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## The “Greenhouse” Effect Explained in Simple Terms

Over the last few years I’ve written lots of articles relating to the inappropriately-named “greenhouse” effect and covered some topics in great depth. I’ve also seen lots of comments and questions which has helped me understand common confusion and misunderstandings.

This article, with huge apologies to regular long-suffering readers, covers familiar ground in simple terms. It’s a reference article. I’ve referenced other articles and series as places to go to understand a particular topic in more detail.

One of the challenges of writing a short simple explanation is it opens you up to the criticism of having omitted important technical details that you left out in order to keep it short. Remember this is the simple version..

### Preamble

First of all, the “greenhouse” effect is not AGW. In maths, physics, engineering and other hard sciences, one block is built upon another block. AGW is built upon the “greenhouse” effect. If AGW is wrong, it doesn’t invalidate the greenhouse effect. If the greenhouse effect is wrong, it does invalidate AGW.

The greenhouse effect is built on very basic physics, proven for 100 years or so, that is not in any dispute in scientific circles. Fantasy climate blogs of course do dispute it.

Second, common experience of linearity in everyday life cause many people to question how a tiny proportion of “radiatively-active” molecules can have such a profound effect. Common experience is not a useful guide. Non-linearity is the norm in real science. Since the enlightenment at least, scientists have measured things rather than just assumed consequences based on everyday experience.

### The Elements of the “Greenhouse” Effect

Atmospheric Absorption

1. The “radiatively-active” gases in the atmosphere:

• water vapor
• CO2
• CH4
• N2O
• O3
• and others

absorb radiation from the surface and transfer this energy via collision to the local atmosphere. Oxygen and nitrogen absorb such a tiny amount of terrestrial radiation that even though they constitute an overwhelming proportion of the atmosphere their radiative influence is insignificant (note 1).

How do we know all this? It’s basic spectroscopy, as detailed in exciting journals like the Journal of Quantitative Spectroscopy and Radiative Transfer over many decades. Shine radiation of a specific wavelength through a gas and measure the absorption. Simple stuff and irrefutable.

Atmospheric Emission

2. The “radiatively-active” gases in the atmosphere also emit radiation. Gases that absorb at a wavelength also emit at that wavelength. Gases that don’t absorb at that wavelength don’t emit at that wavelength. This is a consequence of Kirchhoff’s law.

The intensity of emission of radiation from a local portion of the atmosphere is set by the atmospheric emissivity and the temperature.

Convection

3. The transfer of heat within the troposphere is mostly by convection. The sun heats the surface of the earth through the (mostly) transparent atmosphere (note 2). The temperature profile, known as the “lapse rate”, is around 6K/km in the tropics. The lapse rate is principally determined by non-radiative factors – as a parcel of air ascends it expands into the lower pressure and cools during that expansion (note 3).

The important point is that the atmosphere is cooler the higher you go (within the troposphere).

Energy Balance

4. The overall energy in the climate system is determined by the absorbed solar radiation and the emitted radiation from the climate system. The absorbed solar radiation – globally annually averaged – is approximately 240 W/m² (note 4). Unsurprisingly, the emitted radiation from the climate system is also (globally annually averaged) approximately 240 W/m². Any change in this and the climate is cooling or warming.

Emission to Space

5. Most of the emission of radiation to space by the climate system is from the atmosphere, not from the surface of the earth. This is a key element of the “greenhouse” effect. The intensity of emission depends on the local atmosphere. So the temperature of the atmosphere from which the emission originates determines the amount of radiation.

If the place of emission of radiation – on average – moves upward for some reason then the intensity decreases. Why? Because it is cooler the higher up you go in the troposphere. Likewise, if the place of emission – on average – moves downward for some reason, then the intensity increases (note 5).

More GHGs

6. If we add more radiatively-active gases (like water vapor and CO2) then the atmosphere becomes more “opaque” to terrestrial radiation and the consequence is the emission to space from the atmosphere moves higher up (on average). Higher up is colder. See note 6.

So this reduces the intensity of emission of radiation, which reduces the outgoing radiation, which therefore adds energy into the climate system. And so the climate system warms (see note 7).

That’s it!

It’s as simple as that. The end.

### A Few Common Questions

There are almost 315,000 individual absorption lines for CO2 recorded in the HITRAN database. Some absorption lines are stronger than others. At the strongest point of absorption – 14.98 μm (667.5 cm-1), 95% of radiation is absorbed in only 1m of the atmosphere (at standard temperature and pressure at the surface). That’s pretty impressive.

By contrast, from 570 – 600 cm-1 (16.7 – 17.5 μm) and 730 – 770 cm-1 (13.0 – 13.7 μm) the CO2 absorption through the atmosphere is nowhere near “saturated”. It’s more like 30% absorbed through a 1km path.

You can see the complexity of these results in many graphs in Atmospheric Radiation and the “Greenhouse” Effect – Part Nine – calculations of CO2 transmittance vs wavelength in the atmosphere using the 300,000 absorption lines from the HITRAN database, and see also Part Eight – interesting actual absorption values of CO2 in the atmosphere from Grant Petty’s book

The complete result combining absorption and emission is calculated in Visualizing Atmospheric Radiation – Part Seven – CO2 increases – changes to TOA in flux and spectrum as CO2 concentration is increased

CO2 Can’t Absorb Anything of Note Because it is Only .04% of the Atmosphere

See the point above. Many spectroscopy professionals have measured the absorptivity of CO2. It has a huge variability in absorption, but the most impressive is that 95% of 14.98 μm radiation is absorbed in just 1m. How can that happen? Are spectroscopy professionals charlatans? You need evidence, not incredulity. Science involves measuring things and this has definitely been done. See the HITRAN database.

Water Vapor Overwhelms CO2

This is an interesting point, although not correct when we consider energy balance for the climate. See Visualizing Atmospheric Radiation – Part Four – Water Vapor – results of surface (downward) radiation and upward radiation at TOA as water vapor is changed.

The key point behind all the detail is that the top of atmosphere radiation change (as CO2 changes) is the important one. The surface change (forcing) from increasing CO2 is not important, is definitely much weaker and is often insignificant. Surface radiation changes from CO2 will, in many cases, be overwhelmed by water vapor.

Water vapor does not overwhelm CO2 high up in the atmosphere because there is very little water vapor there – and the radiative effect of water vapor is dramatically impacted by its concentration, due to the “water vapor continuum”.

The Calculation of the “Greenhouse” Effect is based on “Average Surface Temperature” and there is No Such Thing

Simplified calculations of the “greenhouse” effect use some averages to make some points. They help to create a conceptual model.

Real calculations, using the equations of radiative transfer, don’t use an “average” surface temperature and don’t rely on a 33K “greenhouse” effect. Would the temperature decrease 33K if all of the GHGs were removed from the atmosphere? Almost certainly not. Because of feedbacks. We don’t know the effect of all of the feedbacks. But would the climate be colder? Definitely.

See The Rotational Effect – why the rotation of the earth has absolutely no effect on climate, or so a parody article explains..

The Second Law of Thermodynamics Prohibits the Greenhouse Effect, or so some Physicists Demonstrated..

See The Three Body Problem – a simple example with three bodies to demonstrate how a “with atmosphere” earth vs a “without atmosphere earth” will generate different equilibrium temperatures. Please review the entropy calculations and explain (you will be the first) where they are wrong or perhaps, or perhaps explain why entropy doesn’t matter (and revolutionize the field).

See Gerlich & Tscheuschner for the switch and bait routine by this operatic duo.

And see Kramm & Dlugi On Dodging the “Greenhouse” Bullet – Kramm & Dlugi demonstrate that the “greenhouse” effect doesn’t exist by writing a few words in a conclusion but carefully dodging the actual main point throughout their entire paper. However, they do recover Kepler’s laws and point out a few errors in a few websites. And note that one of the authors kindly showed up to comment on this article but never answered the important question asked of him. Probably just too busy.. Kramm & Dlugi also helpfully (unintentionally) explain that G&T were wrong, see Kramm & Dlugi On Illuminating the Confusion of the Unclear – Kramm & Dlugi step up as skeptics of the “greenhouse” effect, fans of Gerlich & Tscheuschner and yet clarify that colder atmospheric radiation is absorbed by the warmer earth..

And for more on that exciting subject, see Confusion over the Basics under the sub-heading The Second Law of Thermodynamics.

Feedbacks overwhelm the Greenhouse Effect

This is a totally different question. The “greenhouse” effect is the “greenhouse” effect. If the effect of more CO2 is totally countered by some feedback then that will be wonderful. But that is actually nothing to do with the “greenhouse” effect. It would be a consequence of increasing temperature.

As noted in the preamble, it is important to separate out the different building blocks in understanding climate.

Miskolczi proved that the Greenhouse Effect has no Effect

Miskolczi claimed that the greenhouse effect was true. He also claimed that more CO2 was balanced out by a corresponding decrease in water vapor. See the Miskolczi series for a tedious refutation of his paper that was based on imaginary laws of thermodynamics and questionable experimental evidence.

Once again, it is important to be able to separate out two ideas. Is the greenhouse effect false? Or is the greenhouse effect true but wiped out by a feedback?

If you don’t care, so long as you get the right result you will be in ‘good’ company (well, you will join an extremely large company of people). But this blog is about science. Not wishful thinking. Don’t mix the two up..

Convection “Short-Circuits” the Greenhouse Effect

Let’s assume that regardless of the amount of energy arriving at the earth’s surface, that the lapse rate stays constant and so the more heat arriving, the more heat leaves. That is, the temperature profile stays constant. (It’s a questionable assumption that also impacts the AGW question).

It doesn’t change the fact that with more GHGs, the radiation to space will be from a higher altitude. A higher altitude will be colder. Less radiation to space and so the climate warms – even with this “short-circuit”.

In a climate without convection, the surface temperature will start off higher, and the GHG effect from doubling CO2 will be higher. See Radiative Atmospheres with no Convection.

In summary, this isn’t an argument against the greenhouse effect, this is possibly an argument about feedbacks. The issue about feedbacks is a critical question in AGW, not a critical question for the “greenhouse” effect. Who can say whether the lapse rate will be constant in a warmer world?

### Notes

Note 1 – An important exception is O2 absorbing solar radiation high up above the troposphere (lower atmosphere). But O2 does not absorb significant amounts of terrestrial radiation.

Note 2 – 99% of solar radiation has a wavelength <4μm. In these wavelengths, actually about 1/3 of solar radiation is absorbed in the atmosphere. By contrast, most of the terrestrial radiation, with a wavelength >4μm, is absorbed in the atmosphere.

Note 3 – see:

Density, Stability and Motion in Fluids – some basics about instability
Potential Temperature – explaining “potential temperature” and why the “potential temperature” increases with altitude
Temperature Profile in the Atmosphere – The Lapse Rate – lots more about the temperature profile in the atmosphere

Note 4 – see Earth’s Energy Budget – a series on the basics of the energy budget

Note 5 – the “place of emission” is a useful conceptual tool but in reality the emission of radiation takes place from everywhere between the surface and the stratosphere. See Visualizing Atmospheric Radiation – Part Three – Average Height of Emission – the complex subject of where the TOA radiation originated from, what is the “Average Height of Emission” and other questions.

Also, take a look at the complete series: Visualizing Atmospheric Radiation.

Note 6 – the balance between emission and absorption are found in the equations of radiative transfer. These are derived from fundamental physics – see Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations – the equations of radiative transfer including the plane parallel assumption and it’s nothing to do with blackbodies. The fundamental physics is not just proven in the lab, spectral measurements at top of atmosphere and the surface match the calculated values using the radiative transfer equations – see Theory and Experiment – Atmospheric Radiation – real values of total flux and spectra compared with the theory.

Also, take a look at the complete series: Atmospheric Radiation and the “Greenhouse” Effect

Note 7 – this calculation is under the assumption of “all other things being equal”. Of course, in the real climate system, all other things are not equal. However, to understand an effect “pre-feedback” we need to separate it from the responses to the system.

### 591 Responses

1. I’ll play Devil’s Advocate. You’re too quick on step 5 (I know note 5 goes into more detail but that’s too much of a jump from a few words to lenghty posts).

Most of what you say has meaning only because of step 5. Please elaborate here, in a concise manner 8) – otherwise one could say that yes, the emission is from a higher altitude but there is more CO2 around so maybe the two effects balance each other somehow.

There is also the issue that even if the troposphere temperature decreases with height, the emission is determined by the temperature of the individual molecules.

Another point that need explanation – as CO2 is heavier than N2, H2O and O2 one might expect it to accumulate nearer to the ground. Has that got any contribution to give to atmospheric temperatures?

The trouble as usual is where to stop the simplification. One can easily understand the troposphere for example by assuming only convection is important. That’s too much of a simplification if the ground and the stratosphere are added. And so on and so forth.

• omnologos,

There is no such thing as the temperature of individual molecules. Temperature is a statistical property. In simpler terms, a bulk property.

• You’re right of course. I should have perhaps spoken about the energy of the individual molecule. But I might be wrong again 😉

Anyway, the underlying question is that in your post you are assuming all molecules to be always in energy balance with the air surrounding them. This is of course just a simplification, and the issue is if the actual underlying mechanism need be added to understand the greenhouse effect to a large extent, in the free atmosphere.

• Omnologos,

The time scale for reaching the local thermal equilibrium within troposphere is nanoseconds, i.e. billionths of one second. Therefore it’s always safe to assume that the molecular energies are distributed according to the Boltzmann distribution and corresponding distributions for other forms of energy than translational motion.

• omnologos,

Pekka is pointing out that all gases are at the same temperature in a “local part” of the atmosphere. See the section “Local Thermodynamic Equilibrium” in Planck, Stefan-Boltzmann, Kirchhoff and LTE.

This is true in the troposphere. As the air thins out well above the troposphere it eventually stops being true.

• More like 10 microseconds for vibrational equilibration, but it doesn’t really make a difference for this argument because the radiative lifetime of vibrationally excited molecules is of the order of 1 s.

Prekka is right for rotational equilibration

• Eli,

I based my estimate on the line width of the pressure broadened absorption and emission lines. My reasoning seems to be erroneous in not taking into account the influence of the more frequent changes in the rotational state and of the fact that those changes are sufficient for broadening the line that corresponds to a transition from a specific vibrational-rotational state to another specific vibrational-rotational state.

As you state it’s irrelevant for the main argument whether the time scale is 1ns or 10µs., but I’ll remember this point in the future.

I have presented similarly erroneous comments before, but again in connections where this doesn’t affect the main point made.

• Omnologos,

From the point of view of a qualitative description little is missing from the step 5, but the unavoidability of the rise in the height of emission is perhaps best seen by an additional argument.

In another comment below I have discussed the role of the stratosphere. From such arguments we can conclude that it’s best to look at the fluxes near the tropopause. An altitude around 17 km might be most appropriate, when one value is used for all locations. (Alternatively a higher altitude could be chosen for tropics and a lower for higher latitudes.)

At such an altitude the up and down fluxes are nearly equal for the wavelengths of strongest absorption peaks of CO2 but at other wavelengths the downwards IR is much weaker than upwards.

When we look down from 17 km, the atmosphere looks very opaque at wavelengths of CO2 absorption peaks. All radiation that reaches the altitude comes from only little lower. At other IR wavelengths the opacity varies. In the atmospheric IR window even the surface is visible. Radiation that reaches the altitude comes from a distance that’s the smaller the more opaque the atmosphere is. Increasing opacity by adding CO2 is guaranteed to shorten the distance, i.e. to move the average height of emission up. That’s true even for the atmospheric IR window, because the share of radiation from the atmosphere goes up relative to that from the surface.

Mixing of gases is so strong throughout the whole atmosphere up to the turbopause at about 100 km altitude that the concentrations do not depend much on the altitude. Beyond that the share of lighter molecules starts to increase with altitude and that of the heavier ones like CO2 to decrease. It takes time for the increase in CO2 concentration to reach fully higher stratosphere, but within the troposphere and lower stratosphere only shorter term variability up to maximally a few years has a more local nature. Antarctic atmosphere follows Northern hemisphere with such a delay.

2. Well done

However, think you missed the most important point with respect to saturation of CO2, that as the concentration increases, the level at which the atmosphere emits in the CO2 bending region rises to a colder level because the emission is trapped higher in the atmosphere Thus, until the level of emission to space hits the stratosphere where T increases with altitude, increasing CO2 will always slow down emission.

The pressure broadening and addition lines argument is actually irrelevant on a practical level.

• Even the case of stratosphere is more complex. The energy balance of the upper stratosphere is very weakly coupled to lower stratosphere and still less with the troposphere.

Additional CO2 will lead to a drop in the temperature of the upper stratosphere and that drop cancels almost totally the increase in radiation out of the stratosphere. That must be the case, because the stratosphere is stratified, i.e. has very little vertical mixing of air, and because the mean free path of the the dominant part of emitted IR is short even in the upper stratosphere. For these reasons the total energy transfer between upper stratosphere and lower parts of the atmosphere is so weak that local heating by solar radiation must be essentially equal to (net) emission of IR up through stratopause..

Thus Increasing CO2 increases emission out form the troposphere, but has only little influence in either direction for the emission from the stratosphere. A very small part of the increased upwards IR at a fixed altitude near tropopause is, however, compensated by increased downwards emission from the stratosphere. This leads also to a small change in net emission through the stratopause.

• Increasing CO2 DECREASES the emission from the troposphere. A simple way to think about this is that the concentration at the level at which the atmosphere emits is constant, because if concentration increased, then radiation would be trapped at the current emission level and would have to occur higher up, and if it decreased, emission would come from a lower level

• Eli,

Stupid of me to write as I wrote. You may notice that I have made the same argument you present elsewhere in this thread.

I was really thinking only about what happens for the radiation from stratosphere and how that extends a small effect down to tropopause.

3. I would also add to your Note 2 that most of the energy from the sun gets absorbed first into the top layers of the ocean – a lot of people seem to be surprised that surface heat anomaly has not raised much but the ocean one did.

4. on June 26, 2014 at 1:25 pm | Reply DeWitt Payne

SoD,

You might consider adding a sentence or two to the Energy Balance paragraph on changing ocean heat content as a measure of the TOA radiative energy balance.

5. Let’s assume that regardless of the amount of energy arriving at the earth’s surface, that the lapse rate stays constant and so the more heat arriving, the more heat leaves. That is, the temperature profile stays constant. (It’s a questionable assumption that also impacts the AGW question).

It doesn’t change the fact that with more GHGs, the radiation to space will be from a higher altitude. A higher altitude will be colder. Less radiation to space and so the climate warms – even with this “short-circuit”.

I dont know how the lapse rate can stay constant with more GHGs. Must a higher altitude always be colder? And how much colder? Say: if it is 100 m higher? My question is about latent heat I think.

• on June 26, 2014 at 2:46 pm | Reply DeWitt Payne

The temperature profile cannot remain constant. A radiative imbalance initially affects the atmosphere more than it does the surface. Even if you are talking about a new steady state after a step change increase in CO2, the atmosphere must still be warmer than it was before or there will still be a radiative imbalance. The lapse rate, that is the slope of the temperature profile may remain the same, but the temperature at every altitude in the troposphere will be higher.

Decreasing the lapse rate while holding the surface temperature constant to increase the temperature at the height of emission doesn’t work either. That would create a radiative imbalance at the surface. Downward emission to the surface is also higher with a lower lapse rate.

• Nobodyknows

I dont know how the lapse rate can stay constant with more GHGs.

The lapse rate is the rate of change of temperature with height. The primary factor that determines this is called the “adiabatic lapse rate”, which is all due to the expansion of air (as it rises) into the lower pressure that you find at higher altitudes. When air expands it cools, a consequence of the first law of thermodynamics (work done = internal energy lost).

When we consider moist air the argument is the same but the calculation is more involved. Now we have to include the energy from latent heat as water vapor condenses out. This reduces the temperature drop with altitude.

See the links in note 3 for more detail.

• If the temperature gradient in the troposphere at more CO2 remains constant (correctly!), then the question remains as it comes to the increase in surface temperature.

The answer is simple: the altidude of the tropopause rises.

• Ebel,

The behavior of the altitude of the tropopause is related, but not exactly the same thing or quantitatively determining for the outcome as both the altitude and the temperature of tropopause may change. In the real atmosphere determining the height of the tropopause is also difficult, because the transition from the troposphere to the stratosphere is not sharp. Simple models are misleading when they produce a sharp well defined tropopause.

Most of the changes that result in radiative forcing and/or surface warming occur deep inside the troposphere, not at the tropopause.

6. Sorry. I lost quotion marks in the two first paragraphs. [Moderator note – fixed that ]

7. Thanks for the replies. As I said I’m playing the Devil’s Advocare role. I think the looking-downward example by Pekka makes the situation quite clear. Also the mention of the turbopause.

8. Thank you for your replay De Witt.
“The lapse rate, that is the slope of the temperature profile may remain the same, but the temperature at every altitude in the troposphere will be higher.”

This puzzels me. If most of the TOA LR come from emission hight between 2 and 10 km, then the temperatures at that hights should cause most of the energy out. Then the stratospheric temperature should not contribute so much.

• on June 26, 2014 at 5:04 pm | Reply DeWitt Payne

You do understand that mathematically a straight line can be described by two independent parameters, the slope and the intercept. By independent, I mean that you can change one without affecting the other. Let’s say the slope of the temperature profile, lapse rate, is 6 K/km and the surface temp is 300K. That means that at an altitude of 10km, the temperature will be 240K. Lapse rate by convention is a positive number even though the slope is negative. If we then increase the surface temperature by 1K and keep the lapse rate constant, the temperature at 10km will be 241K.

The stratospheric temperature doesn’t contribute much. There’s not very much air above the tropopause and there’s hardly any water vapor. So the major radiative gases are ozone and carbon dioxide. Increasing ozone warms the stratosphere because it increases the absorption of incident solar radiation while increasing carbon dioxide cools it because it increases emission much more than it increases absorption.

9. Thank you very much !!
Re: “Water Vapor Overwhelms CO2”

Is it legit to suggest that since H2O vapour concentration is related to temp, and it never surpasses reasonable supersaturated levels because it freezes, rains out… that H2O is a pretty constant – inflexible – GHG? It seems that it is always destined to be more of a “feedback”, than a “driver”.

Is that fair? Too simplistic?

• The ansatz is that relative humidity is constant. Since air temperature increases, the absolute humidity increases

• Manwichstick: Too simplistic. Water vapor concentration does not vary with temperature – the maximum amount of water air can hold AT EQUILIBRIUM varies with temperature. The only time water vapor varies with temperature is when the air is saturated – and that only occurs when liquid water and water vapor are both present. Such a true equilibrium only occurs in clouds (where moist air is rising) and very near the surface of liquid water. I believe that the air even a few meters above the surface of the ocean is typically is only 80-85% saturated (but would appreciate being corrected if the real figure is higher.) So mixing plays a big role in atmospheric water vapor concentration that many often ignore. Furthermore, the rate of evaporation can depend more on wind speed than temperature. (The air in contact with the ocean surface is saturated, so the evaporation rate depends on how fast saturated air is transported away from the surface.) It certainly appears likely that absolute humidity will increase with temperature, but whether this increase will be rapid enough to maintain constant relative humidity is another question.

Large seasonal changes in temperature are accompanied by large changes in absolute humidity. I’d be interested in any good information showing how RELATIVE humidity varies with seasonal changes in temperature.

• manwichstick,

We definitely should think of water vapor as a feedback, even though it is the most important “greenhouse” gas.

A very readable and good summary of some basics is found in this free paper:
Water Vapor Feedback and Global Warming, I.M. Held & B.J. Soden, Annual Review of Energy and the Environment (2000).
They discuss the critical point that the large scale circulation determines the amount of water vapor above the boundary layer.

I noticed today that Isaac Held (lead author of that paper) has a new blog article on water vapor – 47. Relative humidity over the oceans which is very interesting. Isaac Held is always worth reading. He has made a big contribution to climate system dynamics over the last 35 years or so.

There is a series on water vapor on this blog – Clouds and Water Vapor. A comment from Part Seven – Upper Tropospheric Models & Measurement:

One of the key points is that the response of water vapor in the planetary boundary layer (the bottom layer of the atmosphere) is a lot easier to understand than the response in the “free troposphere”. But how water vapor changes in the free troposphere is the important question. And the water vapor concentration in the free troposphere is dependent on the global circulation, making it dependent on the massive complexity of atmospheric dynamics.

• Thanks for the extra references, and Frank, Eli, SoD.
Thanks for correcting me my absolute/relative humidity issue. I wasn’t choosing my words carefully – but you folks seemed to know what I was actually trying to ask.

10. SOD wrote: “At the strongest point of absorption – 14.98 μm (667.5 cm-1), 95% of radiation is absorbed in only 1m of the atmosphere (at standard temperature and pressure at the surface). That’s pretty impressive.”

However, at the tropopause (ca 100 mb), 5% of 14.98 um radiation passes through 10 m of atmosphere. In the stratosphere, (1-10 mb), 5% of 14.98 um radiation passes through 100-1000 m (0.1-1 km) of atmosphere. If you go high enough, even the strongest lines are not “saturated” and doubling CO2 will decrease the rate at which photons escape to space. So saturation by itself is actually irrelevant. (The temperature about one tau into the atmosphere from space is an issue and the temperature doesn’t fall at this altitude for the strongest lines.)

Furthermore, absorption is only part of the story: Twice as many CO2 molecules in the atmosphere emit twice as many photons towards space. The rate of radiative cooling to space is a complicated function of the number of emitting GHGs (primarily CO2 high in the atmosphere), the temperature of the GHGs (which effects the photon emission rate) and the probability of that an emitted photon escapes to space.

• Frank,
The mean free path at the center of the line does not change that fast, because the line width depends on the pressure.

Based on some results that I have calculated the mean free path at the center of the peaks is only 1m at the altitude of 13 km. High in the stratosphere at 50 km the strongest peak leads a mean free path of only 10m at the center of that peak. What changes much more is the width of the peaks and the mean optical depth between the adjacent peaks whose energies are slightly different due to rotational states. When some peaks get lower by a factor of 10 only, the valleys between the peaks drop by a factor of around 10000.

These results are from an old calculation, and I don’t have proper documentation about those calculations, but I think that Voigt profile was used in obtaining those results.

These numbers come from this picture that I have presented before in another thread.

• Peeka: Thanks for the correction. Averaged over a 0.1 um of so, I’m about right; but if you increase the resolution to 0.001 um, I’m wrong. I guess I should have rounded 14.98 um to 15.0 um instead of copying and pasting SOD’s number. Fortunately, that doesn’t change my main points – if you go high enough, doubling CO2 will make a difference in the probability of escaping to space – which is only part of the radiative cooling story.

What happens when a GHG molecule is high enough that an excited state is unlikely to be “relaxed” by a collision? Does the probability of emission of a photon depend on the lifetime of the excited state (rather than B(lamba,T), since there is no T). Is there a simple formula for lifetime (which I probably have forgotten)?

It seems to me that B(lamba,T) is telling us something (similar a Boltzmann distribution) about the equilibrium fraction of molecules with an energy E=hc/lamba above ground state at a temperature T. Actually, I’m not sure whether M-B or B-E statistics apply to this problem. The derivations I have seen don’t approach the situation from a “number of excited molecules times rate of emission per molecule” perspective.

• Frank,

Where the change is line shape gets important is in understanding stratosphere. As lines are very narrow and strong at the center radiation from much lower altitudes passes mostly trough with little absorption, but radiation emitted in the upper stratosphere that’s much warmer than lower stratosphere and the tropopause cannot heat effectively lower altitudes. Radiation from those altitudes gets blocked effectively by layers just below, and has only very little influence further down.

The temperature profile of the stratosphere would be significantly different, if the line shape would not change with altitude.

• on June 27, 2014 at 12:33 am DeWitt Payne

Frank,

What happens when a GHG molecule is high enough that an excited state is unlikely to be “relaxed” by a collision? Does the probability of emission of a photon depend on the lifetime of the excited state (rather than B(lamba,T), since there is no T). Is there a simple formula for lifetime (which I probably have forgotten)?

The intensity of emission is determined by the number of molecules in the excited state, not the lifetime of an individual molecule in the excited state. The Einstein A21 coefficient, which has units of sec-1, multiplied by the number of molecules in the excited state in a unit volume is the emission intensity in that volume. The M-B equation tells you what fraction of molecules at a given temperature will have an energy greater than some level.

However, local thermodynamic equilibrium no longer holds when the probability that an excited molecule decays by radiation rather than collision gets too high. And too high isn’t very high. Without LTE, M-B statistics may no longer apply. Equipartition may not apply. The rate of emission may then depend on the rate of absorption of radiation because collisions are simply too rare to excite many molecules. An individual molecule could have a kinetic energy well above the energy level of an emitted photon, but without collisions, that energy can’t be translated into the vibrational state. I think. QM can be very strange.

• What happens when a GHG molecule is high enough that an excited state is unlikely to be “relaxed” by a collision?

That is called the ionosphere and it is not in local thermodynamic equilibrium. Mostly what happens is that there is so much far UV that most of the molecules are dissociated. Go high enough and there is more O than O2 and lots of ions (N2 is the hardest to crack)

• Frank: Twice as many CO2 molecules in the atmosphere emit twice as many photons towards space.

Yeahbut – the rate of emission varies as the fourth power of temperature AND if you raise the effective altitude of emission, the pressure is lower, which means both fewer molecules AND as Pekka points out, squeezes the line shape effectively raising the absorption coefficient integrated over a line. Thus the total emission rate from the atmosphere decreases.

• on June 29, 2014 at 2:13 pm DeWitt Payne

Eli,

squeezes the line shape effectively raising the absorption coefficient integrated over a line.

I may be misinterpreting what you wrote, but doesn’t narrowing the line decrease total absorption? A Dirac δ function line would absorb no radiation because there’s no radiation to absorb for a line with zero width even though the integral of the the absorption line is finite, specifically, equal to one.

• DeWitt,

Narrower lines increase transmission of wavelengths far enough from the center of the line. Radiation that originates much lower in the atmosphere is therefore absorbed less, but radiation from above or not far below gets absorbed more strongly. As the same lineshape applies to both emission and absorption at nearby altitudes, the narrowing of the lines means that radiation originating high up in the stratosphere is very ofter absorbed not far from the point of emission.

• on June 29, 2014 at 8:00 pm DeWitt Payne

Pekka,

Yes, I understand that. But does the total flux integrated over the line decrease or increase? My thought was that it decreased because an emission line isn’t a δ function. Of course the line width doesn’t go to zero either.

• DeWitt,
It’s likely to increase in the upper stratosphere, because it’s so much warmer than the effective radiative temperature of the lower part of the atmosphere.

• on June 29, 2014 at 10:10 pm DeWitt Payne

Pekka,

I didn’t phrase that correctly. Does the integral of the absorption coefficient increase or decrease as line width decreases?

• DeWitt,

If I’m understanding your question correctly.. the line strength is the total absorption under the curve. So when you integrate the absorption, a(w) vs wavenumber, w you get, S, the line strength (S is the parameter name in HITRAN).

So as the pressure decreases and the line width narrows, the peak of absorption is higher and the trough is lower.

I’m going from memory here, and it’s been a while. Did I understand the question? If so I can go and check my memory against a reference..

• Integral of the absorption coefficient does not change from line broadening. The integral of absorption of a layer of finite thickness decreases for radiation of smooth spectrum, but increases for radiation that’s has an intensity peak nearly as narrow as the absorption peak at the same wavelength.

• Eli: The change in emission caused by a rising characteristic emission level appears to be minor compared with the doubling of emission associated with doubling CO2.

The density of the atmosphere drops by a factor of two about every 5.6 km, about 1.25% every 100 m. So, if the characteristic emission altitude rises 100 m, emission will be reduced 1.25% due to the decreased number of CO2 molecules at that higher altitude.

The temperature falls about 0.65 degK for every 100 m the characteristic emission level rises. At the blackbody equivalent temperature of 255 degK, an 0.65 degK decrease is an 0.25% decrease. Since emission varies with the fourth power of temperature, a 100 m increase in the characteristic emission altitude produces a 1% decrease in emission due to falling temperature.

Therefore, doubling CO2 doubles the number of photons emitted minus about 2.25% for every 100 m the characteristic emission altitude rises.

I couldn’t locate a reliable value for the rise in the of the characteristic emission level caused by doubling CO2. WIth a lapse rate of 6.5 degK/km, however, a no-feedbacks climate sensitivity of 1 degC for 2XCO2 is equivalent to less than 200 m change in altitude. In that case, the lower density and temperature of CO2 at a higher characteristic emission altitude would reduce emission from doubled CO2 by about 4%, reducing 200% emission from doubled CO2 to 192% emission. So, doubling CO2 does nearly double emission of photons by CO2 at the characteristic emission altitude.

The radiative forcing from 2XCO2 appears to be the result of two nearly offsetting factors: a) halving the mean free path of most photons absorbed by CO2 and b) doubling of emission from CO2. The 4% change in emission from lower density and temperature at a higher characteristic emission altitude becomes important only because these factors are nearly offsetting.

• The narrowing of the emission lines is probably also unimportant given a small rise in the characteristic emission altitude. As best I can tell, the total flux of photons emitted from each emission line depends only on the A21 emission coefficient and the fraction of GHG molecules in a particular vibrational (and rotational) excited state. When LTE exists, the fraction of molecules in an excited state depends on temperature. The fact that some GHGs are traveling away from or towards the direction of emission effects the width of the line, but not the number of photons emitted by a particular transition. The fact that some GHGs are in the process of colliding with another molecule when they are emitting a photon also effects the width of the line, but not the number of emitted photons.

The narrowing of lines does effect the distance an emitted photon travels before it is reabsorbed.

• Frank,

I’m probably not telling you anything you don’t know..

The conceptual approach is always a good idea.. but there are significant non-linearities in absorption and emission – and these are integrated over wavenumber and height..

The conceptual approach must be tested against the actual calculations from the equations of radiative transfer.

Doing a set of calculations might be a better idea – I can produce some numbers if that will help, x vs y.

• on June 30, 2014 at 1:48 pm DeWitt Payne

Frank,

The total flux from a gray or black body increases with the fourth power of its temperature. That isn’t true for an individual emission line or for a non-gray spectrum. The rate of change of flux with temperature is dependent on the frequency. At very low frequencies, for example, the intensity is linear with temperature. For frequency, the dependence on temperature is 1/(exp(hν/kT)-1). Using T^4 is at best a crude approximation.

• SOD and DeWItt:

SOD wrote” “Frank, I’m probably not telling you anything you don’t know… The conceptual approach is always a good idea.. but there are significant non-linearities in absorption and emission – and these are integrated over wavenumber and height.”

I could have something wrong. I wrote the above comments because the primary effect of doubling CO2 is to produce twice as many photons from CO2 that travel half as far. This is my personal “big picture” to a first approximation. The lower temperature, lower density and narrower line width at a higher characteristic emission altitude cause a very modest reduction in this doubled emission – a reduction that would be almost negligible (assuming my calculations were correct) if increased absorption didn’t nearly offset increased emission. Another refinement would be to recognize that not all photons emitted by CO2 travel an average of half as far. Those reaching space or the earth (as DLR) don’t. IMO, there is too little emphasis on the big picture – twice as many photons traveling half as far – and too much emphasis on factors that reduce emission slightly. These factors are important to the GHE, but only because the primary factors mostly offset each other.

Perhaps some of my math or physic is wrong. I certainly haven’t properly dealt with the fact that each wavelength has a different characteristic emission altitude and that this altitude is not always in the troposphere where the lapse rate is constant. The effect of temperature on the flux at each individual wavelength does vary with 1/(exp(hv/kT)-1), but the difference between T and T+0.65 degK (100 meters higher) using this formula is 1.5% at 200 degK, 1.0% at 255 degK and 0.72% at 300 degK and 15 um. So T^4 appears to have been a reasonable approximation. I can’t find anything that suggests that my big picture – twice as many photons traveling half as far – is grossly wrong.

(I find the GHE much easily to understand using the Schwarzschild eqn. than any of the above.)

• Frank,

Here is the mental picture I have of doubling CO2 (from fig. 12 in Visualizing Atmospheric Radiation – Part Seven – CO2 increases:

That’s why I can’t do arithmetic on it. For the highly absorbing/highly emissive center of the band the effect is tiny. For the less absorbing side bands the effect is significant.

In my head I can’t average out a large set of numbers where the factor is exp(-something(v) x something doubling). As soon as people start explaining how to do it, my head starts hurting and I have to drink coffee or alcohol, depending on the time of day.

It’s why I like Matlab. I don’t have to manipulate complex stuff in my head. Instead I can run some numbers quickly, produce graphs, change a factor, produce more graphs. It’s a beautiful thing.

So if you can turn the real maths of this absorption & emission model into something conceptually simple that is wonderful. Lacking the mental maths skills I don’t see how it’s possible.

11. There is really no dispute of there being a radiatively induced GHE by rational people. The argument is over how much net ‘enhancement’ to the GHE would or should result from additional GHGs.

• on June 27, 2014 at 1:33 am | Reply Climate Weenie

Right. Some of us like to point out the exaggerations of extent, effect, and impacts.

Perhaps it’s time for the science of joy.

12. My English is not so good to write the following text directly into English – so largely machine translation. For safety, I cling to the German text.

1) There is no saturation: Where is strongly absorbed, is also strongly emits – it’s just a high radiation transport resistance.
2) The oft-mentioned 33 K are only a lower limit of the greenhouse effect (Hölder’s inequality). On closer calculating higher values come out (Kramm about 115 K, 158 K Gerlich). The precise values for certain assumptions are misinterpreted as evidence to the contrary.
3) The decrease in intensity with higher = colder goes past the change in the temperature profile. With more CO2 the pressure of the tropopause decreases (the tropopause is higher) – Schwarzschild criterion. The longer troposphere at approximately constant temperature gradient leads to a larger temperature difference between surface and tropopause. This greater temperature difference is distributed to about 1/4 to the increase in surface temperature and 3/4 to the decrease in stratospheric temperature.
4) Due to the convection of the greenhouse gas, the mixing ratio to about 100 km altitude is constant. Without convection, the greenhouse gas would according to the molecular weight segregate something – but the time of separation is large compared to the circulation time.
5) If none of the isothermal conditions are present, also the distribution of the excited states are not in accordance with the local gas temperature – independent of the lifetime. Nevertheless, you can assign the distribution of the excited states by a Boltzmann temperature. With frequent collisions (LTE) but the difference is gas temperature and Boltzmann temperature well below 1 K.
7) The approximate constancy of relative humidity is due to the circulation and raining. To top rises to 100% saturated air and rains from on temperature decrease. In the ascended air therefore decreases the absolute humidity – while sinking so the relative humidity decreases. Due to the turbulence, the moist air rising and sinking dry air combine to create an average.
8) At the same width of the absorption lines the absorption length is proportional to the pressure. Wherein indication of the absorption length as the pressure difference, therefore, the pressure difference is independent of pressure. The mean pressure difference is approximately independent of the width of the lines.
9) At the tropopause is approximately the whole heat transport upward as radiation transport. In the layers below the tropopause of the total heat transport remains about the same – but because of the increasing radiation transport resistance (higher density) decreases proportion, which is transported beam, so that the convective transport increases.

1.) Es gibt keine Sättigung: Dort wo stark absorbiert wird, wird auch stark emittiert – es ist nur ein hoher Strahlungstransportwiderstand.
2.) Die oft genannten 33 K sind nur ein unterer Grenzwert des Treibhauseffektes (Höldersche Ungleichung). Bei genauerer Berechnung kommen höhere Werte heraus (Kramm ca. 115 K, Gerlich 158 K). Die genaueren Werte für bestimmte Annahmen werden fälschlich als Gegenbeweis interpretiert.
3.) Die Intensitätsabnahme mit höher = kälter geht an der Änderung des Temperaturprofils vorbei. Mit mehr CO2 sinkt der Tropopausendruck (die Tropopause wird höher) – Schwarzschild-Kriterium. Die längere Troposphäre führt bei annähernd konstanten Temperaturgradienten zu einer größeren Temperaturdifferenz zwischen Oberfläche und Tropopause. Diese größere Temperaturdifferenz verteilt sich zu ca. 1/4 auf die Zunahme der Oberflächentemperatur und zu 3/4 auf die Abnahme der Stratosphärentemperatur.
4.) Durch die Konvektion der Treibhausgase ist das Mischungsverhältnis bis etwa 100 km Höhe konstant. Ohne Konvektion würden sich die Treibhausgase entsprechend dem Molekulargewicht etwas entmischen – aber die Entmischungszeit ist groß gegenüber der Zirkulationszeit.
5.) Wenn keine isothermen Verhältnisse vorliegen, ist auch die Verteilung der angeregten Zustände nicht entsprechend der lokalen Gastemperatur – unabhängig von der Lebensdauer. Trotzdem kann man der Verteilung der angeregten Zustände nach Boltzmann eine Temperatur zuordnen. Bei häufigen Kollisionen (LTE) ist aber der Unterschied Gastemperatur und Boltzmanntemperatur weit unter 1 K.
6.) Die latente Wärme einschließlich Kondensationswärme führt nur zur Abnahme des trockenadiabatischen Gradienten (etwa 9,8 K/km) auf den feuchtadiabatischen Gradienten (etwa 6,5 K/km).
7.) Die näherungsweise Konstanz der relativen Luftfeuchtigkeit ist Folge der Zirkulation und Abregnen. Nach oben steigt 100% gesättigte Luft auf und regnet bei Temperaturabnahme ab. Bei der aufgestiegenen Luft nimmt deshalb die absolute Feuchtigkeit ab – beim Absinken sinkt also die relative Feuchtigkeit. Durch die Turbulenz vermischen sich die feuchte aufsteigende Luft und die absinkende trockenere Luft zu einem Mittelwert.
8.) Bei gleicher Breite der Absorptionslinien ist die Absorptionslänge proportional zum Druck. Bei Angabe der Absorptionslänge als Druckdifferenz ist deshalb die Druckdifferenz unabhängig vom Druck. Die mittlere Druckdifferenz ist näherungsweise unabhängig von der Breite der Linien.
9.) An der Tropopause ist näherungsweise der ganze Wärmetransport nach oben als Strahlungstransport. In den Schichten darunter bleibt der Gesamtwärmetransport etwa gleich – wegen des zunehmenden Strahlungstransportwiderstandes (höhere Dichte) sinkt aber Anteil, der strahlend transportiert wird, so daß der konvektive Transport zunimmt.

MfG

13. Just one point: While the CO2 is definitely a GHG and there is a “greenhouse” effect, there is considerable question as to how much CO2 has contributed to historic warming. The Mt. Pinatubo eruption showed, I believe, that CO2 is not as potent a GHG as was previously believe. ERBE readings done in the weeks and months after the eruption showed that a doubling of CO2 would see approximately a 1 C increase in global temperatures. That is below the IPCC’s business as usual scenario and of little concern.

• That is about what you get without any feedbacks.

• Alan: The Mt. Pinatubo eruption can be used to estimate ECS from fast feedbacks (clouds, water vapor) in the absence of slow feedbacks (ice-albedo). However, the answer one obtains depends on the model one uses to analyze the data and what data one uses. Should the temperature data be corrected for ENSO? The first post below reviews the literature (with links) and describes the results from a more comprehensive model (which appears credible, but which certainly isn’t peer-reviewed). The second updated post obtains a most likely ECS of 1.15 degC and can’t properly fit the data if ECS is outside the range 0.8-1.4 degC. The whole analysis is worth reading, especially if you want to understand why it is hard to get definitive results from climate data.

http://rankexploits.com/musings/2012/pinatubo-climate-sensitivity-and-two-dogs-that-didnt-bark-in-the-night/

http://rankexploits.com/musings/2012/pinatubo-climate-sensitivity-revisited/

Now that papers like Otto (2013) have made low values of ECS more likely, perhaps an analysis similar to this one can get through peer review.

14. Let’s assume that regardless of the amount of energy arriving at the earth’s surface, that the lapse rate stays constant and so the more heat arriving, the more heat leaves. That is, the temperature profile stays constant. (It’s a questionable assumption that also impacts the AGW question).

It doesn’t change the fact that with more GHGs, the radiation to space will be from a higher altitude. A higher altitude will be colder. Less radiation to space and so the climate warms – even with this “short-circuit”.

This seems to me to be a non-sequitur.

If convection dominates over radiative transfer as the primary heat transfer mechanism in the troposphere, an increase in radiative forcing from GHGs would still be overwhelmed by the dominant and stronger negative-feedback of increased convection.

Likewise, if an electrical circuit containing a resistor is “short-circuited,” increasing the resistance of the resistor will make no difference to the unimpeded flow of current via the “short-circuit.”

Who can say whether the lapse rate will be constant in a warmer world?

The lapse rate dT/dh = -g/Cp is dependent only upon gravity and atmospheric heat capacity at constant pressure. If anything, adding GHGs would increase Cp and thus decrease the lapse rate to cause cooling. A warmer world would increase evaporation and force lapse rates toward the steeper wet adiabatic to cause compensatory negative-feedback cooling.

with more GHGs, the radiation to space will be from a higher altitude. A higher altitude will be colder…

What altitude are you saying is the ERL? The tropopause is essentially isothermal, so an increase in average radiative altitude would make no difference in radiating temperature. The troposphere is dominated by the negative-feedback of increased convection, so an increase of radiative forcing from a higher ERL would be overwhelmed by negative-feedback from increased convection.

• Paul,

These are all good questions to ask.

If convection dominates over radiative transfer as the primary heat transfer mechanism in the troposphere, an increase in radiative forcing from GHGs would still be overwhelmed by the dominant and stronger negative-feedback of increased convection.

First of all, the “base case” is that of increases in a GHG like CO2 without any feedbacks. So let’s say – in hypothetical case A – that the lapse rate stays exactly the same. In case A, the radiation altitude has moved up.

We’ll come back to your early points later, so let’s pick up your final point, which impacts on this one:

What altitude are you saying is the ERL? The tropopause is essentially isothermal, so an increase in average radiative altitude would make no difference in radiating temperature. The troposphere is dominated by the negative-feedback of increased convection, so an increase of radiative forcing from a higher ERL would be overwhelmed by negative-feedback from increased convection.

Here’s what I said in note 5: ‘ the “place of emission” is a useful conceptual tool but in reality the emission of radiation takes place from everywhere between the surface and the stratosphere. See Visualizing Atmospheric Radiation – Part Three – Average Height of Emission – the complex subject of where the TOA radiation originated from, what is the “Average Height of Emission” and other questions..’

Overall the atmosphere emits from a lot of different altitudes. These altitudes are highly wavelength dependent. Nearly all of the emission is well below the tropopause, in the troposphere. A tiny portion is from the most “saturated” part of the CO2 band in the stratosphere.

So basically almost all of the “greenhouse” effect is from the troposphere, and all of the effect of doubling CO2 is from the troposphere.

This is a “no feedback” case. Obviously the climate has feedbacks so this brings me to the point you have about convection, which I’ll pick up in the next comment.

• “A tiny portion is from the most “saturated” part of the CO2 band in the stratosphere.”

As already written – there is no saturation, not even for a tiny portion.

• Paul,

..If convection dominates over radiative transfer as the primary heat transfer mechanism in the troposphere, an increase in radiative forcing from GHGs would still be overwhelmed by the dominant and stronger negative-feedback of increased convection..
A warmer world would increase evaporation and force lapse rates toward the steeper wet adiabatic to cause compensatory negative-feedback cooling..

According to a random site, cp for air changes from 1.009 at -100’C to 1.005 at -50’C. So that’s about 0.01% per ‘C over that range.
Then from -50’C to +40’C it is constant.

I think we can work on the basis that cp is effectively constant over the temperature changes that might come from increases in CO2.

The dry adiabatic lapse rate will not be changing.
The moist adiabatic lapse rate will be changing if there is more moisture (or less moisture)

Here is a graphic of moist potential temperature (from Potential Temperature), annually averaged:

We can see that in the tropical troposphere the atmosphere is very effectively convecting to the moist adiabatic lapse rate (that’s what the vertical lines of moist potential temperature tell us).

So let’s consider an increase in CO2. On day one the temperature profile is still the same. The emission of radiation to space drops. Nothing happens to the temperature immediately but over the coming days, weeks and months the troposphere warms.

Why would it stop? Because the atmosphere and surface warm up until we reach a new equilibrium point. This has to happen via an increase in temperature. Without an increase in temperature there is no feedback to stop or reduce any climate warming.

So we are on the subject of feedbacks. Well, that’s a lot more complicated and not the subject of this article. But of course it is the interesting subject.

From what you argue we might expect three things:

1. Higher temperatures – which leads to more emission of radiation from the surface and the atmosphere. This is a negative feedback which reduces the impact of more CO2.

2. More water vapor – which leads to a “tropical hot spot” as the lapse rate is lower (ie less cooling per km of altitude) and so the temperature higher up in the atmosphere increases more than the temperature lower down. This is a negative feedback which reduces the impact of more CO2.

3. More water vapor – which is a “greenhouse” gas, aka “radiatively-active” gas, which moves the altitude of emission to space higher in the atmosphere which reduces radiation to space. This is a positive feedback which increases the impact of more CO2.

Is this what you were arguing for, or only 1&2? Surely you must also be arguing for 3 as well?

Of course, many people have used GCMs to try to calculate these different effects, and compared GCMs. Here’s an example from Quantifying Climate Feedbacks Using Radiative Kernels, by Brian Soden et al (2008). One of the coauthors is Isaac Held, both were referenced earlier in the comment thread:

I’m not saying this answer is correct. But it would be a difficult case to make that more water vapor would have an overall negative feedback.

• And just a further comment as the lapse rate always seems to lead to confused discussions.

Saying that the dry adiabatic lapse rate would remain the same is not the same thing as saying that the environmental lapse rate would remain the same in dry areas.

The adiabatic lapse rate only tells us how the temperature changes when convection happens relatively quickly. See the earlier cited articles in note 3 for more on this.

A reasonable case can be made for an “actual” (=environmental) lapse rate staying close to the moist adiabatic lapse rate in the tropics, because that is what we already observe in practice. There is lots of convection, and lots of very strong convection, in the tropics.

In the mid-latitudes and polar regions it is a different story.

Calculating how air expands and therefore loses heat is an easy calculation. Calculating how air moves around in the climate is a completely different story.

• Thank you for this answer SoD.
I think there has been too much confusion in discussions of climate change because the question of negative feedback of vater vapor has been avoided. And I dont say that there is no positive feedback!

• on June 27, 2014 at 1:28 pm Climate Weenie

>>> 2. More water vapor – which leads to a “tropical hot spot”

So, by most examinations, warming has occurred, but the “tropical hot spot” has not. Evidently there are some shortcomings to this understanding.

• on June 27, 2014 at 3:04 pm | Reply DeWitt Payne

Paul,

If convection dominates over radiative transfer as the primary heat transfer mechanism in the troposphere, an increase in radiative forcing from GHGs would still be overwhelmed by the dominant and stronger negative-feedback of increased convection.

Dominates is a rather strong word. The net flows from the surface to the atmosphere and space are ~100W/m² by convection and ~~60 W/m² by radiation, with ~40 W/m² of the radiation going directly to space. Your shorted resistor analogy is incorrect. The question I would like you to answer is: How is convection going to increase if the temperature doesn’t increase?

• A resistor is a bad analogy, because convection is more like a diode (or perhaps a zener diode). A diode lets current pass trough with little resistance as soon a threshold voltage is exceeded, but not when the voltage is in the wrong direction. Similarly convection grows with little resistance when temperature drops with altitude faster than the adiabatic lapse rate, but stops when that’s not the case.

• Pekka, a diode with significant leakage current would be even better. If you use an idealized diode you get trapped into the Gerlach argument about the second law not permitting any energy flow downwards in the atmosphere.

15. Your base premise is wrong, SoD. After that, no appeal to ‘basic physics’ (or the ‘authority’ of Grant Petty, for that matter), no theoretically conjured up warming mechanisms and no calculations made will help you accomplish anything but a reinforcement of your original misconceptions about how the world works.

The earth system warms from the surface UP, not from some hypothesized atmospheric layer in radiative imbalance DOWN. The lapse rate is set from the surface and climbs up with convection bringing the surface heat into and up through the troposphere. The heat moves through the earth system in this way: sun >> surface >> troposphere >> out through the ToA. That’s why we always and only see this progression: surface temps up >> troposphere temps up (how? mostly from the transfer of latent heat through evaporation>condensation)* >> OLR at ToA up (why? from the increased temp and humidity/emissivity through the column below). Never the opposite.

*http://oceanworld.tamu.edu/resources/oceanography-book/Images/rain_ANN.png

The earth system, rather than ‘holding incoming energy back’, radiates whatever it needs to radiate to balance the incoming from the sun, from whatever level(s) convection brings the surface heat up to, from where the ‘radiatively active gases’ radiate it back out to space. (Without their presence, this would be a problem and the earth would be a much hotter and unstable place.) And that’s it. The surface temp is already set, by the incoming radiation from the sun in conjunction with the mass (heat capacity & weight) of the atmosphere on top of the solar-heated surface. Everything after that is just a result of temperature/temperature distribution and emissivity, always having to catch up with variations in solar input. The strange ‘effective (or average) emission height’ hypothesis is nothing but an attempt to turn everything we know about how the world works on its head. It is a result, not a cause of temperature (and emissivity) distribution.

This is what happens when good old meteorology and climatology become ‘high-jacked’ by radiative physicists insisting that their particular field of study in fact rules the roost.

I’m not disputing the radiative properties of the gases in question. Not even the temperature effect they might have in a closed glass box lab exeriment, by forcing reduced temp gradients away from the externally heated surface. What I’m saying is that, in the open surface/atmosphere system, they don’t matter. Well, they matter. The analogy simply fails. They do and can not determine the mean temp of the global surface. At all. Other (and equally ‘basic’) physical processes entirely do that.

The atmosphere does not owe its ability to warm to the presence of ‘radiatively active gases’. It owes its ability to cool adequately to the ‘radiatively active gases’. Because it warms by convection from the surface and cools by radiation to space. Increase its emissivity and you increase its abiity to cool. Basically, the only gas radiating to space from the troposphere is H2O, CO2 almost not at all:

And H2O in the atmosphere would, in purely radiative terms, cool the surface underneath, by absorbing and reflecting a massive portion (close to 80%) of the incoming solar radiation before it can ever be absorbed by the surface. Tropical rainforest areas are several degrees cooler in mean annual temps than tropical/subtropical desert areas. Even as they lie on average more directly under the sun, experience much smaller temp fluctuations than the deserts and would (according to ‘theory’) receive much, much more ‘back radiation’ from the moist atmosphere …

Now, this is how the atmosphere really makes the surface warmer than solar radiative equilibrium (because it sure does):

1) It has a mass and therefore a heat capacity. This means it is able to warm. It does so by being directly convectively coupled with the solar-heated surface below it. Regardless of whether that atmosphere contains radiatively active gases or not, it will warm – conductively>convectively. The atmosphere is able to warm. Space isn’t. Therefore the atmosphere sets up a temperature gradient away from the solar-heated surface that has a finite (sub-max) steepness. Space doesn’t. The atmosphere thus INSULATES the surface. Energy is not able to escape the surface as fast as it’s coming in before it has warmed to a higher mean temperature than before the atmosphere was put in place.

2) It has a mass and therefore a weight (it’s in a gravity field, after all). It therefore exerts a pressure on the solar-heated surface above 0. Unlike space. This pressure makes it harder for energy to escape the surface than without such pressure AT EQUAL TEMPERATURE in two ways: i) it suppresses the evaporation rate, and ii) it suppresses (upward) buoyant acceleration of heated surface air. (The second point here is more subtle and complex than the first one, because it also needs to take into account an atmospheric density distribution factor to work, but that’s for another day. It is still the main reason why the surface of Venus is so hot.)

This is why Mars’s atmosphere is not capable of warming the surface to a mean global temp above pure radiative equilibrium with the sun (S-B) (rather the opposite), even with 95% CO2 in it. It is far from massive enough …

• on June 27, 2014 at 3:09 pm | Reply DeWitt Payne

Sorry, can’t resist.

The atmosphere thus INSULATES the surface. Energy is not able to escape the surface as fast as it’s coming in before it has warmed to a higher mean temperature than before the atmosphere was put in place.

You do understand that that statement is complete nonsense. No, I guess you don’t. Loschmidt was right about the lapse rate, but for the wrong reason.

• You’re a funny guy, DeWitt. No ‘Mr Know-It-All’ arrogance at all.

You don’t ‘get’ common insulation, is that it, Payne …?

• Not much point me having a discussion with Kristian.

Readers interested in Kristian’s point of view, please review our discussion in Visualizing Atmospheric Radiation – Part Three – Average Height of Emission.

When someone replies to this question:

But just to help me, why not tell me what you think. Same surface temperature, same atmospheric temperature profile, more GHGs: what happens to OLR?

– in the way that Kristian did you realize there is no point.
When Kristian comes forward with a textbook (never), or says “oh, I got that wrong”, then we can pick up the discussion. Until then..

• I’m not trying to have a ‘discussion’ with you, SoD. There’s no point in that, I agree …

When someone like SoD simply doesn’t understand that the question asked is completely confined within the framework of his interpretation of how the world works and then doesn’t like the answer, then any sane person would see there is no point trying to have a normal ‘on-the-ball’ discussion about how the surface of the earth is warmed beyond pure solar radiative equilibrium.

I will have no revelations here, that’s for sure. Only reiterations of purely theoretical concepts never even remotely verified by real-world observations.

• on June 28, 2014 at 9:35 pm DeWitt Payne

Kristian,

Replace SoD by Kristian in the second paragraph and you have the real situation. Once again proving that irony always increases. I believe there’s even a relevant biblical quote:

…first cast out the beam out of thine own eye; and then shalt thou see clearly to cast out the mote out of thy brother’s eye.

Matthew 7:5 KJV.

While I’m on the subject: What’s the physical mechanism that causes an atmosphere to INSULATE a planetary surface when radiation is the only means for energy to flow to space? Oh, wait, I know the answer: An atmosphere that can absorb and emit radiation in the wavelength range emitted by the surface, is more transparent to incoming solar radiation than radiation emitted from the surface and where the temperature decreases with altitude. That’s the greenhouse effect in a nutshell.

• DeWitt Payne,

“What’s the physical mechanism that causes an atmosphere to INSULATE a planetary surface when radiation is the only means for energy to flow to space?”

The mechanism is the radiation being the only means for energy to flow to space. Only the radiatively active gases, the so-called greenhouse gases, can (effectively) cool the atmosphere, by radiating to space. The surface is easily cooled non-radiatively by the atmosphere (low thermal resistance at the surface/atmosphere interface), but only the atmospheric ‘greenhouse gases’ can effectively transfer the gained atmospheric energy to space (high thermal resistance at the TOA). The bulk of the atmosphere (N2, O2) insulates the surface by having a very high thermal resistance to space.

• Edim,

A totally wrong answer. Without GHGs the surface would radiate directly to the space and would be much colder (like 30C colder) than it is,

You wrote about the radiation from the atmosphere, but no radiation from the atmosphere would be needed to make the surface cold in absence of down-welling IR radiation from the atmosphere to the surface.

• on July 1, 2014 at 2:11 pm DeWitt Payne

Edim,

The question was rhetorical, as I didn’t expect an answer from Kristian, and I answered it myself in the comment. Apparently you only read the first sentence.

• on June 28, 2014 at 7:30 pm | Reply Climate Weenie

Kristian,

I tend to agree that Meteorology is run roughshod by some ( not necessarily anyone here ) who exaggerate the effects of radiative forcing.

But I would challenge you reflect upon two truths.

1. Convection tends to cool the surface and warm aloft, but only radiance ( by definition ) can remove the surplus energy received from the sun to restore balance to earth. The only way that convection can help restore balance is if it also invokes some process which changes radiance ( increase albdeo, lower average cloud height, change in water vapor profile, etc. ) as SoD has laid out.

2. ‘Radiation Fog’ occurs when one GHG ( water vapor ) is reduced, allowing more intense surface cooling. What is the corollary of this?

16. Sorry, the first link got deactivated by the asterisk. Here it is:

17. Kristian wrote: [Energy flows] sun >> surface >> troposphere >> out through the ToA.

When temperature is stable, the power flux through each step (>>) needs to be equal. What happens if we slow down the “troposphere >> out through the ToA” step with additional GHGs while all of the others remain the same? Won’t the troposphere warm? Won’t this eventually slow down convection (which depends on the lapse rate) and increase DLR? NET radiation does flow through the system in the direction you indicate, but the fluxes are two-way in all cases. (The flux from the earth to the sun and space to the earth are both negligible, of course.)

• No, I’m talking what IN THE DATA FROM THE REAL WORLD, Frank.

It’s nice to have theoretical armchair hypotheses, ideas about how the world should work. If what they’re claiming is happening can’t be observed in the real-world data, then it’s not science. It’s pseudo-science. An unsubstantiated claim.

What works in the lab doesn’t necessarily work in the large-scale Earth system. The two are not analogous.

You have to check it empirically first. +CO2 >> +T. Tropospheric warming >> surface warming. Where’s the signal?

• Kristian: In the real world, I can heat a pot of water from the bottom with or without a lid to reduce convective loss of heat from the top of the water. Will the presence of a lid change the amount of time it takes for the pot to boil? Or better still, see if a lid effects the equilibrium temperature of the water in the pot when when the heat source adjusted so that the temperature is near boiling.

If you know the answer for a pot of water, you know the answer for the atmosphere. So don’t duck the question I asked: “When temperature is stable, the power flux through each step (>>) needs to be equal. What happens if we slow down the “troposphere >> out through the ToA” step with additional GHGs while all of the others remain the same?”

Fundamental principles of science that can be tested or accurately measured in the laboratory apply to the atmosphere. The molecules and photons involved don’t know whether they are in a laboratory or the atmosphere, they will behave the same in either location under the same conditions. Therefore there is no doubt that a sudden doubling of CO2 will reduce the flux of OLR at the TOA. Like pot of water at equilibrium, slowing heat loss from the top will result in warming below.

Weather forecasts demonstrate that we can make models that do a reasonable job of representing the physics of heat transfer in the atmosphere. The radiative transfer calculations performed by forecasting models tell us how much the temperature will fall each night in the absence of SWR and how much that drop will be moderated by DLR from low clouds. Those forecasts tell us where convection will carry water vapor aloft producing clouds and rain. We also understand that those forecasts start failing about a week into the future because we don’t know the current state of the atmosphere accurately enough. If we make slight changes in the initialization conditions of the model (within the error of our knowledge of those conditions), those small changes will alter next week’s weather forecast.

There are numerous problems with converting weather forecasting models into AOGCMs capable of predicting future climate. Weather forecasting models have been tested against observations, but the important predictions of AOGCMs can’t be confirmed by observations. There are numerous parameters in AOGCMs that can’t be accurately calculated laboratory physics or unambiguously determined from observations. Those parameters dramatically influence feedbacks and climate sensitivity. There is no way for us to know which set of parameters (if any) will produce the best representation of our climate after CO2 has doubled or tripled. Given that we are effectively halfway to 2XCO2, it sure looks like the parameter sets used by the models used by the IPCC are over-estimating feedbacks or under-estimating natural variability. Climate models also fail to accurately reproduce phenomena like the MJO or ENSO. You could, if you want, characterize belief in the predictions of such models as pseudo-science. It makes far more sense to challenge religious belief in projections of CAGW than SOD’s scientific explanation of the greenhouse effect.

“…At the strongest point of absorption – 14.98 μm (667.5 cm-1), 95% of radiation is absorbed in only 1m of the atmosphere….”

A skeptic I debate says in response:
“..Modtran shows that only 4% of the 14.98 micron radiation is lost in 1 meter.. Modtran models 70 km as the TOA. At that point Modtran shows that the 14.98-micron radiation has lost 70% or its radiation.”

Can you comment on the his 4% vs your 95%? Did you maybe mean 10 meters?
Also, for practical purposes, the radiative TOA is somewhere near the tropopause, isn’t it?

• bobmaginnis,

The absorption value at a specific wavelength depends on the pressure and temperature (primarily pressure because this varies so much more than temperature as we go up through the atmosphere).

The calculation I gave is at surface pressure (1000 mbar) and temperature.

At 100mbar the story is quite different, here is a comparison from Grant Petty’s book, reproduced in Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Eight:

Onto Modtran. If your debating friend is using the online Modtran calculator, it’s important to understand that Modtran calculates absorption and emission.

Likewise, if you take a look at the calculations I have produced using a line by line model, say in Visualizing Atmospheric Radiation – Part Two you can see that the 15 μm (667 cm-1) radiance is still “alive and kicking” no matter how far you go up through the atmosphere.

E.g., figure 3:

– where the bottom curve is at 23km, the top curve is the surface.

The atmosphere emits strongly if it absorbs strongly. But it emits at the local temperature of the atmosphere.

To calculate absorption you need to know the line strength, the concentration of the absorbing gas and the path length. Then the calculation is quite straightforward, from the Beer-Lambert law.

Perhaps you can ask your debating partner for what they think is the line strength and their source. Or what Modtran tool they are using and are they sure they are calculating absorption only.

Radiative TOA. Well this is just a reference point for a calculation. If you want to know what the spectrum or the flux is at a given height in the atmosphere you need to state the height (and also you will need to know some other conditions like surface temperature etc).

• Ebel,

Very nice graph, thanks.

• on June 28, 2014 at 3:04 pm | Reply DeWitt Payne

bobmaginnis,

There’s another tool on the web for looking at absorption spectra: http://www.spectralcalc.com. There’s lots more you can do there, but you have to subscribe to do that. Lets look at the transmittance spectrum for a 1 m path length at 400 ppmv CO2 at a total pressure of 1013.25mbar and 296K. It looks to me like the transmittance at the peak is about 1%, not 96%.

Now lets look at MODTRAN under similar conditions, observation looking down, US 1976 Standard Atmosphere, 400 ppmv CO2 and a surface temperature of 296K and 668 cm-1. But instead of just looking at radiance, lets go to the raw data and look at transmittance as well. The units are height in km, radiance in W/(m² cm-1). Transmittance is a dimensionless number with a range of 0 to 1.

0.000 0.449 0.31468
0.002 0.449 0.16111
0.040 0.449 0.00000
70.00 0.221 0.00000

0.002km, 2m, is the smallest increment in altitude that produces a change in the data. We can see two things, a viewing height in MODTRAN of 0m isn’t actually 0m. The second is that the transmittance at 2 m is about half what it was at a nominal 0m. If we convert to absorbance by taking the log of the ratio of the transmittances we get a value of -0.2907. Dividing by 2 to get absorbance/m it’s -0.14535, or a transmittance of 0.716, which is a fair bit smaller than 0.96. Transmittance is 0 to five decimal places at an altitude of 40m. Using the calculated absorbance/m, the actual viewing height at a nominal 0.000km is actually more like 0.0034km.

70km is not the TOA for MODTRAN. It goes to 100km. 70km is just the default observation height. So the skeptic you debate is, shall we say, somewhat unskilled in the use of MODTRAN and spectrophotometry in general.

• MODTRAN parameterizes the absorption of bands, rather than doing a complete line-by-line calculation. The MOD part refers to moderate spectral resolution, just as the HI in HITRAN refers to high resolution. It is averaging over all of the sharp features in the real spectrum. It does a good job overall in a really short time. You can read several SoD articles in the time it takes to complete one line-by-line run.

19. Basically, run a way warming hasn’t happened. I still think that the heat is released, rather than retained. If the models that were run were correct, I don’t think there would be any doubt on my part at all. However, the models have failed. Now, that would mean that the feedback is wrong in the way you envision it. …. Lately, however, and the fact that SoD does present the ideas with evidence, rather than just shouting me down, What if that is correct and there is something else that is as large or larger than the feedback from co2. Going back to 1997/98 ‘all things being equal’ it should be much warmer under that scenario, but it isn’t. CO2 levels have certainty exceed all estimates. So the question becomes why, if your analysis is correct why hasn’t the temperature increased ? Would we be facing much colder temperatures?
I also found Pekka’s comment about the resistor interesting because in electrical systems you have a capacitor as well. Combing the two gives a different graph than just a straight line.

• As Fermi said about extraterrestrials, where are they? That, of course is the problem with invoking mysterious stuff.

• on June 29, 2014 at 12:59 pm | Reply Climate Weenie

It depends on which aspect of ‘the models’ you are referring to.

The problem doesn’t appear to lie with radiative forcing.
The warming we observe in the satellite era is roughly consistent with the RF we calculate.

The problem is that the GHG growth rate ( with which the RF and warming rates corresponds ) is less than the low end scenario of previous IPCC reports.

This low rate can then be exceeded by natural variability more frequently and for longer periods than the high rates.

Global warming is real, but exaggerated.

20. Kristian,

We have a few rules on this blog. One is civility. Another one is not just repeating the same points.

Repetition – it’s frustrating when no one answers your question the way you want it answered. Maybe people have just missed your point or ignored you or haven’t understood what you are really getting at. However, at the discretion of the moderator, continual repetition may be snipped or just deleted to avoid a discussion being hijacked or just made less interesting to other readers.

If you find your future comments are not appearing then you will know why (the amazing insight of your science?)

It’s clear what you think.

I haven’t got much patience for people who have no interest in science, no interest in learning, make ridiculous claims as a substitute for learning and start insulting people when they keep pointing out the basic flaws in your claims.

Proven experimental science is not armchair science because you don’t like it, or don’t like the conclusions it leads to.

Probably time to claim victory and move on to other locations that need your real world science insights.

21. Thanks very much for such a clear and simple run through of the greenhouse effect, very helpful for non-experts to read and to link to.

22. @ Eli Rabett:

“”However, think you missed the most important point with respect to saturation of CO2, that as the concentration increases, the level at which the atmosphere emits in the CO2 bending region rises to a colder level because the emission is trapped higher in the atmosphere…””

Ok, we heard about that. But if there is more emission (and absorbtion!) of IR radiation at this “higher level”, so this level could reach the temperature of the level bevore, level at, lets say 300ppm CO2?

• on June 29, 2014 at 2:16 pm | Reply DeWitt Payne

Yes.

But since we assume that the lapse rate doesn’t change much, the atmosphere below that level extending all the way to the surface must warm as well.

23. […] reading comprehension skills than others. In the end, we each believe what we choose to believe. The “Greenhouse” Effect Explained in Simple Terms | The Science of Doom __________________ 98 Boxster 2.5L BSX #129 PCA National DE Instructor PCA Zone 8 DE/Time Trial […]

24. According to the observations (RATPAC) the trend is positive in the troposphere and almost the same (constant temperature gradient), negative in the stratosphere.

• According to that picture all models predict a significant lapse rate change in the troposphere. The observations shown are not conclusive. They seem to be consistent with the models, but also with no change in the lapse rate. The central values of the observation indicate a smaller change than predicted by the models and also a change that does not extend as high up before being affected by the stronger opposite trend of the stratosphere.

• The lapse rate in the troposphere almost does not depend on the concentration of greenhouse gases. At best, the water vapor content change something – but then that’s not the effect as a greenhouse gas, but the heat of condensation.

In the stratosphere ozone affects, otherwise the observation would be almost vertical.

There is therefore a discrepancy between observation and models, ie the model assumptions are not sufficient or excessive.

• Ebel,

According to the present understanding the negative lapse rate feedback is an essential part of the feedbacks as shown the the picture of this comment

https://scienceofdoom.com/2014/06/26/the-greenhouse-effect-explained-in-simple-terms/#comment-66524

Lapse rate feedback means that the lapse rate is reduced, when surface temperature rises. The effect looks small on the scale of your picture, but that’s an artifact of the scale more suitable for understanding the stratosphere.

25. Thanks guys, but I do not really understand.

The height of the tropopause is already and almost determined by the heat content below. The differences between high and low latitudes are well known.
If the temperature at the tropopause changes by 1K, so the hight varies by about 150m and the temperature at the surface also changes by 1K. The lapsrate remains constant. Or is there a quicker T change at 200 hPa?

• The dry lapse rate is very close to a constant, but the moist lapse rate decreases when temperature rises and the absolute humidity increases. The actual average lapse rate is between these cases. According to the climate models the average lapse rate does also decrease, which is a natural result as it’s affected by both the dry and moist lapse rates.

Decreasing lapse rate means that upper troposphere warms at a fixed altitude faster than the surface. As the moist lapse rate applies mainly to tropics, the extra warming of the upper troposphere is expected to lead to the the “tropical hot spot”. If that prediction is not correct as some observations suggest, that would mean that models fail in calculating those details of circulation that determine the tropical average lapse rate.

• on June 30, 2014 at 6:22 pm | Reply DeWitt Payne

As the moist lapse rate applies mainly to tropics, the extra warming of the upper troposphere is expected to lead to the the “tropical hot spot”. If that prediction is not correct as some observations suggest, that would mean that models fail in calculating those details of circulation that determine the tropical average lapse rate.

That’s not entirely surprising if you think about it. Moisture saturated updrafts are very localized. The updraft area is much, much smaller than the size of the grid blocks used in climate models. Most of the mass of air that goes up comes back down in the same area. The moisture in the updraft condenses and rains out, so the downward moving air is drier. The turbulence associated with air movement causes mixing that lowers the average humidity to less than saturated. Less than saturated means that you can’t use the moist lapse rate, which is 100% RH at all altitudes, or the dry lapse rate. It’s somewhere in between.

Climate models cannot calculate this because they don’t have fine enough resolution. So it’s calculated using simplified approximations, or parameterized in the parlance. As I remember, the parameterizations are based on short term behavior, which is fairly well characterized by observations with weather balloons and other measurements. Basically it’s a radiative/convective calculation where the temperature profile is forced to remain within certain bounds. But it’s not at all clear that long term behavior, years to decades, is accurately modeled.

Some would even argue that you couldn’t calculate it even if you had orders of magnitude faster computers because at very high resolution, the problem ceases to be constrained by boundary conditions and becomes an initial value constrained problem, or a weather model, and chaos, literally, ensues.

• DeWItt: Do cloud resolving models what the environmental lapse rate should be in tropical regions?

• on June 30, 2014 at 10:22 pm DeWitt Payne

Frank,

There are no cloud resolving climate models. Anything with a resolution that fine is a weather model, so it’s not clear that producing the correct environmental lapse rate in a weather model means much for predicting the evolution of the tropical hot spot.

26. on June 30, 2014 at 7:09 pm | Reply Peter O'Donnell Offenhartz

I have a question about SOD’s “note 5” posted back on June 26th. Satellite observations and MODTRAN agree that the effective emission temperature of the main CO2 band is approximately -50C. This is also the temperature of the tropopause at temperate and equatorial latitudes. I assume this means that most emission in this band occurs at or near the tropopause at 10km, i.e., this is the altitude where the optical density of the atmosphere in this band (looking downward from space) is about unity. However, while it is true that

“If the place of emission of radiation – on average – moves upward for some reason then the intensity decreases. Why? Because it is cooler the higher up you go in the troposphere. Likewise, if the place of emission – on average – moves downward for some reason, then the intensity increases (note 5).”

It simply isn’t true that it’s “coooler the higher up you go” when you’re in or above the tropopause, which is apparently where the emission in the main CO2 band occurs.

• on June 30, 2014 at 7:52 pm | Reply DeWitt Payne

If you look at the effective emission temperature for the bottom of the CO2 well for each atmosphere and compare it to the temperature profile for each atmosphere (Show Raw Data), you will find that the temperature corresponds to an altitude below the tropopause in every case. So the temperature is still decreasing with altitude. The emission in the center of the band does come from the stratosphere. But the stratosphere equlibrates rapidly so that very soon after a step change in CO2, the stratosphere will cool enough that the emission from the stratosphere will be equal to the emission before the step change.

• Peter

Here is a graph of calculated values from a typical tropical cloud free atmosphere, originally posted here:

The graph shows the cumulative flux reaching TOA from each altitude. We can see that about 250 out of the 273 W/m2 reaching TOA in this case comes from a height below 13 km. In this case the tropopause will be about 17km.

The reason why the simple calculation (calculating the temperature from the total emission and looking up the height of that temperature) doesn’t work is a significant proportion of radiation comes from the surface and the highly emissive water vapor in the boundary layer within the first km above the surface.

The other way to get a conceptual grasp of the problem – the atmospheric emissivity & absorptivity at the tropopause is relatively low, mainly because there’s (almost) no water vapor there. (I could calculate it if I fired up the old calculations). So there’s no way that the tropopause could be “the place of emission for the climate system”.

• on June 30, 2014 at 10:25 pm Peter O'Donnell Offenhartz

I didn’t say “that the tropopause could be ‘the place of emission for the climate system'”. I did imply that the tropopause could be the place of emission for the CO2 band (not the peak, which arises in the stratosphere). I assume that’s why the effective emission temperature in this band is circa -50C.

• Peter,

Sorry I didn’t read your question properly. In the center of the CO2 band, the emission to space is from the stratosphere. In another part of the band, the emission to space is from the tropopause, and in other parts of the band, the emission to space is from the troposphere.

The CO2 band is very wide.

• on June 30, 2014 at 11:12 pm Peter O'Donnell Offenhartz

Agreed.

• on July 1, 2014 at 1:56 am DeWitt Payne

Peter,

I assume that’s why the effective emission temperature in this band is circa -50C.

Well, no. The temperature is ~220K because from the TOA, nearly all the emission at the bottom of the well comes from the stratosphere, not the tropopause. Here’s a plot of brightness temperature from 600-800cm-1 from two altitudes looking down, 70km and 17km and 400 and 800 ppmv CO2. The temperature at 17 km is 194.8K. The minimum temperature for 17km looking down is 195.20K at 400ppmv CO2 and 195.04K at 800ppmv CO2. Obviously the temperature at the peak of the CO2 band, the minimum temperature changes very little because the radiation orginates very close to the observation height.

For 70km looking down, the brightness temperature at 668cm-1 is 246.99K (36.78km, linear interpolation) at 400ppmv CO2 and 250.15K (38.23km) at 800ppmv CO2.

• on July 1, 2014 at 4:23 am Peter O'Donnell Offenhartz

Alas, I am color blind, and cannot make sense of your plots. In addition I am in the process of moving to summer quarters. I will have a closer look at your data when I am settled.

It is indeed the entire 600-800 cm-1 band that is of interest. My question has to do with the altitude at which the optical density of the atmosphere (looking down) at any wavelength is of order unity, per theory. [This is where the mean free photonic path becomes infinite.] It makes a great difference whether this altitude is above or below the tropical tropopause.

• on July 1, 2014 at 5:41 am DeWitt Payne

I’ve modified the graph to include markers on the lines rather than just color.

It makes a great difference whether this altitude is above or below the tropical tropopause.

Not really. Band wing emission will always originate below the tropopause, as can be seen by the brightness temperature. Emission from the center of the band will always be from the stratosphere if the observation height is high enough.

The reason that radiative forcing is defined at the tropopause is because emission from the stratosphere doesn’t count. It won’t change after the stratosphere is allowed to cool to the new steady state. It looks different in MODTRAN because you can’t change the temperature profile in the stratosphere in MODTRAN. Well, you can, but not in the web implementation that everyone uses because it’s free.

• on July 5, 2014 at 1:12 am Peter O'Donnell Offenhartz

Thanks for putting marking labels on your graph for colorblind people like me.

I don’t think the “effective emission temperature” at lower altitudes is particularly meaningful. After all, we are saying is that at relatively low altitudes the atmosphere is optically thick in the CO2 band, thus reducing the magnitude of the emission. However, at 70 km, the translation between emission intensity and effective emission temperature is real: The emission intensity really does tell us about the temperature of the source emission. According to Pierrehumbert’s book, emission occurs (for any given band) at the altitude where the optical density (looking downward) is of order unity; or, in other words, the altitude where the photonic mean free path approaches infinity.

Insofar as I can tell, in the broad CO2 band, much or most of this emission has a source temperature close to that of the equitorial tropopause. It is otherwise difficult to explain why the measured (and calculated) emission temperature is so low, ca -50C.

• Peter,

First a technical comment. The optical depth of one from TOA to a particular altitude does not tell about an mean free path approaching infinity. It tells directly that the fraction 1/e or about 37% of radiation that enters this layer from below exits the atmosphere. In case of uniform density the mean free path is equal to the thickness of the layer that has the mean free path of one.

It’s true that the effective radiative temperature of radiation well inside the CO2 absorption peak does not change much with concentration, but it does change at the edge of the peak. What happens at the edges of the CO2 peak is, what leads to the radiative forcing. That the effect comes from the edges explains also the approximately logarithmic dependence of the forcing on the concentration.

• The changes to the edges of the CO2 peak are of minor importance. The main effect is where it is already the absorption length is important.

The logarithm function is only an approximation – not a law.

• on July 1, 2014 at 9:06 pm | Reply Climate Weenie

Here’s a great image to visualize various bands and emissions:

http://cimss.ssec.wisc.edu/goes/rt/viewdata.php?product=gcall

the 14.3 um band ( sorry, don’t do wavenumbers ) is toward the middle of the CO2 and notice no clouds are discernable – because the emissions are from the stratosphere.

Stepping away from the center and more and more lower features are apparent.

Sorry, dunno what this appears like to the color blind ( the red-to-yellow-to-blue-to-grey may appear as a mess, but the distinctness of the clouds is the feature ).

• on July 1, 2014 at 11:26 pm DeWitt Payne

Wavenumbers are easy. That’s why they’re often used instead of frequency in Hz or radians/sec. Just convert the wavelength to cm and divide into 1. For example, 15μm is 0.0015 cm. 1/0.0015 = 666.67cm-1.

• on July 1, 2014 at 9:08 pm | Reply Climate Weenie

And the Western view:

http://cimss.ssec.wisc.edu/goes/rt/viewdata.php?product=gw_all

demonstrates the hot, and relatively dry signal, allowing the surface to appear through many of the bands obscured by cloud and water vapor in the east.

27. The brightness temperature of the radiation into space is the result of the average temperature in the absorption length. In the middle of the 15 micron band of the absorption length is very short and only goes up into the warm ozone layer – which leads to the summit. In the diagram, the absorption length is specified as the pressure difference across a vertical layer with the absorption length:

The height of the tropopause depends from the intensity of the heat flux and the concentration of greenhouse gases in the stratosphere (Schwarzschild criterion). That is why at the equator to the higher tropopause.

An increase in the surface temperature by 3 K, an increase of the old temperature levels by about 500 m. In this area, the humidity is higher and the lapse rate is lower. On the other side, above the atmosphere colder and thus the humidity lower, bringing the lapse rate further approaches the trockenadiabatischen value. The mean temperature is almost the same everywhere, whether on water or dry land because of pressure equalization via winds. Even in the height of the local temperature gradient approaches the average value.

Die Helligkeitstemperatur der Strahlung ins All ist die Folge der mittleren Temperatur in der Absorptionslänge. In der Mitte der 15 µm-Bande ist die Absorptionslänge besonders kurz und reicht nur bis in die warme Ozonschicht – was zur Spitze führt. Im Diagramm ist die Absorptionslänge als Druckdifferenz über einer vertikalen Schicht mit der Absorptionslänge angegeben:

Die Höhe der Tropopause hängt von der Intensität des Wärmestroms und der Konzentration der Treibhausgase in der Stratosphäre ab (Schwarzschild-Kriterium). Deswegen ist am Äquator die Tropopause höher.

Eine Steigerung der Oberflächentemperatur um 3 K bedeutet einen Anstieg des alten Temperaturniveaus um ca. 500 m. In diesem Bereich ist die Luftfeuchtigkeit höher und damit der Temperaturgradient niedriger. Auf der anderen Seite ist oben die Atmosphäre kälter und damit die Luftfeuchtigkeit geringerer, womit sich der Temperaturgradient weiter dem trockenadiabatischen Wert nähert. Der mittlere Temperaturgradient ist wegen des Druckausgleichs über Winde fast überall gleich, ob über Wasser oder trockenem Land. Auch in der Höhe nähert sich der örtliche Temperaturgradient dem durchschnittlichem Wert.

MfG

28. on July 7, 2014 at 10:00 pm | Reply DeWitt Payne

SoD,

the inappropriately-named “greenhouse” effect

And again I beg to differ. The planet is just a greenhouse with perfect insulation. A glass covered greenhouse will have a higher temperature during the day than a greenhouse with LWIR transparent windows. How much higher depends on how well insulated it is and the type of glass used. Triple glazed low-e glass will make a very large difference.

• DeWitt: A greenhouse prevent some losses by convection, but the atmosphere does not. Maybe you can say that the stratosphere behaves like a greenhouse, but putting more CO2 into the stratosphere causes cooling. IMO, the name is “inappropriate” because it causes more confusion than enlightenment. A reliable list of useful and misleading parallels between greenhouses and GHGs might be useful.

• on July 7, 2014 at 11:39 pm DeWitt Payne

Frank,

Any enclosed space decreases energy loss from the inside of the space by convection. But unless the walls are transparent to incident solar radiation it can’t be a greenhouse. See Pekka’s comment below.

The stratosphere is a reverse greenhouse because it’s more transparent to LW radiation than to SW radiation. That makes the temperature increase with altitude. The single layer non-reflective atmosphere model works the same way. If the atmosphere layer is more transparent to SW than to LW radiation, the surface is warmer than the atmosphere. In the opposite case, the atmosphere is warmer than the surface.

• on July 8, 2014 at 1:19 am DeWitt Payne

Frank,

A greenhouse prevent some losses by convection, but the atmosphere does not.

Umm, the atmosphere cannot lose energy by convection or conduction to space. The vacuum of space is perfect insulation for conduction and convection.

The efficiency of a greenhouse is highly dependent on the quality of its insulation, but the radiative emission characteristics of the glazing is still significant. For economic reasons, though, most people build greenhouses with the cheapest possible glazing, single layer polyethylene film. It’s only if you’re trying to grow tropical plants in Minnesota, say, (or putting windows on your house) that more expensive solutions are needed.

• DeWItt: I picked the stratosphere because convection is unimportant there, not because it is warmed by ozone. I’ve always thought Pekka’s definition of a GHG was the right one – a gas that interferes more with outgoing than with incoming radiation.

Does a greenhouse interfere more with outgoing than incoming radiation? The glass does block LWR, but an additional critical factor is that the emissivity of the glass is less than the ground. Comparing the emissivity of a solid and a gas gets fairly tricky. I think we learn a lot from throughly understanding the principles behind greenhouses and GHGs. Unfortunately, it is human nature to sometimes take what we have learned about one situation and confidently apply it to another situation without the careful analysis one might use with a totally new situation. Or worse, confidently apply faulty understanding from one situation to another. I appreciate that our host and many commenters take the time to attempt to minimize the dissemination of faulty information and I hope many lurkers can tell the difference.

• on July 8, 2014 at 7:55 pm DeWitt Payne

Frank,

You cannot ignore the spectral properties of the stratosphere to make a point about lack of convection. It’s not lack of convection that makes the temperature increase with altitude. The increase of temperature with altitude causes the lack of convection.

Unless it’s coated, the emissivity of glass in the LW IR is nearly one. Kirchhoff’s Law requires it. A low-e coating increases reflectivity in the LW IR. Since the sum of reflectivity and absorptivity/emissivity must be one, emissivity is reduced in the coating. If the outer surface of the glass is cooler than the interior surface of the greenhouse, then the temperature of the surface inside the greenhouse exposed to the sun must increase compared to a greenhouse with an IR transparent cover, much like the surface of the Earth when CO2 in the atmosphere is increased. This is easily proven experimentally, Wood 1909 notwithstanding.

Because even a single glass layer has a non-zero R value, the exterior surface of the glass will be cooler than than the interior when the sun is shining. A low emissivity coating on the glass will increase the effect.

Obviously, the increase in temperature will not be as much as it would be if the walls of the greenhouse were unable to conduct heat away by conduction/convection to the outside. But it will be increased compared to a greenhouse with an IR transparent cover. Better insulation increases the temperature further.

• DeWitt: To analyze the situation properly, I need to account for the SWR flux reaching the ground under the greenhouse, the LWR flux from the ground to the inside of the glass, the LWR flux from the glass to the ground, thermal flux though the glass, and the LWR radiative loss from the outside of the glass. Unless I’m on the moon, I also need to include the DLR flux from the atmosphere to the outside of the greenhouse. Then I need to solve for three unknown temperatures: ground, inside of the glass, and outside of the glass. To do so, I have energy balance at the ground, the inside of the glass, the outside of the glass, and the law for thermal diffusion through the glass which relies on the temperature gradient through the glass.

How analogous is this situation to the atmospheric GHE? It does remind me of a slab atmosphere models for the greenhouse effect (which are good physics exercises, but too different from reality to avoid misconceptions). WIth the simplest slab atmospheres, you assume both surfaces of the slab have the same temperature and you don’t worry about the flux through the slab. Since you mention R values for glass, I assume that you are concerned about the flux through the glass in a greenhouse.

• DeWitt,

I understand your point. But I’m running with Frank’s point of view (July 7, 2014 at 10:19 pm).

• on July 8, 2014 at 1:45 am DeWitt Payne

They coat the surfaces exposed to vacuum of a glass dewar flask with silver for a reason. The low emissivity of the silver coating plays an important part in reducing the rate of heat transfer to or from the contents of the flask.

Yet more proof, if any was needed, that Wood’s 1909 experiment, regardless of his credentials as an experimental physicist, was badly flawed. Absent all the recent hoopla about Wood’s experiment, I seriously doubt you would refer to the greenhouse appellation as inappropriate. His note, which was promptly rebutted without dispute by no less an authority than Charles Greeley Abbot, never should have been resurrected as being definitive.

• As in ‘..the name is “inappropriate” because it causes more confusion than enlightenment..

Whether or not it should is another question that isn’t very interesting to me. Better to explain how radiatively-active gases absorb and emit radiation than get into discussions about parallels. Just my opinion, not trying to convert anyone.

• When we discuss the GHE of the Earth system, it’s of little or no value to go into the details of an actual greenhouse.

How glazing reduces heat losses is important in designing energy efficient windows, but that’s a very different issue. (It’s, however, significant that my IR thermometer gives a reading of about +18C when I direct it to my windows of triple glazing in winter with -15C outdoor temperature. The heat losses from my house would be very much higher if the windows were transparent to IR.)

29. The fundamental point shared by a greenhouse and the atmosphere is that they let solar radiation to enter more freely than they let energy to escape. That justifies the expression for me, further details are of lesser significance.

• Pekka, this is the nearest definition to a ‘greenhouse effect’ that I’ve read here, but you’ve wrongly included radiative gasses in your dialogue.

This dialogue causes confusion because to “let energy to escape.” doesn’t describe the ‘form’ of energy that is permitted “to escape” when any ‘radiative gas’ that is active in the LW emission frequency eventually loses energy as it escapes Earth’s atmosphere. This is ‘energy transmission’ (whatever the rate).

No. Any ‘greenhouse effect’ effectively ‘confines energy’ within a ‘given boundary’. ‘Radiative gasses’ that are active within the LW spectra are ‘unbounded’ and not ‘confined’. Thus, don’t equate to the criteria for a ‘greenhouse’ label.

However, the ‘change of state’ that H2O undergoes does fall into this ‘greenhouse’ category. I would draw your attention to the atmospheric hydrological cycle, with an emphasis on ‘latency’.

Best regards, Ray Dart.

• on July 8, 2014 at 3:08 am DeWitt Payne

Ray,

Any ‘greenhouse effect’ effectively ‘confines energy’ within a ‘given boundary’.

No, it doesn’t. Energy isn’t confined by either a greenhouse or the Earth’s atmosphere. A greenhouse has energy flowing in and out constantly. The flow in goes to zero when the sun goes down, but the energy flow out never stops as long as the temperature inside is higher than the temperature outside.

• “No, it doesn’t. Energy isn’t confined by either a greenhouse or the Earth’s atmosphere.”

The ‘hydrological cycle’ does/is, and this is where ‘latent heat’ is stored. If you’d read my post in its full context this should be aparent.

“A greenhouse has energy flowing in and out constantly.”

Yes, but not in a ‘latent’ form.

“The flow in goes to zero when the sun goes down, but the energy flow out never stops as long as the temperature inside is higher than the temperature outside.”

Please expand on this! You seem to be discussing ‘radiation’, not ‘latent heat’ bounded by a ‘greenhouse’.

Best regards, Ray.

• on July 9, 2014 at 4:00 pm DeWitt Payne

Ray,

I don’t understand why you think that latent heat is in some way deserving of special consideration. The average water vapor content in a column of the atmosphere is equivalent to a layer of liquid water 2.5 cm deep. Annual rainfall averages 100 cm. The atmosphere cannot lose any energy to space except by radiation. But it does this constantly, so the energy is only contained in the sense that the lake on a river behind a dam is contained. It’s constantly flowing in and out.

The walls of a greenhouse can lose energy by conduction and convection as well as radiation. If the loss is great enough, the temperature in the greenhouse will drop enough to cause condensation of some of the water vapor. That energy is not ‘contained’ by the greenhouse.

• “I don’t understand why you think that latent heat is in some way deserving of special consideration.”

Because ‘latent heat’ is bounded by a ‘greenhouse’, where ‘radiation’ (a less efficient ‘energy transport’ mode) isn’t bounded. Thus, where ‘latent heat’ is active, a greater ‘bulk’ of energy transport is possible.

“The average water vapor content in a column of the atmosphere is equivalent to a layer of liquid water 2.5 cm deep. Annual rainfall averages 100 cm.”

I don’t know where your data comes from, but Trenberth et al offer a best average ‘global’ anual precipitation rate of ~1m (that’s about 1 ton of water per year for every square metre of Earth’s surface). However, this only relates to ‘~residency time’ indication and not a ‘latent heat activity’ indicator as the group suggest. The ‘ice : water : vapour’ total atmospheric content may well indicate the latent heat property of Earth’s atmosphere (this changes diurnally), but precipitation doesn’t indicate this (only the sustainability of H2O in Earth’s atmosphere).

“The atmosphere cannot lose any energy to space except by radiation. But it does this constantly, so the energy is only contained in the sense that the lake on a river behind a dam is contained. It’s constantly flowing in and out.”

Exactly. The ‘greenhouse’ loses heat by ‘radiation’, which is less efficient than convective transport of latent heat. Thus, heat builds its flux density ‘within’ the greenhouse.

The ‘lake, river, dam’ scenario only teaches ‘some’ smoothing of a ‘dynamic’ system. It’s not really a good scenario to show this.

“The walls of a greenhouse can lose energy by conduction and convection as well as radiation. If the loss is great enough, the temperature in the greenhouse will drop enough to cause condensation of some of the water vapor. That energy is not ‘contained’ by the greenhouse.”

No, but the ‘latent heat’ is ‘absent’ outside the greenhouse!!!

When we observe Earth’s atmosphere, the altitude of the greenhouse is high enough to radiate thermal energy from the phase change of latent heat from the lower ‘atmosphere/ocean surface’ directly to space.

Best regards, Ray.

30. SOD, this thread doesn’t make sense. What happened to ‘latent transport’???

Best regards, Ray.

• on July 8, 2014 at 3:12 am | Reply DeWitt Payne

Latent heat transport is still convection. Water vapor must be transported by physical movement of moist air to where it condenses. Correct me if I’m wrong, but I believe that there was a paragraph in the post above titled Convection.

31. “Emission to Space

5. Most of the emission of radiation to space by the climate system is from the atmosphere, not from the surface of the earth. This is a key element of the “greenhouse” effect”

Can you explain this image

http://www.nasa.gov/mission_pages/GOES-P/news/infrared-image.html

• DocMartyn,

In Note 5 I said this:

the “place of emission” is a useful conceptual tool but in reality the emission of radiation takes place from everywhere between the surface and the stratosphere. See Visualizing Atmospheric Radiation – Part Three – Average Height of Emissionthe complex subject of where the TOA radiation originated from, what is the “Average Height of Emission” and other questions.

Note that some of the radiation reaching TOA comes from the surface. The reason is that the absorption in the atmosphere depends very strongly on wavelength. 8-12 μm is commonly called the atmospheric window – for the reason that in this wavelength band the surface emission mostly gets through.

The image you provided is at 10.6 μm – where the atmospheric absorption is very low. That’s why it is a heavily used band – you can “see” the surface quite well.

Here is a spectrum shown in Theory and Experiment – Atmospheric Radiation – originally from the textbook Atmospheric Radiation: Theoretical Basis, Goody & Yung (1989)

– Note the measured and theoretical curves are offset for easier comparison.

Zooming in a section and adding the blackbody emission curves for different temperatures for reference:

It should be clear that surface radiation is mostly getting through in 8-12 μm, but elsewhere the radiation is a lower intensity and therefore is coming from the atmosphere.

In the link cited in note 5 of the article I provided a calculation which broke down the TOA radiation by wavenumber (for one particular temperature and humidity profile):

– Note that the right axis is wavenumber – 10 μm is 1000 cm-1 in “wavenumber” – and the left axis is height in km.

We see that in the “atmospheric window” between 800 cm-1 to 1200 cm-1 the surface transmits almost “straight through” (62% of surface flux makes it straight through to the top of atmosphere in this wavenumber range). A small component comes from around the center of the CO2 band (667 cm-1) from the top layer. The rest mostly comes from the “wings” of the CO2 band and where the water vapor absorption is not so strong, around 400 cm-1.

• So can I take it that when you stated:-
“Most of the emission of radiation to space by the climate system is from the atmosphere,”

What you really meant was that

“Approximately XX%of the emission of radiation to space by the climate system is from the atmosphere, and 100-XX% of the emission of radiation to space come from the land/ocean surface via the atmospheric window”

Calculating 100-XX% is non-trivial for a cloudy, rotating planet with a tilted axis but from the 8-13 um window it is at least 50% and probably up to 70%.

• DocMartyn,

Here are three images taken at three different wavelengths:

You can see the 6.7 μm image looks completely different from the 10.7 μm image. The 6.7 μm is “seeing” water vapor in the atmosphere.

• DocMartyn ,

And the page you linked to provides this information, e.g.:

Pop quiz: why, when CO2’s peak absorption is at 15.0 μm is this satellite channel to measure CO2 at 13.3 μm?

• “Pop quiz: why, when CO2′s peak absorption is at 15.0 μm is this satellite channel to measure CO2 at 13.3 μm?”

There is essentially no absorbance of H2O after 12.5 μm, but the CO2 peak at 14.5 μm is saturated. So you need to be somewhere between 12.5 and 14.5μm.

• Determining the precise share of OLR emission that originates from the surface is quite difficult. The 2009 paper of Trenberth, Fasullo, and Kiehl: Earth’s Global Energy Budget gave the estimate of 40 W/m^2 or 17% while a later estimate of Costa and Shine is only half of that with an uncertainty of 20%. The upper limit is thus 10% of OLR. 90-93% of OLR originates in the atmosphere according to this estimate.

• I am a big fan of evolution and just love how things change with time.

In an eyeblink, just 7 years, the Earth energy budget has evolved

Love how Wild et al., just give up on the atmospheric window;

A mature field with all the fundamentals known to 3 standard deviations and physics all solved.

• DocMartyn,

I’m not sure what point you are trying to make.

Hopefully you are now clear why the 10.6 μm image of the earth shows most features of the surface whereas the 6.5 μm image shows the atmosphere?

On the atmospheric window, it is a curiosity value, that is, the value itself is not used in any important calculations. The subject is covered in Kiehl & Trenberth and the Atmospheric Windowhow much radiation escapes to space through the “atmospheric window”, why its value isn’t that important but the complete story anyway.

It’s also covered by a paper which is cited in the above article: Outgoing Longwave Radiation due to Directly Transmitted Surface Emission, Costa & Shine (2012). They note that previous estimates have been ad hoc and make an attempt to calculate this curiosity value.

I looked at your graphic of energy balance progression.

Here’s what Kiehl and Trenberth said in their 1997 paper:

We know everything accurately to 3 standard deviations

Oh no, they didn’t. They said things like:

..The values from our study are listed in Table 1 and are discussed in more detail in the following sections. There is considerable variation for any given flux of energy. For example, values for the net surface shortwave flux range from 154 to 174 W m−2..

..Mean values of the total solar irradiance have varied in different satellite missions from about 1365 to 1373 W m−2 (see National Academy of Sciences 1994 for a review; also Ardanuy et al. 1992)..

..We do not explicitly include the effects of aerosols in the shortwave budget calculations because aerosol optical properties vary greatly due to chemical composition. Thus it is problematic to include them in a global budget..

..Our surface shortwave absorbed flux of 168 W m−2 agrees quite well with the majority of values near 170 W m−2 in Table 1. Recently, results from three observational studies (Cess et al. 1995; Ramanathan et al. 1995; Pilewskie and Valero 1995) suggest that clouds may absorb significantly more shortwave radiation than is accounted for in model calculations (such as the models employed in the present study). These results suggest that the cloudy sky absorption may be approximately 20– 25 W m−2 greater than models predict..

..Gleckler and Weare (1995) estimate zonal mean errors in bulk latent heat fluxes of at least ±25 W m−2, and a large portion of this is likely to be systematic (arising from the exchange coefficient, and biases in surface wind speed, moisture gradients, and sea surface and air temperatures..

• I’m with you Pekka. Latent heat doesn’t have any prefference for temperature, only its environment. It just ‘adds/subtracts to it’ (dependant on the environmental scenario).

Best regards, Ray.

32. , SoD

Just for the record I will ask you one last time.

Where did you get the idea that Gerlich & Tscheuschner said that the warmer Earth cannot absorb radiation from a colder atmosphere?

“see Kramm & Dlugi On Illuminating the Confusion of the Unclear – Kramm & Dlugi step up as skeptics of the “greenhouse” effect, fans of Gerlich & Tscheuschner and yet clarify that colder atmospheric radiation is absorbed by the warmer earth.”

Perhaps it is because you don’t know the difference between energy and heat!

The G&T paper is freely available on the web.

You were not the only one to misread their paper.

Eli Rabett made the same mistake and went on in his silly paper to say that HEAT travels from the colder atmosphere to the warmer Earth.
To make this error it is about as gross as it gets. .

In their reply to Rabett et al G&T make it quite clear that the radiation is absorbed.
Whats so difficult to understand about energy transfer between a hot and a colder object……

energy can be transferred in both directions
Heat can only be transferred in one direction always from hotter to colder object.
If you find that difficult to grasp then this simple reminder will keep you right.
Heat has the ability to do work in the given situation.

.
.

• Bryan,

The differentiation that you make is wrong. When atmosphere emits radiation, the energy is taken from the heat of the atmosphere. When that radiation is absorbed by the surface it adds to the heat of the surface.

When we have a process that takes heat from one place and adds it to another, it makes sense to say that heat is transferred.

Classical thermodynamics is a formal theory that considers heat only on the level of net heat transfer. When discussion is restricted in that way, heat is always transferred from warmer to colder, but today’s physics knows more and allows for discussing further details. Classical thermodynamics is a very restricted theory, it does not make sense to exclude all the additional knowledge. It’s a very unfortunate habit of many skeptics to claim that what was known more than 100 yeas ago should be taken as more reliable than the present knowledge forgetting all the theoretical development and all the evidence collected since those pioneers of science.

The paper of G&T has been discussed sufficiently before and found totally worthless. Most of that material is still openly available. Is that not enough?

• Pekka says that heat is…….

“always transferred from warmer to colder,”
In classical physics

” but today’s physics knows more and allows for discussing further details.”

Perhaps you could point the readers to some ‘new’ physics textbook that claims that heat can be spontaneously transferred from colder to warmer objects.

I have not come across any and I doubt you have either.

It looks a bit suspicious that the ‘greenhouse theory’ advocates seem to depend on ‘new unpublished physics’ to support their conjecture.

You have yet to show that anything in the G&T paper is incorrect and to dismiss it as worthless is merely empty rhetoric.
Now as the pause in global temperatures rise is almost 20 years, despite increasing atmospheric CO2 , perhaps a little more skepticism would be helpful .

• Bryan,

All (or at least almost all) textbooks of heat transfer tell about the present way of considering physics. SoD has presented numerous copies of such textbook pages on this site.

That G&T is incorrect has been shown so many times that doing it once more is totally pointless. SoD has done it here. I wrote a couple of comments about that at Climate Etc, similar evidence has been presented on really many other web sites as well.

• Pekka says

“All (or at least almost all) textbooks of heat transfer tell about the present way of considering physics.”

That should be easy then, to supply the readers with the name of ONE physics textbook to answer my question above

“Perhaps you could point the readers to some ‘new’ physics textbook that claims that heat can be spontaneously transferred from colder to warmer objects.”

I don’t think you can!

• Bryan,

I’m sure that you know, what I’m talking about as you were the first commenter on this thread

https://scienceofdoom.com/2010/10/07/amazing-things-we-find-in-textbooks-the-real-second-law-of-thermodynamics/

If you cannot see, how the examples prove you wrong, it’s your problem.

• Readers will note that Pekka cannot name one physics textbook that states
that heat can be spontaneously transferred from colder to warmer objects.
No amount of equivocation on his part can hide the dead end that he has arrived at.

How can anyone take seriously a conjecture that relies on the the impossible spontaneous transfer of heat from a colder to a hotter object?

Now just in case any reader thinks that Pekka’s viewpoint is seriously considered let me assure you that;

Every physics textbook and physics department on this planet unanimously agree with Clausius that heat cannot be spontaneously transferred from colder to warmer objects.

• Welcome back, Bryan! No discussion of the inappropriately-named “greenhouse” effect would be complete without your much loved self-parodying style.

In concise summary I say that the surface of the earth absorbs energy from the atmosphere and this must change its temperature compared with if it did not absorb it. Bryan says this cannot happen but will never say which bit is wrong:
1. absorption of energy from a colder atmosphere by a warmer earth
2. energy not lost or destroyed (1st law of thermodynamics)
3. energy increasing internal energy (“”)
4. increase in internal energy changing temperature

Bryan has maybe forgotten that I scanned the pages of 6 heat transfer textbooks demonstrating that radiation from colder bodies was absorbed by warmer bodies.

Here is the article: Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamicssimple but necessary – pages from six heat transfer textbooks to confirm the stuff that so many people dispute.

Here is a quote from Bryan in that long fascinating exchange:

SoD

I think its you who’s having a laugh.

You must have searched through hundreds of textbooks to find the few above who have been a bit careless with their thermodynamic definitions.

I think the authors would be appalled at your use of their lack of clarity namely to prove that heat flows from a colder object to an object at a higher temperature.

What happened to the 99.999% of textbooks which give correct definitions that you must have discarded to pick your chosen sample.

We notice that there are no Physics Textbooks in your sample.

Why do you think that is?

I don’t know what you hope to gain by misleading people who might not know enough to see through a bogus scam..

Interested readers should follow the whole thread through, including the book put forward by Bryan.. well, it brought a smile to my face..

Of course, if the radiation from the colder body was absorbed by the hotter body (as Bryan now “clearly states”) then people who believe in the first law of thermodynamics would expect it to change the temperature of that body compared with if that energy was not absorbed.

And thus, interested readers should follow what Bryan says about that energy. And his enthusiasm for “answering” the question of what happens to that energy when absorbed. Specifically, what is the equation for temperature change? Why so difficult to get an answer for a simple equation.

I only know one relevant equation for temperature change of a body with a specific heat capacity,cp and a mass, m. It says if a body emits 5J and gains 3J then we can calculate the temperature change, ΔT = (5-3)/m.cp

In the case where the 3J comes from a colder body where does that leave Bryan? And so, we painfully come to realize, this is why the question never gets answered by Bryan. No equation for temperature change is ever provided.

I assume that Bryan does not know the equation for temperature change of a body. I have pressed Bryan at length many times on this subject, e.g., in this question and this question.

No answer on the equation for temperature change.

Instead, lots of implication that our ideas somehow violate the second law of thermodynamics without actually explaining how.

Then we look back a little further in history on this blog we find something else. Despite his apparent current viewpoint, Bryan seemed convinced that the warmer body could not absorb energy from the hotter body, or that the amount was wrong, or “it depends” but never sure exactly why..

Scienceofdoom and others think that the hot surface has no option but to absorb a photon from the cold surface.
I think a lot of this radiation is in fact scattered from the hot surface and is not absorbed.

Proving the “backradiation” if it exists is very smal and certainly nothing remotely like 300W/m2.

In this comment I tried to pin Bryan down on a very specific point:

When 10um radiation from a -10′C body reaches a 0′C surface how much of this 10um radiation is absorbed compared with the scenario of 10um radiation from a +10′C body.

As usual he was confused:

..Well again it depends on the surface..

So no wonder we are all confused about Bryan.

The second law of thermodynamics is neatly confronted via the equation for entropy.

The Three Body Problema simple example with three bodies to demonstrate how a “with atmosphere” earth vs a “without atmosphere earth” will generate different equilibrium temperatures.

Here is an extract:

So if we take bucket A full of water at 80°C and bucket B full of water at 10°C, Science of Doom is saying that bucket A will heat up because of bucket B? Right! That’s ridiculous and climate science is absurd!

Yes, if anyone was saying that it would be ridiculous. I agree. To take one example from many, in The Real Second Law of Thermodynamics I said:

Put a hold and cold body together and they tend to come to the same temperature, not move apart in temperature.

Of course, it could be that I am inconsistent in my application of this principle..

– so instead I pose a problem of calculating entropy in the case where the atmosphere does change the temperature of the earth compared with a no atmosphere case.

And I ask for readers who – like Bryan – think it is wrong. Just do the entropy calculation and demonstrate that my example has violated the second law of thermodynamics.

Bryan has not yet provided his proof, despite his many comments on that article.

I assume that he doesn’t know what entropy is.

I assume that all the people who are convinced the “greenhouse” effect is a violation of the second law of thermodynamics don’t know what entropy is either.

Otherwise they could just show up and point out where I made a mistake in my entropy calculation and the whole thing would be sorted out.

• SoD,Pekka and De Witt promote alarmist end of world scenarios.
There’s nothing new in that.

http://abhota.info/end2.htm

The trouble is by mutual self support they have created for themselves (here on this site) an alternative universe where heat (they claim) is spontaneously transferred from colder to warmer objects.

Any careful reader will note that SoD has not answered my question about the basis of his mistaken interpretation of the G&T paper.
I cannot see much point in SoDs comments since they are based on a mistaken interpretation….. they are as GG pointed out in his reply ….’vacuous’.

I had a chemistry teacher who thought that mass and weight were the same thing.
Indeed chemists still use the term atomic weight.

Now if it was important for the greenhouse theory that mass and weight were identical then SoD could no doubt find several chemistry textbooks where a sloppy writing style might give this impression.

If you want to know the correct definition of words like heat,energy,mass and weight you must look up a physics textbook.

Surely we can all agree on that!

Thats why SoD and Pekka are now ‘hung up’ and ‘out to dry’

Because they cannot name even one physics textbook that supports their interpretation that photons from a colder object absorbed by a warmer object can be called heat.

The more intense and wider frequency photon stream from the hot object cannot be separated from the energy stream from the colder.
Heat transferred is the difference between the two streams and is always from hotter to colder.
Zemansky summed it up rather well in his classic textbook ‘heat and thermodynamics’ when writing about radiative transfer.

‘The difference between the energy absorbed by a body and that emitted is called heat and is always from a higher to a lower temperature.’

• Photons are not heat, photons get their energy from heat and their energy becomes heat when they are absorbed. That’s, how they transfer heat.

Classical Thermodynamics is an axiomatic mathematical theory that describes well a limited set of phenomena. It was originally not derived from anything; it’s an ad hoc theory that gets its justification as physical theory from the agreement with observations in its limited range of applicability. It’s even presently a very useful theory, but it’s not the most fundamental theory of physics as it can be derived from more fundamental theories through statistical mechanics with some help from quantum mechanics. The deeper nature of heat is also described by the more fundamental theories.

Arguing against the more fundamental theories based on the very limited Classical Thermodynamics and its axiomatic definition of heat is ridiculous.

The semantic question of what we should refer to by the word heat would be irrelevant in absence of the further violation of logic based on picking one definition to state that heat can flow only from hot to cold, and then switching to another definition to “prove” explicit falsehoods about real physics.

• Why is the nature of heat transfer discussed mainly in textbooks of heat transfer rather than in textbooks of physics? The basic answer is simple

– It’s a central question for heat transfer, not for general physics.

Another essential point is that

– Discussing heat transfer in both directions separately rather than the net effect only is of interest almost solely for radiative heat transfer.

What physics textbooks write about radiative heat transfer is in full agreement with what textbooks of heat transfer write, but few of the physics textbooks go even as far as calculating the radiative heat transfer between two bodies. Physics textbooks discuss, how a surface or gas absorbs, transmits, reflects, or scatters radiation. They tell also what’s the source or sink of the energy of the photons. How to use that knowledge in calculation of heat transfer is left to the textbooks of heat transfer.

• on July 15, 2014 at 2:18 pm DeWitt Payne

Bryan,

SoD,Pekka and De Witt promote alarmist end of world scenarios.
There’s nothing new in that.

Weak attempt at deflection by creating a straw man. Please cite an example. Simply stating that there is a greenhouse effect and that, all other things being equal, increased CO2 will cause an increase in temperature doesn’t qualify as either alarmist or an end of the world scenario.

As for the rest, please show in detail how any of what we’ve written results in an actual violation of the Second Law by causing the entropy of the system to decrease using equations and numbers not hand waving, incorrectly, about heat.

Speaking of which:

You keep using that word. I do not think it means what you think it means.

Inigo Montoya, The Princess Bride

33. on July 14, 2014 at 6:20 pm | Reply DeWitt Payne

SoD and Frank,

To continue flogging the deceased equine: If ‘greenhouse effect’ is an inappropriate name because you think it might be confusing somehow, please suggest some other name that would make sense to the general public. I can’t think of anything.

DeSaussure called his glass covered insulated box a hotbox. Somehow, the hotbox effect, doesn’t have the same ring. Not to mention that it’s unlikely that anyone but a student of the history of science would get the reference.

• DeWitt,

I like the name: Inappropriately-named “greenhouse” effect
It’s snappy, concise, and straight away tells everyone what you stand for.

Alternatives: The phenomenon previously known as the greenhouse effect
Slightly more uptempo pop reference.

The effect of radiatively-active gases in moving the emission of radiation to space to a higher colder altitude
Losing its snappiness slightly.

If you had discovered it we would have called it: The DeWitt effect

Most physics phenomena doesn’t have a description, instead they are named after the person who got their publication out slightly ahead of (or after but with more fanfare and panache) the person who really discovered it.

• DeWitt

Please note that Pekka in effect now concedes that no physics textbook can be found where the definition of heat will allow heat to be transferred spontaneously from a colder to a warmer object.

If you wish to check the second law and entropy statements there are several freely available on the web.

Here is a convenient one

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw2.html#c2

• Bryan,

Do you now tell that you agree fully with textbook physics, and that you have no objection to the standard description except that you wish to use the word heat only as it’s defined in Classical Thermodynamics?

If that’s the case, we can all end this discussion noting that the only disagreement is on the semantics.

• Pekka

I always try to keep within the framework of standard physics.
My use the word heat is in line with any physics textbook that I know of.

Of course anyone can redefine a word and with a full explanation of what they mean by the word they can be understood.
However using words like heat mass energy force and so on in an unconventional way (without explanation) can only lead to confusion.

I cant see any point in that approach .

“If that’s the case, we can all end this discussion noting that the only disagreement is on the semantics.”

• on July 15, 2014 at 9:07 pm DeWitt Payne

Bryan,

Of course he does. No one ever said it did. You stand the logic on its head to claim that we do. There is nothing in the physics of the greenhouse effect that violates the Second Law. Heat flows from the hotter Sun to the colder Earth. Heat then flows from the Earth’s surface to the colder atmosphere by radiation and convection and to space from the atmosphere and from the atmosphere to space by radiation.

But, the temperature of the surface of the Earth is dependent on the temperature of the atmosphere. If the temperature of the atmosphere increases, so must the temperature of the surface of the Earth to maintain the flow of heat out to match the flow in from the Sun. But heat continues to flow from warmer to colder, never the other way around.

The simplest example is two infinite parallel planes separated by a vacuum at temperature T1 and T2. If T1 = T2, then no heat flows, by the classical definition of heat. If a flow of heat is imposed between the two planes, then the temperatures are not independent. I can’t find the character Q with a dot over it to represent heat flux so I’ll use F and ignore emissivity.

F = σ(T1^4-T2^4)

If F = 0, then T1 = T2

If F ≠ 0 then T1 = (F/σ + T2^4)^¼

If F is positive then T1 > T2

Increasing T2 therefore must increase T1, but heat never flows from the plane at T2 to the plane at T1. And nobody ever said it did. The heat flowing into the plane at T1 causing the flux F between the planes is the source of the temperature increase of T1.

• All Bryan is doing with his very poor job on sophistry is pointing out that textbooks say E1-E2 > 0 – which everyone else states over and over, no one is confused, we all agree (note 1).

Byran tries to pretend we are saying E1-E2 can be any value, including < 0.

We are not, but you have to sympathize with Bryan, misrepresentation is his only way out.

Unfortunately, apart from the people unable to add up, Bryan’s problem is clearly obvious.

If E1 = 5, E2 = 3, what is the loss in energy by the hotter body?

I say 5 – 3 = 2

Bryan says 5 – 3 = 5

Note 1: E1 = energy emitted by the hotter body; E2 = energy absorbed from the colder body by the hotter body

According to Bryan, the first law of thermodynamics is wrong.

Energy disappears after being absorbed.

If Bryan was interested in physics he would explain this point. If G&T were interested in physics they would explain this point.

Expect more bluster.

• DeWitt

Only some of the proponents of the greenhouse theory make the mistake of violating the second law.

For instance in the comic paper by Halpern et al one section is devoted to saying the greenhouse theory does not violate the second law.
To illustrate they show HEAT moving from colder atmosphere to warmer Earth surface.
They did not say that their use of the word is in any way unusual not realising that they proved the the very point that G&T were making.

Now to make your points without making the gross mistake of getting your definitions mixed up is not too hard if you have been educated properly.

I recommended that SoD read ……

Thermal Radiation heat Transfer, by Siegel & Howell (freely downloadable set of files on the NASA , 3 volumes 1968-1971, NASA Ref SP-164. Check http://ntrs.nasa.gov/search.jsp

Or the lecture notes by Rodrigo Cabellero often recommended by yourself, again freely available.

These authors can make their points without getting their definitions confused.

The Halpern paper in effect proposed that the atmosphere was like an electric blanket with its own heat source – the greenhouse gasses.

Cabellero on the other hand was describing something more like a blanket or the insulating effect of the atmosphere.

The Cabellero approach is much more plausible .
But plausible does not mean it is how the actual atmosphere works but that is a whole new topic.

SoD obviously does not intend to give evidence to support his mistaken interpretation of G&T in that they believe that radiation from a colder body cannot be absorbed by a warmer one.

Instead he seems much more comfortable attacking the bogus proposal that he thinks they should have said.
He has dug a massive hole for himself and yet keeps digging!

• Bryan,

You should finally understand that there’s nothing unusual (let alone wrong) in discussing heat transfer in the way textbooks of heat transfer do. You may prefer the definition of Classical Thermodynamics, but you have no justification for expecting that others would do the same. You are not the highest authority on the meaning of words. There’s nothing “comic” in the discussion of Halpern et al. There’s nothing contrary the the standard use of the words. They do also explain in detail and correctly, what the issue is.

The different energy fluxes of the Earth system can be grouped in many different correct ways. In some of these ways the downwelling radiation forms it’s own component, and is then the largest energy flux that heats the surface. That’s a correct description. Another correct description combines first the downwelling and upwelling IR fluxes as a single net flux that’s upwards.

Both of the above alternatives are correct, and neither can be used as an argument against the correctness of the other.

You may have read correct texts of atmospheric physics, but it’s clear that you have not understood them.

You make the assertion

The Halpern et al paper in effect proposed that the atmosphere was like an electric blanket with its own heat source – the greenhouse gasses.

It seems obvious that you haven’t understood at all, what they write. Please tell, where their argument is based on an own heat source for the atmosphere. (It’s clear that there’s nothing like that in their argument, that’s just a strawman presented by you.)

Debunking a paper that lacks internal logic as the G&T paper does, is somewhat problematic. To me the most serious fault of the G&T paper is that it does not even try to justify it’s conclusions. It just declares them implying that they follow in some miraculous way from the matter covered by the paper, while no such connection is presented or is possible to present. (It’s not possible, because the conclusions are wrong.) It’s difficult to make specific arguments against something that does exist, it’s only possible to observe that the conclusions lack all justification. If someone disagrees, he must be able to tell step by step, how the logic works, when the paper doesn’t even try. Halpern et al find some specific erroneous details, but those are not the worst problem of the paper, only the beginning.

34. on July 14, 2014 at 11:56 pm | Reply DeWitt Payne

Most physics phenomena doesn’t have a description, instead they are named after the person who got their publication out slightly ahead of (or after but with more fanfare and panache) the person who really discovered it.

In that case, we should call it the Arrhenius Effect, as he was the first one to propose that ΔF = αln(C/Co). Or perhaps the Fourier Effect as he was the first to publish that the Earth’s surface was warmer than it should be, given its distance from the sun and likened the process to de Saussure’s hotbox.

• on July 15, 2014 at 12:02 am | Reply DeWitt Payne

According to Wikipedia, Alexander Graham Bell was the first to use a greenhouse as an analogy.

• John Nielsen-Gammon wrote a series of posts arguing for the name Tyndall Gas Effect.

Needless to say, greenhouse effect remains the name in use.

• Pekka

“Please tell, where their argument is based on an own heat source for the atmosphere”

Because it claims it emits heat to the warmer Earth surface.

Its a bit tiresome going over the same points again
We have already agreed that the second law forbids this.
This is not within the framework of physics.

You have failed to provide a physics textbook to support your claims.

You seem to accept sloppy descriptions from Halpern because ‘you have guessed what they really meant to say’

The Halpern comment was about the second law, so sloppy argumentation is not permissible.

Now in certain contexts convenient labels may be given.
For instance the use of the term centrifugal force is often used.
Its quite all right in a certain context.

However if the discussion is about real forces on a satellite (say) then to say that centrifugal force is a real force rather than pseudo force would be mistaken.

You say
“To me the most serious fault of the G&T paper is that it does not even try to justify it’s conclusions. It just declares them ”

Yet you do not give any examples, you just make a declaration!

• That atmosphere emits radiation requires only that it’s heated by some mechanism. All climate scientists agree that the main source of energy is solar radiation, which heats mainly the surface (or ocean below the surface). The atmosphere is heated mainly by the surface by all forms of heat transfer (mainly radiation, convection and latent heat transfer). There’s nothing that contradicts that in the paper of Halpern et al or in any of the posts or comments of SoD, DeWitt or myself.

Your claim on the above is totally false and lacks all justification.

The role of various textbooks has been explained to you tens of times.

Actually I just read the reply of G&T to the Halpern et al paper. In that reply they tell that their original paper was not supposed to be anything more that a debuttal of false simplifications. I.e., they write in that reply that they haven’t even tried to claim that there’s something wrong in the standard theory itself, only in some simplifications selected by themselves. It’s funny that they say in their reply that they have never claimed that the standard theory is false.

Every simplification is false in some obvious way – otherwise they were not simplifications. They cannot be judged on that basis alone. They should be judged on their educational value. I would agree that some of the commonly presented simplifications have not been particularly good on that basis as they may lead to false ideas about the GHE.

The way the paper of G&T is commonly understood is, however, not in agreement with their reply.

• Checking the claims of the original paper of G&T as stated in the abstract we find:

By showing that (a) there are no common physical laws between the warming phenomenon in glass houses and the fictitious atmospheric greenhouse effects,

no common physical laws is too strong, but otherwise I might agree on that

(b) there are no calculations to determine an average surface temperature of a planet,

That’s clearly against the indisputable facts, as it’s easy to find very many such calculations, how accurate they are is another question discussed recently on this site as well.

(c) the frequently mentioned difference of 33 C is a meaningless number calculated wrongly,

Again too strong. The number is based in part on arbitrary assumptions concerning the albedo, but their claim is highly exaggerated and misleading.

(d) the formulas of cavity radiation are used inappropriately,

I don’t think that formulas of cavity radiation are used at all. Thus it’s difficult to see, how they are used inapproriately.

(e) the assumption of a radiative balance is unphysical,

The assumption of radiative balance of the Earth system as whole is highly physical within some limits, otherwise balance of radiation is important for some subsystems like the stratosphere, but less for others

(f) thermal conductivity and friction must not be set to zero, the atmospheric greenhouse conjecture is falsified.

Thermal conductivity is insignificant in most cases, but important in some, most importantly the skin of the ocean.

The final comment lacks justification in the paper. The same error applies to the chapter of conclusions.

In this final comment they present the main conclusion picked from the paper. That’s, however, the particular conclusion that’s presented without even attempt to justify it, as the paper does not study the actual existing theory of atmospheric physics, it finds errors in the strawmen built by the authors.

• on July 16, 2014 at 1:33 pm DeWitt Payne

Bryan,

Are you denying that the atmosphere emits radiation?

If not, then are you denying that the surface of the Earth absorbs about 98% of that radiation?

If so, what then happens to the radiation that is reflected by the surface and why isn’t it measured by spectrophotometers, IR thermometers and pyrgeometers?

Pekka,

G&T’s argument about glass greenhouses is based almost solely on Wood, 1909. That note flew in the face of all previous experimental work starting with de Saussure. You can tell G&T are theoretical physicists because anyone with a strong experimental background would see all the flaws in the experimental design and the inadequate description of the experiment itself. The main flaw being that the covers were not switched between boxes to see if the boxes were identical with regard to insulation and thermometer placement.

The paper was rebutted shortly thereafter and Wood did not defend his results or conclusions. Therefore, G&T’s statement that there are no physical laws in common between the atmospheric greenhouse and a glass greenhouse was not proven.

In fact, de Saussure’s multi-layer hot box works much the same as the atmospheric greenhouse effect, or at least the gray atmosphere simplification, complete with a temperature difference between the glass layers.

G&T’s reference to the cavity radiation formula means the Planck equation for the emission of a black body. A Hohlraum, or cavity radiator, is the best way to approximate a black body. By incorrect, I’m pretty sure they mean using the Stefan-Boltzmann equation. That equation assumes constant emissivity at all wavelengths, which isn’t true of the atmosphere and is only a good approximation of the surface. But, as usual, this is a straw man as the S-B equation is only used to calculate an effective temperature from the integral of the actual emission spectrum. The effective temperature is then just another way of stating the total flux.

35. DeWitt Payne says
Bryan,

“Are you denying that the atmosphere emits radiation?”

No

“If not, then are you denying that the surface of the Earth absorbs about 98% of that radiation?”

Not sure about that particular detail.

“If so, what then happens to the radiation that is reflected by the surface and why isn’t it measured by spectrophotometers, IR thermometers and pyrgeometers?”

DeWitt this is about the best summary of what pyrgeometers measure.

• on July 22, 2014 at 3:08 am | Reply DeWitt Payne

Bryan,

That article isn’t very good. It’s clearly written to confirm your incorrect opinion that pyrgeometers do not in fact measure integrated radiant flux over the wavelength range allowed to reach the detector. They do. But they measure it by difference rather than as an absolute because it isn’t possible to have a reference temperature of 0K. If the reference mass could be maintained at absolute zero, a pyrgeometer would, in fact, measure absolute total radiant flux.

A pyrgeometer or an IR thermometer measures heat flow, in the classical thermodynamic sense of heat, from or to a high thermal conductivity mass of known temperature through an insulating layer using the voltage from a thermopile with junctions on both sides of the insulating layer as the measure of the temperature difference across the layer. Any heat flow through the layer will cause a temperature difference across the layer and thus the voltage to be different from zero. It will be positive or negative depending on whether heat is flowing in or out. The greater the difference in effective temperature, the greater the heat flow. If the voltage is zero, then there is no heat flow and the radiant flux of the field of view impinging on the detector is the same as the flux leaving the detector. The device is calibrated using a cavity radiator or hohlraum, which is an extremely good approximation of a black body.

You like to focus on the negatives of pyrgeometers, which, like any other instrument, are not perfect. But IR spectrophotometers give results that are in agreement with results from pyrgeometers and with theoretically calculated spectra within experimental error. IR thermometers work on exactly the same principle as a pyrgeometer. IR thermometers can correctly measure temperature when the emissivity of the object being measured is known, so there is no good reason to doubt the validity of pyrgeometric measurements.

So yes, Virginia, the atmosphere does radiate in the thermal IR. Pyrgeometers, IR thermometers and IR spectrophotometers all measure this radiation, as they would measure the radiation from a solid object, and the results agree with theory. The difference is that IR thermometers and pyrgeometers integrate over a large spectral range while IR spectrophotometers resolve the distribution of the radiation by frequency or wavelength.

Since atmospheric radiation does vary strongly with wavelengths, if it were reflected by the surface to any significant extent, it would be obvious from the spectrum of radiation observed from the surface. But it isn’t observed because, for most of the Earth’s surface, reflection is insignificant.

• DeWitt,

Emissivity suddenly comes into sharp focus when members of the illuminati come to realize that the emission of radiation from the climate system can’t possibly be the same as the emission of radiation from the earth’s surface. After all that talk about averaging temperatures not working, it turns out that you don’t need to average anything, only sum the energy. There is no flawed assumption.

And if the emitted radiation by the climate system is much greater than emitted radiation from the surface, that’s actually the “greenhouse” effect. It can’t be, it’s forbidden by theory..

Damn, what to do?

Unless.. the emissivity was much lower than anyone said.. that must be it.

Ah, I bet no one has measured it, they just assumed it. Oh, measurements. Lots of measurements.

Ah, they must be based on a flawed instrument.

Ah ha, the instrument relies on physics theory (unlike say “thermometers”, “magflow meters” and all instruments in use in process plants which magically convert the measured parameter into an output without relying on “physics theory”).

You can’t make this stuff up.

I questioned a few things – like here:

..Of course, many people do tedious experiments and write them up in dull papers. But why bother reading them? Remember just a few weeks of writing blog articles can save minutes of research time.

What is the source of the first graph? The one with the footnote that doesn’t match the title?

For anyone with a passing interest in actual science the fact that contributor ‘Max’ has found a graph with a totally different value from decades of study by hundreds of experienced researchers would be worth checking..

And then:

..A person with a spreadsheet can write any value they want for thermal emission of radiation from the surface and then balance it with a downward component from the atmosphere.

What has Wayne Jackson demonstrated? That if you type an invented value of upward emission in and then divide it by the Stefan-Boltzmann equation it gives you an invented value of emissivity.

Well, as the blog owner has explained “some people” doubt the measured values of radiation. That’s pretty convincing evidence..

And a bit later:

Here is a list of MODIS products. MYD11C3 says “Land Surface Temperature & Emissivity”

This is a “land” product. The data sets for this product are all related to land – MODIS/Aqua Land Surface Temperature/Emissivity Monthly L3 Global 0.05Deg CMG.

It seem that your commenter Max added the footer that stated it was ocean emissivity. Well, up to him or the blog owner to provide his source..

Then someone told me a few home truths:

S.o.D’s sarcasm looks like an attempt to cover for the MODIS team’s poor data representation, and to distract attention from the glaring errors in the Trenberth-Keihl energy budget.

And it just keeps getting better from there. I couldn’t make up stuff this good.

36. SOD, Pekka, DeWitt, or Bryan:

There appears to be a dilemma that none of you want to specify in your stale debate. An electric field is a vector quantity. When the electric field from two charges overlap, we usually perform the vector addition and refer to a single resulting field.

Radiative flux is a vector quantity. When a downward and an upward radiative flux “pass through” each other, we argue about whether they cancel. When we discuss the heat transferred by the radiation, we cancel (ie. perform the vector addition). However, we know that the photons passing past each other do not cancel – they certainly aren’t vectors. Those photons can be counted when the reach a detector (facing up or down) through the energy they carry, so the energy fluxes don’t cancel as they pass through each other. Is there a consistent way to know when one should and shouldn’t add these vector quantities?

• Frank

Waves (EM or otherwise) pass through each other unchanged.
Only at a specific point will opposing waves interfere and cause a resultant value.

• Frank,

The photons are effectively independent, because they are incoherent, meaning that the phases of the oscillating electric and magnetic fields are independent. Considering photons emitted at different locations independent is an extremely good approximation for the IR radiation in the atmosphere.

A well known example of the situation, where the radiation is coherent is a laser, where the ‘s’ refers to stimulated. Stimulated radiation is coherent, i.e. locked to the phase of the radiation that is stimulating further emission.

Coefficients found from the HITRAN database contain also the coefficient for stimulated emission. Applying that coefficient in the appropriate formula confirms that stimulated emission is negligible in comparison with spontaneous emission for IR in the atmosphere.

The same facts can be concluded from the theory of Quantum Electrodynamics, which is built on the Feynman diagrams. All simple Feynman diagrams correspond to incoherent photons, but more complex ones represent processes that involve two or more photons. These more complex diagrams tell about phenomena were considering photons independent is insufficient. Quantitatively these corrections are negligible in situations relevant for IR radiation in the atmosphere.

In addition to lasers, much that we know about radiowaves is strongly influenced by coherence. Considering radiowaves in terms of incoherent photons can not explain anything we are accustomed to attach to them.

• on August 4, 2014 at 4:43 pm | Reply DeWitt Payne

Frank,

Radiative flux is a vector quantity.

No, it isn’t. That’s the mistake that G&T make when they criticize energy balance diagrams. The surface defining the flux has an orientation, but the flux through that surface is not a vector. It’s a scalar.

• DeWitt: FWIW, Wikipedia identifies eight fluxes (including radiation and energy) vector quantities.

http://en.wikipedia.org/wiki/Flux

Forces and fields follow the rules of vector addition (and cancel). Opposing fluxes of photons and matter do not cancel, so it makes some sense to treat them as scalars. However, flux in more than one dimension is described with vector notation.

• Frank,

Flux density can be a vector or scalar. But it can only be a vector if you are describing the flux density at a specific point and time specified by the investigator and calculated by integrating the intensity over the entire sphere surrounding the point. When you integrate over a hemisphere defined by a plane selected by the investigator, the result is a scalar not a vector.

Wikipedia Spectral flux density:

Scalar definition of flux density – ‘hemispheric flux density’

The scalar approach defines flux density as a scalar-valued function of a direction and sense in space prescribed by the investigator at a point prescribed by the investigator. [emphasis added]

Vector definition of flux density – ‘full spherical flux density’

The vector approach defines flux density as a vector at a point of space and time prescribed by the investigator.[emphasis added]

What we refer to as DLR and OLR, for example, are scalar quantities, not vectors. The same goes for convection. The actual angular sense is lost, not preserved, during the integration. The scalar value could, in principle, be caused by a high intensity narrow beam or a high intensity from a narrow angle like incident sunlight or from a diffuse source like a cloud covered sky.

37. For most scientific subjects of wide importance, it’s not hard to find review articles explaining the principles involved and summarizing current knowledge.

Maybe I have not looked in the right places but I have not found such a review article dealing with the greenhouse effect. Do they exist? Where should I look?

• Martin A,

I’ve tried in the past to find ones on Planck’s law, Stefan Boltzmann relationship and other fundamental heat transfer, but the best I was able to do was get The Theory of Heat Radiation by Max Planck published in 1914.

It seems that the review papers for stuff well-proven 100 years ago will be in some archives of an august library and not on Google scholar.

For the “greenhouse” effect, the best I can offer are the papers which explained the “radiative-convective” effect – that is, the thermal structure of the atmosphere:

Thermal equilibrium of the atmosphere with a convective adjustment, Manabe & Strickler (1964)

Thermal equilibrium of the atmosphere with a given distribution of relative humidity, Manabe & Wetherald (1967)

Climate Modeling through Radiative Convective Methods, Ramanathan & Coakley (1978)

– the above papers are all free

There is an excellent review article which was made by Ramanathan on invitation:
Trace Gas Greenhouse Effect and Global Warming, V. Ramanathan, Royal Swedish Academy of Sciences (1998)

– sadly this is behind a paywall, but I’ll happily email it to anyone who wants to read it.

38. Sod,

I would love to receive that article

Thanks!

I would be grateful for the Ramanathan review article. I’m assuming that you have my email address.

40. on August 8, 2014 at 4:41 am | Reply Chic Bowdrie

I’m troubled by definitions of the greenhouse effect based on the presence of IR absorbing gases. Clearly the planet would be warmer simply due to the presence of N2 alone compared to no atmosphere. IMO, one should consider alternative possible outcomes resulting from the addition of IR absorbing gases. The most obvious would be the current atmosphere. CO2 doubling or an equivalent projected increase in IR active gases describes another outcome sometimes called an enhanced greenhouse effect. You include this latter outcome as part 6 of the general greenhouse effect where you state “If we add more radiatively-active gases (like water vapor and CO2) then the atmosphere becomes more ‘opaque’ to terrestrial radiation and the consequence is the emission to space from the atmosphere moves higher up (on average). Higher up is colder.”

In note 6 you add “It [a constant lapse rate] doesn’t change the fact that with more GHGs, the radiation to space will be from a higher altitude.”

Where has that fact been established? More IR absorbing gases seems as likely to radiate more to space than less. Your explanation given in The Earth’s Energy Budget – Part Three does not reference any supporting data.

• Chic,

Clearly the planet would be warmer simply due to the presence of N2 alone compared to no atmosphere.

With an N2 atmosphere the surface would be radiating directly to space. The atmosphere would not be interacting with this radiation, it would not be absorbing or emitting.

Therefore, the outgoing longwave radiation of approximately 240 W/m2 would be from the surface.

Or if we want to avoid averages, the OLR of 240 x 5.1×1014 x 86,400 x 365 J/year would be from the surface.

Correct or not?

In note 6 you add “It [a constant lapse rate] doesn’t change the fact that with more GHGs, the radiation to space will be from a higher altitude.”

Where has that fact been established? More IR absorbing gases seems as likely to radiate more to space than less. Your explanation given in The Earth’s Energy Budget – Part Three does not reference any supporting data.

Using the equations of radiative transfer. If we take the case “pre-feedback” then for any given surface temperature and lapse rate we can calculate the changes.

I did the calculation in Visualizing Atmospheric Radiation – Part Seven – CO2 increases.

The code for this is shown in Part Five – The Code – this is just a numerical integration of the equations of radiative transfer.

Obviously I wasn’t the first. I did these calculations so that we could probe into the details and understand at what wavelengths the largest effects were found. And also so the code could be inspected and questioned.

The first paper that really nailed the problem was Thermal equilibrium of the atmosphere with a given distribution of relative humidity, Manabe & Wetherald (1967).
A later more comprehensive paper is Climate Modeling through Radiative Convective Methods, Ramanathan & Coakley (1978).

Note that my calculations are not a GCM, they just calculate how a change in CO2 causes a reduction in OLR for a given temperature profile.

• on August 8, 2014 at 4:09 pm Chic Bowdrie

SOD,

“Therefore, the outgoing longwave radiation of approximately 240 W/m2 would be from the surface. Or if we want to avoid averages . . . .”

A N2 atmosphere would still warm due to conduction and convection. There would be more solar insolation without clouds, possibly 340 W/m2 depending on albedo. Because there is no cooling at altitude, the surface would be much warmer than 255 K depending the magnitude of temperature swings between day and night. So on average an inert atmosphere would be warmer than no atmosphere, but not as warm as our current atmosphere.

“Note that my calculations are not a GCM, they just calculate how a change in CO2 causes a reduction in OLR for a given temperature profile.”

Your calculations are impressive, but I’m still left wanting for evidence that the calculations reflect reality. This from your Ramanathan and Coakley (R&C) reference page 482:

“In judging the importance of [radiative-convective model] results it should be remembered that they represent only the sensitivity of the model climate to perturbations in the atmospheric constituents and that the sensitivity of the model may not reflect the climate sensitivity of the actual earth-atmosphere system. For the CO2-climate problem, however, the increase in Ts due to an increase in CO2 computed by the radiative-convective model is within 20% of the increase in Ts computed by a three-dimensional GCM.”

With GCMs missing the mark by so much, how can they be used to validate R&C’s model?

Also in the literature references, I found no evidence indicating more IR absorbing gases raise an effective emission height. I’m going through Visualizing Atmospheric Radiation – Part Seven – CO2 increases to see if I can find it there.

• Chic,

A N2 atmosphere would still warm due to conduction and convection.

Interesting. Let’s go back to the key point and confirm your thinking:

Where is the emission to space from (with an N2 atmosphere)?
What is the globally averaged OLR?

• on August 8, 2014 at 5:52 pm Chic Bowdrie

sod,

Emission to space from the surface, of course. However, the surface must warm unless it reflects 100% of incoming solar. I assumed the other extreme, 100% absorption or an average of 340 W/m2. OLR would also be 340 W/m2 by definition of in equaling out.

The key point here is that some conduction will occur and the only way that energy can escape is reverse conduction. Convection will oppose that process. So a quasi-equilibrium will develop at altitude with an average global surface temperature that can only be speculated by estimating the daily and latitudinal extremes.

• on August 8, 2014 at 8:53 pm DeWitt Payne

Chic,

If we started out at, say, the temperature of the cosmic microwave background, 2.75K, and turned the sun on, the presence of an atmosphere would cause the surface to have a lower temperature than for no atmosphere until the atmosphere reached steady state. The atmosphere can’t be warmer than the warmest spot on the surface. The warmest spot on the surface could also not be warmer than if there were no atmosphere. So once steady state is achieved, no further energy will be transferred from the surface to the atmosphere and vice versa, assuming no convection. If you allow convection, the equator wouldn’t be as hot and the poles wouldn’t be as cold, but the average would be only slightly higher.

Nitrogen isn’t perfectly transparent to IR because of collision induced absorption. But the absorptivity is so low and at such long wavelength that it wouldn’t make much difference.

• on August 9, 2014 at 6:18 am Chic Bowdrie

DeWitt,

If you want to start with a 3K planet and turn the sun on, then let’s do that for both cases—no atmosphere and a N2 atmosphere devoid of any IR activity. Assume both planets have the same albedo = 1. The sun will continue to warm the surfaces of both planets until they become hot enough to radiate on average the same energy as they receive. I call this a quasi-steady state because the planet’s surfaces are constantly heated and cooled as they rotate and revolve around the sun. Because some of the energy absorbed by the surface of the N2 planet is conducted into the atmosphere and convected upwards, this atmosphere will continue to warm until reaching an average temperature profile similar to Earth’s dry lapse rate. Because this N2 atmosphere provides some insulation not possible with a no atmosphere planet, the average global temperature of the N2 planet will be greater than the no atmosphere planet. You cannot assume no convection when a planet has an atmosphere.

“The atmosphere can’t be warmer than the warmest spot on the surface.”

On average, of course. But every night, the atmosphere will be warming the cooler surface by conduction.

“The warmest spot on the surface could also not be warmer than if there were no atmosphere.”

I can’t agree with that. The no-atmosphere surface does not have the “back” conduction provided by the atmosphere.

• on August 9, 2014 at 7:07 pm Chic Bowdrie

Assume both planets have no albedo or at least anything other than 1. Sorry.

• Chick,

The N2 atmosphere cannot affect the surface temperature in any other way than by transferring a little heat from one spot of the surface to another and by storing a little heat, when the surface is hot, and later release the same amount to the surface, when the surface is cooler. It cannot influence the effective radiative temperature of the surface as virtually all radiation is absorbed by the surface and emitted by the surface.

Thus the surface emits as much at it absorbs. This requirement of energy balance determines the surface temperature. N2 atmosphere cannot changed that much, as the only effect comes from the lesser temporal and spatial variability of the surface temperature. With the T^4 behavior of the Stefan-Boltzmann law the same average T^4 leads to slightly warmer average T, when the variability is less.

Lesser variability means that the maximum temperature is reduced and the minimum temperature increased.

• on August 9, 2014 at 8:30 pm Chic Bowdrie

Pekka,

“With the T^4 behavior of the Stefan-Boltzmann law the same average T^4 leads to slightly warmer average T, when the variability is less. Lesser variability means that the maximum temperature is reduced and the minimum temperature increased.”

This much I understand and agree with totally. Which is why I don’t understand why you think an atmosphere, albeit inert, “cannot influence the effective radiative temperature of the surface ….” If any radiation absorbed by the surface is conducted, then the atmosphere will convect it upwards. Once that energy is elevated it won’t readily be returned to the surface. Conduction is too slow and convection isn’t driven in that direction. Assuming realistic thermal conductivities, it may take some time, but eventually the atmosphere will establish a lapse rate with a warmer average temperature than a similar planet with no atmosphere.

Another way to look at it is both planets will eventually radiate the same amount as received. But the atmosphere planet has that extra heat capacity due to the atmosphere. It has to be saturated before the final pseudo- or quasi- equilibrium is reached.

• Chic,
The atmosphere cannot convect or conduct heat upwards for long, because the heat cannot go anywhere from a non-emitting atmosphere and the heat capacity of the atmosphere itself is small. All the heat that goes up must soon come back down. There’s no net transfer of energy either up or down over a full year.

The surface is heated only by the sun, and cooled with the total power of emission that corresponds to the surface temperature. A N2 atmosphere changes that very little. Therefore the surface is would have nearly the same temperature as without any atmosphere. The only significant additional energy flow to the Earth surface is the downwelling radiation from the atmosphere, but N2 does emit such radiation to a significant degree leading to a cold surface (near 0C with zero albedo).

• on August 9, 2014 at 9:19 pm Chic Bowdrie

Pekka,

“The atmosphere cannot convect or conduct heat upwards for long, because the heat cannot go anywhere from a non-emitting atmosphere and the heat capacity of the atmosphere itself is small.”

The atmosphere only needs a non-zero heat capacity to gain heat. Once heated, the N2 near the surface must rise. It isn’t coming back as fast as it goes up because of the dynamics of convection.

I don’t understand the context of “no net transfer of energy either up or down over a full year.

We may have to start putting some numbers up or we’ll end up at an impasse.

First let’s try a thought experiment. If I irradiate ice in a vacuum with IR, will it melt?

• on August 9, 2014 at 9:21 pm Chic Bowdrie

More appropriately, will it sublime?

41. […] that there have been more than 50 posts on this topic (post your comments on those instead). See The “Greenhouse” Effect Explained in Simple Terms and On Uses of A 4 x 2: Arrhenius, The Last 15 years of Temperature History and […]

42. […] wrote The “Greenhouse” Effect Explained in Simple Terms to make it simple, yet not too simple. But that article relies on (and references) many basics […]

43. […] wrote The “Greenhouse” Effect Explained in Simple Terms to make it simple, yet not too simple. But that article relies on (and references) many basics […]

44. “If we add more radiatively-active gases (like water vapor and CO2) then the atmosphere becomes more “opaque” to terrestrial radiation and the consequence is the emission to space from the atmosphere moves higher up (on average). Higher up is colder”

OK, I’ll stick my neck out.
I have the same problem that Peter O’Donnell Offenhartz alluded to earlier.
I have used MODTRAN to view the spectra for Tropical and Arctic Summer atmospheres (didn’t look at the others).
http://climatemodels.uchicago.edu/modtran/modtran.html

This does seem to indicate that the Effective Radiating Level of radiation coming from the bottom of the ‘CO2 well’, around wavenumber 666 is in the stratosphere, a region where temperature increases with altitude.
For example, look down from 70km and observe the bottom of the ‘CO2 well’, around wavenumber 666. This, presumably, radiation from the Effective Radiating Level(ERL).
Estimate its temperature by reference to nearby the plank curves.
Now look down from progressively lower and lower altitudes. What happens? When we get to 40km, we see that the bottom of the CO2 well appears to be emanating from somewhere ‘cooler’ than it was when viewed from 70km.
If the temperature at ERL was decreasing with altitude, it should appear to radiate from somewhere warmer as we descend.
The argument based on ERL is very easy to understand and I have used this argument myself, after reading your blog. But it does rely on the ERL being in the troposphere, and it doesn’t appear to be.

Or, have I done something wrong?

• on May 15, 2015 at 2:02 pm | Reply DeWitt Payne

By convention, forcing is measured at the tropopause. In fact, you are also supposed to allow the stratosphere to equilibrate. Adding CO2 will cool the stratosphere. But the contribution from the stratosphere is small enough that if you measure the change in upward radiation (looking down) at the tropopause it’s pretty close. Looking down from the top of the atmosphere, emission from the center of the CO2 band will increase if you increase CO2. But if the stratosphere were allowed to cool, it wouldn’t change very much.

When you change the temperature in MODTRAN, you only change the temperature in the first ten kilometers above the surface. If you measure the difference in upward radiation starting at the surface and moving upward, the difference increases, reaches a maximum and then decreases. If the stratosphere were allowed to equilibrate, it would reach a maximum and then not change much with altitude.

• MikeB: You are exactly correct. If temperature didn’t drop with altitude, the GHE wouldn’t exist. As radiation (I_0) of a given wavelength passes an incremental distance (ds) through the atmosphere, the change (dI) is caused by both emission (first term) and absorption (second term) of the Schwarzschild eqn:

dI/ds = n*o*B(lamba,T) – n*o*I_0
dI/ds = n*o*[B(lamba,T) – I_0]

where n is the density of GHG and o is its absorption cross-section. Outward radiation (I_0) comes from below – produced by a B(lamba,T) term usually with a higher T. So the term in brackets is usually negative and increasing n decreases the outward flux. Above the tropopause, however, temperature increases with altitude and the ERL for the strongest CO2 emission bands is in the stratosphere.

MODTRAN is numerical integrating the Schwarzschild eqn upward and downward through the atmosphere over all wavelengths. If you look down from the tropopause (which varies with latitude) and then the TOA (70 km), you will see that OLR increases a few W/m2 above the tropopause. (Globally OLR decreases from about 390 W/m2 to 240 W/m2 with increasing altitude, so the increase due to emission from the stratosphere is pretty small.)

• MikeB,

If CO2 only absorbed across wavenumbers 666 – 668 cm-1 with such a huge absorption then more CO2 would cause less warming – for exactly the reasons you describe.

But CO2 absorbs across a much much wider range, with most of that absorption being many orders of magnitude lower. Consequently, the emission to space from most of the CO2 band is from the troposphere.

Take a look at Visualizing Atmospheric Radiation – Part Seven – CO2 increases. Hopefully the bigger picture as explained makes sense. But feel free to ask further questions.

45. SoD
Thank you for the reply. If you repeat the experiment you will see that whole base of ‘CO2 well’, from wavenumber 640 to 680 (as far as I can tell) is already in a region where temperatures increase with increasing altitude.
I was ignoring the central ‘blip’ at wavenumber 666 although I think an upward blip like this is also an indication of temperature inversion.

This is not to say that increasing CO2 levels will not widen the CO2 absorption band in the ‘wings’. Modtran also shows a reduction in upward IR heat flux with increasing CO2. That is accepted. The only point I am making is that the argument for the greenhouse effect based on the ERL raising to a level which is cooler doesn’t seem to be supported. But, it is not supported by evidence and I have never seen any statement or reference saying that the ERL is in the troposphere.
So, I think the argument from this perspective is, for the want of a better word, wrong. I think we are back to ‘back radiation’ providing an additional heating flux on the ground.

• on May 18, 2015 at 1:26 pm | Reply DeWitt Payne

The effective level of radiation emission is a mathematical construct. You take the average global emission, 239W/m², convert it to temperature and use a standard atmospheric temperature profile to convert temperature to altitude. That’s 254.8K. Using the 1976 U.S. Standard Atmosphere profile with a surface temperature of 288.2K, that’s 5.14km. An increase in surface temperature of 1K would raise that level to 5.29km.

The other thing you have to remember is that most atmospheric emission comes from water vapor, and that comes from the lower troposphere. The scale height for water vapor is ~2km. That means that 86.5% of all the water vapor in the atmosphere is below an altitude of 4km. Adding CO2 has no immediate effect on water vapor emission and, if you allow the stratosphere to equilibrate, no effect on emission at the CO2 band peak. But it does increase the effective emission level in the band wings.

• MikeB:

..The only point I am making is that the argument for the greenhouse effect based on the ERL raising to a level which is cooler doesn’t seem to be supported. But, it is not supported by evidence and I have never seen any statement or reference saying that the ERL is in the troposphere.
So, I think the argument from this perspective is, for the want of a better word, wrong. I think we are back to ‘back radiation’ providing an additional heating flux on the ground..

If you arbitrarily define the CO2 band to be 650-690 cm-1 then you are correct. If you define the CO2 band to be the whole band then you are not correct.

In this calculated graphic (from Visualizing Atmospheric Radiation – Part Seven – CO2 increases) you can see that for a given surface temperature the TOA flux is reduced due to doubling CO2, and you can see which wavenumbers provide the effect:

In this graphic, from Visualizing Atmospheric Radiation – Part Three – Average Height of Emission you can see where the TOA radiation is emitted from as a function of both wavenumber and altitude:

[The 0-1500 is wavenumbers in cm-1 and the 0-20 is altitude in km]

• SOD and MikeB: The difference between the 280 and 560 ppm of CO2 lines might be clearer.

By defining the TOA as 50 hPa (19 km) in the above graphs, SOD is missing the changes to TOA caused by some of the stratosphere. Using the MODTRAN calculator (US standard atmosphere, no clouds, 400 ppm CO2 and NO other GHGs), I find that the altitude with the minimum value for OLR is 20 km. OLR rises 0.5 W/m2 by 30 km and 1.6 W/m2 by 70 W/m2. The biggest visible change is associated with the narrow 666 cm-1 line in the middle of the band, but the effective radiation temperature of the whole band rises from just below 220 degK at 20 km to just above 220 degK. Doubling CO2 to 800 ppm decreases OLR by 3.0 W/m2 at 70 km and 3.6 W/m2 at 20 km. (I suspect that MODTRAN, like SOD, calculates equilibrium temperatures for the stratosphere.)

Most of the action is in the troposphere. 400 ppm of CO2 alone can reduce OLR by about 100 W/m2 between the surface and 12 km, reduces another 1.6 W/m2 between 12 and 20 km, and adds back 1.6 W/m2 from 20 to 70 km (from the relatively warm stratosphere). Doubling CO2 reduces OLR whether the TOA is at 50 hPa (19 km) or 70 km.

Caveat: When I added the default amount of other GHGs (including stratospheric O3, the change in OLR between 20 km and 70 km was only 0.1 W/m2, not 1.6 W/m2. In a US standard atmosphere with no GHGs except stratospheric O3, OLR decreases steadily with altitude beginning above 12 km – because the surface of the earth is warmer than the stratosphere. The colder atmosphere between doesn’t absorb or emit significantly at these wavelengths (even when the default tropospheric ozone is included). So, stratospheric CO2 and O3 have opposite effects on OLR that nearly cancel.

• on May 19, 2015 at 7:38 pm DeWitt Payne

Frank,

MODTRAN keeps the temperature above 13km constant. In reality, an increase in CO2 will cause the stratosphere to cool rapidly, in weeks to a few months, and there will be little or no increase in emission with altitude above the tropopause. That’s why the official protocol for calculating radiative forcing from ghg’s requires that the stratosphere be allowed to equilibrate while the troposphere is held constant.

• MikeB,

Another point – as I stated in Note 5:

The “place of emission” is a useful conceptual tool but in reality the emission of radiation takes place from everywhere between the surface and the stratosphere. See Visualizing Atmospheric Radiation – Part Three – Average Height of Emission – the complex subject of where the TOA radiation originated from, what is the “Average Height of Emission” and other questions.

Conceptual tools have their uses. They have their limitations as well.

If this conceptual tool has lost its usefulness, instead just think about OLR reducing due to more GHGs.

Visualizing Atmospheric Radiation – Part Seven – CO2 increases shows how the “no feedback” result of doubling CO2 is calculated and which wavenumbers make the largest contribution.

The maths is the result. Graphs and explanations are ways to help us grasp it.

I found the subject difficult to grasp until I was able to carry out the calculations myself. I realize this is out of reach for most people – you need a lot of time and a tool like Matlab to be able to carry out the intensive numerical calculations required by the equations of radiative transfer.

But I am more than happy to produce the results in different ways for interested commenters..

• on May 20, 2015 at 8:43 pm Chic Bowdrie

Scienceofdoom,

“Conceptual tools have their uses. They have their limitations as well.”

Very true. The problem I see with concepts like ERL and a MODRAN tool is the emphasis on a static atmosphere, where a snapshot of the atmosphere in one century can be compared one in another; where you have time to calculate the effect of a change in CO2 and speculate on the amount of energy that might be represented by those calculations. In reality, energy fluxes through the atmosphere are constantly changing. During the day, CO2 and H2O molecules absorb radiation at the Earth’s surface and transfer that energy to the bulk air where it rises higher up in the atmosphere to eventually be transferred back again as radiation to space. What is needed is a conceptual tool that reflects what is happening in real time, not calculated differences between snapshots in time.

46. on May 18, 2015 at 8:51 pm | Reply Peter O'Donnell Offenhartz

@DeWitt Payne

I’m not sure you have answered MikeB’s question. You provide a “scale height” for water vapor and another height for ALL infrared-absorbing gases taken together. How about a scale height for carbon dioxide by itself.

47. on May 18, 2015 at 11:12 pm | Reply DeWitt Payne

CO2 is a well mixed gas, unlike water vapor. The scale height is the same as for the other well mixed gases, ~8km. Therefore at 4km for the 1976 U.S. Standard atmosphere, the pressure is 701mbar, meaning 70% of the atmosphere and by extension the CO2 in the atmosphere is above that altitude, compared to 13.5% of the water vapor. And again, most of the atmospheric emission to space comes from water vapor, so all the emission to space from water vapor is from below 4km, That also doesn’t include the emission through the atmospheric window directly from the surface.

• DeWitt: You didn’t account for the fact that there is much less CO2 (.04% or 400 ppm) in the atmosphere than water vapor (about 1% or 10,000 ppm) near the surface. Atmospheric pressure is 10,000 kg/m2. At 4 km, you are still beneath about 4.2 kg of the 6 kg (70%) of CO2 above every m2 of the surface. At 4 km, you are still beneath 8 kg of the 62 kg (13.5%) of water vapor over every m2 of surface. (This is after converting ppmv to ppmw. There are about 4-fold times as many water vapor molecules as CO2 molecules above 4 km.) You also have to factor in the difference in cross-section.

If you remove all of the other GHGs from the MODTRAN calculator for the US standard atmosphere, water vapor emission for 100-400 cm-1 has an effective radiation temperature of 230-240 degK (8 km) and above 1300 cm-1 of 240-260 degK (6 km). The lowest effect radiation temperature for CO2 is 220 degK (10 km)

• on May 19, 2015 at 2:29 am DeWitt Payne

My numbers say 25kg water vapor/m², i.e. 25mm of liquid water, not 62kg or 62mm. But you have a point. Water vapor absorbs strongly below 400cm-1 which is still well within the thermal IR range. Still, if we set CO2, methane and tropospheric ozone to zero, atmospheric transmittance increases rapidly with altitude. Tropical atmosphere (for maximum water vapor content), looking up

0km 0.163
2km 0.348
4km 0.490
8km 0.711

The troposphere in the tropics stops at about 17km altitude.

48. Chic Bowdrie wrote: “The problem I see with concepts like ERL and a MODRAN tool is the emphasis on a static atmosphere, where a snapshot of the atmosphere in one century can be compared one in another; where you have time to calculate the effect of a change in CO2 and speculate on the amount of energy that might be represented by those calculations. In reality, energy fluxes through the atmosphere are constantly changing. During the day, CO2 and H2O molecules absorb radiation at the Earth’s surface and transfer that energy to the bulk air where it rises higher up in the atmosphere to eventually be transferred back again as radiation to space. What is needed is a conceptual tool that reflects what is happening in real time, not calculated differences between snapshots in time.”

That is what climate models, cloud-resolving models and re-analysis of weather data to do for us – but then you are buried in data from which it is difficult to abstract important concepts. Satellites are continuously monitoring the changes in OLR and reflected SWR. Their data says that climate models do a good job of representing OLR from clear skies (the GHE and water-vapor plus lapse rate feedback) and a poor job with clouds and ice-albedo feedback.

• on May 23, 2015 at 6:40 pm | Reply Chic Bowdrie

Frank,

Thank you for the response and the paper which AFAICT says, “the gain factors of longwave CRF obtained from most of the CMIP models have positive values, in contrast to the small negative values obtained from satellite observations.” Hopefully someone in the climate model community got the word.

Meanwhile a novice like me continues frustrated over the use of spectrograms showing that IR absorbing gases absorb some of the radiation at some point after it leaves the surface and emit it somewhere higher up. This technique is used to quantify the amount of extra energy that is “trapped” due to increasing CO2. All that is left is to associate the trapped energy with some amount of temperature increase. Most of the climate debate is over the magnitude of the temperature change. Scientists who look into all the actual energy transfer scenarios involved, ie conduction, evaporation, radiation, and convection, surmise that temperature gradients and energy fluxes are not all determined by radiation physics. IOW, it is not enough for models to accurately predict OLR from atmospheric composition and temperature at altitude. There has to be some consideration for how the energy got there in the first place. I know that’s what the GCMs are supposed to include, but there’s more wrong with them than just cloud radiative forcing.

This is not meant to be a criticism of SoD’s work exhaustive work here. I’m just not able to understand how doubling CO2 can lead to lower “spectral emitted power” at 50 hPa as indicated in the response to MikeB above. At that elevation if CO2 isn’t radiating at least as much, what else is?

• Chic wrote: “Scientists who look into all the actual energy transfer scenarios involved, ie conduction, evaporation, radiation, and convection, surmise that temperature gradients and energy fluxes are not all determined by radiation physics. IOW, it is not enough for models to accurately predict OLR from atmospheric composition and temperature at altitude.”

When I first came to SOD, I used to hate all of the emphasis on radiation and the lack of discussion about convection. Surface temperature is controlled by both. My biases made it hard to always remember that the only way for energy to enter or leave the planet is by radiation. Convection merely redistributes the energy that is present. We can calculate the amount of heat transported by radiation, but not by convection. Physics only tells us when an unstable lapse rate will permit buoyancy-driven vertical convection, not how much energy will be carried upwards by it. It doesn’t tell us why the earth’s average lapse rate is 6.5 K/km, instead of some other value. So SOD can (and does) tell us far more about radiation than convection.

I used to dream that winds over the ocean could blow faster, causing more evaporation and transporting more heat away from the surface. Unfortunately, the more heat that is carried aloft from the surface, the smaller the likelihood of an unstable lapse rate. The heat un the upper troposphere has to escape to space before more can arrive.

Chic wrote: “Hopefully someone in the climate model community got the word.”

One of the co-authors of the above PNAS paper is Manabe, one of the original developers of the GFDL climate model (and co-author of the first paper on “radiative-convective equilibrium”). He may be the driving force behind the Coupled Model Intercomparison Project (CMIP) including experiments on seasonal change. It is hard to believe they don’t recognize these problems, whether they discuss them fully or not. Policymakers desperately need useful projections and have invested billions(?) in climate models. Complete public candor has its drawbacks.

49. on May 24, 2015 at 10:03 pm | Reply Chic Bowdrie

Frank, your comment reminded me of my first foray into the “deep” science of climate change. That was right here at ScienceofDoom. Now I dream about being reincarnated as a climate scientist. So many questions, so little time.

You said, “We can calculate the amount of heat transported by radiation, but not by convection.”

Is that strictly true or is it just a lot more difficult to set up and solve the differential equations involved? I haven’t delved into that yet, but my understanding is that an unstable lapse rate is caused by the heat produced at the surface when the sun comes up. That heat does not wait for a signal from the TOA to see if there’s room for more heat up there. Are we on the same page with that? Also, I’m pretty sure the lapse rate is calculable from thermodynamic principles.

• Chic,

You said, “We can calculate the amount of heat transported by radiation, but not by convection.”

Is that strictly true or is it just a lot more difficult to set up and solve the differential equations involved? I haven’t delved into that yet, but my understanding is that an unstable lapse rate is caused by the heat produced at the surface when the sun comes up. That heat does not wait for a signal from the TOA to see if there’s room for more heat up there. Are we on the same page with that?

We can write an equation for heat transported by radiation. We have some data required – the temperature profile, the concentration of GHGs and their absorption properties.

The equation can be solved (a numerical solution can be obtained given the data noted above).

We can write an equation for heat transported by convection.

But we can’t solve it. Take a look at Turbulence, Closure and Parameterization.

..Also, I’m pretty sure the lapse rate is calculable from thermodynamic principles.

This is partly correct.

The lapse rate under convection can be calculated. See Potential Temperature. In the tropics, due to the large amount of convection (even though a larger area has subsiding air than has ascending air), the temperature profile roughly matches this calculated lapse rate.

Outside the tropics there is much less convection and the lapse rate often doesn’t match the calculated lapse rate under convection.

• on May 26, 2015 at 4:40 am Chic Bowdrie

SoD,

Sorry I didn’t catch your response until now. You write,

“We can write an equation for heat transported by convection. But we can’t solve it.”

Does that include numerical methods? If you can write an equation, a numerical method solution should be possible. Or is it similar to the turbulence problem that is too massive even for today’s computers?

• Chic Bowdrie,

Does that include numerical methods? If you can write an equation, a numerical method solution should be possible. Or is it similar to the turbulence problem that is too massive even for today’s computers?

If you take a look at the article I linked, Turbulence, Closure and Parameterization, you will see the answer to your questions:

1. Does that include numerical methods? – Yes
2. If you can write an equation, a numerical method solution should be possible – Not necessarily
3. Or is it similar to the turbulence problem that is too massive even for today’s computers? – It is exactly “the turbulence problem that is too massive for today’s computers”. And tomorrow’s. And probably the computers in 2035.

• Chic commented: “That heat does not wait for a signal from the TOA to see if there’s room for more heat up there.”

Upward buoyancy-driven convection only occurs when the rising air is warmer – and therefore less dense – than the air it displaces – even after it has expanded (under the reduced pressure) and cooled. The “signal to continue rising” is the temperature of the air immediately above a rising parcel of air.

Dry and moist adiabatic lapse rates can be calculated from first principles: https://scienceofdoom.com/2011/06/12/paradigm-shifts-in-convection-and-water-vapor/. However, these simply tell you when or where the lapse rate is unstable – it doesn’t tell you how much latent heat (and simple heat?) will move upwards through an unstable region before it stops being unstable.

Lapse rates in the real world are highly irregular: perturbed by the diurnal cycle (ground, with higher emissivity, cools at night faster than the air immediately above) and perturbed in temperate zones by colliding masses of warmer and colder air. Radiosondes are launched twice a day from several hundred(?) sites around the globe and the altitude vs temperature profiles or “soundings” (posted on the web somewhere) are used to initialize weather prediction programs. The are often displayed on Skew-T diagrams.

FWIW, I have asked and searched for an explanation for why the earth’s average environmental lapse rate in the troposphere is 6.5 K/km and never gotten one. We can calculate how much an increase in absolute humidity will change the lapse rate in a world with 2XCO2 (to my knowledge, the only recognized form of lapse rate feedback), but I can’t see an unambiguous reason why that humidity-induced change must be a change from today’s value, 6.5 K/km. If the Hadley circulation were to speed up in a 2XCO2 world, stronger trade winds would produce more evaporation and the ITCZ would carry more heat to the top of the troposphere, reducing the overall lapse rate. However, only radiation can carry that heat from the upper troposphere to space and 2XCO2 slows this process. Furthermore, the amount of air ascending and descending must be equal, and descent requires increasing density by radiative cooling. If a faster Hadley circulation carried heat HIGHER, heat could escape more easily. Convection in tropical regions does reach far higher than in temperate ones, and I’ve tried reading (without comprehension) papers explaining why the top of the convective region is located where it is.

50. on May 25, 2015 at 12:22 pm | Reply Chic Bowdrie

Frank,

That is why a simple quantitative dynamic model of the processes we are discussing would be helpful. Saying what would happen is really just giving our opinion of what should happen. You seem convinced that 2xCO2 slows radiation to space. Other than something generated from a computer model. where is the data showing that is true?

Your interpretation is based on an assumption that more GHGs will make emission to space occur higher up in the atmosphere. SoD makes the same assumption in this post. I would like to see that data as well.

My understanding is that 6.5 K/km is a result of surface evaporation and cloud formation. The latent heat is removed from the surface and released at altitude forcing a reduction in the dry lapse rate.

• on May 25, 2015 at 1:45 pm | Reply DeWitt Payne

Chic,

Radiative transfer is about as bullet proof as anything in climate science can get. The properties of the absorption/emission lines of H2O, CO2 and the other radiatively active ghg’s have been measured in the lab and in the field. They have also been calculated using quantum mechanics from first principles, ab initio. The database containing this information, HITRAN, was originally started by the Air Force as a research tool for imaging and heat seeking missiles. The radiative transfer equation is about as solid as F=ma. Calculated atmospheric emission spectra have been compared many times to measured spectra with agreement on the order of 1%. The same theory is used to calculate atmospheric temperature profiles from satellite measurements of microwave emission from oxygen near 60 GHz.

If you’re really serious, you should buy Grant Petty’s A First Course in Atmospheric Radiation. It’s only $36 direct from the publisher. There are simple one dimensional radiative convective models, but they don’t actually calculate convection, you set a limit for convective stability. You can start out with about any surface temperature and air temperature and they will converge to a temperature profile that looks very much like the standard average profile based on measurements. Beyond that, you must have a full bore Air Ocean General Circulation Computer Model. There is no in between conceptual model. The general assumption is that the current atmospheric profiles for different regions happen for a good reason even if we can’t say why with high confidence and that a perturbation of a few degrees won’t change them much. • on May 25, 2015 at 7:12 pm | Reply Chic Bowdrie DeWitt, Thanks for the info on microwave emission and suggestion to get my own copy of Petty’s book. However, I’m not challenging the results from radiative transfer based models which accurately predict OLR. But AFAIK these are point-in-time measurements. Do these models account for the net input and output over a 24 hour period or longer? This may be the objective of GCMs, but 1) they aren’t accurate and 2) they are too complicated for any conceptual learning. At least, I would have to retire and make it a full-time project learning how they work. Also, I wouldn’t call a model a radiative-convective one if it can’t calculate the contribution of convection in real time. Isn’t that contribution most crucial when the static or stable atmosphere is perturbed? Which happens essentially around the clock and causes a lot more than a few degrees of perturbation. • Chic wrote: “Also, I wouldn’t call a model a radiative-convective one if it can’t calculate the contribution of convection in real time.” By specifying the composition, temperature and pressure, one can calculate (from first principles and measured absorption cross-sections) the OLR and DLR flux into and out of any layer of atmosphere thin enough that these inputs can be treated as constants. An average of 240 W/m2 of post albedo SWR passes through the atmosphere with about 162 W/m2 reaching the surface. If average temperature is assume to be constant with time (steady-state), we know that OLR – DLR + convected heat (latent plus simple) must equal the inward flux of SWR at that altitude (a value between 162 and 240 W/m2 which increases with altitude). We can calculate OLR-DLR for any input temperature profile (say one from observations). Convection must be providing the rest of the needed outward flux of heat. That is radiative-convective equilibrium (at least as I understand it). If the temperature at a particular location is not assumed to be in a steady state, then the temperature will rise or fall when convection doesn’t deliver the appropriate upward flux. Falling local temperature tends to: 1) promote convection from below (via a less stable lapse rate from below), 2) suppress upward convection from the location, and 3) suppress radiative cooling. Rising local temperature does the opposite. These tendencies gradually restore local radiative-convective equilibrium. So the concept can be applied to steady state and non-steady state situations. At altitudes where the atmosphere is too optically thick for the necessary outward flux to be maintained by solely by net outward LWR, the lapse rate with be determined by convection. When the atmosphere becomes optically thin enough that convection is no longer needed, we say that atmosphere is in a radiative equilibrium and has lapse rate that can be calculated from first principles. The tropopause marks the altitude where convection is no longer needed. • Chic wrote: “My understanding is that 6.5 K/km is a result of surface evaporation and cloud formation. The latent heat is removed from the surface and released at altitude forcing a reduction in the dry lapse rate.” If relative humidity over the oceans ever reached 100%, there would be no more surface evaporation. So evaporation itself is limited by the rate at which convection carries moist air away from the surface. In fact, there is a very thin adhering layer of air above the ocean where the relative humidity is 100% (equilibrium), and the rate of evaporation is limited by transport out of this thin layer. For this reason, winds are at least as important as SST for evaporation rate. Turbulent convection associated with surface winds mixes a boundary layer (1-2 km thick?) of the atmosphere with some air from above, producing a fairly homogeneous layer with about 80% relative humidity and sometimes boundary layer clouds at the top. Convection raises parcels of air from the boundary layer into the “free troposphere” until clouds form and rain falls. However, during this process some drier descending air is turbulently mixed into the rising air (“entrainment”). 6.5 K/km is the end result of a very complicated process involving dry and moist lapse rates that are not perfectly adiabatic. The large grid cells in climate models can’t reproduce these processes from first principles, so they must be parameterized. I bought Grant Petty’s book on Atmospheric Radiation and his other book on Atmospheric Thermodynamics. Purely physics and chemistry of the atmosphere. Neither climate change nor the IPCC are mentioned in the index. Equivalent texts can easily run$100+ and not be as good.

• on May 25, 2015 at 8:09 pm DeWitt Payne

Chic,

I don’t think you have a grasp of the complexity of what you’re asking. Frank mentioned atmospheric boundary layer above. That’s only a small part of the atmosphere. Here’s an article about the structure of the ABL and its diurnal variation. You can also look at SURFRAD data to see how EM radiation levels and meteorological data vary daily at nine different US sites. We have data, we don’t, and aren’t likely to ever get, a simple conceptual model.

51. on May 26, 2015 at 12:34 am | Reply Chic Bowdrie

Frank,

Evaporation above a body of water would produce less dense air capable of generating its own convection even with an otherwise stable temperature profile (no wind or excess solar insolation etc.). Cooler denser and less humid air will displace the saturated air. Otherwise your brief description of a “very complicated process” seems straight forward. I’m curious why you noted that climate change and IPCC are not mentioned in Petty’s books.

In your later comment you write, “Convection must be providing the rest of the needed outward flux of heat. That is radiative-convective equilibrium (at least as I understand it).” This is the reason a dynamic model would be helpful, so we can understand the radiative-convective dance. There is no static equilibrium in the real world. A dynamic model can eliminate the wind, etc. and just focus on a column of air subjected to a daily dose of sun. The initial conditions will eventually transition into final values that repeat every day. This should allow greater insight into when, where, and how much convection contributes to the heat transfer process.

DeWitt,

I don’t know what I don’t know, but I do know atmospheric physics is complex. I’m just not satisfied with hand-waving conceptual explanations such as Frank and I discussed. Thanks again for the references, in case I start working on my own dynamic model.

• Chic,

Saturated air at 30C contains 4.2% water vapor. That reduces the density 1.6% from that of dry air. The same reduction is obtained by raising the temperature by about 5C. The heat used to evaporate water from 0% to 4.2%, would raise the air temperature by nearly 100C. Thus evaporation is not a important factor on driving convection, rather the contrary. In practice the differences in humidity are much smaller, but the ratio remains the same.

From the point of view of modeling it’s important to notice that radiative heat transfer can be calculated reliably and reasonably accurately, when the state of the atmosphere is known. The relationship between the temperature profile and the strength of convection goes in the way that small changes in the temperature profile lead to large changes in the strength of convection when the adiabatic lapse rate is locally exceeded. Combining these two observations we can conclude that we can calculate most directly
– the temperature profile and

Furthermore we can estimate the total energy transfer from the power of heating of the surface.

The strength of convection can then be deduced by subtracting radiative energy transfer from the total energy transfer. Trying to do it in other ways is bound to be much less accurate than this approach.

As has already been discussed, the lapse rate is well determined by known physics, wherever vertical convection takes place. Diurnal and other temporal variation as well as spatial variability at all scales from tens of meters to thousands of kilometers are another major complication as is also horizontal mixing and all other weather related phenomena. This kind of complexity seems to lead to the typical value of 6.5 C/km for the environmental lapse rate, but deriving the value by a model calculation is surely very difficult, perhaps impossible based on present knowledge.

• on May 26, 2015 at 6:33 pm Chic Bowdrie

Hi Pekka.

Are you sure the air temperature has to change during evaporation? If so it wouldn’t be called latent heat. Evaporation takes heat out of the surface. The energy is transferred, but there’s no change in T. It’s like a phase change. I don’t do climate science, but if there was any doubt about a principle as basic as this, I would be headed to the lab to experiment right now.

You are in agreement with SoD on the impossibility of modelling convection dynamically if that is what you are getting at. I’m not interested in calculating lapse rates or OLR as a function of atmosphere composition. I’m proposing to use the parameters we know, DSR, OLR, atmosphere composition, heat capacities, etc. with the equations we know to produce a dynamic model that illustrates how energy is transferred through the atmosphere. Static models are used all the time to explain how radiation works. It’s time to demonstrate, not just explain, how convection, evaporation, and conduction work in conjunction with radiation. Ignorance is not bliss.

• Chic,

The temperature need not change in evaporation, but the point I tried to make is that temperature gradients are much much more important in driving convection than the smaller mass of H2O as compared with N2 and O2.

The other point I tried to make is that one order of looking at the main variables is likely to be more productive than another. Specifically it’s very difficult to calculate convection from the equations of fluid dynamics like the Navier-Stokes equation, while constraining the strength of convection is possible based on estimated energy fluxes.

I propose that you read the latest blog post of Isaac Held and in particular his paper on the importance of understanding the basics in climate modeling. (Held gave the link to this paper in his answer to SoD.)

Held has written many posts where he discussed, what can be learned from general principles and simple models. That’s an important approach, but many essential issues remain poorly understood. The Earth system is inherently complex, and many of its features may be impossible to understand in any other way than using large complex models. That’s unfortunate, but that may be true.

• on May 26, 2015 at 9:13 pm Chic Bowdrie

Pekka,

It’s not a matter of whether humidity or temperature is more important. Any perturbation of an equilibrium temperature profile may result in convection. I think I follow what you’re saying about energy flux constraining convection, but it’s no consolation if my ultimate goal is understanding the process rather than having an ax to grind. I look forward to reading the comments of Isaac Held. Thanks.

• Chic wrote: “I’m curious why you noted that climate change and IPCC are not mentioned in Petty’s books.”

Climate change and the IPCC are politicized. Get your atmospheric physics and chemistry from someone like Petty who doesn’t over-simplify or distort so that we can all understand/fear the potential looming danger.

SOD once recommend a book called “Elementary Climate Physics”. It has a worked example with a rising characteristic emission level predicting 18 degC warming from 2XCO2! I still haven’t figured out what assumptions permitted this travesty. I also have an aversion to presenting models of the earth with isothermal optically thick layers, which magically emit blackbody radiation from both sides, but don’t have any mechanism (i.e. a temperature gradient) that allows energy to pass through the layer. I have enough trouble figuring out what to believe without having to doubt my reference books.

When you move from atmospheric physics and chemistry to the more complex subject of climate change, you can’t avoid the possibility of bias.

Unfortunately, material from the skeptics is even less reliable.

Grant Petty has nothing to sell – except his passion for clearly explaining atmospheric physics and chemistry and a great price.

• Chic wrote: “I’m just not satisfied with hand-waving conceptual explanations such as Frank and I discussed.

“Hand-waving” hurts, but perhaps it is accurate given the limitations of fluid mechanics. I’d love to read a better explanation, if you ever find one. I struggled with the incompatibility of radiation (W/m2) and convection (degK/km). FWIW, you can track convection of latent heat in the real world by the amount of precipitation that falls. The global average is roughly 1 m/year (1 m3/m2/yr) and that can be converting into W/m2 (with some uncertainty since some comes down as ice, rather than liquid water.) IIRC, the KT energy balance diagram gets its value for latent heat from such a calculation. Simply heat is a fudge factor (“consistent with other data) that makes the radiative imbalance agree with estimates from ARGO)

• on May 27, 2015 at 10:11 am Chic Bowdrie

I hope you aren’t taking the hand-waving remark as a criticism. Everybody does it and it is one of the ways we learn from each other. Not everyone has the time or expertise to do the elaborate maths that SoD provides.

I should probably just leave it at that, but then you suggest convection has units of degK/km. That is a temperature distribution. Convection moves energy through the atmosphere and therefore has to have units of W. Precipitation would account for some convection, but wouldn’t there still be much convection occurring even during periods of clear skies?

52. on May 27, 2015 at 12:00 am | Reply Chic Bowdrie

From Isaac Held’s “The Gap Between Simulation and Understanding in Climate Modelling:”

“Conceptual research versus hierarchy development. A theoretically inclined researcher might design and build a model for a particular purpose and then discard the model. The model is not intended, in many cases, to have life of its own, but is, rather, a temporary expedient. In the limiting case, the model is not fully described and the result not fully reproducible. Or, an existing model might be used in the same way, but with the focus on the concept, not on the model itself. I refer to this as conceptual research. Much of the best work with comprehensive models can be classified as conceptual, as can, for example, much of the paleoclimatic research with computationally efficient climate models of intermediate complexity. In this context the model is a useful tool that helps one think about the system and search for ways in which to interpret observations.”

“The health of climate theory/modeling in the coming decades is threatened by a growing gap between high-end simulations and idealized theoretical work. In order to fill this gap, research with a hierarchy of models is needed. But, to be successful, this work must progress toward two goals simultaneously. It must, on the one hand, make contact with the high-end simulations and improve the comprehensive model development process; otherwise, it is irrelevant to that process, and, therefore, to all of the important applications that are built on our ability to simulate. On the other hand, it must proceed more systematically toward the creation of a hierarchy of lasting value, providing a solid framework within which our understanding of the climate system, and that of future generations, is embedded.”

My desire for a conceptual dynamic model illustrating the synergism between convection and radiation is probably at the bottom end of Helm’s hierarchy of models. Somewhere somebody must have been there, done that. However, it would certainly be an improvement over the static radiation models front and center in most presentations of the greenhouse effect.

• Chic,

My desire for a conceptual dynamic model illustrating the synergism between convection and radiation is probably at the bottom end of Helm’s hierarchy of models. Somewhere somebody must have been there, done that. However, it would certainly be an improvement over the static radiation models front and center in most presentations of the greenhouse effect.

[Isaac Held is the author, not Helm].

The challenge is modeling convection – movement of the atmosphere. Lack of success in modeling this turbulent atmosphere isn’t because it isn’t seen critically important or because somehow climate science thinks “radiation is it”.

Convection has way more focus in climate science than calculations of radiation. Interested people reading basic explanations of the greenhouse effect will obviously see a lot about radiation because it – along with the lapse rate – are the key determinants of the basic greenhouse effect.

Actual climate science research efforts have a completely different focus from presentations of the building blocks of basic science to the public and enthusiastic climate science amateurs.

Radiative-convective models – such as Thermal equilibrium of the atmosphere with a given distribution of relative humidity, Manabe & Wetherald (1967) and Climate Modeling through Radiative-Convective Models Ramanathan & Coakley (1978) – are perhaps the simplest step forward. These rely on the observation that when the temperature profile reaches a certain point, convection is initiated. In the tropics the calculated adiabatic lapse rate is quite close to the actual measured lapse rate.

All models of atmospheric circulation require parameterizations. There are many thousands of papers on the subject of atmospheric circulation. Held himself has made some great contributions. You can see the kind of work in a paper like Nonlinear axially symmetric circulations in a nearly inviscid atmosphere, Held & Hou (1980), or from another well-known author in Baroclinic Instability, Pierrehumbert & Swanson (1995). These papers -and pretty much every paper on atmospheric circulation – unfortunately jump straight into intensive maths.

If you are interested in gaining understanding of the topic I recommend An Introduction to Dynamic Meteorology, James R. Holton. First I recommend reading a more introductory book on fluid dynamics that derives and explains the Navier-Stokes equations. I think Holton’s book is excellent although it is a little too difficult for me. I was never able to really get an understanding of conservation of potential vorticity although I am determined to give it another go.

• on May 27, 2015 at 10:17 am Chic Bowdrie

Good stuff. Of course I’m interested in gaining understanding. I just don’t have the time right now. If you have, please give it another go. Your blog is a great classroom.

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57. In another article – The Debate is Over – 99% of Scientists believe Gravity and the Heliocentric Solar System so therefore.., Bob Armstrong asked (and I relocated the question here):

Yes gravity has very explicit equations including that light blue shifts , ie : gets hotter , descending in a gravitational well .

GHG has NONE quantitatively explaining why the bottoms of atmospheres are hotter than their radiative equilibrium as see from outside . All that is required is a simple differential quantitatively explaining the trapping of thermal energy by a spectral filter .

On the other hand , explicit simple calculations based on the laws of gravity have been presented which impressively match atmospheric temperature profiles .

• on August 1, 2017 at 4:03 am | Reply DeWitt Payne

Oh, puhleeze. Not that again. If it were just gravity, the surface wouldn’t stay hotter.

• That’s exactly why it can . The gravitational energy cannot be ignored . It must be accounted somehow or you have a massive hole in your theory .

And , Mike , there was a computation in a comment on WUWT which calculated that same 9.8k/km . But , as I say below , my priorities are to make succinct notation to express these relationships run on current and future chips so someone else would have to motivate taking a closer look .

• Bob Armstrong wrote: “simple calculations based on the laws of gravity have been presented which impressively match atmospheric temperature profiles ”

If it were gravity, the deep ocean would be warmer than the surface ocean, there would be no stratosphere, and the atmospheric lapse rate would be 9.8 K/km. The only thing impressive about a theory that matches exactly zero observational data is that anyone would believe it.

• Deep ocean temperature is the “anomaly” to be explained . ( It’s interesting that it is quite close to the gray body temperature in our orbit . ) Of course unlike land where the temperature gradient with depth is consistent , you have massive convection from the poles to the tropics carrying the cold water along the bottom .

It was only after showing the absurdity of Venus’s extreme bottom of atmosphere temperature could be explained as a spectral effect , http://climateconferences.heartland.org/robert-armstrong-iccc9-panel-18/ , that comments by HockeySctick and others on WUWT with calculations got thru to me that gravity cannot be left out of the equations and in fact fills the gap between the radiative equilibrium temperatures and the bottom of atmosphere temperatures .

More recently , the fact that gravity causes light to blue shift “settles” the case for me . As Alan Guth , among others , explains very simply , gravity computes as a negative energy . It must be balanced in the thermal equations .

My focus is on implementing “the shortest path from the chip to the math” in my CoSy Forth based APL . So I’ll wait for someone interested in implementing these computations in CoSy to really look closely at them .

However , it remains the case that fundamental simple calculations based on gravitational energy have been presented which convincingly match atmospheric “lapse rates” . Despite all the verbiage , I have yet to see what should be a simple experimentally testable differential explaining the asymmetric “trapping” of energy by some electromagnetic phenomenon .

[ nb : took me a couple of times , but I can appreciate the maintaining coherence of particular subject threads . I’ll try to emulate that on my nascent http://cosy.com/Science/ComputationalEarthPhysics.html .
However , title of the other thread : “… 99% of Scientists believe Gravity … ” almost demands the observation ” except in Climate Science ” .

• Bob Armstrong,

No matter how much gobbledegook you spew, it does not change the fact that the theory you promote does not agree with the observational facts. Whereas theories based on actual physics, such as the laws of Thermodynamics, do agree with the facts.

• Please point me to the specific equation by which one computes the “trapping” of thermal , ie : kinetic , energy on Venus to more than 25 times the energy density the Sun supplies to it’s orbit ( its surface temperature to ~ 2.25 times the gray body temperature its orbit ) . Actually the ratio is much greater given Venus’s extreme reflectivity over the peak of the solar spectrum .

Perhaps Ironically , I think it might have been Pikka here who steered me to TiNOX , which I used in this slide to show youall are claiming Venus creates an even higher solar heat gain than humanity has yet produced .

Just show me ( us ) that calculation and we’ll become believers . Otherwise you are just engaging in untestable word waving no how many irrelevant equations you include in your . Alternatively show us an experimental demonstration of the trapping . I , in fact , am working to establish a prize for best “YouTube” demo of relevant basic physics .

On the other hand , youall are ignoring gravity . That’s inexcusable . Please provide the equations by which it has no effect on the temperature profile of atmospheres altho it does on light .

• Bob Armstrong wrote: “Please point me to the specific equation by which one computes the “trapping” of thermal , ie : kinetic , energy on Venus”

That is a doubly ignorant request. Is there a “specific equation” used to design a bridge? Of course not. And there is no such thing as trapping of thermal energy. To actually understand the processes that do occur, you need to invest some time and effort. Here is a good place to start: https://scienceofdoom.com/roadmap/atmospheric-radiation-and-the-greenhouse-effect/

Bob Armstrong wrote: “Please provide the equations by which it has no effect on the temperature profile of atmospheres …”

Well, that one is actually easy: dS = dq_rev / dT.

Bob Armstrong wrote: “Just show me ( us ) that calculation and we’ll become believers . … On the other hand , youall are ignoring gravity .”

You are obviously ignoring all that SoD has posted here, so it seems I am wasting my time. So you now go on my “ignore” list.

• Could you please point out what the terms in your dS = dq_rev / dT in SI parameters are so I can apply it to some simple case to demonstrate the trapping of energy ?

I know that SoD has presented a LOT of material . In a sense , I just want a little . A handful of APL expressions gets us from the parameters of the Sun to the ~ 278.6 +- 2.3 from perihelion to aphelion for a gray ball in our orbit .

The expression at http://cosy.com/Science/warm.htm#EqTempEq for arbitrary spectra will give the radiative equilibrium temperature for our measured absorptivity=emissivity spectrum as seen from space — and that produces the 255K value when fed that crude hypothesis .

What I’m looking for is simply the next expression which gets from that balance to our surface temperature , somehow overcoming the constraint of the Divergence theorem that the average interior temperature has to match the equilibrium calculated for the spectrum as seen from outside .

But , in any case , you are still left with how to incorporate the gravitational energy gradient into your system . Somehow you have to account for gravity having a heating , blue shift , effect on radiation but , by your ignoring of it , not on matter . Please explain that .

• Bob Armstrong,

That is actually a somewhat reasonable comment, so I will suspend your being on my ignore list.

You wrote: “Could you please point out what the terms in your dS = dq_rev / dT in SI parameters are”.

S is entropy, q_rev is reversible heat transfer, T is absolute temperature, the d’s indicate derivatives. You can learn how to use it from any book on thermodynamics, in association with the Second Law. You will find it a lot of work to do that.

“so I can apply it to some simple case to demonstrate the trapping of energy?”

Since everything tends towards equilibrium, there is really no such things as the trapping of energy. Best you can do is to slow down the approach to equilibrium.

You wrote: “I know that SoD has presented a LOT of material . In a sense , I just want a little.”

In other words, you don’t want to do the work required for understanding, even when someone like SoD goes through a lot of trouble to make it as easy as possible. I can’t help you there.

You wrote: “somehow overcoming the constraint of the Divergence theorem that the average interior temperature has to match the equilibrium calculated for the spectrum as seen from outside.”

There ain’t no such thing. Off hand, I can’t imagine what it is that you think demonstrates that. Pour hot coffee into a thermos bottle and put ice water into a second, identical bottle. The surface temperatures will be virtually identical, even though the interior temperatures are very different.

You wrote: “But, in any case , you are still left with how to incorporate the gravitational energy gradient into your system.”

Gravity has no significant effect on conductive or radiant energy transfer. It does affect convective transfer. SoD discusses that: As I recall, he starts with radiant transfer only, then considers the effect of convection.

You wrote: “Somehow you have to account for gravity having a heating …”

It doesn’t. Heating requires a transfer of energy and therefore a source of energy. Gravity is not a source of energy.

I am thinking that my first sentence above was a mistake.

• Could you define “q_rev is reversible heat transfer” ? “reversible” makes it sound like a symmetric operation and the your explanation makes it sound like simply a conversion between temperature and entropy .

Where is the equation of the form T1 = f[ T0 ] * dx , such that T1 > T0 ? The differential need to be in space , not temperature .

And you are simply wrong about the divergence theorem .

Sorry , I have a party to prepare for : http://cosy.com/y17/CoSyMidSummerMela2017.html . So I’m signing out for now . I’m still not convinced that any of youall really understand how to calculate the temperature of a billiard ball under a sun lamp . That requires agreement on the computation for arbitrary spectra at http://cosy.com/Science/warm.htm#EqTempEq , and even that basic computation seems to be impossible to get agreement on . That would be a starting point .

And the gravitational influence , ie : energy , which MUST be balanced in the equations is NOT trivial .

• I didn’t read carefully to the end of your comment :
“”You wrote: “Somehow you have to account for gravity having a heating …”

It doesn’t. Heating requires a transfer of energy and therefore a source of energy. Gravity is not a source of energy. ”

That’s absurd . And I’d refer you to the appendix on Alan Guth’s Inflationary Universe book explaining why gravity computes as a negative energy — which is exactly what is needed to balance the temperature gradient with the “ToA” radiative balance . The simplest computation is the 9.8K % km mentioned earlier . I’ve got other priorities than to explore the nuances , think mass density must enter into it , but that energy MUST be accounted for .

Further , I’ve been reading Huseyin Yilmaz’s 1965 Relativity and Modern Physics which cites the then recent Pound Rebka experiment ( google it ) demonstrating the blue shifting of light by gravity .

• on August 3, 2017 at 6:46 pm DeWitt Payne

Bob Armstrong,

Gravity and heat capacity at constant pressure determine the adiabatic lapse rate, Γ_d = g/C_p. By convention, this is a positive number even though temperature decreases with altitude in the troposphere. However, gravity does not determine the temperature. That depends on the rate of incident solar energy transfer to the surface and the rate of loss of energy from the surface by radiation and convection. When the sun goes down at night, it gets cooler and when it comes back up, it gets warmer. The gravitational acceleration and the heat capacity, however, don’t change. Gravity alone also does not explain why the temperature in the stratosphere increases with altitude. Absorption of incident solar radiation by oxygen and ozone and emission of radiation by ozone and CO2, however, does.

• DeWitt , I’m not familiar with your notation , but definitely the notion that gravity effects only pressure is the standard GHG paradigm assumption . However it won’t satisfy the total energy balance with the spectral radiative equilibrium required by the Divergence Theorem . Incidentally I think it is interesting that the temperature goes down from the gray body ~ 279 as one enters the top of the atmosphere . While I don’t like asserting things I can’t compute , I would imagine that’s where the “darkness” of our IR spectrum shows its effect . And , as you point out the absorption of , particularly UV by O2 & O3 create a heated layer .

I do wish I had more time to deal with these issues in more detail , but it’s the language to express the computations which is my business . But a point of APLs is that it is rather trivial to map the equations , ie : differentials , over a sphere , or even a spherical shell of voxels .

I’ll repeat , tho , that I just want to see the one dimensional differential , in x , “trapping” energy by , from everything I know , symmetric electromagnetic forces . I think it quite useful in thinking about the problem to think of it horizontally rather than vertically . I’ve made a diagram which I’ve never totally fleshed out , but which does capture the fundamental claim . The total spectral filter function of the atmosphere can simply be collapsed as a product across all the layers . The hypothesis is that the average of the temperatures at b and c is greater than the average at a and d .

So , show me the equations .

• DeWitt , Let me also comment that in the case of Venus where the GHG hypothesis is not tenable both because the bottom of atmosphere temperature is so extremely beyond even the greatest solar gain that humanity has managed to create , but because only ~ 3% of solar radiation even reaches the surface , the dark side surface temperature does stay hot even thru a , what is it , half-year night .

• Bob Armstrong,

You wrote: “the notion that gravity effects only pressure is the standard GHG paradigm assumption.”

It is much more than that. It is basic physics, specifically the Second Law of Thermodynamics.

You wrote: “So , show me the equations.”

Why? You are obviously too lazy to take the trouble to understand them. If you weren’t, you could find them on this web site, with just a little effort.

• Once again you are exactly wrong . It is the GHG hypothesis which repeatedly has be stated to violate thermodynamics . And when you presented an equation it was irrelevant not being a function over distance — which is the critical claim . And as for the requirement for computable equations , I created this slide for my Heartland presentation :

I have pointed very specifically at the experimentally testable equation for equilibrium temperature for arbitrary spectra at http://cosy.com/Science/warm.htm#EqTempEq but have not been able to get either agreement , disagreement , or request for clarification . It is an absolutely fundamental non-optional computation if you are going to make a computable trail from the parameters of the sun to our observed temperature . It’s not a request to wander the web once again for the testable equations I have never found . It’s a specific testable foundation upon which we either can reach mutual understanding or it’s not science — and actually , despite differences , I believe SoD is one of the few blogs , which like Anthony Watts WUWT is ultimately seeking honest understanding .

So , Mike , There’s a specific simple equation in terms of the ratio of dot products of source , object and sink spectra . Is it right or wrong ?

• on August 4, 2017 at 3:11 pm DeWitt Payne

Bob Armstrong,

I’ve done the radiative transfer calculations for the atmosphere near the surface of Venus. The difference between the upwelling IR from the surface and the downwelling IR from the atmosphere is on the order of 15 W/m², very close to what you would expect from the global average of the small amount of solar energy that reaches the surface. It’s about as much as you would get on the Earth with heavy cloud cover during the day. The pictures from the Venera probe were taken using available light.

The Venusian atmosphere is almost, but not quite, opaque in the spectrum of EM radiation emitted by the surface. It would be less opaque and the surface temperature lower if there weren’t small amounts of water vapor and sulfuric acid in the atmosphere. Those compounds fill in the holes in the spectrum.

Gravitation alone also doesn’t explain where all the CO2 in the Venusian atmosphere came from.

• I don’t see how gravitation would have anything to do with where the CO2 came from . I do wonder why it’s there and so dense .

But I just want the equations . Do you have a link to your calculations ? Maybe we can sort some of this out . But my time is limited . My priority is http://CoSy.com , the language .

Perhaps I’ll get back to it Monday . My first step is to pull up the calculations of radiant energy density corresponding to the ~ 750K temperature .

• and Bob also replied in that article, his comment relocated to here:

. I’ll just note that if gravity is not included , one of the 2 macroscopic forces and their associated energies is left out of the equations . — of which I find none on your GHG explained page .

58. Over in another article – The Debate is Over – 99% of Scientists believe Gravity and the Heliocentric Solar System so therefore.., mark4asp posted a comment. I’ve moved it here for the reason explained in that article:

CO2 is not responsible for 24% of the GHGE as you claimed at least once here in this blog. I doubt you can scientifically show how much of the GHGE is due to CO2 and how much to water. The atmospheric ratio of water:CO2 varies from 95:1 (equator) to 1:1 (poles). But polar GHGE is small because the land temperature is low. Equatorial GHGE is higher and it certainly cannot be 24% due to CO2. Because water is a far more efficient GHG than CO2, and there is far more water. CO2 has 3 absorption bands. Water has dozens.

Your proofs are not indisputable when they are based on unverified assumptions; as all climate models are.

PS: unverified = not shown scientifically.

I noticed a fascinating thing about Google. When one Googles “GHGE water” and clicks on images. There are none. All the leading images omit water. That’s believer ‘science’ for you!

59. Bob: The idea that gravity creates a lapse is based on the ASSUMPTION that the sum of kinetic and potential energy of gas molecules is always a constant. This is incorrect. If potential and kinetic energy were distributed evenly throughout the atmosphere, the lower atmosphere would be enriched in gases with high MW (CO2 and Ar).

Here in the troposphere, between collisions a gas molecule falls under the influence of gravity for a few picoseconds before it collides with another molecule. The increase in kinetic energy from falling is ridiculously small compared with its average kinetic energy and is followed by a collision that could leave it with almost no kinetic energy or twice as much. There is no way to keep the kinetic energy added to a falling molecule from being exchanged with all neighboring molecules. There is a website where you can watch gravity “being turned on” in a group of colliding molecules, see them gain kinetic energy as they fall and then see that kinetic energy immediately diffuse throughout the gas, leaving an isothermal gas with a density gradient.

At the edge of space, collisions are rare and the amount of kinetic energy that can be gained or lost between collisions can be significant. Above about 100 km, the atmosphere is enriched in lighter gases. However, the concept of a thermodynamic temperature (proportional to mean kinetic energy) doesn’t even exist at these altitudes, so the concept of a lapse rate is dubious. In the troposphere, however, we know that convection must be responsible for creating the lapse rate because there is no isotopic enrichment.

• Thank you Frank . I only have time for a brief comment , but all you say with respect to the mixing of different weight molecules strikes me as a good description and is what I’d expect . ( Again , give me the equations ( and a motivation ) and I’ll see how succinctly they can be expressed in CoSy . )

I would only comment that I don’t think convection is intrinsic in the thermal gradient . It’s powered by the thermalization of the radiation occurring at the bottom . Were the absorption to occur at the top , which must be to some extent true of the high altitude UV absorption , there would be no convection , but still the temperature.pressure gradient — as there is as you continue into the solid earth .

• Bob: Although others here have disagreed with me, I am of the opinion that the lapse rate we observe depends on the mechanism that transfers heat vertically the fastest through the atmosphere. In the post below, SOD calculates the lapse rate for a graybody model of the atmosphere in pure-radiative equilibrium with no convection. It varies linearly with optical density, which increase exponentially as one approaches the surface.

https://scienceofdoom.com/2010/08/08/vanishing-nets/

Fastest Mechanism: Heat transfer by convection with unstable lapse rate. Result: linear lapse rate = Cp/g.

Next fastest: Radiative equilibration. Linear with OD.

Next fastest: Thermal diffusion by collision. Result: Isothermal. See Feynman for example.

Slowest: Molecular diffusion. Lapse Rate also = Cp/g, but with isotopic fractionation.

Others correctly note that temperature is not defined when collisions are infrequent, so they object to molecule diffusion. When you have a single molecule in a box with elastic walls, its kinetic energy varies linearly with height, but it doesn’t have a temperature.

• Frank , have even less time .

But one quick question to know what I’m reading :
Does “graybody” really mean “gray” , ie : flat spectrum across the whole spectrum , ie : Absorptivity = Emissivity = k
across the whole spectrum so that it drops out of the equation and the equilibrium temperature simply corresponds to the total radiant power impinging on the body , ie : the same temperature as a black body , ie : ~ 278.6 +- 2.3 from peri- to ap- helion in our orbit ?

Or is it the sort of screwy inexcusably crude step function with one constant AE fraction across the peak of the solar spectrum and a value of 1.0 in the longer wavelengths ? Is it this bit of confusing uselessness which produces the extreme 255K when you plug in 0.7 AE , in CoSy working right at the x86 register level ,

.7 .25 2_f ^f 278.6 _f *f |>| 254.83

Please tell me “gray” means fully flat spectrum . That’s the only value useful in computations .

• I think others need to explain why that simple change in velocity ceases as average path length decreases , ie : pressure increases .

Here’s a link to one of HockeySchtick’s analyses : http://hockeyschtick.blogspot.com/2014/11/derivation-of-entire-33c-greenhouse.html . There are some proponents of the notion who are as mathematically incompetent as much I’ve seen on the GHG side , but Hockey seems quite competent and has references for the “thermo-gravitational” effect going back to Maxwell .

( I think it bizarre that now days important scientific references are to people going by names like “HockeySchtick” and “ScienceOfDoom” . )

Happy MidSummer !

• on August 5, 2017 at 2:05 am DeWitt Payne

Frank,

It’s g/C_p.

I don’t know what you mean by having ‘molecular diffusion’ different from thermal diffusion. If you don’t allow collisions, you still have no temperature gradient at equilibrium, assuming you start with a Boltzmann kinetic energy distribution. The kinetic energy distribution will still be the same for a given atom or molecule at all altitudes, i.e. isothermal. By the way, if temperature is undefined, then so is heat capacity.

As Wikipedia puts it:

If several systems are free of adiabatic walls between each other, but are jointly isolated from the rest of the world, then they reach a state of multiple contact equilibrium, and they have a common temperature, a total internal energy, and a total entropy.[7][8][9][10] Amongst intensive variables, this is a unique property of temperature. It holds even in the presence of long-range forces. (That is, there is no “force” that can maintain temperature discrepancies.) For example, in a system in thermodynamic equilibrium in a vertical gravitational field, the pressure on the top wall is less than that on the bottom wall, but the temperature is the same everywhere.[my emphasis]

https://en.wikipedia.org/wiki/Thermodynamic_equilibrium

• Bob: Invest the time to read SOD’s post about the lapse rate that would be found with pure radiative equilibrium. It involves a “gray atmosphere” (not gray-body, I was rushed too) with equal absorbance at all wavelengths of outgoing thermal radiation (and no absorption of incoming SWR). This is also found in basic textbooks.

The average lapse rate we observe is the product of SEVERAL mechanisms of heat transfer, mostly radiation and convection. They are summarized by the phrase “radiative-convective equilibrium”: In locations where purely radiative equilibrium produces lapse rates that are unstable to convection, the average lapse rate is near the moist adiabatic lapse rate. As best I can tell, those locations are the troposphere in the tropics and subtropics and seasonally parts of the temperate zone. These heat transfer processes are so much faster than thermal and molecular diffusion that the latter aren’t responsible for what we observe. (The argument for a gravitationally-derived lapse rate is based on molecule diffusion, and ignore the much faster processes of convection, radiation and collision.)

The post below discusses the temperature of the atmosphere would be if all the air in any location were compressed to 1 atm (potential temperature) and all of the water vapor condensed (moist potential temperature). Regions of the atmosphere that are well mixed vertically by convection have the same moist potential temperature with altitude. The atmosphere is stably-stratified when the moist potential temperature rises with altitude.

https://scienceofdoom.com/2012/02/12/potential-temperature/

• Jeez , I still got cleaning to do & a beer run .
But , I’ll try to find that post .
I’ll repeat tho , that I’m only interested in the single equation which causes an asymmetric trapping of energy by spectral effects — one which would work as lawfully in colored glass as a convective medium , one which works sideways a well as up and down .

The only differential I know of for this general case is the Schwarzschild . Under what parameter domain does it trap a higher energy density on the side away from a source ?

BTW : Anthony Watts is also very dismissive on the gravitational explanation , but allows rational discussion . And ironically it’s in comment there that , having found NO computable explanation of Venus’s 25 times greater bottom of atmosphere energy density than that the Sun supplies to it’s orbit , I saw the explicit equations of HockeySchtick and others which provide remarkably good first order fit to the atmospheric profiles .

I guess that defines a paradigm — when both sides rule out an obvious , non-optional parameter .

• DeWitt asked: “I don’t know what you mean by having ‘molecular diffusion’ different from thermal diffusion.”

Imagine two atmospheric layers of neon at T1 initially resting on top of a layer of argon at a lower temperature T2. I’ve picked gases that don’t absorb thermal radiation. Postulate no convection or radiation. Thermal diffusion brings them to a uniform temperature. Molecular diffusion provides molecular homogeneity. I believe – but I’m far from sure – that the former process operates faster than the latter. If it didn’t, would equilibrium be isothermal?

Now think about solid. They have thermal diffusion, but no molecular diffusion.

• on August 5, 2017 at 2:39 pm DeWitt Payne

I believe you should use the term ‘conduction’ instead of thermal diffusion to avoid confusion when talking about heat transfer. There is, in fact, thermal diffusion of interstitial or otherwise loosely bound components of solids, like copper in gold, but it isn’t heat transfer. Conduction is indeed faster than diffusion, especially in good conductors like metals. Conduction leads to thermal equilibrium even with dissimilar materials in contact with each other, like your example of neon floating on argon.

The main problem, however, is that there will be turbulence in any large system with sufficient density to allow local thermal equilibrium, like a planetary atmosphere. Turbulent mixing, sometimes called eddy diffusion, is orders of magnitude faster in a gas or liquid than conduction. That’s also why the different components of the atmosphere are well mixed below the turbopause at ~100km altitude for the Earth. The only place conduction dominates heat transfer in the atmosphere is the very thin layer at the surface.

Assuming no turbulence, your neon/argon system would reach thermal equilibrium long before it reaches homogeneity.

• DeWitt: “Assuming no turbulence, your neon/argon system would reach thermal equilibrium long before it reaches homogeneity.”

Good, I think we agree on the basics. Extra credit question: How does one calculate the ratio of the thermal conductivity to molecular diffusion? (I presume the units have the same dimensions.)

I understand that turbulent mixing occurs above certain Reynolds number and that the scale of fluid flow in the ocean or atmosphere produces turbulent flow perpendicular to any surface. I didn’t know that the phrase eddy diffusion referred to the same phenomena. I understood that that boundary layer was the extent of the turbulent layer near the surface, but the free atmosphere above is turbulent (see clouds) for reasons I don’t understand as clearly. X-ray crystallographers used to send experiments into space so they could grow large protein crystals in the absence of turbulence in microgravity.

• Frank,

You wrote: “I understand that turbulent mixing occurs above certain Reynolds number … I understood that that boundary layer was the extent of the turbulent layer near the surface, but the free atmosphere above is turbulent (see clouds) for reasons I don’t understand as clearly.

In the first sentence, you actually gave the answer to the second sentence. The Reynolds number is V*L/nu where V is velocity, L is the length scale of the system, and nu is kinematic viscosity (see note below). Since L is huge in systems like the atmosphere and ocean, flow is everywhere turbulent, except for above about 100 km.

Convective stability does not prevent turbulent diffusion, but it weakens it by suppressing the largest eddies. Convective instability amplifies the largest eddies and produces large convective cells that greatly enhance vertical transport over what would otherwise occur from turbulent diffusion. The boundary layer is the region in which such convective cells occur.

Note: The kinematic viscosity, nu, is the dynamic viscosity (often given in units of poise) divided by density. It has units like m^/s, the same as the diffusion coefficient. In simple versions of the kinetic theory of gases, it is numerical the same as the diffusion coefficient, but in reality it is somewhat different. In liquids, they are very different.

Frank asked: “How does one calculate the ratio of the thermal conductivity to molecular diffusion? (I presume the units have the same dimensions.)”

The thermal diffusivity, alpha, is related to the the thermal conductivity, k, by
alpha = k / (rho*CP)
where rho is density and CP is heat capacity. Alpha has the same units as diffusion coefficient and in gases they have similar values.

• on August 5, 2017 at 6:22 pm DeWitt Payne

Mike M.,

I stand corrected. Concentration (not homogeneity in a gravitational field) and thermal equilibrium would take about the same amount of time in a column of gas. For any large column height, that would probably be a very long time.

• Frank wrote: “If potential and kinetic energy were distributed evenly throughout the atmosphere, the lower atmosphere would be enriched in gases with high MW (CO2 and Ar).”

There seems to be a subtle but important error here. Imagine a tall, perfectly insulated tube containing a gas mixture, say Helium and Argon. At thermal equilibrium, the temperature will be the same everywhere in the tube and independent of height. But the mixing ratio of Argon will be greater at the bottom of the tube than at the top.

• on August 5, 2017 at 6:04 pm DeWitt Payne

Mike M.,

A key caveat is ‘in the absence of turbulence’. That condition turns out to be very difficult to meet in the real world for anything larger than a few cubic centimeters and a short time. Perfect insulation is also thought experiment territory, not the real world.

However, very high g in a centrifuge can be used for isotope separation, so it might be possible to devise an experiment that wouldn’t be possible at 1g to show that acceleration alone does not create a temperature gradient. I doubt anyone would want to bother doing it, though.

60. Thank you for moving my comment here. Climate scientists say CO2 is responsible for 24% of the GHGE. So if the total effect is 33C, 8C of that is due to CO2. So climate scientists say. Impossible. Because the Water vapour : Carbon dioxide ratio at the tropics is about 80 : 1. Water is a more powerful GHG. Sensible people believe the CO2 contribution to the GHGE is between 5% and 10%. Which is what I think. I wrote a blog to explain why. The net effect is that CAGW is cancelled due to lack of interest from CO2. https://greenfallacies.blogspot.co.uk/2017/08/greenhouse-gas-effect-ghge-is.html

• mark4asp wrote: “Carbon dioxide ratio at the tropics is about 80 : 1. Water is a more powerful GHG.”

The mixing ratios of absorbing gas at the surface is irrelevant. What matters is the amounts in the upper troposphere (around 5 km) since that is the source of most emission to space. The amount of water vapor at such altitudes is dramatically less than at the surface.

• on August 5, 2017 at 7:21 pm | Reply DeWitt Payne

mark4asp,

If you want to see where the 24% comes from, go to

http://climatemodels.uchicago.edu/modtran/

This site has been recently modified to cover a wider spectral range so that the calculated emission is much closer to the value calculated using the Stefan-Boltzmann equation.

Looking at the Tropical Atmosphere and setting the height to zero, looking down, upward IR heat flux is 445.252 W/m². That’s pretty close to 448.29 W/m² calculated from the S-B equation using T = 299.7K and an emissivity of 0.98. If I zero out everything and set the altitude to 100km looking down, the upward flux is 442.112 W/m². So everything you can’t set to zero reduces the upward flux by 3.14W/m². Add back everything else and the upward flux is 298.017 W/m². The total greenhouse effect is 445.252-298.017 = 147.215 W/m².

Setting just CO2 to zero gives an upward flux of 329.072 W/m². 329.072 – 298.017 = 31.055 W/m². 31.055 / 147.215 * 100 = 21.1%, not 5-10%. I’m not going to go through the exercise of looking at all the atmospheres with and without clouds and averaging, but you should get the picture. 24% is a reasonable approximation.

61. At equilibrium, temperature is independent of altitude.

Let’s imagine that we have a non-condensable gas in a very tall, perfectly insulated tube. There is no flow of energy in or out. Start the gas out with a vertical temperature distribution matching the dry adiabatic lapse rate. Some people claim that the gas is then at equilibrium, but they are mistaken.

At equilibrium, entropy is a maximum. Now imagine we reversibly transfer an amount of heat, Q, from the gas near the bottom of the tube, where the temperature is TH, to the gas near the top, where the temperature is TL. The net change in entropy is

delta_S = (Q/TL) – (Q/TH).

Since TH > TL, delta_S is positive. Therefore this process is spontaneous. Heat will flow until the temperature is everywhere the same, at which point we have equilibrium.

• You are neglecting the change in gravitational energy .

• A change in gravitational energy is work. Exchanging energy as work has no effect on entropy. If it were not so, we could build perpetual motion machines. My calculation only “neglects” gravity in the same sense as it “neglects” the Dow Jones Average.

• No, he’s not.

An isothermal column of gas under the influence of gravity has constant total (kinetic plus vibrational/rotational plus gravitational potential) energy per unit volume with altitude. The increase in gravitational potential energy of the molecules with altitude is matched by the decrease in the number of molecules per unit volume with altitude.

For the argument that a gravitation induced temperature gradient would violate the Second Law see:

http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/

or any of several articles by Robert G. Brown at WUWT.

• DeWitt,

You wrote: “An isothermal column of gas under the influence of gravity has constant total (kinetic plus vibrational/rotational plus gravitational potential) energy per unit volume with altitude.”

I feel a bit bad about disagreeing with you when you are agreeing with me, but that is not right. The pressure decreases exponentially with height while the gravitational potential energy per molecule decreases linearly. So the two effects can not cancel. Also, two gases can have the same molecular weight but different heat capacities, in which case the condition you describe can not apply for both gases.

At your link, Willis also makes an error in his nice Stygia example in that it is the Second Law rather than the First Law that would be violated in that example.

There is no requirement of equal energy. The uniform temperature is all due to the Second Law. You can prove that either using entropy, as I did above, or by showing that it any other result allows for perpetual motion, as Willis did.

• on August 6, 2017 at 2:56 am DeWitt Payne

Mike M.,

I wonder why nobody pointed that out in the comments? Oh, well.

Yes, it’s the Second Law, not the First. Energy would be conserved because the whole Stygian atmosphere would cool if you were able to extract energy from a gravitationally created temperature gradient. But the Second Law says you can’t do that.

62. A few comments from the article “The Debate is Over – 99% of Scientists believe Gravity and the Heliocentric Solar System so therefore..” moved here..

I’m tired of what was Global Cooling 1970’s, that became Global Warming
and now Climate Change. Climate is synonymous with change and so the phrase is meaningless. It is used to inflict guilt on a population that has no choice but to burn fossil fuels. This is thanks to a science community that suppresses alternative energy supplies. This is reality if you can handle it?

Sure. And the oil companies suppressed a carburetor that would get 100 miles per gallon. Not.

Crystal radios were powered by the RF energy of the broadcast signal, not some magical source of power. They only work with amplitude modulated signals and then only if you’re fairly close to the transmitter and have a good antenna.

If you’re so confident, build a working model.

I hope it’s clear that I have substantial respect for a lot presented on this blog ( including the effort at developing “building blocks” for various sub-topics ) .

But DeWitt’s answer to the notion of “free” energy :
“If you’re so confident, build a working model.”
prompts my response : If you’re so confident some spectral effect can “trap” energy , show us an experimental demonstration of it .

BTW : I remember cleaning and poking around with a cat’s whisker on a germanium crystal to get weak headphone audio from 50kw Chicago stations 40km away .

Bob Armstrong challenged: “If you’re so confident some spectral effect can “trap” energy , show us an experimental demonstration of it.”

Given the average temperature of the surface of the Earth, it should emit an average of about 390 W/m2. Spacecraft show that only 240 W/m2 of thermal IR is leaving the top of the atmosphere. Laboratory measurements show us that GHGs are the only gases that can change this flux by absorption and emission (not “trapping”). GHGs in the atmosphere slow down the rate at which heat escapes the planet – a form of insulation – thereby allowing it to be warmer than it would be otherwise.

For the same reason, increasing GHGs will cause some warming. Readers of this site know that the appropriate question is “How much?”, not “Will it?”

Here I go feeding the creature said to live under bridges. *sigh*

It’s easy. Take a well insulated cardboard box with one end open. Paint the bottom black and cover the other four sides with reflective film, a silver space blanket with the aluminized side out is what I used. I used a 12″ square piece of galvanized steel painted flat black with the thermocouple on the back on the bottom of the box. Cover it with thin polyethlyene film, cling wrap will do. Point it at the sun and measure the temperature of the bottom. Now replace the cling wrap with glass and repeat (or build two boxes). The bottom temperature will be higher with the glass cover because the glass ‘traps’ the IR radiation from the bottom while the cling wrap is nearly transparent to thermal IR.

Here’s the data from an experiment using three layers of cover on a 12 inch cube with thermocouples on the bottom and each cover layer plus ambient air temperature. The layers are glass four inches above the bottom, polycarbonate in the middle and acrylic on the top:

These are what the boxes look like:

The insulation is six inches of fiberglass in a box constructed of foamed plastic. The temperature would be higher with the sun directly overhead. This data was collected in the winter with a relatively low sun angle so there’s substantial circulation in the box and the bottom plate is a lot warmer on the high side compared to the low side. Drying the box first helps too.

Maybe some day I’ll be motivated to repeat the experiment using resistance heating instead of sunlight so I can accurately measure the energy input with the box facing down to minimize internal convection.

The bottom of a glass covered box cools more slowly than a cling wrap covered box too, at least until frost or dew forms on the cling wrap and changes its transmission characteristics. That, by the way, is the reason why it’s not worth buying low IR transmission polyethylene for greenhouses.

• SoD,

Thanks for fixing the image links and moving the posts to the correct thread, not to mention getting my post out of moderation. Photobucket wants an exorbitant amount of money to allow third party hosting.

• sorry I missed that not seeing the images .

It’s awfully complicated to figure out what the radiative balance is , but at least it’s a step towards analyzable experiment — and far from the geometry of spheres at issue . Can you put equations to your experiment ?

Here’s the experiment I did when first diverted into this quagmire : http://cosy.com/Science/warm.htm#PingPong .

• Complicated??? It’s bone simple compared to the atmosphere. You have transmittivity, reflectivity and absorptivity for the cover layers for the solar spectrum and for the thermal IR. Assume at least 90% transmissivity for the incident radiation and 10% for the thermal IR. A large amount of the loss of incident radiation flux will be due to reflection, not absorption since the layers are relatively thin. A first approximation would be to ignore absorption of incident radiation for calculating the layer temperature.

The structure of the solar spectrum, then, isn’t important, only the integrated energy over the spectral range. In round numbers that would be ~1,000W/m² at the top. For the bottom, assume absorptivity/emissivity of one for all wavelengths, which makes the emission spectrum identical to a blackbody. If you can’t come up with the equations yourself, you can’t possibly deal with atmospheric radiation.

It’s in Petty’s book, sort of. He ignores reflectivity for his simple single layer, non-reflecting atmosphere model. Unfortunately, Amazon in their Look Inside feature leaves out the critical pages. This source (see 7.3.2 about half way down) leaves out a lot of features. Most of those left out may not be significant, however.

Note that for a transparent layer with an index of refraction different from the surrounding atmosphere, you get some reflection from each interface even with normal incidence and internal absorptivity of zero.

The reason I haven’t done much with equations is that I don’t have good numbers for the heat transfer to and through the insulation. As you can see from the graph, it takes quite a long time to reach anything close to steady state. Electrical resistance heating would be a lot simpler. You wouldn’t have to worry about clouds and tracking the sun and you could let it run for a long time.

63. ! EXPERIMENT !

We all apparently agree on at least an incredibly crude estimate of the radiative equilibrium temperature of the Earth , ~ 255K , altho I have yet to see any agreement on the generalization in terms of a ratio of dot products of spectra times total radiative flux . ( It almost makes me think that nobody here knows what a dot product is , but that’s impossible . )

The issue is , and I have yet to see either equation or experiment showing how our surface can be hotter than that equilibrium .

Just scrolling thru these responses I see neither equation nor experiment .

I , in a very fundamental totally classic sense , don’t care about planets or atmospheres . Just show me the equations by which the interior of a ball is held to a higher temperature than its radiative equilibrium .

Then we can start talking quantitative testable theoretically based physics .

• There are none so deaf as those who will not hear, none so blind as those who will not see.

The information has been given to you, but apparently not exactly in the form you claim to want and you won’t accept anything less.

Line-by-line radiative transfer programs reproduce observed atmospheric spectra to within 1% or so. See, for example, here:

https://www.atmos-chem-phys.net/13/6687/2013/acp-13-6687-2013.html

Now I give up on an apparently hopeless case.

• Can’t you just point me to the central equation ? I can implement it from there in terms of its parameters .

From my side , I can’t get you to agree to the computation for the equilibrium temperature of a radiantly heated colored ball .

I’ll repeat that I come to this as an APL programmer for whom a half dozen expressions gets us to the temperature of a gray ball in our obit — in fact to a gray ball surrounded by any arbitrary map of radiant power over the entire celestial sphere .

One more expression implements the parameter of source and object spectra . Even at that point it seems impossible to reach mutual agreement .

Now I just simply want the next expression in terms of the next critical parameter which quantitatively explains how the interior is held hotter than that radiative equilibrium despite the Divergence Theorem .

• Bob,

I’ve pointed you to this before. Here it is again:

Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations

I look forward to your Groundhog Day comments requesting this equation in the future.

• Jeez I don’t have time for this now because I got lots to do preparing for presenting CoSy at http://euro.theforth.net/ . But unambiguously settling this science is very important . And to me that means being able to implement the equations .

I still don’t understand why I can’t get simple agreement on the computation for a ball of arbitrary spectrum . Surely that’s a necessary first step to any further computations . Can we have agreement on calculating the equilibrium temperature of a simple single color painted ball ?

I take it you are asserting that the “secret sauce” for trapping energy is in the Schwarzschild differential with some set of parameters . I’ll have to leave that until I get back . However , I don’t find it promising when you comment :

“To solve the equation in practical terms we need to know:
>>> the temperature (vs height) in the atmosphere <<<
the concentration of each absorber vs height
the absorption characteristics of each absorber vs wavelength

When the temperature profile is exactly what is the parameter to be calculated .

Finally , while I'll agree that DeWitt's experiment is "bone" simple compared to the atmosphere , I think it is far from being a simple analog of the the heat flows at issue . ( And the historic experiments like Ritchie's ( see bottom of http://cosy.com/#PlanetaryPhysics ) are brilliant in their simplicity and symmetry . )

I have posed , above , a 1 dimensional analog of the heat trapping hypothesis : https://scienceofdoom.com/2014/06/26/the-greenhouse-effect-explained-in-simple-terms/?replytocom=120217#comment-119774 .

That sort of setup with a heat source on one side and sink on the other with a symmetric filter – surface – filter sandwich in between is far cleaner , quantifiable and unambiguous .

Anyway , I guess you all will be happy that I'll be out of here for at least a month .

• Bob Armstrong wrote: “Can we have agreement on calculating the equilibrium temperature of a simple single color painted ball ?”

If you mean a ball illuminated by the sun and surrounded by space, then there is no such equilibrium temperature. There is, however, a steady state temperature. And we all agree on how to compute it.

The difference between equilibrium and steady state is vitally important. Unfortunately, people often refer to steady state as equilibrium. Sloppy, but most of the time is it clear enough when equilibrium does not actually mean equilibrium.

An atmosphere would not alter an equilibrium temperature, but that is irrelevant. An atmosphere can certainly alter the steady state temperature.

• Over a cycle a conserved quantity like energy comes to a constant mean . These sorts of comments make me wonder if “climate scientists” have even the equivalent of an undergraduate degree in applied math and physics — focused on heat transfer — which should a requirement for anybody presuming expertise in the field .

The equilibrium temperature of a gray ball in our orbit , given a sun about 5780 . is about 278.6 +- 2.3 from peri- to ap-helion .

The idea that that is not meaningful is testably false .

My continuing perception is that this is a field which is in fact a branch of applied physics , but is taught as if a “social science” based on theoryless observation and measurement .

No other field of applied physics has been so spectacularly stagnate :

• Bob Armstrong,

You wrote: “Over a cycle a conserved quantity like energy comes to a constant mean . ”

Although often sloppy in language (like most physicists, in this case) they understand the distinction between steady state and equilibrium. You obviously don’t. And they understand the distinction between closed an open systems. You obviously don’t.

You wrote: “These sorts of comments make me wonder if “climate scientists” have even the equivalent of an undergraduate degree in applied math and physics ”

They do. You don’t.

Time for me to take the advice of Confucius: “He who knows not and knows not that he knows not is a fool – shun him”

• I assume you can calculate that 278.6 temperature and the 2.3 variation around the orbit .

Yes or No .

Only then is there any point in going further .

Again , I am interested in the computable basic physics . So far I have seen no evidence that you know how to calculate anything quantifiably experimentally testable .

Your reference to physicists in the third person indicates that you yourself aren’t one so don’t have the education to meaningfully discuss these issues .

• Bob Armstrong: Imagine a rotating spherical spacecraft of radius r with a highly conduction surface that gives the surface a uniform temperature T. The emissivity of the spacecraft’s surface is e, and its absorptivity for solar radiation is a. Emissivity = absorptivity at any given wavelength, but the spacecraft absorbs visible light from the sun and emits thermal IR, so a and e can be different. Incoming solar is S (W/m2). At steady state, incoming and outgoing energy must be equal. The relevant equations can be solved for T.

(a*S)*(Pi*r^2) = (eoT^4)*(4*Pi*r^2)

aS/4 = eoT^4

What went wrong with your experiment with ping pong balls? 1) You thought you were changing only absorptivity when you painted the balls different colors. You forgot that you were also changing emissivity. 2) You provided a second route for heat to escape from your balls: conduction to air which is moved away by any wind or convection.

This is why DeWitt carefully insulated his boxes: If you want the IR transparency of the top of the box to make a big difference, you need to make sure the only way heat gets out is through the top of the box. Experiments that fail are often improperly insulated. Unless we know a, e, and S for DeWitt’s boxes, we can’t calculate the inside steady state temperature. Even if we did, no insulation is perfect. In the real world, there will be some heat flux through the insulation and another term on the right hand side of the equation.

What would happen if the emissivity of our spacecraft decreased by 2%? The temperature would need to rise about 0.5%. (1.005^4 = 1.02) If the steady state temperature happened to be 255 K, it would rise 1.3 K. Doubling CO2 will reduce the emissivity of the planet by about 2%

Our planet is quite different from an idealized spacecraft. But we think of Earth as a simple spacecraft with T = 288, e = 0.61 and a = 0.7, what would happen if we removed the atmosphere and e rose to almost 1? The answer would depend on what happened to absorptivity. The standard assumption is to keep absorptivity constant.

• Frank,

A major problem for calculation is convection inside the box. That leads to non-uniform temperature of the bottom and the cover layers. That makes it very difficult to calculate energy balances, especially at the bottom where the effect is the largest because the temperature is highest. And yet, if you calculate the energy balance of the interior layers using only the emission of the layer above and below based on the measured temperature and an emissivity of one, it comes out fairly close.

That’s why someday I might like to use resistance heating and support the box with the opening down. That should minimize convection as it’s hottest at the top.

Looking at the glass layer closest to the bottom and using data collected over 36 minutes after the box had reached more or less steady state (after the data in the graph above was collected), the temperature of the bottom averaged 147.4°C, the glass layer was 120.49°C and the polycarbonate layer above the glass was 91.4°C. Using the Stefan-Boltzmann equation and an emissivity of one, the emissions from each layer were 1774, 1362 and 1002 W/m². Adding the emissions from the bottom and polycarbonate layers gives 2776 W/m². The glass layer emits from both sides, so it’s total emission is 2 * 1362 or 2724 W/m². That’s less than 2% different. I consider that to be fortuitous, considering the unknowns, rather than accurate. Note that the incident solar radiation before passing through any of the layers is probably less than 1000 W/m².

• My equation at http://cosy.com/Science/warm.htm#EqTempEq is in terms of Total spectra but it requires understanding of arguably the most important computation in all of science , the “dot” product .
I find it inconceivable that people here don’t know what that is , but when repeatedly these crude computations are presented rather than the general computation for arbitrary spectra , I must wonder .

Nothing went “wrong” with my ping pong ball experiment . It essentially confirmed my grade school nerd understanding of a question in a boy’s science book back in the Eisenhower admin while Gore was out on the playground learning how to win class president :

A white ( light gray ) rock and a black ( dark gray ) rock are sitting in the desert sun . Which gets hotter ?

The answer which is essentially the insight of the Ritchie experiment which became formalized by Kirchhoff ( and Stewart and others ) in the 1850s is that they come to the same temperature .

The ping pong balls are a lot closer analog to our ball in space heated on one side by the 5 millionths of the celestial sphere of the Sun’s disk , radiating to cold temperatures in all others than DeWitt’s . In fact the improvements are pretty obvious and could be made quite comparable with a radiant heat source at one end of a cryogenic vacuum chamber .

I now understand the equations of balance much better than I did a decade ago . That was a Mr Wizard style first cut experiment that confirmed my ball park understanding .

My equation simply to arbitrary source and sink power spectra and arbitrary object absorptivity==emissivity spectrum , ie : color . When fed the parameters which produce the endlessly parroted 255K meme , so does it — because it is simply a generalization in modern executable notation of that computation .

But again , no one here seems to know how to do the most basic computations over total spectra . I would love it for someone to prove me wrong because then we could start having a meaningful Quantitative analytical discussion .

64. Bob,

Hard to point out your many mistakes. Not worth going through them in detail because I’ve done it before.

..no one here seems to know how to do the most basic computations over total spectra..

1. If you go to the link where I sent you, and use the equation I gave you – remember you requested it – which is across the whole spectra, you will find that increasing GHGs higher in the atmosphere reduces OLR (outgoing longwave radiation).

You probably don’t understand the equation but it is derived from first principles.

Just integrate eqn 16 (in the article) across all wavelengths.

Next time you say (some variant of) “no one has given me an equation” “no one knows how to compute over total spectra” I’ll just delete everything after the first few words with a link to this comment.

a) Prove the equation wrong
b) Prove that the equation, when integrated over all wavelengths, using the absorption lines of CO2, does NOT reduce OLR for more CO2 higher in the atmosphere
c) Admit you have no idea what a) and b) actually mean or why they are relevant

2. The sun heats the surface of the earth through the mostly transparent atmosphere. The climate system emits radiation to space from the mostly opaque (colder) atmosphere. More CO2 increases the opacity of the atmosphere for terrestrial radiation, but has almost no effect on the transparency for solar radiation.

The above statement is a simple summary of what you can find in the equation I gave you.

I know you don’t understand any of what I have written. Or the equation I provided.

However, you asked for the equation. You claim *we* have no idea how to calculate full spectrum radiation. The equation provides the proof that we do. And your insistence otherwise demonstrates your confusion over physics basics.

Future repetitions of your confusion are very likely to be dramatically shortened and redirected to this comment.

• Can’t we start with something simple and testable ? That’s the way physics works . Surely it should be trivial for us to agree on the computation of a simple solid asymmetrically irradiated uniformly colored ball , eg : a billiard ball under a sun lamp .

If we cannot Quantitatively do that — which is clearly — or should be — an undergraduate level homework assignment , there is no point in going further . We are not able to compute anything and the utter quantitative stagnation pointed out by Nir Shaviv will endure .

I will repeat something I have cited before . I have gained great respect for Lavoisier who turned alchemy into chemistry by bringing his accounting background to bare insisting on a complete quantitative “audit trail” from inputs to outputs . Here we seek an “audit trail” from the output of the Sun to our surface temperature to at least the 4th decimal place , but we are stopped dead a the simple issue of mean temperature of a solid ball much less one with a semi-transparent cover .

Rather than simply providing the simple computations for bodies of arbitrary color showing that you do know how to compute equilibria for even simple situations , you revert to your untestable complex final case .

That’s not how physics is done . You prove and demonstrate your quantitative understanding at each step along the way . As I point out in the slide below , Griffiths spends 280 pages out of 550 on electrostatics before moving on to dynamics . I continue to see no evidence here that anyone here actually knows how to compute anything beyond the most basic Stefan-Boltzmann .

As I said , when I get back from Europe , I will find time to implement Schwarzschild and see if there is any range of parameters which “trap” energy on the side away from the source . That single testable differential and a demonstration of it is all that is needed to quantitatively settle the issue .

• Bob,

Here you are going around telling everyone how climate science doesn’t understand radiation basics.

Because Bob is confused therefore climate science as a whole has made a massive mistake.

I’ve given you the equations you asked for. They are simple. Absorption. Emission. Put them together. Rearrange. It’s only equation #16 in the steps that you have to get to. It’s not difficult to follow the steps if you understand the basics of radiation and of calculus.

You only have 3 options. It’s clear you are on option c at the moment.

That is fine – most visitors to the blog don’t understand these equations either. I didn’t before I got textbooks out of the university library and then bought a couple more textbooks.

What is fascinating is your monumental ability to believe that your ignorance of basics means climate science is wrong.

I started with a different mindset – I assumed that people who had spent their lives studying atmospheric physics, professors of physics in august institutions, probably knew more than me and at least I should try and read their textbooks before pronouncing the discipline to be completely failed.

At least you are on common ground with many other visitors here. Their lack of understanding of physics basics also leads them to pronounce climate science as a failed discipline.

• Sorry I don’t find equation #16 .
In my construction of the audit chain from the Sun’s temperature , disk and orbital geometry it’s the about the 5th or 6th . First you calculate the Stefan-Boltzmann gray body temperature in the orbit , ~ 278.6 +- 2.3 from apsis to apsis in ours , then you construct the Planck function , one more line , then compute the equilibrium for a colored body , one more line .

I don’t see how you can skip that step . If it is equation #16 , please point out that equation so I can compare it with mine and we can either have agreement , or an experimentally testable difference .

I find it hard to believe that anybody in the field has so much trouble presenting such a simple testable computation . Without it , you literally can’t compute the temperature of a billiard ball under a sun lamp .

And if you can’t do that , what’s the point in going further ?

• Bob, you asked for the equations. I have given them.

You say:

Sorry I don’t find equation #16 .

Good for you. Do you find eqn 1? Eqn 2? If I put my hands over my eyes I can miss all of them as well.

I wish I had answered my exams like you do. Much less preparation would have been needed.

====
Q. Prove eqn 16 is the result of applying the Beer-Lambert law of absorption and the Planck law of emission to the OLR at top of atmosphere.

A (in exam). Sorry I don’t find eqn 16.

Bob-world result: Congratulations – 1st class honors.
====

Bob, you are still currently on option c. You are trying to move to option a – which would look better for you – by “doing a Bryan” as we say here.

Prove the equation wrong (option a) means:
i) find the flawed equation, or
ii) demonstrate that one equation doesn’t lead to another

It is super easy. It is like doing an exam where someone has already given you all the steps.

I say super easy. What I mean is, someone who has read and understood a textbook on radiation and heat transfer would find it super-easy.

Having just listened to another history book, Nixon was advised on Vietnam:

Claim victory and leave

My recommendation to you.

• I didn’t ask anything about atmospheric attenuation . If you cannot calculate the radiative equilibrium , ie : balance , temperature of a radiantly heated ball in a vacuum you are stopped dead in your quantitative analysis .

How do you calculate the temperature of a radiantly heated colored ball in a vacuum ?

• Bob,

Just for you. Here is the equation you can’t find, from Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations:

• Let me add , whatever is going on in the distribution of temperature in the “thick layer of paint” of the atmosphere , that mean HAS to equal the radiative equilibrium determined solely by the lumped spectrum as seen from outside .

I continue to simply ask for the computation of that equilibrium for arbitrary spectra .

• I continue to simply ask for the computation of that equilibrium for arbitrary spectra .

The mathematical operation is called integration. I assume you’ve heard of it. It’s done numerically because there isn’t an analytic solution. The flux over many small spectral ranges covering the range of interest are summed. Instrumentally, it’s done using an instrument that has constant sensitivity over the range of interest and whose input is filtered to only cover that range. For measuring the IR energy emitted by the atmosphere to the surface from 4.5-100µm, you can use an instrument called a pyrgeometer. The design is much like that for a non-contact IR thermometer but it has a 180 degree field of view and covers a wider spectral range.

https://en.wikipedia.org/wiki/Pyrgeometer

• My god , have any of you taken a look at my math background ? Here’s a link to notice I just received from one half of the tek informing CoSy : https://kx.com/blog/machine-learning-kdb-k-nn-classification-pattern-recognition/ .

So you want to talk integrals : give us the definition of a dot product as an integral . Show us you can compute the mean temperature of the lumped planet and atmosphere . The remaining 3% unexplained variance is detail .

• Bob,

Please stop using the term ‘equilibrium’ when you mean ‘steady state.’ The Earth’s atmosphere and surface isn’t at equilibrium and never will be as long as the sun is shining.

You’re math background is utterly irrelevant in this case. In fact it probably hurts as it gives you false confidence that you understand the physics, which you clearly don’t if you think equilibrium and steady state are the same thing. It also seems to blind you to what’s actually going on because it doesn’t fit your paradigm.

Atmospheric transmission calculation programs, like LBLRTM, don’t use matrix math. Whether I can give a definition of a dot product as an integral (I can look it up, but why bother) has nothing to do with whether I understand atmospheric transmission of radiation far better than you. If you want to express Schwarzschild’s equation in vector terms and then solve it (which you’ll have to do numerically rather than analytically), go right ahead. But don’t expect us to do it for you.

• Bob,

I would also like to point out that the only reason you’re getting responses at all at this point is that I, for one, don’t want people who read this for information to think that I’m ignoring you because you’ve found a gaping hole in radiative transfer theory. You haven’t and there isn’t.

• Calculating the equilibrium of the lumped surface and atmosphere ( Incropera el al , would call it a control surface ) is a gaping hole ?

You are right . It’s undergraduate stuff . Without it you cannot possibly compute the temperature of a planet in terms of the energy it receives from the Sun .

You quibble about “steady state” versus “equilibrium” ? Then give us the calculation for the steady state .

To paraphrase my friend Professor Cork Hayden , youall will do anything other than compute an experimentally testable value .

Can’t you see why that leads to the utter stagnation Nir Shaviv graphs ?

But I guess that leads to endless careers — that is until a Trump gets elected to clean out the non-performing swamp .

I guess we all agree this is totally pointless . Youall literally don’t know how to calculate the temperature of a billiard ball under a sun lamp but will go on claiming to be able to say something meaningful — if untestable and without experimental demonstration — about the temperatures of planets .

Wow .

Bye .

• Can someone help me understand something. I believe I found the elusive formula that Bob is looking for. One thing I noticed though is that the diameter of the receiving body sends to be irrelevant. Is that accurate? In my mind, larger receiving body would equate to more heat gained. Or is it related to the fact the equation I found is assuming an albedo of 0?

• Brad , that’s accurate . A point and a sphere are equivalent . in these computations .
A lot of physics is in terms of basic surfaces of high symmetry : planes , cylinders , spheres . Somebody’s name is attached to the essential set but I can’t find it right now .

The diameter of the body is irrelevant.

There are two key points for non-trivial problems of this type:

1. Emissivity/absorptivity is usually different at different temperatures – especially when you look at something like the absorptivity of solar radiation at wavelength 0-4um (corresponding to a temperature around 5800K) versus the emissivity of terrestrial radiation at wavelengths 4-100um (corresponding to temperatures around 300K).

An example is snow which is highly reflective to solar radiation – around 50% to 80% depending on how fresh it is. But highly absorbing, and therefore, highly emitting, for terrestrial radiation – around 95%.

If you assume absorptivity = emissivity for the incident (heating) radiation and for the outgoing (cooling) radiation then you get a nice simple answer. But that doesn’t apply to the sun-earth problem (see note at end).

2. The surface of the earth is the “layer” absorbing the solar radiation and the atmosphere is the “layer” emitting the outgoing radiation. The temperature difference between the two is critical.
Note that in reality the emission to space of terrestrial radiation takes place in some proportion from the surface, and then from different heights in the atmosphere.

If we take a simple version of reality and say emission to space takes place from 5km above the surface, and the average lapse rate (temperature profile) is 6K/km then we have a temperature difference of 30K between the surface being heated and the “surface” that is radiating to space.

Both of these points I have tried in previous articles to explain to Bob but it is all wasted words (so I didn’t really try in comments this article).

Note: an important point that confuses newcomers to this problem (and many examples of confusion can be seen in many previous articles) emissivity = absorptivity at a given wavelength. So for a given body for say 10um radiation, emissivity = absorptivity. Likewise at any other wavelength.

Absorptivity = proportion of radiation absorbed (vs reflected)

This does not mean that emissivity = absorptivity at two completely different wavelengths. And when you heat the earth from a 5800K object (the effective surface temperature of the sun) with wavelengths 0-4um while the earth is only 300K and therefore emitting with wavelengths 4-100um you can’t assume that emissivity = absorptivity. It isn’t.

When you say that “for a given body at 10um (don’t have the fancy symbols on phone) that emissivity= absorptivity, is that sporadically for a blackbody?

If not just let me know, I obviously have some reading to do.

• The definition of a blackbody is emissivity = 1 at all wavelengths, absorptivity = 1 at all wavelengths.

Emissivity(λ) = absorptivity(λ), where λ = wavelength.
This is true for all bodies.

In general, emissivity at λ1 is not equal to emissivity at λ2.

I explain more about this and some other points in Planck, Stefan-Boltzmann, Kirchhoff and LTE. That article specifically covers some basic areas commonly mangled by “contrarians” who don’t understand radiation basics but of course have great confidence that they do.

Most general textbooks on heat transfer and radiation cover black bodies and then derive formulas for “gray bodies”. Gray bodies are mythical but useful concepts that have constant emissivity (and therefore absorptivity) over all wavelengths. This is the usual method of textbooks – start with simple, then go to more complicated and more general.

Once you introduce varying emissivity with wavelength – which is actually what we find in reality for just about everything – the maths gets more complicated and you can only usually solve the problem numerically. Derivations and solutions to these kind of problems are found in many textbooks on atmospheric radiation.

You can read the series Visualizing Atmospheric Radiation and see lots of detail.

Visualizing the solution to the problem is difficult (I try to provide some perspectives on visualizing through that series).

– Across wavelengths you have strongly varying emissivity and absorptivity – by many orders of magnitude.
– Vertically through the atmosphere the temperature drops by around 100K
– Vertically through the atmosphere the pressure drops by several orders of magnitude (so the concentration of “greenhouse” gases drops by this same factor)
– Vertically through the atmosphere CO2 is well-mixed but water vapor is not

• scienceofdoom,

Thanks for the info. Will read that first link you provided. My comprehension on the matter is still lacking. When you say emissivity=absorption for any body, yet earth absorbs a particular visible wavelength, let’s say 3um but doesn’t emit 3um at the same rate. Anyway, thanks again. Will look into the other info you have provided.

• Definitely the https://scienceofdoom.com/2010/10/24/planck-stefan-boltzmann-kirchhoff-and-lte/ has some useful info , and reference to Siegel & Howell : Thermal Radiation Heat Transfer is one I’ll likely purchase . But much seems nuance when a case like Venus , where the surface energy density is more than 25 time that which the Sun supplies to the orbit , and , I think it was mentioned here that only about 3% of that radiant energy reaches the surface . 90% is reflected in its upper atmosphere .

By the way , just to show there is some common understanding here , the Planck thermal radiation function given on that page is

{[ l ; T ] ( ( 2 * h * c ^ 2 ) % l ^ 5 ) % ( _exp ( h * c ) % l * boltz * T ) – 1 }

Expressed in K on my http://cosy.com/Science/ColoredBalls.html page .

And Kirchhoff’s “law” says essentially that absorption and emission are just to directions thru the same “filter” . I like to lump them as a parameter ae , a function of wavelength . At any wavelength they are identical . but generally they they vary as has been said here at different wavelengths giving an object its color . The “coupling” between a power spectrum like the Sun’s and an object of a fixed color is given by the dot product ( the sum of the products at each wavelength ) of the two spectra .

For a flat spectrum , ie : gray , ae , being constant drops out of the equation so a gray object , no matter how light or dark , comes to the same temperature as a black , determined by the total impinging power .

That’s why the gray body temperature , ~ 279K in our orbit is a constant against which the effect of changes in ae spectrum can be calculated .

I do hope everybody agrees with all the fundamental stuff said here . I can’t see anything at all controversial or non-classic .

Emissivity is a factor, actually a ratio – it is between 0 to 1.
Likewise for absorptivity.

Emission is a value, usually in W/m2. This is dependent on the temperature of the emitting body and the emissivity.

Absorption is a value, usually in W/m2. This is NOT dependent on the temperature of the absorbing body, it is dependent on the incident radiation x absorptivity.

It’s quite easy for emission at 10um to be different for absorption at 10um, even though of course emissivity = absorptivity at 10um. Incident radiation on the surface at 10um = 0 for example (all the solar radiation is really in the range 0-4um). If the body is anything like 250K-350K it will be emitting quite strongly at 10um.

Understanding:

a) emissivity = absorptivity
b) they are wavelength dependent
c) emission at any wavelength does not have to equal absorption because of the above explanation

– is critical to making sense out of the basics of radiation. Most “climate basics critics” confuse and mangle these essential uncontroversial ideas. Emissivity is not emission. Absorption is not absorptivity.

A little more about these points in Basics – Emissivity and the Stefan Boltzmann Equation.

• SoD,
Thanks. Think it’s starting to click after a bit more reading. First hangup I had was intermingling emission with emissivity and absorption with absorptivity. The second issue I had that finally resolved was that i was trying to reconcile why difficulty of the sun had to equal absorptivity of the earth surface, when that isn’t what is implied by emissivity= absorptivity. Finally clicked when I saw the Ritchie experiment in that the ability of a particular surface to absorb equates to its ability to emit.

• Bob,
So now my question to you is why does a gray object having a constant ae for all wavelengths mean that it drops out of the equation when calculating temp of an object irradiated by a great source? I see no proof of that. I see your equations where you’ve demonstrated that temp of a black body according to sb is nearly identical to (t^4 x portion of source in sky)^1/4. But just because sb constant doesn’t have a large impact on the final number doesn’t mean:

1) that your shortcut is right or
2) that you can drop out all other constants such as an ae that isn’t equal to 1

Given an ae of anything other than 1 drastically changes the number received by sb, and thus shows that ae can only be ignored when it is 1. If you don’t think that’s accurate let’s choose a nearly perfectly reflective body, not a perfect white so still considered gray. According to my understanding of what you are saying, if 99.9% of the radiation were reflected the sphere would still end up the same temp as a black body. Is that correct?

• It’s exactly because of that result of Ritchie’s brilliant experiment — and is the basic fact I learned literally working thru boy’s science books as a kid in the 1950s . A dark object heats up quicker , but it also cools quicker . The equilibrium is only determined by the correlation between the object ae spectrum and the source spectrum versus the sink spectrum .

As , I believe DeWitt pointed out , snow has a very high reflectivity over the peak of the solar power spectrum . however it is much more absorptive and equally emissive over the longer wavelengths at the peak of the Planck thermal function at the < 300K temperatures of objects in our orbit . And that < 300K thermal spectrum is radiating to the 0.999995 of the celestial sphere at ~ 3K .

So the equilibrium temperature of a snowball Earth could well be low enough to keep it a snowball .

That's a rough picture of the situation .

• Bob,
Maybe I didn’t word my question very well. Hypothetically, here is the scenario.

2 identical sized objects (p1 and 02), orbiting equidistant from the same source of radiation. p1 is black body, p2 is gray with an absorptivity of .001. According to what I’ve read on your page, you are saying that the temp on p1 and p2 are going to be equal. Am I understanding you correctly?

• There is a singularity at 0 , as it looks like you recognize .

But , Yes . That’s what Ritchie’s experiment confirmed — an many since . I’m sure I must go thru the math somewhere on CoSy .

Consider what that ball looks like . It’s nearly totally reflective — to all wavelengths . Real shiny or real white . I think MgO is the whitest .

Heat’s slow getting in , and slow getting out .

All these basic physics relationships are eminently testable . That’s why I’m pushing a Ritchie Prize for best “YouTube” demo of any of these essential foundations : http://cosy.com/#PlanetaryPhysics .

• Bob,
My hypothetical and the Ritchie experiment are not the same. The key difference is that the emissivity of the source stayed the constant I my scenario while in the Ritchie experiment it changed.

In Ritchie here are my observations. The source of heat is the same, but the emitting surfaces are different in their emissivity. And while the receiving surfaces are equidistant, they are once again different in their absorptivity. For the sake of calculation I’m going to assign some arbitrary values. Let’s say his dark surface is aeD=.9 and his polished surface is aeP=.2. The reason the temps stay even on each side, as I think you pointed out on one of your pages, is because eD * aP = eP * aD. However, I hypothesize that the results would have been very different had the source had matching surfaces (whether both P or both D) or likewise if both receiving surfaces were the same.

For that matter, by your logic, the center cube could have been spun around matching up the ae surfaces (dark emitting too dark, polished emitting to polished) and the temp would have remain unchanged.

• Brad , you are right the Ritchie experiment is sort of the complement of the situation where the surfaces are the different but the source the same .

See http://cosy.com/Science/RadiativeBalanceGraphSummary.html for the issues in terms of spectra .

SoD’s simplification ( other than it is simplified to disks rather than the actual spherical geometry ) is that it ignores that those absorptivity and emissivity values are over different sections of the spectrum . They represent the hypothetical spectrum labeled “alarmist” in this graph :

I’ve already pointed to the computations of a and e as dot products over the two relevant spectra : the Sun’s power spectrum from which we receive heat versus the rest of the celestial sphere over which we radiate it from our thermailzed temperature to the cold of space .

I should note that browsing CoSy you’ll find that early on I found the equilibrium temp by quantitative search . ( The K code is on CoSy somewhere . ) It was only later , about the time I learned Martin Hertzberg came to generally the same understanding , that I realized the equilibrium , in terms of energy , boils down to the ( ratio of the dot products with the source versus the sink ) times the total energy impinging on the point — or sphere .

It seems an astoundingly hard bit of computation to get across , but you can see how it is a straightforward generalization of the “Alarmist” worse case giving 254.9 in the graph .

BTW : A simple way to understand why size doesn’t matter for the radiated object’s temperature : Consider 2 points sitting near each other . Now stick them together . Are they going to change temperature just because they are now right next to each other ?

For more reference, also some basics also at The Earth’s Energy Budget – Part One.

I think you have caught up. Once the absorptivity and emissivity are different, there isn’t a simple answer. Of course, at the same wavelength absorptivity = emissivity. With the earth being illuminated by radiation centered on 0.4um but the terrestrial radiation being centered on 10um we can’t assume that absorptivity of solar radiation equals emissivity of terrestrial radiation. And they are not.

Ein = αR.πr2, where R = incident solar radiation, α = absorptivity, ε = emissivity, r = radius of sphere
Eout = εσT4.4πr2 where σ = 5.67×10-8

and the factor of 4 is explained in the above article (incident solar radiation from a long way away strikes an area of pi.r^2, emission comes out from the whole surface area of the earth – 4.pi.r^2)

So solving for T = (αR/4εσ)1/4

If α = 0.7 (this means an albedo of 0.3) and ε = 1 (roughly a black body emission), and if you put in R = 1360 W/m2 (what we measure by satellites above the atmosphere) you get T = 255K

If α=ε then T = (R/4σ)1/4

And if you put in R = 1370 W/m2 you get T = 278K and this is completely independent of the actual value of α. But only for α=ε.

If some people never understand that emissivity is not equal to absorptivity at completely different wavelengths, they can never understand the basics. And I can’t help them.

66. Bob,

ae value of the sun should be 1 in your table, not .7

Given the suns surface temp and observable irradiance across the spectrum, it is nearly identical to what would be expected of a black body radiator.

• Bob,
Also, I may be wrong but I think the ae of the sun is irrelevant to the situation. Whether it is an ae of 1 or an ae of .7 doesn’t change how much we message is hitting us. It would change the amount of radiation we would expect to be receiving at a given temperature but it would have no bearing on what our actual measured irradiance from the sun is.

• It’s not the ae of the Sun which matters , it’s the received power spectrum . However , it’s the e , emission , side of the Sun’s spectrum which , multiplied by the Planck function , determines the actual spectrum it emits and we receive .

• The assumption producing the 255K value is that the Earth has an absorptivity of 0.7 over the Sun’s peak but radiates line a black body at the equilibrium temperature , Those are the numbers SoD used in his post , https://scienceofdoom.com/2014/06/26/the-greenhouse-effect-explained-in-simple-terms/?replytocom=120463#comment-120460 , above .

I just use the Planck thermal power spectrum for about 5800K for the solar spectrum . See the Heartland presentation for details . It would be nice to use a measured outside-the-atmosphere spectrum , but what would be far more important , and I think it would make things a lot clearer would be an ae spectrum for the Earth ( and atmosphere ) from outside . If somebody hands me a table of the spectrum , I’ll run the dot products — or they can do it themselves . It would be nice to see what the equilibrium temperature for an actual ToA ae spectrum is rather than the crude step function which produces the 255K value — far too crude for use in arguing about the 4th decimal place variations in estimated surface temperature this is all about .

But I’ve got better things to do than look around for such a table or scrape one from a graph . Right now , I’ve got to make sure I’ve got everything I need to head down to Denver to get to the Vienna Forth conference .

• Bob,
SoD used an emissivity for the sun of 1 and an activity for the earth of .7. In the chart that you shared from your site, the only ae value that doesn’t change is the sun’s, which you have pegged at .7. My point is either the table is wrong or the math that arrives at the answers is wrong since it would take an e of 1 and an a of .7 too arrive at 255. Your table obviously shows a value of 255 though. So is it labeled wrong and if not, how did you arrive at that number mathematically?

Also, I realize my hypothetical situation never was answered. Do you think each block would have stayed the same power had Ritchie turned the center emitting block?

Also I believe the disk formula is perfectly correct given a sphere projected onto another object is seen as a circle, not a sphere. At least that’s my take on why that is the correct way.

• Yes the disk formula is correct . But my APL computations are the start of a model of an arbitrary spectral map over the whole sphere .

I’m not sure what you are hung up on . I have said I use — as do any of these sorts of computations — a Planck thermal radiation function as an approximation of the Sun’s spectrum , that is , an ae of 1 times over the whole spectrum .

It is the ae spectrum of the Earth that is plotted and produces the difference from the gray body temperature . I am doing exactly the same computation but making it explicit that the 0.7 is over the solar spectrum and the 1.0 over the longer wave lengths .

Actually , I believe that’s a way the experiment was performed : rotating the center cube so dark faces dark , and light – light . Check the links I include . They are the main references I know of .

In any case , that’s a reason to replicate the experiment and get lots of YouTube traffic . That apparatus pictured was standard physics demonstration apparatus probably until the 1930s or ’40s .

• I read this essay sometime ago when it was recommended by Judith Curry. I had some issues with it back then but now, I’ve found out more, I have even more issues. The first sentence of the article is wrong, because the greenhouse effect, as described by climate alarmists, certainly is AGW. The basic physics were never proven. Not 100 years ago; not today. The basic physics underlying the climate models have always been _assumed_.

Given I disagree with you, I should first start by saying where I _agree_. I agree that certain gases, let’s call them GHG, absorb radiation at certain frequencies. I agree that H2O, CO2, O3, N2O, CH4 are such gases. I rewrote the order in terms of importance. I’m glad to see you consider H2O a GHG. Most alarmists I try to talk to refuse to mention water beyond dismissing it. The claim “CO2 does all the forcing”, so only CO2 is to be discussed!

Here, below, is my understanding YOUR more model, written at the molecular scale. I’m somewhat sorry I have to thought read you like this and put words in your mouth. But climate alarmists all refuse to talk about “The Science”, claiming it was long ago “settled”. Then comes what I think really happens.

[What I think you think]
The surface (earth and sea) warm due to absorption of much sunlight. The same surface slowly cools by emitting black body radiation (BBR) a much longer wavelengths (and lower frequencies). The cooling radiation has much less energy than the warming sunlight. So there must be a lot more of it. Because the sum total of energy gained and lost by the surface must be the same. Most of this energy is lost as black body radiation. When heading out to space, it is intercepted by GHG, which absorbs radiation at distinct wavelengths characteristic of certain electron transition modes [centered at 2.7, 4.3, 15 micrometers for CO2]. The same molecule then retransmits the radiation but equally in every direction. You call this “trapping heat”.

[What I think is wrong with that]
I disagree with two things here. (1) The same molecule retransmits the energy it absorbs. I’m not saying it can’t happen. I’m saying it’s not the general case. Will Happer says that there is a distinct time gap between absorption and retransmission. That this gap is generally over 100 milliseconds. What can happen in that time? 100 million molecular collisions can happen. A typical gas molecule collides with about 1 billion other molecules each second. On each collision they share energy with each other. The higher energy molecule will share more of it’s energy with the lower energy molecule than the converse. So the net effect is to bring all the molecules energetically close to each other. As soon as one molecule gains energy, it will be lost by sharing before that energy can be ‘retransmitted’. LWIR is absorbed but it is _thermalized_. (2) Previously we agreed that O2 and N2 are not “GHG”. But O2 (certainly) can transmit microwave energy and it does (even though it may have no significant absorption at the black body LWIR wavelengths emitted by the surface). We know O2 can transmit microwave because the satellites measuring the temperature of the atmosphere (giving RSS and UAH temperature series). Specific microwave emissions of O2 are measured and this is used to determine atmospheric temperature. In my model the CO2 gets a one-time BBR / LWIR absorption. The majority of retransmission will not be by CO2 but by O2, N2, and H2O which are the vast majority of atmospheric gases. There are many papers, on for example, O2 emission going back from the 1960s.

• mark4asp,

In reply to your recent earlier comment of August 1, 2017 at 11:24 pm I pointed you to the article that provides equations – Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations..

You could write an argument in words about why the tensile strength of steel wasn’t important in steel reinforced concrete and it might sound quite convincing to an amateur.

But the test of physics ideas is in deriving and proving equations, and then showing those equations demonstrating reality and providing predictions of what happens if we change factor x and factor y.

I provide the equations. So:

1. Go and demonstrate the equation wrong.
2. Or demonstrate that the solution to the equation doesn’t match reality.
3. Or demonstrate that the solution to the equation doesn’t result in more CO2 = more surface warming (all other things being equal).

So far on this blog many people have shown up and waved hands and not a single person has taken me up on the actual equation of radiative transfer.

No one.

That should be a useful demonstration of how little people who doubt the radiative effect of CO2 actually understand about atmospheric physics and radiation.

67. Bob,
According to this chart you shared here, you have an ae for the sun of .7. That’s why I bring it up again. ae for the sun should be 1. That is what SoD used above in his calculation. An absorptivity of .7 was used for earth, an emissivity of 1 used for the sun.

Everything I can find on the Ritchie Apparatus, which sadly isn’t much, says that opposite colored faces were directed toward each other. Again, I hypothesize that rotating the device such that similar colored faces are directed toward each other will provide a big change in temperature. To me, it’s easy to understand why a low emissivity paired with a high absorptivity would equate to pairing a high emissivity with a low absorptivity. I simply can’t understand the logic though that says that when I rotate the device the high emissivity paired with the high absorptivity is going to stay the same temp as low absorptivity and low emissivity. But again, that’s my hypothesis at the moment. Look forward to making a device to illustrate such.

http://physics.kenyon.edu/EarlyApparatus/Thermodynamics/Ritchie's_Apparatus/Ritchie's_Apparatus.html

You say it would make things clearer to have an ae spectrum for the earth and atmosphere from outside. Is that not the .7 that is used as an average for earth? The actual ae is going to vary second to second given the observers location, rotation of the earth, cloud cover, ground cover, etc. There isn’t one simple spectrum to be analyzed, it’s a continually moving target, that from my understanding, approximates to .7.

Also, have you considered the other implications of your claim that any gray body in space will be the same temp as a black body in space? Do you wear black clothing outside in the summer? According to your claim it shouldn’t matter. The gray body clothing (no matter how white it is it will never be an a=0) will all be the same temp. If that is in line with your hypothesis please let me know, I can very easily demonstrate your hypothesis to be false.

Lastly, this discussion isn’t in regards to the accuracy of modeling climate change through absorptivity and emmisivity. The discussion and blog post is about GHG and how they work and the implications of the greenhouse gas effect. You seem to be taking the position that because the equations for radiative temperature transfer aren’t anywhere near accurate to the 4th decimal place then that means the GHG effect is false. That’d be like me saying that since my odometer only shows whole numbers on my car that means that I’m not actually driving because to get from point a to point b involves linear, continuous movement, not jumps of whole mile increments as my odometer claims.

The Ritchie apparatus has a fundamental problem. While lamp black does indeed have an absorptivity of close to one for the solar spectrum as well as for the thermal IR at 300K, it’s not at all clear that the white paint also has low absorptivity in the thermal IR range. If the identity of the white pigment were known, it would be possible to look it up.

You are correct that Bob’s table is bogus. The sun does not have an emissivity of 0.7. The Earth also does not have the same absorptivity for the solar and thermal wavelength ranges. The emissivity of 0.61 comes from a non-physical calculation that assumes the emitting temperature is 288.4K, but the emission power is equivalent to a blackbody at 255K (~239W/m²).

0.61 * 0.0000000567 * 288.4^4 = 239.3 W/m²

The value of 0.61 is not a constant. It depends, among other things, on the CO2 concentration in the atmosphere. Increase CO2 and the value would decrease.

The Earth does not have a blackbody emission spectrum and only about 20% of its total emission to space comes directly from the surface. The rest comes from different altitudes in the atmosphere, where it’s colder than the surface.

• DeWitt,
How does not knowing the ae of the painted surface impact the results? As I see it, the painted surfaces are going to have the same ae as each other, the black surfaces are going to have the same ae value as each other, and the values for the painted and black surfaces will be different from each other. I understand not having the exact ae for the white surface news we can’t quantitatively hypothesize what the temp will land, the construction of the experiment will properly demonstrate that a(of paint) * e(of black) is equivalent to a(of black) * e(of painted). Also I think it will prove when reversed that the temperatures ae(of painted) will not be equivalent to ae(of black).

I am hoping to set up a device to show this using a lamp black surface paired with a highly polished metallic mirror surface.

• Figure out the equations behind this . Then we’ll be on common ground ;

• Continue reading on one of the links to the references to Ritchie’s experiment , I think RIchmyer . It describes an experiment answering objections to Ritchie”s .

The geometry of the system is important. In the limiting case of infinite opaque parallel planes, the emissivity isn’t important as long as it’s not identically zero, which is impossible in the real world. The photon ‘gas’ between the planes will have a blackbody spectrum and intensity at the temperature of the walls, which will be equal assuming no heat in or out by other than the radiation between the planes. That’s because any photon not absorbed will be reflected back to the other side. That will continue until it’s eventually absorbed. That gives you the same number density of photons as if they were 100% absorbed, which is also not possible in the real world.

That’s why an insulated cavity with a small hole for observation is such a good approximation of a black body. It’s the effective equivalent of near infinite parallel planes. If the emissivity of the walls is low, then even a small hole could be a significant heat leak. But if there is no hole, there is no leak.

In the Ritchie apparatus, if the emitting/absorbing area is large, the spacing is small and the surfaces are parallel, the emissivity of the surfaces won’t make a difference either.

• Bob,
Figure out equations for what exactly?

• The “solar gain” and temperature profile of the layers of TiNOX . I see no evidence that anyone here could get a job as an engineer in that company .

• Bob,
Everything I’ve read on Ritchie’s experiment is as I have stated. That the temps stay in balance between the 2 chambers when opposite colored faces are facing each other. The implication is given, although I can’t find it listed anywhere that I can link to, that the negative of the experiment (like colored faces facing each other) caused the temperature to no longer stay constant. The temperature went up on the side with the lamp black faces. Consequently, the image that I find that leads me to believe this says details and raining from the experiment can be found in Richmeyer “Introduction you Physics”. Trying to find that I see you’ve left a review on Amazon for it. Maybe you’d be so kind to take some picture of page 197 for me.

• Bob,
I’m not sure why you feel it’s necessary for individuals to provide you equations when they have already done so. I’ve read a half dozen or so insightful, knowledgeable blog articles as well as plenty of commentary that I could find the info if I wanted to. Common sense tells me that even if someone does come in and give you exactly what you want you’ll simply dismiss it. After all, you think that radiative heat transfer equations that involve the sb coefficient as well as differing values of ae are unnecessary. That is the equation that would be used to demonstrate what is happening, so again, you’ll just dismiss it and go on asking how to calculate the temperature of a gray ball being irradiated (even though I’ve demonstrated your website calculations to be wrong).

While I’m not a physicist and don’t have the experience or education that many of the individuals here do, it’s really not difficult to understand what is happening with TiNOX. The outer quartz layer has high transmissivity to radiation in the visible spectrum. This allows the light from the sun to simply pass through (mostly except for some blue light that gets bounced around and can exit out the top due to refraction). The visible spectrum light is absorbed by the copper. The copper of course emits radiation but does so at a much longer wavelength. The quartz doesn’t have the same transmissivity for this wavelength though. In fact, glass and quartz are going to be highly reflective to IR. So the IR becomes trapped, not completely, but mostly.

That’s it in a simple nutshell, but if course this is a problem if you believe that a gray object heated by a radiator will arrive at the same temp as a black object. So I’d say, if anything, the task is on you to show how your equations fit this observation. After all, your equation says that this panel would arrive at the same temp as any other object, so why doesn’t your equation fit the observation?

Yes, that’s it. Or that’s part of it anyway. If T_1 = T_2 then no heat is transferred. If you add up the emitted and reflected photons, you get the same photon flux from each surface as for an emissivity of one. The quick and dirty way to see that is to have one surface with ε = 1 and the other with ε = 0.1 with both at the same temperature, T. The surface with &epsilon = 0.1 only emits 0.1σT^4, but it reflects 0.9σT^4 from the other surface. No heat is transferred because 0.1σT^4 from the other surface is absorbed.

• I left out a semicolon after the &epsilon. Now, though, you know how to insert special characters in HTML.

• Regarding my last post made minutes ago. Because of this thermalization effect, on “retransmission” the frequencies will have changed. Unless something is emitting at its absorption frequencies, CO2 can’t absorb a second time because the radiation centered at 2.7, 4.3, 15 micrometers is not longer there [unless O2, N2, H2O emit close to those frequencies (not likely!)]. That basically ruins your model, does it not?

By “retransmission” here I just mean transmission by atmospheric gas, not retransmission by CO2 as the model would have it.

• mark4asp,

Can you entertain the idea that you don’t have an idea how to do heat transfer equations? Or that you have no idea about atmospheric radiation?

Radiation absorbed is thermalized. The energy absorbed is transferred by collision into the local atmosphere within a few ns.

Random invented ideas by novices would “ruin a model” if they had a relationship to physics.

Go read the derivation of the equation of radiative transfer – it is based on fundamental physics and has been well-known since 1950 (a physics Noble prize winner developed this solution).

But what do Nobel prize winners like Subrahmanyan Chandrasekhar, or for that matter atmospheric physics textbooks like Atmospheric Physics: Theoretical Basis by RM Goody (1964) actually know?

• Mark4asp,
I’m sure someone will correct me if I’m wrong but this is my limited understanding on the subject.

Co2, and other gases, have these narrow absorption bands because that wavelength oh photon has the right amount of energy to move them to an excited state, not too much, not too little. Consequently, when these same molecules move from excited back to ground, they will give up a photon of equal energy as they absorbed, ie the same wavelength.

To me, it sound like you are saying a molecule will absorb at 15 and when it emits it will be at some lower value, and that is simply false. Again, Maybe I’m wrong, I’m sure SoD or someone will correct if so.

68. And again Bob,
Why do you need an ae spectrum for your model? According to your hypothesis, ae has no bearing on where the temp will end up. Remember, a gray body in space will become the same temp as a black body. You shouldn’t need an ae numbers to prove anything in your model. You simply need the distance from the radiator, radiators temperature, radiators radius. Everything else is irrelevant by your hypothesis.

69. Dewitt,
All of that makes perfect sense. I’m still a bit lost as it seems like in your comment that began with “The geometry of the system is important…” that you are saying the Ritchie apparatus experiment would be a failure or inconclusive in its findings perhaps. I also realize I didn’t explain fully how I would set it up so let me know if this would provide conclusive results.

Line a box and lid with 2-3 layers of aluminum foil. Attached and suspended 3-4 inches from the top of the lid would be 3 containers, accessible from the outside. The containers would all identical in size, shape, and basic construction with only the outside color being what made each different. Perhaps cubes could be made of tin duct work (still trying to think of what would be readily accessible and inexpensive). The center cube (radiator) fully wrapped in aluminum foil with the exception of one side which would be painted a flat, lamp oil black. There would also be an aluminum lined center divider to permit radiation transfer between each side of the box.

For the other containers, one would be wrapped similarly to the center which is to say aluminum foil on all side except one which would be painted flat black. The other container would only be wrapped in aluminum foil.

Experiment would go as follows:
With the lid in place and the environment sealed, fill the center container with boiling water and outer containers with room temp, or possibly ice water. Observe and record temperatures periodically to see how the outer vessels temperatures are changing. My guess is going to be that with opposite colored planes facing each other the outer vessels will change at the same rate. My second guess is that if I rerun the experiment but reorient the setup such that vessel 1 (v1) which has a black face is facing the black side of the radiator and vessel 2 (v2) which only is aluminum sides is facing the aluminum side of the radiator then v1 will heat quicker than v2 due to higher ε coefficients in that environment when compared to the v1 side which should have a very low transfer rate due to the low ε

Does that sound accurate or is there something that I am missing?

• Also, do you think vacuum sealing the box would be necessary to help cut down on convection?

When I say geometry is important, I mean the view factor as well as the emissivity determines the rate of heat transfer between objects at different temperatures. In the basic Ritchie apparatus, it looks like you can rotate the three parts. If the ends are parallel, you maximize heat transfer. Any rotation will, I think, slow heat transfer because the apparent surface area viewed is smaller.

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node137.html

Metallized plastic, i.e. the silver looking space blanket material, not the gold, may be better than aluminum foil. You want the metal coated side out. An ohm meter will tell you which side is metal and which side is plastic.

Convection could be a problem, but evacuating your box will not be easy.

70. Dewitt,
Yes, understanding your comments in regards to geometry I’d have to agree. I wouldn’t want any of the vessels to be able to be rotated freely or arbitrarily on an axis, just the ability to install the center box facing one direction or rotated 180 degrees so that all relevant sides were always near parallel.

You have the heat transfer equation. If the faces are parallel, then the view factor should be the same, so that’s out of the equation. Then all you have that’s different is the factor 1/(1/α1 + 1/α2 – 1) for opposite color faces. That factor is the same for both sides, so the rate of heat transfer from the hot central cylinder will be the same and the differential temperature won’t change.

For same color faces it’s 1/(2/α1 – 1) on one side and 1/(2/α2 – 1) for the other. If α1 is greater than α2, then the rate of heat transfer will be higher for that side and the differential temperature will not be constant.
Say α1 is 0.9 and α2 is 0.1. For opposite color faces, the factor is 0.099 on both sides. For same colors facing, the factor is 0.82 for α1 = 0.9 and 0.053 for α2 = 0.1. The rate of heat transfer between the two dark faces is nearly 16 times faster than between the two light colored faces.

What this has to do with the greenhouse effect is not clear.

• DeWitt,
Thanks for your input and I agree completely. My interest in the Ritchie experiment is due to Bob stating that all gray bodies come to the same temp regardless of their ae, in fact he claims gray body will be the same temp as a black body (ae=1). As part of his claim, he states that when like colored faces are directed at each other in the Ritchie experiment that there will still be no temperature differential between the two. You are right that this has little to do with the greenhouse effect outside of proving already established scientific theories and principles. I’ll try to stay on topic.

I have a more detailed reply below, but in short, the Ritchie experiment as described is not definitive and could indeed show no difference with white facing white and black facing black compared to black facing white. It’s even possible for the experiment to show a difference the wrong way depending on the IR emissivity of the paints involved. There are black paints with low IR emissivity and white paints with high emissivity.

I guarantee you would see a difference with carbon black and aluminum foil, though.

• I regards to the experiment and if I’m understanding all that I’ve read correctly, my focus should really be coatings or surfaces with drastically different values of ae in regards to IR. If I were to just use two coatings with differing ae in the solar spectrum they could still have similar, or even reverse values from what would be expected, in the IR spectrum. Given that my radiator will only be around 100C it will be emitting in that IR spectrum.

Also, that is why there are such differing ae values in the table you linked below. The absorptivity has to do with visible spectrum while the emissivity is dealing with IR. Is that accurate?

The absorptivity has to do with visible spectrum while the emissivity is dealing with IR. Is that accurate?

Correct. Well, visible and near IR, where you see most of the energy from sunlight. That’s wavelengths shorter than about 4μm.

The surface emissivity is also related to the temperature seen by an IR thermometer. No home should be without one, by the way. They’re not very expensive for the basic fixed emissivity models. If you have one, you can easily see the effect of emissivity by pointing the IR thermometer at the walls of a pot containing boiling water. You need to be fairly close so the field of view of the IR thermometer is covered by the pot and curvature isn’t important. A shiny stainless steel pot will read much lower than an anodized aluminum or seasoned cast iron pot even though they are at the same temperature.

72. mark4asp,
[moved this response so it appeared in the right place in the thread]

• Since you presented no reply button, here is my reply. You did not answer the point raised by Will Happer, and summarized by me. Here: http://sealevel.info/Happer_UNC_2014-09-08/Another_question.html

You pretend to deal with some criticisms of mainstream GHG hypothesis but you have nothing to say about this one. You don’t even acknowledge the point was made.

Admit SoD, you are in denial.

• mark4asp,

Sorry, but Happer leaves out an important fact that is well known by anyone who has studied molecular spectroscopy. CO2 can also be raised to an excited state by collision with other molecules as well as de-excited, not just by absorption of a photon. At 300K and one atmosphere, about 7% of CO2 molecules are in the first bending mode excited state at any instant whether IR radiation is present or not. You can calculate this easily using the Maxwell-Boltzmann energy distribution and the fact that the first bending mode has a degeneracy of 2, that is the vibration can be in plane or out of plane.

https://climaterx.wordpress.com/2013/04/10/thermal-behavior-of-co2/

The Einstein A21 coefficient gives the rate at which the excited molecules will spontaneously emit. At thermal equilibrium, exactly as many photons are absorbed as are emitted.

https://en.wikipedia.org/wiki/Einstein_coefficients

The Einstein coefficients and other important data, like line broadening, about all the CO2 lines as well as many other molecules can be found in the HITRAN database.

• By the way, the same radiative transfer theory and molecular data that is used to calculate the greenhouse effect is also used to design things like heat-seeking missiles. In fact, the precursor to the HITRAN database was started by the US Air Force for that very reason.

• Collisional excitation and de-excitation of molecules is a feature of the physical principle called microscopic reversibility. At the level of individual molecules, the equations of motion are symmetric with respect to inversion of time.

https://en.wikipedia.org/wiki/Microscopic_reversibility

• Mark4asp and DeWitt: Actually, as I read it, there is no conflict between the consensus, SOD, DeWItt and Will Happer at the link you provided:

http://sealevel.info/Happer_UNC_2014-09-08/Another_question.html

The problem is with the figure that shows absorption (causing vibrational excitation) and re-emission of a photon (causing vibrational relaxation)

A single isolated CO2 molecule would behave as shown. In the lower atmosphere, as DeWitt, Happer, SOD, and the consensus agree, there would be about 1 billion collisions between the absorption event and the emission event, some of which relax the excited state and some of which create a new excited state. Collisions relax “excited CO2” about 13X faster than collisions excite “ground state CO2”, creating an equilibrium where DeWitt’s 7% of CO2’s are excited (though that fraction varies with temperature). When the rate of excitation by collisions is much faster than the rate of excitation by absorption, then we say that local thermodynamic equilibrium (LTE) exists and that the rate of emission depends only on the local temperature, not the local radiation field. (If more photons are emitted by a parcel of gas than absorbed by it, the parcels temperature can drop. However, the number emitted at any instant in time depends on the temperature at that time, not the number absorbed during that instant.)

To create the ILLUSION for the public that the effects of rising GHGs are easy to understand, the consensus often skips over the billion collisions (with collisional relaxation and excitation) between absorption and emission (not re-emission) of a photon. This illusion creates difficulties and mistrust for people like you interested in knowing more than the basics.

How do we know that the consensus is taking into account those billion collisions? In the derivation of Planck’s Law, the Boltzmann distribution (LTE) is used to calculate the fraction of molecules in an excited state. So when you see Planck’s Law or the Planck function being applied, those billion collisions haven’t been ignored. The Planck function is used to calculate radiative transfer of heat in our atmosphere. That process is another one that is often over-simplified into black or gray-body radiation.

• But the problem is that the models have CO2 absorbing at every elevation.

The net effect of the GHG effect is to immediately convert radiation that can be absorbed by CO2 into radiation which can’t be. This pretty much confirms the CO2 saturation effect. The air itself (molecules which collide with the excited CO2) are heated. Absorption by CO2 at the bands it can absorb at is practically a one-off event. After that the thermalized atmosphere can emit radiation. Even O2 and N2 emit microwave radiation. But CO2 will not be absorbing that energy a second time. Re-emitted radiation is now at the wrong frequency.

So the climate models are just wrong. CO2 does its initial absorption close to the ground where the radiation is black body. But that energy will be thermalized. So the CO2 is very limited. After that, the photons which CO2 are just not there. It’s only partially black body as radiation moves further away from the ground. The photons required by CO2 at 2.7, 4.3, 15 micrometer wavelengths have been stripped out.

• mark4asp,

I look forward to your paper rewriting the theory of atmospheric radiation.

Don’t forget to include in your explanation (the reviewers will want to see this) how it is that current theory correctly calculates what we observe? For example, this is from Goody and Yung (the 2nd edition of Atmospheric Radiation: Theoretical Basis, 1989):

There are many more theory and experiment papers around, many images posted in different articles on this site. If the theory is wrong how does it correctly predict the flux values and the spectra of outgoing longwave radiation and downward longwave radiation?

Why do all the textbooks on atmospheric radiation get it wrong?

Perhaps these questions don’t trouble you at all.

• DeWitt Payne: “Happer leaves out an important fact”.

This was implicit in my explanation when I said that O2 and N2 have microwave emissions. But AGW assumes the spectra stay, pretty much, the same because you say: “CO2 absorbs then re-emits”. That statement implies a general case. Happer implies that is the exceptional (and very rare) case. So the wavelength probability distribution will alter.

As for me writing papers. It’s something I’d never do without having a body a data to support it. I believe scientific laws are generalizations made from patterns found in data. That scientific theory is unification and simplification of laws and underlying patterns (revealed by data). In contrast : you AGW people seem to have contempt for data and adore modeling. I do not think you are doing science.

• mark4asp

As for me writing papers. It’s something I’d never do without having a body a data to support it. I believe scientific laws are generalizations made from patterns found in data. That scientific theory is unification and simplification of laws and underlying patterns (revealed by data). In contrast : you AGW people seem to have contempt for data and adore modeling. I do not think you are doing science.

1. You haven’t read a textbook.
2. You haven’t explained why the theory as explained since at least 1950 is wrong. Theory based on the Beer Lambert law of absorption and the Planck law of emission.
3. You haven’t explained why the observations at top of atmosphere and the surface match the theory.

Don’t bother with a paper.

Just write an equation for top of atmosphere radiation as a function of wavelength.

If you understand the subject it should be easy.

Or prove standard theory is wrong. I’ve given you an equation and a derivation.

• Your argument reminds me of: “if it’s not CO2 what else could it be?, therefore it must be CO2”. If you think that’s silly, something along those lines was said by the head UK metrologist 10 years ago. So much for “The Science” guiding policy!

I don’t need to give alternative explanations to to see the flaws in your hypothesis. That’s not how science is done. To disbelieve you, I only need to know of deep flaws in your hypothesis and methods. Here is another criticism worth making. The net effect of GHG is not the sum of absorptions of individual gases. https://twitter.com/GillesnFio/status/906182271719399424 BTW, I don’t expect to win anyone over here. The reason I’m still posting is: your side call me a ‘denier’. I feel I must give an explanation. I find your arguments totally unconvincing. It’s because I find your methods (models all the way down) wrong, and your conclusions (CO2 is > 50% of climate warming) unsupported by science.

• mark4asp,

Your argument reminds me of: “I can’t speak Chinese but your translation of this Chinese literature is completely wrong”.

• mark4asp,

The reason I’m still posting is: your side call me a ‘denier’. I feel I must give an explanation. I find your arguments totally unconvincing. It’s because I find your methods (models all the way down) wrong..

My side? I haven’t called you anything and definitely not a “d—-er” – check The Etiquette and the history of comments in this blog.

I have called you confused and uninformed. That’s because you have no idea about physics. That part is pretty simple. People who have an idea about physics engage with equations. They say interesting physics things like:

– “here is my equation”
– “that equation doesn’t apply in that situation, because..”

..and your conclusions (CO2 is > 50% of climate warming) unsupported by science.

I haven’t claimed that. You must be confusing me with some other blog.

73. Another reply to your argument of “You haven’t explained why …” is you have not explained the Medieval Warm Period nor the Little Ice Age.

• mark4asp,

You are definitely mixing this blog up with ideas in your head, or completely different blogs.

This is a science blog. So provide your equation. Or prove the equation for radiative transfer wrong.

If you can’t engage with scientific ideas then your amazing insights will not be lost (no old comments deleted), but your new comments will be deleted. You can join the small and select group of scientifically illiterate visitors whose self-confidence is unmatched by any actual evidence (equations).

• A science blog eh? Primarily interested in the GHGE (which is why you only wrote one blog on the MWP, perhaps?). So I should search it for what you think of scientists doing GHGE studies: Barrett, Laubereau, Iglev, Lightfoot, and Mamer. No references here to any of those 5 people. Why do your side ignore all these papers?

“Your side”. That’s perfectly justified. No one here mentioned any of these 5 people until I just brought them up. The last two papers are very unsympathetic to CO2 as the main cause of climate change.

Barrett, J 1985, ‘Paper on Spectra of Carbon Dioxide’, Villach Conference, Austria, October 6-19.
Barrett, J 2005, ‘Greenhouse molecules, their spectra and function in the atmosphere’, Energy & Environment, vol. 16, no. 6.
Laubereau, A & Iglev H 2013, ‘On the direct impact of the CO₂ concentration rise to the global warming’, EPL, vol. 104, no. 2.
Lightfoot, HD & Mamer, OA 2014, ‘Calculation of Atmospheric Radiative Forcing (Warming Effect) of Carbon Dioxide at any Concentration. Energy & Environment’, Energy & Environment, vol. 25, no. 8, pp. 1439-1454.

4 papers cited by John Abbot and John Nicol in chapter 19: “Global Warming The Contribution of Carbon Dioxide to” in “Climate Change: The Facts” 2017. .

So yes. There are sides in this debate. Yes :- everyone takes a side even those claiming not to. I suggest cultural cognition for some of the most interesting data on sides in science:

• mark4asp,

Which one of these papers you cite provides an equation for radiative transfer?

Ever since you first arrived with ideas not found in physics textbooks I’ve asked you what is wrong with the equation found in textbooks.

Now you list a bunch of papers you claim to like. So tell me which paper gives the equation for radiative transfer and what it is. You must know.

• The disputed issue here has little to do with radiative transfer equations. At the macro level, you attribute 24% of the GHGE to CO2. I do not. That is the disputed issue. I don’t think you wrote down, anywhere, how you did that; what data and which radiative transfer equations were used.

• mark4asp,

Other than Swarzschild’s equation, there is no simple equation to show how much CO2 contributes to the greenhouse effect. It has to be calculated numerically using an atmospheric radiative transfer program. In fact, I’ve already done that for you above:

https://scienceofdoom.com/2014/06/26/the-greenhouse-effect-explained-in-simple-terms/#comment-119830

What’s your simple equation to show that the contribution from CO2 is significantly different from 24%? You can, of course, like the White Queen in Through the Looking Glass, believe anything you want, but you won’t convince anyone else you’re correct without showing your work.

“There’s no use trying,” she said: “one can’t believe impossible things.” “I daresay you haven’t had much practice,” said the Queen. “When I was your age, I always did it for half-an-hour a day. Why, sometimes I’ve believed as many as six impossible things before breakfast.”

Below is a link to a table of solar absorptivity and IR emissivity for various materials. As you can see, a lot of white pigments have high IR emissivity, so the effect on heat transfer in the Ritchie apparatus from white paint vs black will be small. Obviously the white pigments would have poor performance in a solar water heater.

A high ratio of absorptivity to emissivity means that the material will be hotter when exposed to sunlight than a material with a low ratio, like the white pigments listed in the table. A material with high solar absorptivity and low thermal emissivity is the ideal for a solar water heater. There are several listed in the table including high temperature oxidized stainless steel.

http://www.solarmirror.com/fom/fom-serve/cache/43.html

Note that polished aluminum has a high ratio even though absorptivity and emissivity are both small. Polished aluminum will get quite hot when exposed to sunlight.

Almeco, the manufacturer of TiNOX, has the data to show that it does work.

A car painted black exposed to the sun will be hotter than a car painted white or silver, especially on the painted surface. The interior temperature won’t be as much different because the interior is heated by sunlight through the glass as well as by conduction through the body.

• I’m at the Dyalog APL conference north of Copenhagen with Sweden close across the strait & not much time to get thru email . But I’ve saved the link to the solar mirror table that youall generally DO understand that if ae = k across the spectrum , it drops out of the equation . The ratio is 1.

Yes the Ritchie experiment shows it does not matter which faces are facing each other . There is only a difference in temperature if one of the faces is changed . I pointed out that one of the sources linked goes on to discuss an experiment which overcomes objections to Ritchie by having all surfaces at a uniform temperature . 

• Bob,

ae is rarely constant across the solar and thermal spectral ranges unless the surface is black. Emissivity tends to be high in the thermal IR spectral range unless the surface is highly reflective like a polished metal. As the table I linked above shows, white paint is much more reflective, has lower ae (not that there is much emission from even a 373K surface at those wavelengths) than black paint in the solar spectrum range but is very similar to black paint in the thermal IR range. That explains the results from the Ritchie experiment. The surfaces may have looked different in the visible, but wouldn’t look different in the IR at the same temperature. As I said, if Ritchie had used aluminum or gold foil instead of white paint, there would have been a very large difference.

• “ae is rarely constant across the solar and thermal spectral ranges ”

Of course not . That’s why agreement on the calculation of the equilibrium temperature for any arbitrary ae spectrum is a non-optional first step in any quantitative analysis .

The Ritchie experiment applies to any 2 surfaces with any 2 arbitrary spectra . The objection raised was the “darker” one would be locally hotter . The alternative experiment described following the description of the Ritchie one nk in the Richtmyer , 1928 link , https://archive.org/stream/in.ernet.dli.2015.260663/2015.260663.Introduction-To#page/n210/ , overcomes that issue .

75. Is there not agreement on use of sb for calculating the temperature a body will arrive at given the distance, size, and temperature (or power emitted) of the radiator as well as the absorptivity of the receiving body?

That’s exactly how IR thermometers work. They calculate an effective temperature based on the measured emission from the object of interest. The emission is measured as a temperature difference across a thin thermal resistance to a copper block at a known temperature. The outer surface of the device has a high absorptivity in the thermal spectral range.

You can even point an IR thermometer at the sky away from the sun and measure an effective temperature as long as it’s not too cold so the effective temperature is below the lower limit for the device.

Note, that’s an effective temperature because you don’t necessarily know the emissivity of the object being measured or that the emissivity is constant with wavelength.

You can only use the S-B equation to calculate a real temperature in your example if the emissivity of the source and the sink are constant over the spectral range of the emission and absorption. Otherwise, you have to go back to the Planck equation and integrate.

• I repeat that for a gray , ie : ae == k across the entire spectrum , k drops out of the equation and the temperature is determined exactly be the SB equation based on the total power impinging on the body . In our obit that is about 279.

If ae is less over the peak of the solar power spectrum than over the thermal spectrum for an object around 279 , then the object will be colder . If you plug in an ae of 0.7 over the solar peak and 1.0 over the ~ 279 thermal distribution you will get a value of about 255 .

The formula for arbitrary spectra is the 4th root of the ratios of the dot products of the respective spectra times the gray body temperature as I have repeatedly stated but I have yet to see anybody understands that statement or offer a computable , and therefore experimentally testable alternative .

76. Bob,

Absorptivity doesn’t drop out of the equation. An opaque body with less than ae of 1 means that light is being reflected. A body reflecting light won’t have an increased temperature because of this reflected light.

77. I feel like you’ve probably seen these equations before….

Blackbody temp:

S is flux at surface from radiator:

T = (S / (4 * sb))^1/4

For a S of 1367 (flux from sun on earth) this yields a black body temp of

279, no surprises

When we need to take into consideration that the receiving object has a lower absorptivity (increased reflectivity) then the equation becomes:

T = (a * S / 4 * sb))^1/4

In the case of earth, S=1367, a=.7:

T = 255

So I see from your other reply that you arrive at the same numbers, what I don’t understand is why you say that the absorptivity value drops out an is not needed for the calculation. What am I misunderstanding?

You’re not doing an energy balance. You have to calculate and balance the emission as well as the absorption. For that you need to specify the geometry as well as the spectral characteristics. The average temperature of a spherical body depends on the thermal conductivity and the rotation rate, among other things.

For a thin flat plate perpendicular to the incident radiation and with perfect insulation on the unlit back side, the temperature of the plate is 394K. If the back side of the plate radiates to space and the thermal conductivity is high enough that the temperatures of the front and back are equal, then the temperature is 331K. If the plate is a perfect insulator, then the back side of the plate will be 2.725K and the average temperature will be 198.4K. For a non-rotating sphere with very low thermal conductivity, the unlit side will also be at the temperature of empty space, 2.725K The temperature of the lit side will vary, with the highest temperature at the equator with the sun directly overhead and reducing to 2.725K at the terminator. Integrating the temperature over the whole surface of the sphere gives an average temperature of 158K.

The numbers normally quoted are for an isothermal sphere. That would require either a very small, rapidly rotating sphere with metallic conductivity or superconductivity for a large sphere. For any other set of conditions for the sphere, the average temperature will always be lower than for an isothermal sphere. A mathematical principle called Hölder’s Inequality requires it.

Bob is correct that for a grey superconducting sphere, where the emissivity is constant at all wavelengths, the temperature will always be 279K and the value of the emissivity, as long as it isn’t identically zero, doesn’t matter.

Your calculation of 255K is only correct for an isothermal sphere with an absorptivity for solar radiation of 0.7 and a thermal IR emissivity of 1. You’re correcting the absorption for a lower absorptivity, but you’re not correcting the emission for a lower emissivity.

For a solar absorptivity of A_1 and a thermal emissivity of E_1, and a solar flux of S, the equation is A_1 * S = E_1 * σ * T^4. Then T = (S * σ * A_1/E_1)^1/4. Obviously then, if A_1 = E_1 it drops out and the temperature is the same for any non-zero value of A_1.

What Bob neglects is that the Earth is not gray, particularly the atmosphere, and the emission to space does not all come from the surface at the temperature of the surface.

• DeWitt , I’m not neglecting , I’m seeking agreement on the computational equation for arbitrary source and sink power spectra and object ae spectrum .

Then we can look at the details of temperature thru the “thick layer of paint” represented by the atmosphere . And , as importantly non uniform color ie : maps , over the sphere ,

• DeWitt,
Thanks for the great info and clearing up my misunderstanding.

• The issue is when ae is Konstant across whole spectrum as in the flat line in , I will repeat , this graph :

We get exactly the same 255 number when we put in the assumption of ( 0.7 ; 1.0 ) as in the the step function .
How the hell can we agree that that is correct but the computation for the flat constant gray line be wrong ?!

Jeez .

• Bob,

aeEarth is not 0.61. Period. That number incorrectly assumes that all radiation to space comes from the surface and the atmosphere has no effect. That’s wrong. The emissivity of the Earth is indeed close to one, but most of the emission comes from the atmosphere above the surface where it’s colder than 288.4K. The 254.9K temperature is another questionable number. It assumes that the albedo of the planet, aeSun in your table, would remain the same if the atmosphere were transparent in the IR and all radiation did come from the surface. But if you remove all the water vapor, you also remove clouds, which account for about 2/3 of the albedo. It also assumes an isothermal surface, which wouldn’t happen either.

255K is the effective temperature of the planet as seen from space because the planet must emit close to the amount of radiation it receives or the temperature would go up or down depending on the sign of the imbalance. But the spectrum is not a black or gray body spectrum. It has structure that comes from the water vapor, CO2 and other IR active gases in the atmosphere. But that’s just a calculation. One should talk about the global annual average total flux to space of ~239W/m², not the S-B calculated temperature of ~255K. And the total flux from the surface of ~490W/m², ~390W/m² by radiation and ~100W/m² by convection.

The point is that we can and do measure the flux from the surface , both IR (look up SURFRAD) and convective (look up EBEX-2000) and the flux to space (CERES, for example). The flux from the surface is much higher than the flux to space. The measurements aren’t precise enough to close the balance to the less than 1W/m² that is thought to be the current imbalance at the top of the atmosphere, but they’re in the ballpark. The greenhouse effect is the difference between the upward flux at the surface and the upward flux to space. It’s not just theoretical, it’s been measured.

• “aeEarth is not 0.61. Period. ”
Absolutely RIGHT !

But that is what its , I’ll call it : aShort % eLong ratio would have to be to have its surface temperature explained by its radiative equilibrium .

The hypothesis is false .

Earth’s surface temperature , much less Venus’s for which the claim is a quantitative absurdity by an order of magnitude , cannot be explained by a spectral phenomenon .

Are you implicitly agreeing with the ratio of dot products computation and that a flat gray spectrum drops out of the equation — simple high school algebra ?

• Bob,

The only thing you’re correct about is that a black or gray body with a constant absorptivity/emissivity with wavelength exposed to EM radiation will have the same temperature regardless of the the (non-zero) value of the absorptivity/emissivity. It is simple algebra.

The Earth does not have to have an emissivity of 0.61 to radiate only 239 W/m² to space with an average surface temperature of 288K if the radiation to space does not all come from the surface. And it doesn’t. In the tropics only 12.6% of surface radiation is transmitted directly to space because the atmosphere is effectively opaque for 87.4% of the radiation emitted from the surface. But because the pressure and so the number density of absorbers drops with altitude, eventually an altitude is reached where the optical depth drops to 1. That’s where the maximum rate of emission to space happens.

But if you knew anything about emission and absorption spectroscopy, you would already know this. Temperature decreases with altitude in the troposphere where the majority of the emission to space originates. Since at any wavelength, the emission cannot exceed the emission calculated for a black body by the Planck equation, the maximum intensity of emission drops with altitude. That’s why the Earth radiates to space at a much lower rate than the surface radiates to the atmosphere.

You’re also wrong about Venus. The radiative transfer and energy balance calculations work out quite nicely. To all intents, no radiation from the surface directly reaches space. It all comes from much higher up where it’s a lot colder because, just like the Earth, temperature decreases with altitude at pretty close to the calculated adiabatic lapse rate. Sufficient sunlight does reach the surface to drive this process. The Russian Venera probe took pictures of the surface of Venus in the visible spectrum using available light.

I get the impression you think the atmosphere is a thin coat of colored paint on the surface as in the Ritchie apparatus. It isn’t. It has significant depth, which is important for radiative transfer calculations and it’s neither isotropic nor isothermal as a coat of paint would be. MODTRAN, for example, breaks the atmosphere up into about 30 vertical layers for calculation purposes. More layers wouldn’t improve the accuracy of the calculation. Up to 25km, the layers are 1km thick. From 25 to 50km, they’re 5km thick, then there’s a 20km thick layer and a 30km thick layer. That gets you to 100km. The actual atmosphere continues beyond that, but emission and absorption in the thermal IR are negligible above that altitude.

• The fact that we can agree that the temperature of a flat spectrum gray ball is constant , no matter how dark or light , and comes to the same temperature , determined totally by the total energy impinging on it , means at least we have reached a common testable computational foundation . Your agreement to the computation for the ( .7 ; 1.0 ) and .61 ratio required for a ~ 288 equilibrium implies an understanding of the general ratio of dot products of spectra times the gray body temperature , altho it would would be nice to see you explicitly agree or express the general case in your notation .

As I’ve said , if you can point me to a table of the ToA spectrum I’ll run the computation for that spectrum and give a value which should start being more useful and realistic than the crude 255 value . But sitting here in Copenhagen after being with some of my brightest old friends at the https://www.dyalog.com/ APL conference , I’m not going to “screen scrape” a graph .

In any case we seem to be on the verge of speaking classical analytical quantitative physics .

And that’s not a jump to clouds or any other anisotropies . Let’s be sure we quantitatively understand the simple uniform semitransparent case first . And that will be essentially the Schwarzschild differential applied to a uniform sphere with a surface of a single color and transparent layers of single color .

Then , we will have a common quantitative testable foundation to start dealing with non-isotropies .

And thus comes the rub . I’ve got to get on to other things now , but suggest you review the Divergence Theorem and consider its implications for any claim that the energy density of any interior layer of a sphere is different than that radiative balance we appear to be converging on agreeing on the computation of .

• Oh, I forgot about clouds. Clouds are completely opaque to radiation from the surface below. Radiation to space starts at the cloud tops for 60% of the Earth’s surface. Needless to say, the cloud tops are colder than the surface, so they radiate less to start with.

• Bob,

The radiative properties of the atmosphere, i.e. the fact that the atmosphere radiates at a higher rate to the surface than it does to space does not violate the Divergence Theorem. It’s not up to me to prove that it doesn’t. It’s up to you to prove that it does.

• DeWitt , you should know by now that only computable count with me . Verbal claims mean nothing . I again refer to Lavoisier for the goal of creating an quantitative audit trail from the output of the Sun to our surface temperature . He converted alchemy to chemistry by that insistence on a quantitative chain .

Are we agreed on the computation of radiative balance for radiantly heated balls in a vacuum ?

• Bob,

Verbal claims mean nothing

Right back at you. As I said, the onus is on you to prove that radiative transfer calculations are incorrect. You haven’t provided any equations or calculations. You have just made completely unsupported assertion. Example:

Earth’s surface temperature , much less Venus’s for which the claim is a quantitative absurdity by an order of magnitude , cannot be explained by a spectral phenomenon.

Prove it.

Measurements agree with calculations, so you can’t. The posts at this site, which you likely haven’t bothered to read since your mind is already made up, are yet more evidence, and they include equations and calculations. I’ve even given you a link to an open source radiative transfer computer program, LBLRTM. You have all the information you should need.

A passing reference to the Divergence Theorem is just more hand waving on your part. The surface temperatures of spheres in vacuum is also almost completely irrelevant to what happens at the surface of a planet with an atmosphere that is not transparent to radiation emitted from the surface, as is the Ritchie experiment, which was fatally flawed.

I won’t bother to respond to anything you post that only makes more unsupported assertions.

By the way, the Socratic Method, which seems to be how you are trying to argue by getting me to agree about spheres in a vacuum, is a big turnoff for me. IMO, Socrates got off way too easy. Bloodless crucifixion, the standard method of Athenian execution at the time, would have been far more appropriate.

78. “Clouds are completely opaque to radiation from the surface below.”

<- Statements like that are why I distrust climate modelers.

• Why’s that? Are they not opaque to OLR?

• So now your are talking about just being “opaque”, when, previously, it was “completely opaque”. Which wavelengths are the opaque to? Is that every single, earth exiting, black body OLR wavelength?! I would not have thought so. That is what I interpret as “completely opaque”.

Do you have a citation to show that wavelengths to which they are completely, relatively, and not at all opaque to?

• Mark,
I’m not DeWitt but I’m sure he has the capacity to backup his statement. Doubt a few minutes of searching reveals a number of papers discussing the opacity of clouds to infrared radiation. Here’s one I found that discusses the difference at varying wavelengths. Clouds are highly opaque to any radiation shorter than RF. While it doesn’t say completely, the percentage passing through was .005%, which is going very near 0. Maybe if someone says something that you think to be wrong or inaccurate you could provide your own reference to why instead of just saying they aren’t to be trusted because they didn’t give you enough information for your liking. I’m rather new to the topics and discussions here but have thoroughly enjoyed the ability to have dialogue with those that know more than myself as well as the challenge of researching unfamiliar topics to challenge my own misconceptions.

• mark4asp,

Nitpicking my own grammar: opaque is an absolute adjective. So completely opaque is incorrect usage because the ‘completely’ is redundant. IOW, there is no difference between ‘completely opaque’ and ‘opaque’.

Go to the MODTRAN site and select the first cloud model and then click on show model output and look at that. The transmittance from 5000 to 4.545 μm is 0.0001. To all intents and purposes, that’s opaque, especially when compared to a transmittance of 0.1261 for the clear sky tropical atmosphere.

The low frequency range of MODTRAN was recently extended to 2 cm-1, ~5,000μm. It turns out that at wavelengths longer than 1000μm there is some slight transmittance. But the emission intensity from the surface is very low at those wavelengths. At 2cm-1 (4999.998μm) the transmittance is 0.10725. The radiance, though is 9.63E-10 W/(cm² steradian cm-1). Peak radiance is 1.22E-05 W/(cm² steradian cm-1) at 17.857μm, four orders of magnitude higher. The transmittance there is 0.00000

http://climatemodels.uchicago.edu/modtran/

• Modtrans is irrelevant. Climate ‘heroes’ like Hansen did not use it (https://wattsupwiththat.com/2014/04/12/a-modtran-mystery/). Catastrophic predictions depend on accelerated forcings due to water vapour feedback. This water vapour feedback is simply written into climate models as parameters. It is not derived from any kind simulation of first principles in the General Circulation Model runs (GCMs). Long tails (in probability distribution charts) of the predictions, show large Equilibrium Climate Sensitivity (ECS), often 3, 4, even 6 times the effect of CO2 alone. These long tail, high ECS, values from the climate models are used to blackmail us into replacing our energy systems with those which do not work (e.g. Energiewende, South Australia). It’s politics not science. I basically object to the use of these climate models, which pretend to be simulations. If it were only about CO2 forcing, there would be no issue. No COP, no IPCC, no Paris. Because CO2, _alone_, obtained by burning fossil fuel, will never be shown to change climate in a harmful way.

• mark4asp,

Now you’re moving the goalpost from your original post in this subthread.

I happen to agree that the long term high positive feedbacks from GCM’s, where a very small radiative imbalance at the TOA leads to a very large temperature change in a relatively short time looks fishy. But that’s a separate point.

• To: DeWitt Payne,

I admit I moved the goalposts but… This is the main point which tipped me over to scepticism. My overriding objection to ‘climate consensus’. 1) Actually impossible projections in the long tail for the ECS distributions. 2) Much of it due to arbitrary inclusion of H2O feedbacks based on no ‘science’ I accept as science. Comparatively: other objections I have are just nitpicking. I thought I’d come clean on it.

• mark4asp,

Have you heard of the term lukewarmer (luckwarmer if you’re Eli)? That’s the term invented by Steven Mosher to describe someone who accepts that increasing CO2 will likely cause some warming, but that the equilibrium climate sensitivity to a doubling of atmospheric CO2 falls in the lower half of the IPCC range, from 1.5 – 3 °C. That’s my position.

• mark4asp wrote: “Statements like that are why I distrust climate modelers.”

As appropriately skeptical scientists, we ALL distrust climate modelers. As Box said, All models are wrong, some are useful. We – YOU – need to know which are USEFUL. (Though they rarely tell the public, even climate modelers know their models are wrong – they have serious disagreements with each other.)

Radiation transfer calculations are only a small subset of the calculations done by climate models. These calculations start with absorption cross-sections measured in the laboratory, including parameters for how the cross-section varies with temperature and pressure. For the most important lines, these measurements have been made by several laboratories and done long before the hysteria about climate change. There seems little reasons to distrust these measurements.

The theoretical framework – the Schwarzschild eqn – that uses these cross-sections (o) is derived from Einstein coefficients for absorption and emission of photons

dI = n*o*B(lambda,T)*dz – n*o*I_0*dz

where dI is the change in radiation of a certain wavelength as it passes an incremental distance, dz, through a medium. That change is produced by absorption (the second term) and emission (the first term). This equation reduces to Beer’s Law at high temperature (where emission can be ignored) and to Planck’s Law (when there is equilibrium between emission and absorption and dI = 0). Many routinely use and are confident in Beer’s and Planck’s Laws without knowing that they are basically corollaries of the Schwarzschild Eqn.

This equation has then been tested in our atmosphere to see if it agrees with what we observe. It does. For example, see this SOD posts (and several others)

When one looks at theory vs observation in the atmosphere, one quickly realizes that the bigger problem is knowing the temperature and composition (n and T in the Schwarzschild eqn) of the atmosphere. One uses radiosonde data for that and finds good agreement. However, AOGCMs MUST calculate n and T for themselves. IMO, there is plenty of room for skepticism about how composition and temperature are calculated (convection, condensation and precipitation) and little room for doubt about how much radiation is transferred through a given increment (dz) of atmosphere. Those calculations do require a computer program to perform numerical integration (say with MODTRAN or HITRAN) and therefore appear “non-transparent”.

Amazingly enough, our host wrote his own program to do numerical integration and demonstrated what theory predicted about complicated confusing situations like “saturation” of bands and overlapping bands.

To reduce computation time, AOGCMs take short-cuts in calculating radiation transfer. An early model comparison project found that some of these shortcuts weren’t accurate enough, but that problem has been fixed.

Nevertheless, it amazes me how many technically competent skeptics don’t even bother to learn how radiation transfer is actually calculated and are ABSOLUTELY SURE that it is being done wrong.

Even worse, those skeptics forget that 60% of the sky is covered with clouds. AOGCMs must predict the type of cloud formed and therefore its optical properties.

IMO, there is plenty of room for controversy about AOGCM’s, but questioning radiation transfer mostly a waste of time. At best, it is a sign of ignorance. At worst, a sign of the “alternative facts”, “fake news”, and conspiracy theories that are taking over the blogosphere.

79. Just for giggles, I dug up my Venusian atmosphere spreadsheet, all 900MB of it, as described here: https://scienceofdoom.com/2010/06/22/venusian-mysteries-part-two/#comment-4367

At a temperature of 730K and an emissivity of 1, the surface radiates upward 16,101.8 W/m². The radiation downward from the atmosphere is 16,096 W/m². The estimated average incident solar radiation is ~17W/m². If you want to duplicate my calculations, you’ll have to subscribe to Spectralcalc yourself. I’m not going to upload a spreadsheet that large. Even in binary form, it’s 250MB.

• So, why use anything else than heat transfer equations? There is no reason to not use the sb-equation for transfer, other than it doesn’t support tour claims. “Opaque” is a term for optics, why is greenhouse-theory he only theory that uses equations for optics to calculate heat and temperature, when all other cases use heat transfer and thermodynamics?
Heat and temperaturen is NOT frequency-dependent, it is power-dependent.
You are describing a thermodynamic system, but don’t include the first law.
Try it:
dU(TSI)=W(4g²)+Q(4*0.0000000567*256⁴)
g=9.78, surface accelereration acc. to NASA
g²=95.7Nm², units of thermal resistance, stress and pressure

Heat transfer works, and thermodynamics, stop fiddling

• Optics and radiation are the same thing. The atmosphere, while mostly transparent too the visible spectrum (we can see through it ) is not transparent to radiation in the non visible spectrum. In fact, to some wavelengths it is opaque. Meaning radiation emitted at lower altitude at that wavelength doesn’t make it through the top of atmosphere.

• Radiative heat transfer is important in engineering as well. See, for example, Modest, Radiative Heat Transfer.

A gravitational field cannot provide a continuous source of energy, which is required to maintain a temperature gradient. That would violate both the First and Second Laws.

80. Of course the equations you use work for transfer, the problem is that you use them to get a reversed, opposite, effect than what heat transfer with the s-b equation produce. There is no way to get an increasing temperature of the heat source with the s-b equation, but you manage to do that with the equations you use. The surface is the source of any, including co2, emission in the atmosphere, and you are claiming that the power of the heat source is increased. From less emission of heat by co2. It is a very, very strange claim.

Another thing, the effective emission at 255K, 240W/m^2, from a sphere, needs a source power of 960W/m^2. To increase the effective emission, you need to increase the emission from the source. According to the inverse square law. A few watts *decrease* in the coldest part of the system(the atmosphere), will not increase the power of the source at 960W/m^2. Doesn´t the inverse square law apply to the greenhouse theory?

81. “A gravitational field cannot provide a continuous source of energy, which is required to maintain a temperature gradient. That would violate both the First and Second Laws.”

This is also a very strange statement. The force of gravity(surface acceleration, 9.78 according to NASA) adds 9.78Nm to the surface continously. It compresses the atmosphere by 95.7Nm^2 every second, while heat emission simultaneously expands it. Pressure, as we know, is closely related to temperature.
To produce a force of 95.7Nm^2 you need 95.7W/m^2 of energy. According to the inverse square law, the source of that energy is 4*95.7Nm^2=383W/m^2, which happens to be equal to heat emission at 287 Kelvin. Weird coincidence.

The effective emission from a perfect blackbody absorbing TSI at 1360.8W/m^2 would be 340.2W/m^2. If we use the first law, we can see that the missing energy of emission is exactly the energy of the force of gravity:
340.2-95.7=244.5W/m^2, equal to an effective emission at 256K.
The first law works, and there is no reason that it shouldn´t. Because gravity does work, you can measure it as easily as putting yourself on a scale.
The difference between heat and source power in a thermodynamic system is the work done, there is no reason why that wouldn´t hold for earth. And we got the mean values of the heat source, the emitted heat and the work done. Since it fits, and also work for Venus and Mars, I think it trumps any calculation where less emitted heat means more emitted heat, like the ones we see here.

It seems like you have blinded yourself in your calculations. Why would less heat in the atmosphere mean more heat in the system. That is contradictionary in itself. There is no example of that anywhere else in physics. It is like you could increase the temperature of a running engine by cooling the exhaust.

• Lifeisthermal,
Do you wear a jacket when it is cold outside? If so, do you feel warmer than you would without it? The rate of emission of your body hasn’t changed. Rather an insulating layer has helped to slow heat transfer to the atmosphere surrounding you, much in the same way that miles of atmosphere slow heat transfer to space. The reason is ultimately because emission from the outer shell of the jacket (top of atmosphere) is much lower than emission from the body (surface) . The net result of this is that the body is warmed, even though it’s heat source has remain unchanged. A warmer body emits more radiation, causing the jacket to warm more and consequently emit more. The body will continue to heat until the rate of absorption of the jacket on the inside is matched by the rate of emission of the jacket on the outside, or you get warm enough that you unzip.

• A jacket is thermal insulation. Thermal insulation *prevents* absorption in cold surroundings(the atmosphere is cold surroundings). Co2 increases absorption in cold surroundings. This is very basic, the gh-analogy with blankets and jackets is as wrong as it can get, but gh-believers don´t get this.

An atmosphere does NOT slow emission from the solid earth. This is very old thermodynamic knowledge. The emission depends on the internal state of the emitter. Only. The atmosphere is the opposite of the internal state of the solid earth.

• I haven’t replied to some of DeWitt’s misunderstandings since I headed to give my Vienna talk , https://t.co/8lVsFIEsJb. But you are right . Gravitational energy CANNOT be left out of the equations , and only with it ( computing as a negative as , eg , Alan Guth clearly explains ) can the total energy equations be balanced .

Gravitational energy MUST be accounted for or else the theory is incomplete .

For all their word waving , NO simple differential , ie : parameter sub-domain of the Schwarzschild equation , ( nor experimental demonstration ) ever been presented explaining how a em spectral effect can trap energy causing the bottoms of atmospheres to be hotter than their tops , whatever their optical properties .

If it’s not quantitative , it’s meaningless .

• lifeisthermal and Bob Armstrong: In our atmosphere, no mass rises without some other mass falling. For the most part, this buoyancy means that the effects of gravity are small and that no net work is being done.

Small differences in atmospheric pressure at a given altitude exist, change with the weather, and produce winds that do little work because viscosity is low. Even the terminal velocity of rain drops is low, meaning that most of their potential energy is dissipated in the atmosphere (and not returned to the surface).

If gravity were important, the lower troposphere would be enriched in higher molecular weight gases compared with the upper troposphere. Fractionation by molecular weight occurs above the turbopause (100 kM), but convection and radiation dominate heat transfer in the troposphere and stratosphere, not gravity.

• I have yet to see ANY computable testable equation for the trapping of heat by an em phenomenon . It just would take a single differential . Nor have I ever seen an experimental demonstration of the phenomenon .

On the other hand , While I’ve not gone thru them thoroughly , ie : implemented them , HockeySchtick at , eg : http://hockeyschtick.blogspot.com/2014/11/derivation-of-entire-33c-greenhouse.html , does provide calculations .

I will simply repeat that gravitational energy MUST be included in total energy equations . If it’s not expressed as heat , how is it expressed ?

Only computable equations count as responsive answers .

• Bob A writes: “I have yet to see ANY computable testable equation for the trapping of heat by an em phenomenon . It just would take a single differential . Nor have I ever seen an experimental demonstration of the phenomenon.”

Observation: An average of 390 W/m2 of upward radiation is emitted by the surface, but only 240 W/m2 is escaping to space. This is not “trapping of heat” – it is better described as a resistance to outward heat flow by radiation – analogous to insulation.

Bob A writes: “On the other hand , While I’ve not gone thru them thoroughly , ie : implemented them , HockeySchtick at , eg : http://hockeyschtick.blogspot.com/2014/11/derivation-of-entire-33c-greenhouse.html , does provide calculations.”

It is very difficult to convert a change in radiation flux into a change in temperature such as 33 K. One needs to reach mutual agreement about what the albedo of the Earth would be with no GHE’s in the atmosphere (no clouds? more clouds? same clouds?) and a much colder surface (frozen oceans? more ice caps?). Also agreement is needed on the temperature difference between the poles and equator, the temperature difference between night and day, and the lapse rate on the planet without GHGs. This is a challenging calculation even for the much simpler Moon.

The GHE is 150 W/m2, the difference between average surface upward emission (390 W/m2) and TOA outward emission (240 W/m2). If this difference (which is produced by GHGs in the atmosphere) didn’t exist, the planet would be much colder. That is a GHE.

To a first approximation, 2XCO2 should add another 3.7 W/m2 to that 150 W/m2, but the changes (feedbacks) associate with that process are so complicated that even the IPCC won’t say if ECS is 1.5 K or 4.5 K (and admits a 30% likelihood it could be outside this range).

The GHE and aGHE are simple when expressed in terms of W/m2 and horribly complicated when expressed in terms of degC. Those complications don’t mean the GHE and aGHE don’t exist.

Bob A writes: “I will simply repeat that gravitational energy MUST be included in total energy equations . If it’s not expressed as heat , how is it expressed ?”

There is no such thing as “gravitational energy”, but there is “gravitational potential energy”. It can be converted to other forms of energy when something FALLS. Nothing is falling today. The atoms in our atmosphere fell to Earth 4.5 billion years ago and generated enough heat to melt to cold matter of interstellar space – which is why the Earth has a denser core than mantle. However, that heat dissipated long ago by radiative cooling to space.

• Critical equation please . Then I can implement it , and play with it and understand the parameter space which traps energy . I know of no other relevant differential than the Schwarzschild , and while I have not explored it , I don’t see under what set of parameters it asymmetrically “traps” energy .

“There is no such thing as “gravitational energy” ” is just physically ignorant .
Tell us that apples don’t fall .

• Oh , I see you do agree that apples fall converting potential to kinetic energy , ie : heat .
But you claim that kinetic energy simply dissipates by em radiation — leaving the apple free to float again .

It’s a weird sign of the www age that this debate is in many ways between “Science of Doom” and “HockeySchtick” . He’s got equations . where are yours ?

• on November 2, 2017 at 7:28 pm Peter Offenhartz

It might be helpful to note that the earth’s atmosphere is roughly iso-energetic with altitude, that is, the sum of thermal energy and gravitational potential energy does not usually change much with increasing altitude. Another way of thinking about this is to mathematically concert potential energy to thermal energy; the result is called “potential temperature” and it is roughly independent of altitude.

• Interesting . That’s as one should expect to make the total energy balance equations work w/o mysterious “trapping” of energy by some symmetric em force ( without enabling equation ) .

Note that it was only a while after my Heartland talk showing the quantitative absurdity of Venus’s bottom of atmosphere temperature being explicable by spectrum that discussions on Anthony Watt’s ( who is still in the GHG box ) blog got thru to me that gravity was left out of the energy balance equations — but CANNOT be .

• The atmosphere convects according to the heating by the surface, at 383W/m^2. As you say, gravity compress the atmosphere equally. With a force of 9.78Nm in a point, 95.7Nm^2 on a square meter. To produce a force of that strength on a spherical surface from a source below, the inverse square law tells us that the source power is 4*95.7=383W/m^2.

“convection and radiation dominate heat transfer in the troposphere”

And you think convection is only upwards?
If gravity didn´t contract equal to expansion of heat, the atmosphere blows into space. As I wrote, these calculations work for Mars and Venus as well, and you can´t get surface temperature right on earth without saying that a cold fluid with increasing amounts of dry ice makes it´s own solid heat source emit more power. You chose the least probable cause of surface temperature instead of including all forces and heat.
Surface temperature: 1/2*TSI/(4/3)^2. Equal to the necessary source power of gravity, 4g^2. Do you really think a cold fluid as a heat source is more reasonable?
This is just a fraction of the relationships that can be found with simple principles, like the inverse square law, electric field equations, equation of state for van der Waals fluid, shell theorem with volume etc.

Basic thermodynamics and geometry is enough, but I guess you don´t care. It´s a shame, because thermodynamics is the only 100% consensus science.

• Bob,
If heat and gravity were linked a you imply then as an object heats it would gain gravitational potential. As an object would cool it would lose gravitational energy. So put your apple on a hot plate until it gets hot enough to gain gravitational potential and you might have a basis for your argument. Do you have a testable experiment that demonstrates an object gaining gravitational potential as it heats, or the opposite?

• That’s basically a non sequitur . But , in fact , thru general relativity , it will .

I think the problem the purely em GHG hypothesis has to face is to balance that very easily calculated gravitational energy . It doesn’t disappear and the trade-off of depth in a gravitational “well” and heat is ubiquitous extending to the centers of all planet size bodies , and for that matter , the blue shifting of light itself , confirmed only around 1960 .

• Peter Offenharrz,

Gravitational potential energy is not included in potential temperature, at least not directly. For calculation purposes, parcels are exchanged so there is no net change in gravitational potential energy.

θ ≡ T_1 * (p_0/p_1)^(R/C_p).

Altitude is not included in the equation, only pressure. Maybe you can relate the ratio p_0/p_1 to gravitational potential energy. I haven’t done the math.

• on November 2, 2017 at 10:21 pm Peter O'Donnell Offenhartz

@ DeWitt Payne

What you say is absolutely true, but the idea behind potential temperature is the swap between gravitational potential energy (“mgh”) for thermal energy (“RT”). Remember that P1/P2 = exp(-mgh) for near-isothermal conditions (if memory serves!)

• Bob,
How is my comment irrelevant. You state that gravitational potential and thermal energy are linked. Can you provide any testable theory or experiment that demonstrates this? Maybe an equation, since that is what you always seem to be demanding, that demonstrates a direct link between thermal energy and gravitational potential?

Also, light from the sun is red shifted, not blue shifted, as it approaches earth. Saying it is blue shifted is completely false.

• Right now I’ll have to just point you once again to HockySchtick’s analyses . As I have said , I have other priorities fleshing out CoSy , my reduction of APL in Forth — unless someone motivates my working in that direction .

At least he has equations — and references going back to Maxwell . A number of other people have similar analyses by now also I believe — including here .

I just did the trivial computation of radiative balance for arbitrary spectra which showed the absurdity of the notion that Venus’s surface temperature , 2.25 the gray body temperature in its orbit ( 25 times the energy density the Sun delivers ) could be explained by a spectral effect .

And it continues to be impossible to even get agreement on that trivial essential experimentally testable computation , much less any equation showing how the constraint of the Divergence Theorem for the internal temperature to match that calculated for the effective radiative surface is overcome to explain how an internal temperature is sustained hotter than that radiative balance .

My attitude remains that if you can’t get agreement on how to calculate the ( experimentally testable ) temperature of a billiard ball under a Sun lamp , what’s the point of speculating further . I’ve got other things to do .

• Bob,

You keep asking for equations. I gave them to you last time (and not for the first time) on Aug 28th:

Bob,

I’ve pointed you to this before. Here it is again:

Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations

I look forward to your Groundhog Day comments requesting this equation in the future.

You replied:

Jeez I don’t have time for this now..

..because you are so busy. Etc. Etc. Now you again ask for the equations.

None of the many super-confident people who show up disputing the simple greenhouse effect – this includes you Bob, as a super-exemplar – ever actually point out a flaw with the equations derived from first principles. These equation also predict OLR & DLR flux and spectral values of radiance at the surface and at top of atmosphere.

You can’t point out a flaw. You can’t show that the equations when applied get a different result.

That’s because you don’t understand them, or the very basics of atmospheric radiation. It’s pretty simple. Absorption. Emission. Both spectrally dependent.

But instead of facing up to your lack of understanding you just keeping asking for the equations.

Everyone can see why

With your final comment you tease us yet again:

..I’ve got other things to do

This time let’s say – it’s not you, it’s me.

You can join the illustrious group of …. whose testimony to ignorance is forever preserved but whose further contributions are no longer required.

• SoD,

Thanks. Bob reminds me of another poster whose name also started with a B.

For completeness: He claims he never saw a simple experiment. I posted results of a simple experiment to demonstrate how the bottom of the atmosphere can be warmer than the top solely due to EM radiation. He claims I never gave an equation for the temperature an object exposed to solar radiation. That’s wrong too. His memory is either conveniently short or, more likely, he is just a tr0ll like that other poster.

82. Oh , and Brad : Does putting a coat on a corpse warm it ?
The planets are heated from the outside ( except for a bit of geothermal and human ( not inconsequential in cities ) which is minor ) .

• Bob,
Only if the coat is unique in that it is transparent to the visible spectrum while not transparent to ir spectrum, and the corpse has a visible light spectrum heat source impeding it. My point still stands that without changing the rate of emission, the body can be at a higher temp without changing the heat source.

• I remember now what DeWitt’s comment that was left hanging .

He explicitly refused to commit to what should be an undergraduate level exam question for an equation to calculate the equilibrium temperature of a ball of uniform color ( absorptivity=emissivity spectrum ) radiantly heated by a disk with a given power spectrum .

That experimentally testable computation is exactly what answers your question as to the equilibrium temperature of a dead body under a semitransparent , ie arbitrary absorptivity spectrum , body bag .

In all of this endless and endlessly stagnate word waving , total resistance to any experimentally testable non-optional basic computations .

• Bob,

• Again , the most basic computation for arbitrary source and object spectra is at http://cosy.com/Science/warm.htm#EqTempEq in a freely downloadable , tho somewhat esoteric , APL . But the essential computation is a ratio of dot products which are essential to computations involving spectra ( and most other physics or any quantitative subject . )

Plug in the spectrum of the object as seen from the source(s) and sink(s) and it will give you the equilibrium temperature .

Since even at a density of ~ 3 molecules of CO2 per 10,000 of air it is essentially opaque in its absorption bands within a couple of hundred meters ( at sea level ) , adding more makes only a negligible , altho calculable , change in the color of our atmospheric body bag . and thus a virtually undetectable effect on our mean temperature .

( But , of course , being the source of carbon to carbon based life , the extra molecule per 10k we’ve restored to the atmosphere is visibly greening the planet — and that likely has as much an effect on our equilibrium temperature . )

• Bob Armstrong,

That’s either a trivial problem or exceedingly complex depending on the details. It’s also almost completely irrelevant to the greenhouse effect because that depends on the spectral and physical properties varying with wavelength and altitude.

I also refused because you failed to explain where you were going. That made me suspicious that you were attempting to use the Socratic Method. I simply won’t play that game.

• I would consider it towards the “trivial” side . But you refused to give a testable equation for even that .

And it is the essential equation for determining the equilibrium temperature of a radiantly heated object of arbitrary spectrum . It is a fundamental building block for any further complexities .

And in fact , the change in our spectrum as seen from the outside is the ONLY effect changes in CO2 or any other contributor to our spectrum can have on our mean temperature .

• Bob,

By the way, I’ve pretty much given up reading your posts. I just happened to spot that comment in passing and decided to reply. Don’t expect anything further. If this were my website, you would be banned for wasting bandwidth. You contribute nothing and think you already know everything you need to know.

• Are you aware that “visible” is referring to human eyes? It has nothing to do with heat, which is power dependent. Your eyes does not dictate heat transfer.

• But , of course , power is a function of wavelength and thus the equations need to be in terms of whole spectra . Thus the equations in terms of the dot products expressing the “coupling” between them .

• I would say wavelength is a function of power, since the emissive power, T^4, holds for all emitting bodies, and wavelength is not included. The wavelength shortens as power increase, and I´m talking about natural sources just to be clear.

• By power as a fn of wavelength , I mean the essential E = h ν relationship . But the equilibrium temperature of , eg : a planet lumped with its atmosphere , is a function power spectrum from the Sun’s disk ( which in my computations a http://CoSy.com as simply the Planck thermal spectrum for ~ 5800K ) and the planet’s “color” which is what’s seen from the outside : essentially its surface over the transparent atmospheric wavelengths and its atmosphere over the wavelengths over which it is essentially opaque .

The simple scalar SB equation just holds for gray bodies . Otherwise the “coupling” between the object’s spectrum and it’s sources and sinks determines the a and e which are in the crude scalar simplification .

• Yes, I understood what you meant. E=hv is the theoretical definition of the composition of the energy, how it´s distributed in the spectrum. I just wanted to make it clear that my point of view is the physical reality, and that is what I base my thinking on. In reality, there is an increase or decrease in power, in the source or the heated body, and from that there can be a change in hv.

I have begun to question the concept of “grey” bodies, since I found these results in my simple calculations of gravity and heat. The term “grey” seems to come from the observation of a body which doesn´t emit all absorbed energy as heat. My conclusion is that what is not emitted is work, because that is what the first law says, and I have no reason to not trust the first law. So the term “grey” is pretty useless, it´s better to say heat+work, and be confident that the heat that is not emitted, is work.

• Lifeisthermal,
You stated earlier that opaque only refers to optics. My mistake for thinking you were referring to visual. It still doesn’t change the analogy that visible light passes through the atmosphere relatively undisturbed, while infrared radiation that is emitted by the surface is absorbed depending on particular wavelengths. Do you disagree?

• lifeisthermal,

The term “grey” seems to come from the observation of a body which doesn´t emit all absorbed energy as heat.

Nope. Grey means not black. EM radiation that reaches a surface can be absorbed, transmitted or reflected. Absorptivity, transmissivity and reflectivity must sum to identically one. Period. A blackbody absorbs 100% of the radiation that impinges on a surface and neither reflects nor transmits EM radiation. With the possible exception of a black hole, a true blackbody doesn’t exist in nature. With a grey body, any radiation that is not absorbed must be reflected or transmitted. Note that the Second Law requires that the emissivity of any body must equal its absorptivity. This is even true for gases as long as the gas is in local thermal equilibrium, i.e. the gas particles have sufficient density to have a Maxwell-Boltzmann energy distribution. It’s called Kirchhoff’s law of thermal radiation.

83. lifeisthermal,

Power equals force times velocity. Energy equals force times distance. The average vertical velocity of the atmosphere is zero. Therefore no power or energy is required to maintain the pressure. If you compress a gas into a pressure vessel, there is a rise in temperature. But over time, the temperature returns to ambient by radiation, convection and conduction. No further power is needed to maintain the pressure in the vessel. The atmosphere works the same way. The gravitational field works somewhat like the walls of the pressure vessel.

If you were correct, which you aren’t, where would the energy come from to maintain the atmospheric surface pressure?

• By his claim it also seems that heating and cooling would be a very simple process. Pressure gas and trap it in a vessel and it would constantly produce heat until unpressurized. Conversely, vacuum and seal the same container for a permanent cooling effect. Utility bills would be much better with this kind of physics.

• Emission is what is the simple process, because you only need to care about T^4.
Then you give an example of work being done on a container, which proves my point.

• But it doesn’t stay warmer than it’s surroundings. If i compress a volume of gas into a container and seal it, this means that the heat energy has been condensed as well. This heat energy being condensed equates to the temperature of the pressurized volume being higher than the surrounding environment that it started in. Because of the increased temperature, the pressurized volume will emit at a higher rate than its surrounding, causing the pressurized volume to cool, and consequently emit less. This will happen until the pressurized volume and they surrounding environment are at the same temp (ie rate of absorption = rate of emission). The trapped choline will still be at higher pressure and will remain so until there is a way for it to mix with its surroundings, at which point there will be a temporary drop in temperature.

Pressurized volumes don’t stay heated, they gain temperature that is relayed to the work done to pressurize them. Once the work to increase the pressure stops, so does the increase in temperature. At that point, absorptivity and emissivity rates will determine how quickly the volume returns to equilibrium with its surroundings.

• “But it doesn’t stay warmer than it’s surroundings.”

I agree with this, and everything else in your post.

I think you may have misunderstood me. Gravity puts pressure on the surface continously. Since we are dealing with watts and Nm we can say energy per second. The pressure of 95.7Nm^2 is added each second in the atmosphere, and heat expands it equal to the source power of gravity, 383W/m^2 each second. A steady state. Do you understand how I think?

Ultimately, it doesn´t matter how I think, because the calculations show the equality. My thoughts actually comes from the calculations, because they came first.

• lifeisthermal,

The pressure of 95.7Nm^2 is added each second in the atmosphere

No, it isn’t. Atmospheric pressure is constant when globally averaged. The only way it could increase is for mass to be added to either the planet, increasing g at the surface, or to the atmosphere. Neither is happening.

84. DeWitt Payne,

“Nope. Grey means not black. EM radiation that reaches a surface can be absorbed, transmitted or reflected.”

Yes, yes, I know this. But there is no black bodies in reality, as far as I know, and no body is grey, right?

What I´m talking about is centered around the first law. If some part of the heat is transmitted or reflected, it will include work. Don´t you agree?
Transmission through mass will no doubt have an effect on that mass where mass is displaced to some extent. That is work. Reflection will also include some level force acting on the reflecting mass, that is no secret. Absorption, if not all energy is emitted perfectly, will include work, as the heat moves through mass and loses intensity from displacement of mass.

“Note that the Second Law requires that the emissivity of any body must equal its absorptivity.”

Emissivity seems to be widely misunderstood. Some greenhouse-minds seems to think emissivity can make a body emit less than T^4, which is wrong. Emissivity includes absorptivity in the fact that it is the fraction of heat not absorbed, and of course not emitted.

The amount absorbed heat, according to the stefan-boltzmann law for transfer between bodies, depends on the internal state(T^4) of the lower temperature emitter. But the emission depends only on the internal state. *Only*. The emitted heat from a body is independently related to the temperature of the emitter. This is really, really fascinating, because it is a universal independent relationship. I even think it is the only independent relationship.
Many people seems to think emission is caused by absorption, especially gh-believers. But that is not true, emission is caused by the internal state, and absorption is dependent on emission from the absorber. This means that heat flow is dependent only on itself, independent. This is Prevost´s conclusion from observation, and I believe it has not been questioned.

So, in the universe, we have one universal observation of energy flow, which is heat/light. And we observe one universal force, gravity. Which is indirectly observed by it´s effect on heat/light(wobbles and such). They are always observed together. The most rational, conservative and logic conclusion, is that we are observing cause and effect. Right?

What is not emitted as heat, must be work, because the first law says so. And we know the first law is true, and that gravity does work. Right?

• lifeisthermal,

If some part of the heat is transmitted or reflected, it will include work. Don´t you agree?

No, I don’t. A tiny amount of work could be done by reflection. Photons have momentum, so when they are reflected, there is change in momentum that must be applied to the reflecting object because momentum is conserved. That’s how a solar sail would work. But if the reflecting object doesn’t move, no work is done.

Emissivity seems to be widely misunderstood. Some greenhouse-minds seems to think emissivity can make a body emit less than T^4, which is wrong. Emissivity includes absorptivity in the fact that it is the fraction of heat not absorbed, and of course not emitted.

Wrong again. It’s not just “some greenhouse-minds, it’s basic physics. The Stefan-Boltzmann equation includes an emissivity term.

j* = εσT^4

An object can emit less than σT^4.

From Wikipedia, for one:

A body that does not absorb all incident radiation (sometimes known as a grey body) emits less total energy than a black body and is characterized by an emissivity, ε < 1

https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

You are trying to warp physics to support your conclusions. It doesn’t work that way.

By the way, why no comment on the temperature of, say, a compressed gas cylinder?

• “But if the reflecting object doesn’t move, no work is done.”

Compression, heard of that?

“Wrong again. It’s not just “some greenhouse-minds, it’s basic physics. The Stefan-Boltzmann equation includes an emissivity term.

j* = εσT^4”

I thought you knew thermodynamics. Stefan did not include εσ, neither did Boltzmann. A body emits according to it´s temperature, and you need to read up on this.

The draper point, 798K, how do you fit emissivity into that? That all solids emit equally at the same temperature?

Emissivity is the relation to the heat source, you should know that. Shame on you.

• “I thought you knew thermodynamics. Stefan did not include εσ, neither did Boltzmann. ”

Ooops, sorry. Of course Boltzmann included σ. But ε is still the relation to the heat source.

• I only just read what the draper point is so I’m sure I’ll be corrected if I’m wrong.

From my brief reading of it, it is the point at which all solids will start to emit light in the visible spectrum. This makes no claim of the intensity though. An object with an emissivity of .5 will emit the same wavelengths, but at a reduced amplitude, as an object with an emissivity of 1. Peak of the curve would be at the same wavelength, but much shorter.

• lifeisthermal,

Emissivity is the relation to the heat source, you should know that. Shame on you.

Cite a physics textbook reference for that. If you can’t, then shame on you.

• Emissivity can also change with a change only to the surface. For example, polished aluminum will have lower emissivity than a non-polished aluminum. Material and temperatures unchanged, only difference being the surface. You are wrong to say that emissivity is linked to temperature. Emission (which is to say quantity emitted) is linked to temperature. As the temperature of an object increases, so does the amount that it emits. It’s emissivity however is a constant relative to the particular object being observed.

• lifeisthermal,

By the way, you’re in violation of blog etiquette.

Basic Science is Accepted – This blog accepts the standard field of physics as proven. Arguments which depend on overturning standard physics, e.g. disproving quantum mechanics, are not interesting until such time as a significant part of the physics world has accepted that there is some merit to them.

The moderator reserves the right to just capriciously delete comments which use as their premise that standard textbook physics is plain wrong.

Your interpretation of emissivity, for one, is not standard physics. Now if you can find, say, a physics textbook that supports your view, you might have a point. But you won’t.

• “By the way, you’re in violation of blog etiquette.”

Yes, call for daddy, you coward. It is uncomfortable, huh? Being put in the corner. Standard climate scientist behaviour.

Wonderful for me though, that you show your true colors.

• Resorting to ad hominem now are we? If you don’t like the message, shoot the messenger. How sad.

I’m not the one in the corner. You are, unless you can cite references to support your contentions. You haven’t yet.

It’s easy to demonstrate that different materials have different emissivities. Get an IR thermometer. They’re inexpensive and very useful. Use it to measure the temperature of a shiny stainless steel or polished aluminum pot containing boiling water. Then measure a seasoned cast iron or black anodized aluminum pot also containing boiling water. They will have different apparent temperatures. The shiny pot will have a lower apparent temperature than the other pot.

85. […] that other articles explain the basics. For example – The “Greenhouse” Effect Explained in Simple Terms, which has many links to other in depth […]

86. lifeisthermal,

New policy – please can you answer the two questions over at Two Basic Foundations.

I’ve found in the past that the steady stream of over-confident, but confused, visitors can never bring themselves to answer either of these questions (for obvious reasons).

87. […] in another article, a commenter […]

88. mark4asp, you said, November 5, 2017 at 10:34 am:

Modtrans is irrelevant. Climate ‘heroes’ like Hansen did not use it (https://wattsupwiththat.com/2014/04/12/a-modtran-mystery/). Catastrophic predictions depend on accelerated forcings due to water vapour feedback. This water vapour feedback is simply written into climate models as parameters. It is not derived from any kind simulation of first principles in the General Circulation Model runs (GCMs).

No doubt you didn’t invent this last part yourself, you heard it on the grapevine. But now you’ve made the claim it’s time to produce evidence.

I just wrote an article on this – Water vapour feedback is simply written into climate models as parameters? – please add your proof over there.

Or – this would be a wonderful step – admit your error.

89. Question in regards to a/e, and if posted in an article somewhere here already please link it. I feel like I’ve read it before here somewhere but can’t find it.

When calculating the effective temperature for earth an absorptivity of .7 is used while an emissivity of 1 is used for outgoing. So why the difference? I cab thunk of a couple possibilities but not site if I’m on the right track.

First is that part of that .7 is due to reflection in the atmosphere, which obviously keeps radiation fruition reaching the surface. But this reflection doesn’t reduce the actual absorptivity of the surface. So that is one reason for emissivity of the surface to be higher than total absorptivity of the sun. Second, and i think more importantly, is that just because the surface does reflect some radiation, especially in the visible spectrum, doesn’t mean that its absorptivity in the IR range isn’t high, which would consequently mean high emissivity given that our surface temp equates to emission in the IR range. Am i on the right track with my thinking?

• The difference is caused by the different spectral ranges involved. Solar radiation wavelength is shorter than 5μm while thermal emission is longer than 5μm. Reflection of solar radiation is mainly from clouds and snow and ice covered regions. Reflection from water is a function of the angle of incidence and whether the surface has been effectively roughened by wind action and whether there is a significant concentration of particulate matter near the surface. The reflectivity of the land surface depends on the cover. See the Wikipedia article on albedo, for example: https://en.wikipedia.org/wiki/Albedo

Emissivity at wavelengths longer than 5μm is slightly less than one. MODTRAN, for example, uses a value of 0.971 for the emissivity.

90. “More GHGs
6. If we add more radiatively-active gases (like water vapor and CO2) then the atmosphere becomes more “opaque” to terrestrial radiation and the consequence is the emission to space from the atmosphere moves higher up (on average). Higher up is colder. See note 6.
So this reduces the intensity of emission of radiation, which reduces the outgoing radiation, which therefore adds energy into the climate system. And so the climate system warms (see note 7).”

“So this reduces the intensity of emission of radiation”, Yes
per square meter, not the overall amount.

No. Surely not. Semantics warning please.
Time period specification.
There is a slight reduction when the gases are added which is due to a delay in adjusting to the new setup *. As soon as the GHG level effect is established it is back to situation normal.
The planet can only radiate out the energy received [if sun and the earth otherwise unchanged] hence it has to, repeat has to radiate what comes in sooner or later no matter how much GHG there is.

Reasons,
-One logically, it should be emitting from higher up only because there is a larger amount of energy needing to go out. e.g. hotter sun
-Hence the amount of energy going out is from the surface area of the imputed sphere which is a lot more energy from lower energy molecules for the increase in radius to the new TOA.
-the molecules lower in the atmosphere at the level where the IR could go out unimpeded are still there and they are radiating out increased energy as well.

which therefore adds energy into the climate system, Yes

-This is the tricky bit.
* Different materials have different times in which they release the energy they build up hence the time it takes to start emitting the heat back.
Hence there is an absorption time for GHG and while they are reaching this plateau there is an increase in the energy stored in both the GHG and the atmospheric molecules in general through their kinetic heat which can only easily be released to space through said GHG.

“And so the climate system warms (see note 7).” Yes, the atmosphere we live in warms.
I am glad you did not say the earth warms. The energy coming in is still, longterm and virtually always, the energy going out.

• angech,

..There is a slight reduction when the gases are added which is due to a delay in adjusting to the new setup *. As soon as the GHG level effect is established it is back to situation normal.

The planet can only radiate out the energy received [if sun and the earth otherwise unchanged] hence it has to, repeat has to radiate what comes in sooner or later no matter how much GHG there is..

Your second statement is almost correct. Energy in = energy out in the absence of warming or cooling.

But your overall comment shows that you don’t really understand how heat transfer works.

It is very simple. Try and grasp it before writing anything else.

More GHGs reduce the amount of radiation leaving the planet. What happens next?

1. Energy in is greater than energy out.

2. Therefore, the planet stores energy and the temperature goes up (the first law of thermodynamics)

3. As the temperature increases more radiation is emitted.

4. Eventually energy out is equal to energy in.

5. At this point a new steady state is reached, the planet is now at a higher temperature.

This concept is so simple that I wonder at people who can’t understand it.

• on December 27, 2018 at 5:04 am Morris Walters

“This concept is so simple that I wonder at people who can’t understand it.”

No it really isn’t. What may be simple for you isn’t really all that simple. I’ll point out his conceptual error though. He states it clearly, he just doesn’t understand what it means.

” has to radiate what comes in sooner or later”

Anyway.

• angech,

The reduction in W/m² is far larger than the increase in area caused by the increase in altitude. Thus there is indeed a reduction in the total amount of energy emitted by the planet.

For a doubling of CO2 and thus a forcing of 3.7W/m², the reduction in emission is (1- 235.3/239)*100 = 1.55%. The radius of the planet is 6,371km. On average the emission comes from 5km above the surface, so 6,376km. An increase in area to compensate for that would be an increase of radius of:

0.9845 = (6376/(6376 + x))²

x = 50km

The actual increase in altitude from doubling CO2 is more like 100m.

91. “This concept is so simple that I wonder at people who can’t understand it.”
Nearly as succinct as Dirac.
Who put it thus, ‘ I thought It was a statement of ignorance not a question”
Nonetheless I am trying to learn and your site offers both learning and a chance to ask questions. I will try harder.

• angech,

If you are trying to learn the basics that is good and I don’t want to discourage you.

Your earlier comment suggests some strange idea, unknown in heat transfer.

As always with heat transfer fallacies I have to try and think “what is in this person’s head?

You appear to think – and I’m just trying to help you – that somehow a body/a system has some requirement to emit radiation/transfer heat at a rate determined by steady state requirements.

No.

Bodies emit radiation based only on their temperature.
Bodies conduct heat based only on their respective temperature difference.

If you reduce a body’s radiation somehow then (all other things being equal) the body gains energy. It doesn’t have some immediate “drive/propensity” to radiate at its earlier rate. (There is no law in heat transfer that requires this).

The result – the body gains energy. This gain in energy increases its temperature. An increase in temperature leads to higher radiation (and/or conduction, depending on the situation).

So the key to the problem is that changes in heat transfer balance cause changes in temperature (internal energy). Changes in temperature result in changes in heat transfer.

I don’t know whether I have helped.

• You are helping.
“Your earlier comment suggests some strange idea, unknown in heat transfer”
I am missing part of the puzzle for sure. Perhaps a desire to have less GHG effect?
As you say the physics is well studied.
There seem to be a lot of people in different boats, like myself, but we do not seem to share the same misconceptions, just misconceptions in general.
The discussion at WUWT in particular highlighting the difficulty in getting general agreement in what should be an explainable subject.
I looked at a couple of Willis’s Steel greenhouse ideas,
I still am stuck on the fact that the sun only produces so much energy and for whatever reason the earth radiating out at a higher temperature than the sun puts in seems wrong.
Thank you for the explanation, which like DeWitt’s has taken me somewhere along the line to the answer but unfortunately not through my obtuseness.
Prefer to leave it there at the moment, thanks both, and mull it over some more.

92. DeWitt,
For a doubling of CO2 and thus a forcing of 3.7 W/m², the reduction in emission is (1- 235.3/239)*100 = 1.55%.
OK.

Not sure I have grasped the rest of your maths concept yet.

Technically a reduction in emission initially means a fall in the height of the TOA initially? perhaps it would be by a 100 meters by immediate CO2 doubling as 1.55% is now not going to space initially.
The TOA can only go out as the temperature of the atmosphere heats up due to the reduction in emission.
So you can only have a TOA further out as and after the atmosphere adapts to having more energy in it.
Now as SOD says eventually it is in balance.
Yet once in balance the sun is not giving any more energy in than what comes out.
SOD says the planet is at a higher temperature.
I would say the atmosphere of the planet is at a higher temperature.
The planet is not an engine generating new energy, just temporarily diverting some of the heat before it goes out.
Now the emission of energy to space, total amount, is unchanged, has to be as it is now still just putting out as much as the sun is putting in.
As it was before the CO2 increase.
As it is putting it out at a higher TOA but the same total the intensity per square meter 100 meters further out is less.

50 kilometers and 100 meters are ballparks apart if we are talking about similar things which is a worry.

• angech,
An increase in co2 raises the effective height that radiation can be emitted from in the atmosphere and make it out to space without being absorbed again.

Temperature decreases with height.

Because of decreased temperature the rate of emission is lower.

A lower rate of emission means less energy leaving the system.

Less energy leaving the system drives surface temperatures higher until the surface emission is high enough to offset the decreased rate of emission.

DeWitt probably thought as I did; that you were trying to say that the increased height would equate to a larger radius and thus larger surface area of emission. And that this increased surface area would offset the reduction in flux. His point is that to offset the 3.7 W a doubling would bring with surface area would require an increase of 50km in the radius. However, the change in emission height that is being suffused is about 100 meters. Therefore, the increase in surface area due to a larger radius is not near enough to offset the reduction to outbound longwave radiation.

Same radiation in, less out, increased surface temperature.

• on November 28, 2017 at 2:26 pm nobodysknowledge

“Temperature decreases with height.

Because of decreased temperature the rate of emission is lower.

A lower rate of emission means less energy leaving the system.

Less energy leaving the system drives surface temperatures higher until the surface emission is high enough to offset the decreased rate of emission.”

This is an ideal connection, but it seems not to be right. It has one condition, and it is that the lapse rate is the same for different amounts of CO2. Observations show that the outgoing longwave radiation at TOA hasnt increased the last 40 years. So emission hight is compensated for lapse rate. CO2 is only affecting the distribution of heat in the atmosphere.

• Nobodysknowledge,

Do you agree that the effective height of emission is increased as concentrations of co2, or other greenhouse gases, increase?

• nk,

I don’t think that the measurements of outgoing IR radiation are accurate and precise enough to see the fairly small difference caused by slowly increasing CO2. The current difference should be about 0.5 W/m². That’s about an order of magnitude smaller than the current measurement precision. That’s also about the rate of increase of ocean heat content, which is what would be expected. Not that it’s easy to measure OHC, but that’s probably the current best measure of TOA radiative imbalance.

• nobodysknowledge wrote: “This is an ideal connection, but it seems not to be right. It has one condition, and it is that the lapse rate is the same for different amounts of CO2.”

The calculation of the Planck response makes that condition and several others: that surface temperature increases by the same amount anywhere and that there are no changes in water vapor, surface albedo, or clouds. All of those things do change, resulting in various feedbacks that alter the amount of warming from that expected by the Planck response alone. But feedbacks are not needed for the qualitative argument made above by SoD and others.

nobodysknowledge wrote: “Observations show that the outgoing longwave radiation at TOA hasn`t increased the last 40 years. So emission hight is compensated for lapse rate.”

The greenhouse effect would cause the outgoing long wave radiation to first decrease then come back to whatever is required to restore the balance of radiation in and out. That could be either an increase or decrease, depending on how albedo changes.

nobodysknowledge wrote: “CO2 is only affecting the distribution of heat in the atmosphere.”

It also affects the distribution in the ocean and changes the total energy in the system.

• angech,

SOD says the planet is at a higher temperature.
I would say the atmosphere of the planet is at a higher temperature.

Precision is important when making statements about science. SoD says the surface of the planet is at a higher temperature. The atmosphere can’t warm significantly without the surface warming too.

You have to think of the atmosphere as something like insulation. Adding ghg’s to the atmosphere increases the effective thickness of the insulation by raising the average emission height. The surface can then be warmer while the planet still radiates the same amount of energy to space as when the ghg level was lower.

• on November 29, 2017 at 12:32 pm nobodysknowledge

Brad Schrag: Do you agree that the effective height of emission is increased as concentrations of co2, or other greenhouse gases, increase?
Yes I think it is right.

93. It is extremely difficult to help angech.

Your wife is likes the house cold. You like it hot. Assume the furnace runs at a constant rate and that the pressure inside and out is about the same. If the window is closed, the only way for the heat to escape the house is through the exterior surfaces of the house. Given enough time – warming in the pipeline -, the house will equilibrate at a high temperature.

Now your wife sneaks over and opens the window all the way. Can you agree that the heat can get out much faster? Given enough time, you will yell, “Who the hell opened the window?” It will equilibrate at a low temperature.

You sneak over close the window halfway. Now the exit of heat has been slowed down. There is warming in the pipeline. Given enough time, the house will equilibrate at a moderate temperature.

Furnace – constant
Window – adds no heat, and: is the control knob.

angech thinks they’ve missed something. They did. It’s likely a bunch of tiny little things that net to almost zero effect.

94. JCH [thanks, your example is good]

There is a difference between a house with a furnace inside it [Willis steel greenhouse type problem] and a house with a bonfire outside it.
The example is not answering my needs.

It raises good questions about how heat travels through different substances and at what speed they travel and what actually travels.

In the real world the sun puts a certain amount of energy into the system and after absorbing it and conducting it back to where it radiates out the end result is energy in equals energy out.
The atmosphere is warmer when the amount of energy going back to space has been impeded from getting out but is still losing the same amount of energy to space when it has resorted itself. Presumably this means the radiating surface is still putting out the same energy the intensity is less per square meter because it is now out another 100 meters. The air below is hotter from the surface up but in terms of being able to radiate it out the CO2 or GHG are temporarily trapping that extra amount of heat which takes a slightly longer time to exit. The air being warmer because it is at an associated higher kinetic energy.

“Temperature decreases with height.
Because of decreased temperature the rate of emission is lower.
A lower rate of emission means less energy leaving the system.”
using this logic
means that the hotter the world gets, the higher the emission height goes, the less energy leaves the system therefore if the world got hot enough no energy could leave the system?
I think you would agree that if the world got a lot hotter there would be more energy going to space, not less, after any equilibration period of course.
“An increase in co2 raises the effective height that radiation can be emitted from in the atmosphere and make it out to space without being absorbed again”.
Initially it lowers the ERH [effective radiation height] but after it eqilibrates the height is hiugher but the output of energy the same, SOD says higher as the temperature is higher.
My weird idea is that the temperature of the atmosphere is artificial due to the retained IR/GHG effect and the actual effective radiating temperature, that we see from space, remains unchanged.

• angech,

There is a difference between a house with a furnace inside it [Willis steel greenhouse type problem] and a house with a bonfire outside it.
The example is not answering my needs.

The example (furnace inside) is actually very appropriate.

The sun radiates centered around 0.5μm. The atmosphere is mostly transparent to this radiation so the radiation is absorbed at the surface.

The earth radiates centered around 10μm. The atmosphere is mostly opaque to this radiation so in very simple terms it acts like insulation.

If the atmosphere was equally opaque to solar and terrestrial radiation then it would be like the problem of the house with the bonfire outside it.

• Yes, the atmosphere is also somewhat like the exterior surfaces of a house. I could replace windows with insulation, but most people have dueled with their spouse/kids over the window and understand the open and shut of the matter.

No analogy is perfect, but I used the house to convince a friend of mine that the concept of warming in the pipeline cannot be dismissed as ridiculous, which he was doing. And I have used it often since then. It sort of brings the GHE “home”.

• angech,
you said:

using this logic
means that the hotter the world gets, the higher the emission height goes, the less energy leaves the system therefore if the world got hot enough no energy could leave the system

and conveniently left out:

Less energy leaving the system drives surface temperatures higher until the surface emission is high enough to offset the decreased rate of emission.

It’s going to be a constant shifting change, one way or the other. I suppose if I wanted to entertain your logic of a thought experiment we could set the rate of GHG increase to be exponential and continuous. Then the atmosphere wouldn’t have time to adjust and voila, no heat leaving. But that doesn’t really get us anywhere.

I believe you are accurate in the claim that the effective temperature will basically remain unchanged. I’m a bit fuzzy on the order of what happens next, it’s a bit of a chicken-egg discussion in my head, but this would be my limited understanding.

increase co2 will lead to a change in lapse rate.

the change in lapse rate will increase the effective height of emission

as such, less radiation will be leaving the system, temporarily

the result of less radiation leaving will be a surface warming

a warmer surface will help to restore balance at ToA with the newly adjusted lapse rate

once the earth has arrived again at steady state, the observed temperature from space will still be 255K, but the surface temperature will now be higher than previously.

For an extreme example, look at Venus. It has an effective temperature of 230K, or 158W/m^2. Yet the surface temperature is 730K. So despite the fact that the surface is emitting at a rate of 16100W/m^2 we measure at the top of atmosphere 158W/m^2.

https://scienceofdoom.com/2010/06/12/venusian-mysteries/

As a side note, I’m still learning so I’m sure I’ll get corrected on some of my details.

as I understand it, the change in lapse rate is a second order effect, relatively minor, and if I recall correctly more associated with latent heat than radiative properties. I could be wrong on the latter point.

The first order change from increased CO2 is emission height, not lapse rate.

That height can maybe be conceptualised as a height of constant opacity; ie constant absolute CO2 concentration. Above that height, LW emitted by CO2 goes direct to space, below it it is reabsorbed. Obviously that’s a gross simplification, but it gives you the concept.

As CO2 rises in relative concentration (ppm), the height at which *absolute* concentration (kmol/m3, or equivalent) will rise. The temperature at that height must however remain constant, to keep total radiative balance. So the surface temperature rises, as the lapse rate remains constant, to a first approximation.

There’s a useful diagram here: https://chriscolose.files.wordpress.com/2010/02/gt5.jpg

which is part of a good overview https://chriscolose.wordpress.com/2010/02/18/greenhouse-effect-revisited/

• Isaac Held on the lapse rate feedback:

As a moist parcel of air ascends it cools as it expands and does work against the rest of the atmosphere. If this were the only thing going on, the temperature of the parcel would decrease at 9.8K/km. But once the water vapor in the parcel reaches saturation some of this vapor condenses and releases its latent heat, compensating for some of the cooling (you get about 45K of warming from latent heat release when a typical parcel rises from the tropical surface to the upper troposphere). A warmer parcel contains more water vapor when it becomes saturated, so it condenses more vapor as it rises, and temperature decreases with height more slowly. That is, the moist adiabatic lapse rate, – \partial T/\partial z, decreases with warming

It does seem to be caused by increased latent heat release.

• as I understand it, the change in lapse rate is a second order effect

Per Soden and Held, roughly half the absolute value of the water vapor feedback:

Of course, the interesting thing is that it hasn’t occurred over the satellite era ( the, so far, missing hot spot ).

• on November 29, 2017 at 4:28 pm nobodysknowledge

I find the reference to Isaac Held interesting. It shows a most dramatic change in lapse rate with doubling CO2 (something like 5 degK warmer at 7 km, and 10 degK warmer at 12 km). According to a CMIP3 enemble for the tropical troposphere. I think this shows the high tropospheric hotspot, and then the systematic bias of models.
What I cannot understand with the great warming of the atmosphere up to about 15 km that is shown in the models, is that it should not result in an increase in IR to space.

• on November 29, 2017 at 4:34 pm nobodysknowledge

I am referring to the top right graph in the Held blog referred by verytallguy.

• nk,
I see you responded above that you agree the effective emission height changes. I’ll just reply here though to keep everything together.

At increased elevation the rate of emission is lower. This means the system won’t be in steady state. To get back to steady state, the new emission height has to warm up such that it’s rate of emission matches the incoming. How does that happen?
If we work backwards from effective emission height to surface using the lapse rate, which is K/km, we see that for that outer emission surface to warm and restore balance the surface must be warmer as well. As the surface warms, the lapse rate will eventually warm the layer at effective emission to the point that the system is in steady state.

I’ve been wondering about the order of those steps. If we woke up tomorrow to a doubled co2 from today what would happen? In my mind, the lapse rate would have a kink in it of sorts, that would drive up temperatures on the surface. The warming surface would drive convection which would result the lapse rate as it warms that outer cold layer. But for that outer cold layer to stay warm enough to balance incoming and outgoing ultimately means that the surface must remain warmer. I also think my thought experiment is a bit useless though because it is about sudden and abrupt changes that really don’t happen at all. GHE is a very gradual and slow process, both in its increase and decrease.

• Brad Schrag: “If we woke up tomorrow to a doubled co2 from today what would happen?”

The upper 50-100 meters of the ocean has an enormous heat capacity and readily exchanges heat with the atmosphere. So a step change in CO2 would cause a more gradual change in surface temperature over a decade or two. The atmospheric circulation responds much more rapidly than that (a matter of months, I think), so it should be able to keep up with the change in sea surface temperature.

No doubt a true step function would produce some interesting transients. But the ocean would even a lot of that out.

• What I cannot understand with the great warming of the atmosphere up to about 15 km that is shown in the models, is that it should not result in an increase in IR to space.

Right – that’s what the Lapse Rate feedback represents – negative feedback.

But it would still be less than the radiative forcing imposed by increased IR opacity and less than the modeled positive feedback of water vapor.

95. πREarth2Fsun(1−a)=4πREarth2 σTtop4
Ttop=((Fsun/4)(1−a) σ)1/4. Sorry for the poor attempt at formula posting

-Basically the TOA for emission is determined by the energy coming in and later going out.
-Very basically for an unchanging sun and albedo there seems to be be a specific radius or height for the TOA.
-This should not change whatever the temperature in the greenhouse because all that can escape to space is the same energy as what comes in, and for a rotating sphere averaged out it is immutable?
-When the absorbing body changes characteristics, as in a GHG increase, once at equilibrium the emitting height must be the same as before.

-DeWitt calculates an extra height based on the temperature being warmer.
The problem is that this is the temperature in the house, not the temperature of the emitting surface.
Like JCH and SOD the assumption is that the rise in temperature reflects a real temperature radiating to space but this is not correct.
At the stage the IR can escape to space it can only escape at the rate of the incoming solar energy, no more no less.

This is a different problem to a changing of albedo with clouds as perforce the amount of energy reaching the “surface” is altered, which is not true for a GHG which is merely holding up some of the energy before it releases it thus giving a higher temp but one which does not radiate to space until there are no more GHG particles in the way.

Sorry for the obtuseness, as SOD says “Your earlier comment suggests some strange idea, unknown in heat transfer.”

• -When the absorbing body changes characteristics, as in a GHG increase, once at equilibrium the emitting height must be the same as before.

No.

You can see this very easily by a thougth experiment in which you remove all GHGs and replace them by the same amount of some non-GHGs.

With no long wave active molecules at all in the atmosphere radiation to space would be directly from the solid surface. You agreed, that with GHG there will be some altitudes above the surface, where radiation to space takes place.

So, how is this difference between the two setups to be explained? If any concentration of GHGs should lead to the same radiation heights in steady state, as you suggested, there will be no concentration dependent mechanism at all – it would be just some on/off mechanism. Thus a limit concentration that switches the mechanism on must be there.

Does this sound propable to you: At some certain concentration the emission height for steady state suddenly “jumps” from the surface to some fixed altitudes?

It is quite easy to visualize: If you begin to eat more calories than you burn, you will get fatter until your body due to his heaviness needs so much more energy for muscles, that your burning equals your eating again.
If you are now not doing one of the following things:
a) burning more ( sports )
b) eating less ( diet )
c) undergo some surgery ( fat removal )
your steady state will be one of a fat guy constantly. It will not eventually go back to it’s “old equilibrium” by some magic, where you have been in a slim shape.

Why should an accumulation of energy lead to no final difference, as you suggested in the quote above? Human body fat is no more than representation of accumalated energy due to nutrition. So the analogy is for the first principle not that bad.

• An exceptionally rotund analogy. I’m stealing it.

• angech,

Here is a thought experiment for you.

A large sphere internally heated by 238W/m^2 is surrounded by a single layer shell. The opacity of the shell can be carried from 0-1. 0 being that no radiation makes it through, 1 being that all radiation passes through, .5 being that half passes through and .5 is absorbed, I think you get the picture.

As the heat source for the surface is unchanging, what changes as the transmissivity goes from 1 (everything gets through) to 0 (nothing gets through without being first absorbed by the shell). Specifically, can you describe what happens to:

1) surface temperature
2) shell temperature
3) observed outgoing radiation seen by a satellite orbiting outside the shell

* extra credit for an equation that correctly describes it

• angech,

Just picking one thing:

DeWitt calculates an extra height based on the temperature being warmer.

No. The extra height occurs because the atmosphere has become more opaque, not because the temperature of the atmosphere or the surface has changed, because it hasn’t yet.

96. “ChrisA
-When the absorbing body changes characteristics, as in a GHG increase, once at equilibrium the emitting height must be the same as before.
No.
You can see this very easily by a thought experiment in which you remove all GHGs and replace them by the same amount of some non-GHGs.
With no long wave active molecules at all in the atmosphere radiation to space would be directly from the solid surface.”

That idea was mentioned in the article I took the formula from.
I thought I had “rationalised” it out.
Sorry for the obtuseness.
Back to thinking square one.

Probably fits in with DeWitt
“No. The extra height occurs because the atmosphere has become more opaque, not because the temperature of the atmosphere or the surface has changed, because it hasn’t yet.”

97. […] US Order Oxandrolone in US Order Oxandrolone in US Order Oxandrolone in US Order Oxandrolone in US Order Oxandrolone in US Order Oxandrolone in US Order Oxandrolone in US Order Oxandrolone in US Order Oxandrolone in US […]

98. […] the “greenhouse” effect post your comment in one of the many articles about that, e.g. The “Greenhouse” Effect Explained in Simple Terms. Comments placed here on the science basics will just be […]

99. […] the “greenhouse” effect post your comment in one of the many articles about that, e.g. The “Greenhouse” Effect Explained in Simple Terms. Comments placed here on the science basics will just be […]

100. […] the “greenhouse” effect post your comment in one of the many articles about that, e.g. The “Greenhouse” Effect Explained in Simple Terms. Comments placed here on the science basics will just be […]

101. You have some misguided and some arrogant misconceptions of physics. I wonder, would you be willing to backup these claims in a live debate with an actual physicist?

I can arrange that.

• Thanks for your kind words.

No.

“Actual physicists” can explain the flaws in my articles right here. They don’t even need to provide their “actual physicist” credentials, just make arguments from theory and experiment.

I look forward to it.

• When you are given comments like this, ““The planet can only radiate out the energy received [if sun and the earth otherwise unchanged] hence it has to, repeat has to radiate what comes in sooner or later no matter how much GHG there is..”

That are 100% true with an arrogant reply like this, “Your second statement is almost correct. Energy in = energy out in the absence of warming or cooling.

But your overall comment shows that you don’t really understand how heat transfer works.
It is very simple. Try and grasp it before writing anything else.”

I find it hard to give you kinder words under those circumstances.

102. I became a skeptical of anthropogenic global warming, AGW, caused by a greenhouse gas effect, GHGE, about 3 to 4 years ago. There’s a long story behind the genesis of my skepticism I won’t tell [it fascinates and horrifies me; it will bore the reader]. When I questioned AGW my points were simple:
* What is the GHGE testable hypothesis?
* Who wrote it?
* What are the scientific tests for it?
* Which tests are validated?

After some obfuscation, I found: there is no testable hypothesis for GHGE, no proper tests, no validations. The climate modelers promoting GHGE don’t bother with rigorous science. Even the modelers often resort to ‘argument from authority’, and ‘ad hominem’ against their critics. Just like E-NGO climate activists. It’s as if they take their cue from climate activists! One ‘scientist’ Ken know very well and worked with even told me I was a “fossil fuel shill”. Me, the ex-nuclear power activist (well to be fair to him, he did not know my history).

I discovered the nearest thing to a GHGE hypothesis are 2 papers: Manabe & Wetherald, 1967 and Held & Soden, 2000. But there are no tests, nor proper test validations. Modelers just assumed the GHGE model and declared it ‘settled science’ without it ever going through the normal scientific validation process.

• The GHG theory can be disqualified in one sentence.

“Line spectrum cannot, under any circumstances, increase the temperature of its radiative source.”

• CD Marshall said:
““Line spectrum cannot, under any circumstances, increase the temperature of its radiative source.””

Glad you weren’t anywhere close to the invention of the infrared CO2 laser in the early 1960’s. People like you are impediments to scientific progress. Same goes for mark4asp

• on July 28, 2020 at 7:42 am CD Marshall

Actually it’s a good thing you weren’t there either. A CO2 laser in no way makes that statement incorrect. A laser works by focusing photons into a concentrated beam. Does a magnifying glass increase the potential energy of line spectrum? No it does not. A focus of concentrated energy is not the same thing as increasing the energy greater than its source.

Perpetual energy devices simply do not exist in this universe under our current laws of physics.

• “A laser works by focusing photons into a concentrated beam.”

LOL, you ignoramus. To get to that point a laser needs to convert energy from one wavelength to another. What do you think GHGs do?

• on July 28, 2020 at 7:29 pm DeWitt Payne

“Line spectrum cannot, under any circumstances, increase the temperature of its radiative source.”

That sentence is nonsense. Do you mean that the spectral energy density or brightness temperature of an emission line cannot be greater than the temperature of its radiative source? That would only be true for a thermal source. It’s certainly not true of emission from a laser.

• on July 28, 2020 at 9:24 pm CD Marshall

Going back to its source and increasing the temperature. Lasers require work. Anything can increase it’s source with work. A line spectrum is a specific wl/f and no, it can’t increase its radiating source greater by reflecting back to it.

Energy can transfer both ways but to increase temperature you need to meet requirements this is both on a macro and a micro scale.

• on July 29, 2020 at 12:06 am DeWitt Payne

That’s simply not true. If you enclose a heat source in a perfect reflector, the temperature will increase without limit. A less than perfect reflector will increase the temperature of the source compared with being surrounded by a perfect absorber at a fixed temperature. That’s how the highly reflective space blankets work. They also block convection, but at what we consider normal temperatures, say 300K, radiative heat transfer dominates.

But this is pointless. You have your mind made up and no amount of evidence will change it.

• on July 29, 2020 at 2:41 am CD Marshall

“If you enclose a heat source in a perfect reflector, the temperature will increase without limit”

Nowhere near true unless the reflector can also withstand indefinite pressure. The temperate won’t exceed its heat source; any heat increases beyond its source would only come from increased pressure.

Space blankets restrict convection and a higher reflection equals a lower emissivity.

No magic energy is available on planet Earth and If these notions were remotely true we’d have no energy crisis.

• on July 29, 2020 at 3:24 am Brad Schrag

The sun is the primary heat source for the planet.

Greenhouse gases stand in the way of heat exiting the system from the surface.

They reduce the rate of heat loss from the system, causing it to reach a higher equilibrium temperature with it’s primary heat source, the sun.

This doesn’t require heat to move from atmosphere to surface, it only requires energy to move from atmosphere to surface.

• on July 29, 2020 at 4:50 am CD Marshall

Yes the Sun heats the Earth that is 100% true.
Yes IR is the only means of energy leaving the system, not the main source of heat in the Troposphere. Surface IR is 20% intercepted and 80% emits to space. The thickness of the atmosphere until it gets so thin that light waves can escape freely to space is about, 20km in altitude (quick math). Given that the speed of light is 300,000 kilometers per second, it would take unobstructed light wave 66-67 microseconds to get through the atmosphere (quick math).
IR reactive gases aid the dispersion of heat out of the atmosphere they do not prevent it. How else would homonuclear diatomic molecules cool off? Only other way is the adiabatic lapse rate. IR moves at the speed of light, “prevention” is far less than a second. In fact, a photon can be reflected over 14 thousand times in a second (quick math).
A higher surface temperature (in direct sunlight) means the hotter the surface gets the greater the Troposphere expands to compensate and the emissions to space increases by the fourth power of the absolute temperature of the radiating body to appropriate the Conservation of Energy Law, under that law the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. Delta U=Q-W. Despite sordid arguments, IR reflected energy is not new energy introduced to a system.
The Troposphere expands on the day side and contracts on the night side, because convection has allowed it to cool. Thus if it is cooler, the surface is not HOTTER. Energy does move from surface to the atmosphere and back, but that does not automatically equate higher temperatures when dealing with bosons.
So, “They reduce the rate of heat loss from the system, causing it to reach a higher equilibrium temperature with it’s primary heat source, the sun” would be true if we had solar energy irradiating the entire planet 24 hours a day, but we do not. It’s not the primary heat source of the Sun as much it is the primary heat source of the Sun at the angle of incidence. Maximum temperature from solar irradiation at the Poles is very different from the Equator.

• on July 29, 2020 at 4:51 am CD Marshall

Sorry it took my paragraphs out.

• CD Marshall,

Please go to Two Basic Foundations and answer those two basic questions.

• You are going to have to be more specific than a generalization of an article based on the pre conceived concept you are right about the GHGE.

• It’s a test.

• on July 29, 2020 at 5:49 pm DeWitt Payne

CD Marshall,

I don’t have anything better to do right now so here goes.

Nowhere near true unless the reflector can also withstand indefinite pressure. The temperate won’t exceed its heat source; any heat increases beyond its source would only come from increased pressure.

Pressure from what? Maybe you’re assuming an atmosphere inside the reflector. That doesn’t have to be true. Let’s take a pellet of 238PuO2 and put it in an evacuated highly reflective inside and outside container. There will be a small amount of helium produced from the emitted alpha particles, but the internal pressure will still be negligible for a really long time. So the majority of the heat transfer will be by EM radiation. Are you saying that the temperature of the surface of the pellet in the reflective container will be the same as a free pellet in intergalactic space? If so, you’re not much of a physicist.

You also don’t seem to know how a laser actually works. Here’s some more questions for you: Define population inversion and stimulated emission.

BTW, being a physicist does not confer infallibility. Gerlich and Tscheuschner probably have more publications than you and are just as wrong about the greenhouse effect as you.

• LOL I’m not a physicist. SOD declined that invitation to meet one. Do you really think a physicist has time to comment on blogs? I would have thought the fact that I was not speaking in equations was a dead give a way I was not a physicist.

Nevertheless, thank you for your input I will read it and get back to you.

• on July 30, 2020 at 1:00 am CD Marshall

BTW I will clarify for you, a temperature can only reach its potential heat source. A contained source being warmed can increase greater than the source if it could withstand the expansion of heated air. creating pressure.

Many people misinterpret potential heat source (usually not a physicist). For example, a light bulb, the filament is not the potential heat source (it produces heat), the bulb gets its power from the voltage applied across the circuit, from outside the bulb…so if you can increase the voltage without blowing the bulb or the circuit you can increase the potential energy of the bulb and therefore, the temperature.

• CD Marshall wrote: “I will clarify for you, a temperature can only reach its potential heat source.”

You are correct, but energy in the atmosphere and the surface/ocean comes from the SUN, whose surface is about 6500 K with a peak spectral intensity in the visible range. This means that concentrated (focused) sunlight in solar ovens and non-photovoltaic solar power plants that use sunlight to heat a working fluid can only heat to a theoretical maximum of 6500 K. This is also the theoretical maximum the Earth’s and Venus’s temperature can be driven to by a runaway GHE.

Looking at the inner rocky planets, it should be obvious that the theoretical maximum is irrelevant to what we observe and that the presence of an atmosphere makes the surface of Venus hotter (740 K) hotter than the surface of the sunlit side of non-rotating MERCURY (only 430 degK). This is because the GHGs in Venus’s atmosphere slow down the rate at which the 740 degK surface of Venus radiatively cools to space. Slowing radiative cooling is a form of insulation, whether performed by an atmosphere or a space blanket or a DeWitt’s perfect reflector.

To understand the power of insulation, consider the interior of the Earth which is maintained at 7000 K (roughly the same as the sun’s surface) by a relatively tiny amount of radioactive decay. That decay produces the roughly 0.1 W/m2 of heat that escapes continuously through the surface. (Thermal conductive and the 20-30 K/km temperature gradient allow us to calculate how many W/m2 are escaping through the crust.) Insulation can dramatically raise the temperature inside an object whether that insulation is provided by resistance to thermal conduction of heat in the crust or slowing the escape of thermal IR by GHGs in the atmosphere of Earth and Venus.

You may argue that the colder atmosphere can’t “heat” the surface of the Earth. The surface and the atmosphere are both heated by the sun. In standard models of planetary heat flux – the NET flux – the only flux controlled by thermodynamics – is from hotter to colder: from the sun to the atmosphere and surface, from the warmer surface to the colder atmosphere, and from the surface and atmosphere to space. A lot of power from DLR is absorbed by the surface, but more power is emitted as OLR by the surface.

Pressure alone does not create heat; pressure must move or compress something to do work on it and thereby heat it. The coldest place in the ocean is at the bottom where the pressure is greatest. The same thing is true for deep lakes. In the stratosphere, the temperature rises with altitude even as the pressure is falling.

• Frank thank you for your input. Energy does come from the Sun I agree with you 100%. What strikes the atmosphere of our planet and what punches through the top of the atmosphere are two different variables. The TOA is around 1300-1400 W/m^2 what strikes the surface is based on TOA and the angle of incidence and albedo which is between 940 W/m^2 at the Equatorial Solar Zenith and around 82 W/m^2 at the Poles.

As I stated,
“It’s not the primary heat source of the Sun as much it is the primary heat source of the Sun at the angle of incidence. Maximum temperature from solar irradiation at the Poles is very different from the Equator.”

The maximum temperature obtained on Earth from the Sun is not that of the Sun’s full temperature but at the TOA which around 120 Celsius/393 Kelvin. The Moon’s temps in direct sunlight is roughly 123 Celsius/396 Kelvin.

Venus is claimed to be from GHG I disagree and I have my theories but we’ll leave that for another subject. Heat isn’t being trapped on Earth, so it can’t overrun the system with a higher out of control looping mechanism. The atmosphere is not a solid insulator like the Earth’s outer crust. Earth’s outer insulator prevents the surface from rising to a geologically estimated 120 Celsius at the Equator multiply that by 92 bar atmosphere and lava pools of 2000 Celsius and you got a good idea of the isothermal hell on Venus. Please keep in mind gas (yes under pressure which was of course my original point being gas under pressure) does play some role in temperature. Perfect examples of those are the outer planets with immense pressure creating super heated cores.

“You may argue that the colder atmosphere can’t “heat” the surface of the Earth. The surface and the atmosphere are both heated by the sun.”

The atmosphere has a temperature because it is first heated, certainly true. That source of heat is mainly the Earth’s surface via conduction/convection/advection/lapse rate. The SURFACE of the Earth can be heated to a maximum of the incidence of radiation not directly from the Sun’s main temperature. A Greenhouse will only reach the maximum temperature of the incidence of radiation so don’t put a greenhouse in Antarctica.

For example, one day In direct sunlight my gauge calculated 40C/545.3 W/m^2 at 9:35 AM at 41.08482° N.

Many factors prevents the Earth’s surface from reaching maximum thermal potential from solar irradiance. We couldn’t live on most of this planet if it were not so.

The atmosphere is heated by the surface and the surface is not directly heated by the TOA all the time. As I stated, the Troposphere expands during the day under direct sunlight and cools at night, that cooling by way of IR and the lapse rate.

Now the biggest misconception is that an energy exchange automatically equates an increase in temperature. That is far from being true, heading back to line spectrum re-routed from the atmosphere back to the surface.

These laws are obeyed on the macro and the micro, frequency/wavelength AND energy dictates temperature change. A match is certainly hotter than then say an Arctic day at -20 Celsius, but that match even though higher in frequency and wavelength does not have enough energy to overcome the energy of the system and change the temperature gradient, that same match placed at the middle of the Troposphere at -18 Celsius will also not change the system.

Distinctions not being comprehended between energy, heat and temperature are creating a lot of confusion in the current concepts of climate science.

• Took out my paragraphs again, what’s up with that?

• CD Marshall: There is a lot of handwaving in your 8/2 reply. Perhaps I can simplify:

1) Our planet warms when it absorbs more radiation than it emits, cools when the opposite is true, and remains steady when they are equal. The planet’s radiative imbalance (I) is given by:

I = (S/4)(1-a) – oeT^4

where S is solar irradiation, the factor 4 corrects for geometric factors (the sun irradiates a hemisphere of area 2*Pi*R^2 with an average cosine of angle of incidence of 0.5 and radiates over an area of 4*Pi*R^2), a is albedo, o is the S-B constant, e is the “effective emissivity” of the planet given an average surface T. The planet’s surface temperature has been approximately the same during the Holocene, so there was likely little imbalance when the Industrial Revolution began adding CO2 to the air.

Radiative transfer calculations (which correctly calculate Dave Evans “re-routing feedback) predict that a doubling of CO2 with no other changes will reduce radiative cooling to space by about 3.7 W/m2 if nothing else changes, but will not interfere significantly with incoming SWR. That will make produce an imbalance +3.7 W/m2 and the planet will begin to warm. Unlike your match, the reduced radiative cooling to space from doubled CO2 continues from year to year and century to century. Warming will continue until our climate system emits (LWR) or reflects (SWR) enough more energy to restore balance.

(For the moment, we don’t need to worry about internal transfer of heat between the surface, the ocean and the atmosphere or latitudinally or vertically transfer or minor differences between day and night. The main concern is the energy imbalance or balance across the TOA. Conservation of energy demands that the heat from a radiative imbalance must raise the temperature somewhere below the TOA and it will continue to do so until enough additional heat is emitted or reflected to restore balance. To a first approximation, all of the internals modes of energy transfer will distribute heat about the same way it is distributed today, but everything will warm.)

The climate of the Moon, Venus and Mercury are pretty complicated subjects. The temperature of the Moon changes radically between the two-week long day and night. I simply wanted you to recognize that the surface of Venus is hotter than the sunlit surface of Mercury (which lacks an atmosphere). According to the imbalance equation above, the only way that can occur is by differences in emissivity (e) and albedo (a). And Venus has a much higher albedo than Mercury. So it must have a very low effective emissivity. In other words, very little of the thermal IR emitted by the 740 K surface escapes to space and the atmosphere slows radiative cooling (insulating the planet from radiative heat loss).

I presume you recognize that energy/work and temperature (internal energy) can interconvert and that heat capacity is the conversion factor. Force and energy are different. A force must move an object to do work on it (force times distance) so a force can not be directly converted into energy or temperature. For the same reason, pressure (force per unit area) must change (compress) the volume of something before it is converted to energy or temperature. The bottom of the ocean is not warmed by the weight of several kilometers of water above. The soil under the foundation of a skyscraper is not any warmer than nearby ground due to the weight of the building. And Venus is not hot because of the weight of its atmosphere. It is hot because 740 K is the surface temperature where (S/4)(1-a) = oeT^4. Fundamentally, this is just conservation of energy.

• I’ve probably suggested this before, but you can read “Atmospheric Radiation: Theoretical Basis” by Goody & Yung, Oxford University Press 1989. It’s a textbook. Or “Radiation and Climate” by Vardavas & Taylor, Oxford Science Publications 2007. It’s a textbook.

Or if you want a simpler textbook, try “A First Course in Atmospheric Radiation”, Grant Petty, Sundog Publishing 2006.

Perhaps you aren’t the textbook type of commenter. But not reading the relevant textbooks and claiming that no one has tested the GHE hypothesis is entertaining. Ignorance is bliss.

Finally, I recommend reading the article –

Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations

– that introduces the radiative transfer equations (as simply as possible). So far, commenters with a similar profile to you have never explained what is wrong with these equations, derived from first principles and proven in large numbers of applications.

There’s a reason why, and sharp-eyed readers can probably figure it out.

103. There are good reasons I ask for a testable hypothesis. I want to know all the details of the basic GHGE model. Why is it modeled that way and not another? What of the other possible models:
1. by modeler: Dr David Evans,
2. by scientist Ferenc Miskolczi
(those 2 models are similar to the GHGE model supported by Ken).
3. But there are other alternative models too: which model a GHGE by calculating the mean free path to space, MFPtS, for outgoing longwave radiation (I know of 2 versions of MFPtS models). I’m sure there are as many possible models are there are physicists on earth.

There an estimated 10^e500 possible string “theories”, one or none of them may describe the universe. In comparison there are a mere 10^e80 protons in the universe. Scientists devised the notion of testable hypotheses to deal with problems just like this. Unless we can test a hypothesis against the real world why should be waste time pontificating on it?

If we cannot test a hypothesis, we’d be fools to spend trillions of dollars rejigging our energy systems by replacing all the fossil fuels with more expensive, less reliable renewables (ruinables, as their critics call them).

• Mark

Please go to Two Basic Foundations and answer those two basic questions.

• mark4asp wrote further above: “When I questioned AGW my points were simple:
* What is the GHGE testable hypothesis?
* Who wrote it?
* What are the scientific tests for it?
* Which tests are validated?”

You won’t find a convincing simple explanation for the enhanced greenhouse effect (eGHE = rising GHGs will make the planet warmer), because climate scientists want to provide the public an over-simplified explanation that everyone thinks they understand: “GHGs absorb thermal infrared and “trap heat” in the atmosphere.” Many of us are smart enough to know that GHGs also emit thermal radiation and cool the planet, and we become skeptics like you and I. Unfortunately, the science behind the full explanation is fairly complicated.

For me, the explanation begins with what are commonly called “radiative transfer calculations”, which predict how much radiative energy will travel from point A to point B in an atmosphere, given the composition and temperature of the atmosphere between these points. Literally thousands of absorption and emission lines are involved in these calculations and the process is computerized. In some cases, you also need to be concerned about scattering of radiation, in addition to absorption and emission. Our host referred you to one of his articles on how radiation transfer calculations are performed:

I have five textbooks covering radiation transfer calculations, including the inexpensive and highly-recommended Grant Petty’s “A First Course in Atmospheric Radiation” – a text intended for meteorology students, not climate science. Each text presents radiation transfer using slightly different formalism and it’s tough material to understand. To help people like you understand how radiation transfer calculations are used in climate science, I wrote a Wikipedia article on Schwarzschild’s Equation for Radiative Transfer. It explains where this equation comes from (Einstein coefficients and quantum mechanics) and how it simplifies to familiar science: 1) Beer’s Law for absorption – which is only valid when a powerful light source is used and emission is negligible. 2) Planck’s Law which is only valid when absorption and emission are in equilibrium – an assumption made by Planck when he derived that law. in the derivation of that law. As it turns out, absorption and emission are not in equilibrium in our atmosphere. A quick look at the spectrum emitted by our planet should immediately tell you that our planet certainly doesn’t behave like a black or graybody. So trying to describe our planet in these terms runs into problems. College physics and chemistry deal cover Planck’s and Beer’s Laws without theses limitations to their general applicability, and ignore the more fundamental Schwarzschild’s equation. Since Schwarzschild’s equation is a differential equation whose solution can only be approximated by numerical integration, learning about it doesn’t provide the public with an intuitive understanding of the eGHE. When climate scientists talk about “settled science”, they are discussing radiative transfer calculations. The predictions of radiation transfer calculations have been confirmed by numerous experiments in our atmosphere.

The law of conservation of energy predicts that – if the rate of radiative cooling of our planet is slowed by rising GHGs and if incoming SWR is unchanged – then the resulting radiation imbalance must cause our planet to warm somewhere. It doesn’t tell us where it will warm, how much it will warm or how fast it will warm; but it does guarantee that it will warm.

How much our planet will warm depends on how much radiative cooling to space increases for every degK the planet warms and how reflection of SWR increases (or decreases) with every degK the planet warms. This sum of these two is called the climate feedback parameter (W/m2/K) and can be combined with the radiative forcing for 2XCO2 to calculate warming after CO2 has doubled and incoming and outgoing radiation are in balance. (If you think in terms of an overall planetary climate feedback parameter, you can avoid the messy and unnecessary concept of many feedbacks amplifying of a no-feedbacks climate sensitivity.)

Climate scientists tend to skip over the climate feedback parameter and jump straight to equilibrium climate sensitivity. Climate sensitivity is what climate scientists use AOGCMs to calculate. However, since AOGCMs can’t calculate convection, precipitation, evaporation and cloud formation from first principles of physics and chemistry, AOGCMs must model such processes with parameters that are tuned by a non-rigorous process. So there is no need to believe that the output of such models must be right. In fact, climate models are unable to accurately reproduce the large seasonal changes in emission of OLR and reflection of SWR from clear and cloudy skies that we observe from space every year. Furthermore, the predictions of models are often mutually inconsistent. Our inability to unambiguously quantify the amount of warming that will follow a forcing doesn’t cast doubt on the reliable science (conservation of energy plus radiative transfer calculations) that predicts some warming.

Our host has several posts on Miskolczi. Dave Evans goes wrong when he postulates a “re-routing feedback”. Radiative transfer calculations automatically include re-routing feedback. And, like many, Dave appears to think that the predictions of climate science are fundamentally derived from feedbacks and can be invalidated by handwaving arguments about feedbacks. In reality, AOGCMs are right or wrong because they properly or incorrectly represent the truly fundamental processes in our planet’s weather: radiation, convection, precipitation, evaporation, clouds, ice, etc. If these fundamental elements and processes were correctly described, then AOGCM output would be correct. Feedbacks can be abstracted from the output of an AOGCM and are right or wrong if that AOGCM is right or wrong. Feedbacks are derived from AOGCMs; but AOGCMs aren’t built from feedbacks. IMO, arguing about whether feedbacks are right or wrong is a waste of time and a great source of confusion.

104. “that introduces the radiative transfer equations (as simply as possible). So far, commenters with a similar profile to you have never explained what is wrong with these equations, derived from first principles and proven in large numbers of applications.”

“Much of thermodynamics deals with three closely related concepts: temperature, energy, and heat. Much of students’ difficulty with thermodynamics comes from confusing these three concepts with each other…” -Chapter 1 – Energy in Thermal Physics Daniel V. Schroeder: An Introduction to Thermal Physics

Apparently it’s not just students who struggle with this concept.

• Many skeptics wrongly believe thermal infrared emitted by the colder atmosphere can be absorbed by the warmer surface of the Earth, because that violates the 2LoT. Others wrongly believe that thermal infrared photons absorbed by GHGs in the atmosphere are “re-emited” (instead of the excited states so produced being relaxed by collisions, aka “thermalized”. The Schroeder text you mention covers statistical mechanics, the branch of physics and chemistry explains how large number of colliding molecules and photons following the laws of quantum mechanism produce the macroscopic laws of thermodynamics that were deduced much earlier from empirical observation. Individual photons can travel from the colder atmosphere to the warmer surface because they follow the laws of QM, not thermodynamics, but statistical mechanism shows why the NET flux (heat) of photons between two locations is always from hotter to colder. The net flux in energy flux diagrams produced by climate scientists is always from hotter to colder, but they show two-way fluxes, not net fluxes. When you have a Boltzmann distribution of energy among ground an excited states (one way to define the concept of local thermodynamic equilibrium), then the rate photons are emitted depends only on temperature – which is the case in the atmosphere (but not LED lights). Schroeder’s book can help you learn these challenging subjects. However, the fundamental physics about why GHGs slow radiative cooling to space is described by radiative transfer calculations, which is covered in by the post ScienceofDoom linked above or the Wikipedia article on Schwarzschild’s Equation for Radiative Transfer or Grant Petty’s cheap textbook, “A First Course in Atmospheric Radiation”. Schroeder may not even cover this subjects.

So, show us you aren’t just another transient commenter who likes to stir things up and constantly change the subject, but is unwilling to risk learning anything that could conflict with what you already believe. It is more entertaining to listen to “popes” speaking to today’s equivalent of Flat Earth Societies in highly partisan echo chambers – as if Galileo and the scientific method never existed.

105. on August 9, 2020 at 12:23 pm | Reply nobodysknowledge

Demetris Koutsoyiannis has looked at the hydrological circle, including precipitation. He has found that models do a bad job.
Revisiting global hydrological cycle: Is it intensifying? Demetris Koutsoyiannis. 2020.
” While most of climate impact studies have been based on the assumption that climate models provide plausible predictions (usually termed projections) of future hydroclimate, there is a number of studies that claimed that this cannot be true as, when compared with real data of the recent past (after the predictions were cast) or even earlier data (already known at the time of casting the prediction), prove to be irrelevant with reality (Koutsoyiannis et al., 2008, 2011; Anagnostopoulos et al., 2010). This becomes even worse if we focus on extremes (Tsaknias et al., 2016). Tyralis and Koutsoyiannis (2017) developed a theoretically consistent (Bayesian) methodology to incorporate climate model information within a stochastic framework to improve predictions. However, because of the bad performance of climate models, application of this methodology leads to increased uncertainty or, in the best case, in results that are indifferent with respect to the case were the climate model information is not used at all. In summary, as implied by Kundzewicz and Stakhiv (2010), climate models may be less “ready for prime time” and more ready for “further research”.”

“Comparison of model outputs with reality, as the latter is quantified by the satellite (GCPC) observations, is provided in Figure 17. As expected by the assumptions and speculations mentioned in section 3, climate models predict increase of precipitation after 1990-2000. This hypothetical increase is visible in Figure 17. However, real-world data do not confirm the increase. Noticeable is also the large departure of reality and model outputs in terms of the average global precipitation. All these observations support the claim that climate models dissent from the hydrological reality and they further illustrate the fact that the real-world precipitation has not been intensified according to the IPCC expectations.”

• on August 9, 2020 at 12:24 pm | Reply nobodysknowledge

• on August 9, 2020 at 12:33 pm | Reply nobodysknowledge

This post should have been placed under Models and rainfall. Sorry

• Koutsoyiannis compares model output and “reality”, without considering what how accurately “reality” is measured. Unfortunately, he never calculates the confidence intervals for any trends and obscures much of his data by plotting raw data without taking anomalies to remove the much larger effect of the seasonal cycle.

For example, consider this ERA5 data from 1980s to 2010s (Table 2)

Increase in surface temperature over land: 0.30 K/decade
Increase in dew point over land: 0.15 K/decade
Increase in “surface temperature” over the ocean: 0.14 K/decade
Increase in dew point over the ocean: 0.10 K/decade

Does this mean relative humidity has fallen? Even over the surface of the ocean? How well does reanalysis extract a dew point from the data it processes and how has the reliability of that process changed with time? If we look on to Figure 5. The ERA5 and NCEP reanalyses differ about the specific humidity at 850 mb by 5-10% and by nearly 100% at 300 mb.

In Figure 10, there is a 10-20% difference between rainfall measured over land using gridded rain gauges (CPC) and measured from space (GCPC) or by re-analysis.

In Figure 17, Koutsoyiannis complains that observed global precipitation (GSPC) does not show a significant upward trend (as models do), but the observations are fairly noisy. The big problem is that global precipitation is 10% (8 W/m2!) bigger in models than we observe from space! However, precipitation from the NCEP and ERA5 reanalyses (Figure 10) are closer to the predictions of climate models. Which version of “reality” is the right one?

106. “So the temperature of the atmosphere from which the emission originates determines the amount of radiation”

Hang on, I thought amount of radiation out is fixed at 240 W/m²?

• Nope. When greenhouse gases increase, emission to space decreases until the surface and atmosphere warm up to restore the balance. If emission to space were fixed, then there would be no effect from changes in ghg concentration. Emission from the sun at the top of the atmosphere is nearly constant, but the amount reflected, the albedo, could change if the fraction of the planet covered by clouds changed.

• Output isn’t fixed. When output = input the system is in balance. The output can increase it decrease door to the radiative properties of the atmosphere, which then drive changes in system temperature.

• Brad: You may find it clearer to express this thought in terms of the law of conservation of energy. When incoming and outgoing radiation are equal (in balance) from year to year, the temperature of our climate remains constant. When they are out of balance, the energy difference is added to or subtracted from internal energy, ie temperature. When there is an imbalance, it is warming or cooling somewhere.

We measure a radiative imbalance in terms of W/m2. A +1 W/m2 radiative imbalance is fairly huge – enough to warm the atmosphere surface, and top 50 m of the ocean at a rate of 0.2 degC/y. We sometimes call the top 50 m of the ocean the mixed layer because winds stir the oceans down to roughly 50 m and seasonal warming and cooling of the ocean reach down about this far.

107. I meant that it’s fixed at 240 W/m2.

If there were changes to albedo etc, then sure, temperature would go up or down, but when the system has settled again, the average emissions will be 240 W/m2 (again).

• Change in effective radiative temperature of the emissions to space can drive changes as well, not just a change in albedo.

• Steady state output is constant only if the input is constant, i.e. remains at 240 W/m². A change in albedo, for example, would raise or lower that number ‘permanently’. Albedo can change with temperature. During a glacial period, for example, the albedo increases because ice and snow is more reflective than dirt and plants so the steady state output would be significantly lower than 240 W/m².

• Mark: The sun’s output is nearly fixed at about 1367 W/m2 (except for the solar cycle and periods of weak solar activity like the Maunder and Dalton minima, which are usually small enough to ignore). That solar output is absorbed by a planet with a surface area 4 times larger than the planar disk that intercepts sunlight. That means solar input at the edge of space is nearly fixed at 342 W/m2 averaged over the entire planet. Right now, about 102 W/m2 is reflected and scattered back to space by clouds, aerosols and the surface (including snow and ice), so about 240 W/m2 is absorbed.

On a warmer planet, many are convinced that reflection of sunlight by snow- and ice-covered surfaces is going to decrease, and that storm tracks are going to move polarward and reflect less sunlight. If so, the above reflection of 102 W/m2 is NOT fixed and more than 240 W/m2 could be absorbed.

These cloud and surface changes in reflection of SWR in response to a change in surface temperature are called SWR feedbacks. They are measured in units of W/m2 per degK of surface warming or W/m2/K. If reflection decreases 0.5 W/m2/K (technical +0.5 W/m2/K because that warms the planet), then the current 1 K of surface warming since pre-industrial conditions has decreased reflection of SWR from 102.5 to today’s 102.0 W/m2. That extra 0.5 W/m2 has added to the +2.5 W/m2 from rising CO2.

The key unknown is how much more radiation our planet emits (LWR) and reflects (SWR) as it warms so that balance is restored after a forcing. (This concept is confusingly called climate sensitivity by climate scientists.)

108. Frank, sorry yes, if the snow all melts, then albedo goes down, temperature and W/m2 go up.

That’s not what I meant though. What puzzles me is the notion that there can be “warm” 240 W/m2 radiation and “cool” 240 W/m2 radiation.

Radiation does not have a temperature. Neither does potential energy, latent heat, chemical energy etc.

Temperature is dictated by how much thermal energy there is, which is measured in joules, not W/m2.

If 240 W/m2 are emitted, that works backwards to 240 joules (of thermal energy) being emitted every second for every square metre. It does not matter from where they are emitted, it only matters how many.

• It matters where it is being emitted from because a large portion of those emissions originate from the atmosphere. The atmosphere temperature profile is mostly dictated by the surface temperature and the lapse rate. So when emissions to space have to originate from higher in the atmosphere in order to leave the system then there is a reduction is outgoing emissions. Since outgoing < incoming, the system warms.

• Brad, again, thanks, but the article explains this theory.

The bulk of IR emissions to space are from the atmosphere, no dispute there.

My question is, how and why is there a difference between “cool” 240 W/m2 being emitted from higher up versus “warm” 240 W/m2 being emitted from lower down?

Or are you arguing that with more CO2, only 239.999…9 W/m2 are being emitted, so each second, the number of Joules in the atmosphere is going up by 0.000…1 Joules?

IR (or indeed sunlight) does not have a temperature in and of itself, that must be clear, it’s got to hit something first.

summary here

On a pedantic note, the surface temperature is dictated BY the lapse rate. The average temp of the atmosphere = 255K, just enough to emit the 240 W/m2. Earth rotates fairly quickly and the atmosphere is good at distributing energy, so it averages out and average temp = effective temp. The lapse rate is an independent variable dictated by gravity and specific heat capacity.

• Correct Mark. If you increase co2 the initial response is a decline in emissions from the system. The system isn’t always emitting 240Wm-2, it’s always emitting 240Wm-2 WHEN it’s in steady state. The condition of steady state isn’t a guarantee. Changes to ghg concentration will perturb the steady state and alter the rate of emissions from the system. This is what would cause the system to warm or cool.

• Mark, in regards to your comment about the surface temp and lapse rate, I don’t find that comment to be accurate – much less relevant, as the lapse rate isn’t changing to a first approximation. Surface temps are influenced by radiative properties of the atmosphere, not the lapse rate itself which is simply a description of the change in temperature with altitude.

1. Effective temp of atmosphere (the emitting layer) = average temp of atmosphere (because of relatively fast rotation and mixing) – depends on sunlight etc.

2. Lapse rate depends on gravity and specific heat capacity.

3. So lower layers are warmer than average and upper layers are cooler. The effective-average temp sets the temp of the mid point about half way up and lapse rate says how cold tropopause is and how warm sea level is.

4. Hard surface or ocean surface must be +/- same temp as lowest layer of atmosphere.

• Sorry – I posted this in the wrong place and can’t see how to delete it.
——————————————————————————–
Brad: “If you increase co2 the initial response is a decline in emissions from the system. The system isn’t always emitting 240Wm-2,”

Yes, that is obviously the theory. No need to repeat it.

I just fail to see how INCREASING the amount of gas which can radiate IR to space can REDUCE the amount of IR radiated to space. That’s like arguing if your car has a bigger radiator, the engine will get warmer.

That is the logical step I am stuck on.

• Mark: You asked a great question: How does a flux of energy (power/unit area) become a temperature which is proportional to energy? It doesn’t! An energy flux in W/m2 becomes a RATE of temperature change. You also need to know the depth of the material being heated through its surface area, which give you the volume being heated. Heat capacity (J/volume/K) then allows you to convert a W/m2 flux into a rate of warming K/s. Or you need to know the mass of the material being warmed through its surface are and use J/kg/K for heat capacity. Volume works for warming the ocean. Mass works for warming the atmosphere.

In the case of our planet, we talk about the radiative imbalance created at the TOA by an instantaneous doubling of CO2 (a radiative forcing). As the planet warms in response to reduced radiative cooling to space, OLR will increase and the radiative imbalance will gradually shrink to zero. If you integrate that radiative imbalance (power) over the time, you get energy – which is found in the internal energy of our warmer climate system

109. Brad: “If you increase co2 the initial response is a decline in emissions from the system. The system isn’t always emitting 240Wm-2,”

Yes, that is obviously the theory. No need to repeat it.

I just fail to see how INCREASING the amount of gas which can radiate IR to space can REDUCE the amount of IR radiated to space. That’s like arguing if your car has a bigger radiator, the engine will get warmer.

That is the logical step I am stuck on.

• The analogy I prefer to is that co2 to fog since fog is much more perceptible to us.

Consider for a moment that you are in a foggy area where you only have a few feet of visibility. Fog dissipates with height so as you rise in altitude the fog thins, allowing your visibility range to increase. Eventually you’ll reach a height at which most of what you see is a clear picture of the open sky.

Now, increase the amount of fog. You no longer have as clear of visibility so must travel higher to have an equivalently clear picture of the sky as before.

Now let’s clarify some other points.

Effective temperature is not the avg temperature of the atmosphere. Effective temperature is the temperature you arrive at when you integrate the radiative spectrum of the system as seen from space. This is 240Wm-2, which equates to 255K. This has nothing to do with the avg temperature of the atmosphere. Take the sun for example. It is for all intents and purposes a giant atmosphere. It has an effective temperature of 5800K. That is not it’s act temperature, but rather it’s a blackbody equivalent of the amount of radiation it emits.

The other hurdle you seem stuck on is that more emitters does not mean more emissions. Rate of emissions is highly temperature dependant, so a slight cooling to the effective emissions layer has a dramatic impact on the amount of energy it emits. And remember from the fog analogy, you have to move upward (to cooler temps) to see space as the fog thickens.

I’m also going to point out here, just in case, that the fact that 255K in our atmosphere is close to the mid point is merely coincidental. Consider other system (Venus, Sun, Mars) and you’ll see their effective emissions layer is no where near the center of mass of their atmosphere.

• Mark, another thought in trying to drive this point home – if you have now emitters in a system you also have more absorbers. This is why increasing concentrations doesn’t simply equate to more emissions, as you’ve also increased the number of absorbing particles above the original emitting layer of the atmosphere.

• I just fail to see how INCREASING the amount of gas which can radiate IR to space can REDUCE the amount of IR radiated to space. That’s like arguing if your car has a bigger radiator, the engine will get warmer.

That’s because you don’t understand radiative transfer. Emission originates at the surface, but only a small part of that radiation escapes directly to space. Ghg’s emit and absorb radiation. That atmospheric emission is proportional to the temperature and concentration of the gas. Temperature and concentration (pressure) decrease with altitude. At 400 ppmv and wavelengths near the center of the CO2 band, effectively all radiation emitted from below is absorbed at an altitude of a few meters. That continues as altitude increases until the number density of CO2 molecules (number of molecules/m³ or pressure times length: atm cm e.g.) gets low enough that 50% of the radiation emitted upward can escape to space. That’s the effective emission altitude. The same sort of thing applies to the band wings. The result is that, because temperature in the troposphere decreases with altitude, the intensity of the emitted radiation decreases as the concentration increases because the effective altitude of emission increases.

If you want to see how that works, try going here:

http://climatemodels.uchicago.edu/modtran/

You can vary all sorts of parameters and see what happens.

If you set all the variables except viewing altitude to zero, you get a blackbody emission curve with a power of 380.254 W/m² at a surface temperature of 288.2 K. The emissivity is not one (0.98 IIRC) and some energy at long wavelengths is not included in the integration so it’s less than σT^4. Now let’s add some CO2 to the atmosphere:

10 ppmv 364.554 W/m²
20 ppmv 360.786 W/m²
50 ppmv 356.076 W/m²
100 ppmv 352.622 W/m²
200 ppmv 348.854 W/m²
400 ppmv 345.4 W/m²

Emission decreases non linearly with increasing CO2. The same goes for water, methane, tropospheric ozone and freon.

Note that initially the CO2 absorption band gets wider and deeper, but at about 50 ppmv, the effective emission altitude for the center of the band is in the stratosphere where the temperature increases with altitude so you get a peak in the middle of the dip.

Grant Petty’s textbook isn’t expensive ( in fact, it’s currently on sale for the original price of \$36) and can give you a good grounding in the fundamentals of meteorology and radiative transfer.

• Mark wrote: “I just fail to see how INCREASING the amount of gas which can radiate IR to space can REDUCE the amount of IR radiated to space. That’s like arguing if your car has a bigger radiator, the engine will get warmer.”

I struggle with this problem for a long time. Twice as much CO2 will emit twice as many photons and those photons will travel on the average half as far between absorption and emission. There will be no change in the flux to a first approximation. However, the traveling half as far argument doesn’t apply to photons that would have escaped to space before doubling. I didn’t understand radiative forcing until I learned the proper physics.

The emission and absorption of radiation (photons) is the controlled by the laws of quantum mechanics, which simplify to “Schwarzschild’s equation for radiation transfer” in the stratosphere and troposphere (in the absence of scattering, which is usually negligible for thermal IR, but not visible):

dI = emission – absorption (+/- scattering)
dI = n*o*B(lambda,T)*ds – n*o*I*ds = n*o*[B(lambda,T)-I]*ds

where dI is the change in intensity of radiation at a wavelength lambda (spectral intensity) as it passes an incremental distance ds through any material, I is the spectral intensity of the radiation entering the increment ds, n is the density of molecules absorbing and emitting radiation (GHGs in the case of thermal IR), o is their absorption cross-section at wavelength lambda, I is the spectral intensity of the radiation entering the increment ds, T is the local temperature, and B(lambda,T) is Planck’s function. This equation is used to calculate the forcing from 2XCO2 by integrating the thermal IR emitted by the surface as it travels from the surface to space. DLR is calculated traveling from space to the surface. Individual photons don’t obey the Second Law of Thermodynamics, but the net flux of radiation calculated using Schwarzschild’s equation is always from warmer to colder.

The first time SOD showed me this equation, the “lights went on” and all my confusion about radiative forcing disappeared. I don’t know if the implications of this equation are immediately obvious to others. SOD has written many posts showing the non-intuitive results that are produced by numerically integrating this equation.

When radiation has passed far enough through a homogeneous isothermal atmosphere, eventually absorption will come into equilibrium with emission. dI will be zero and I = B(lambda,T). Planck’s law can be considered to be a corollary of Schwarzschild’s equation and Planck derived his law explicitly assuming radiation where absorption and emission are in equilibrium with “quantized oscillators”. Planck’s law doesn’t apply in the atmosphere, because absorption and emission are not in equilibrium at all wavelengths and altitudes. This is why our planet’s OLR spectrum doesn’t look like that of a blackbody.

If really high intensity radiation enters the ds increment, the emission term is negligible and Schwarzchild’s equation simplifies to the differential form of Beer’s Law. A laboratory spectrometer uses a filament at several thousand degK to produce light so intense that thermal emission is negligible. Using laboratory spectrophotometers causes many people to ignore emission and think only in terms of absorption. Both absorption and emission comparable in the atmosphere.

The implication of Schwarzschild’s equation can be simplified to: The spectral intensity of radiation traveling through an atmosphere is being changed by absorption and emission so it approaches “blackbody intensity” at a RATE proportional to the density of absorbing/emitting molecules and the absorption cross-section (n*o). If you take the most strongly absorbing wavelength for CO2, 90% of the photons emitted upward from the surface (with near blackbody intensity) have been absorbed in the first 1 m – and they have been replaced with an equal number of photons emitted upward by CO2 molecules. If you put a low emissivity coating on the surface so that it emitted no radiation up, upward emission by from CO2 molecules will have produced a blackbody intensity within a few meter of the surface. And if you shone a laser tune to this wavelength upward from the surface, absorption would dominate emission and the upward flux would also have blackbody intensity a few meters above the surface. By the time radiation gets well into the stratosphere, the density of CO2 molecules has been reduced to the point where radiation can travel a long distance between absorption and emission and the temperature can change over that path. If temperature changes, temperature-independent absorption and temperature-dependent emission may no longer be in equilibrium.

Now let’s consider a horizontal layer of atmosphere ds thick in the middle of the troposphere. Photons are emitted from this layer in all directions, but we can break that emission into three parts: a component in the +z direction, a component in the -z direction and a component in horizontal directions that contributes nothing to heating or cooling the planet and can be ignored. (This is called the two-stream approximation.) Emission from our layer of atmosphere contributes equally to dI in the +z direction and the -z direction. Now, what about absorption. Radiation entering this layer is also traveling in all directions and again we only need to consider the components traveling in the +z and -z directions. Since radiation traveling upward was emitted lower in the atmosphere where it is usually warmer, upward I is greater than the local B(lambda,T) emission. Therefore, the [B(lambda,T)-I] term is negative and dI is negative. When you add up all of these dI terms from the surface to the edge of space, the 390 W/m2 upward flux at the surface is reduced to 240 W/m2 (OLR). More importantly, when the concentration of GHG (n) is increased, dI gets more negative, explaining why rising GHGs reduce radiative cooling to space. For radiation traveling in the opposite direction, dI is usually positive and the downward radiative flux increases from zero at the edge of space to about 333 W/m2 at the surface (DLR).

To be completely candid, increasing GHGs (n) also reduces the magnitude of the [B(lambda,T)-I] term, because photons travel a shorter distance when the concentration of GHGs are higher and the temperature difference between the altitude where incoming radiation (I) was emitted and the layer is emitting B(lambda,T) is smaller. For this reason, the wavelengths where photons travel a long distance between emission and absorption make the greatest contribution to radiative forcing.

And if you are really sharp, you will have noticed that there would be no GHE or enhanced GHE (radiative forcing) with an isothermal atmosphere whose temperature didn’t change with altitude. And why rising CO2 cools the stratosphere.

• Frank, that’s all very interesting.

So why is it so easy to reconcile surface temp on Mars, Earth, Moon and Venus without this tweak?

All you need to know is
– distance from Sun
– height of cloud cover and albedo
– mass and height of atmosphere
– specific heat capacity of gases in atmosphere
– strength of gravity
– rotation speed (affects the Moon in particular, as it has no atmosphere to even things out).

You don’t need to worry about whether thermal energy is transmitted by conduction, convection and radiation. It is safe to assume that energy in all its forms does its best to even out everywhere.

(For Earth you have to adjust for latent heat of evaporation as well)

• Mark,

So why is it so easy to reconcile surface temp on Mars, Earth, Moon and Venus without this tweak?

That’s completely bogus. Ignoring radiative transfer in a planetary atmosphere does not explain planetary surface temperature. It has been discussed at length elsewhere on this site.

• Mark: I hope you recognize the Schwarzschild equation as more than “interesting” – as an accurate representation of the physics/chemistry that determines how radiation interacts with GHGs in our atmosphere. This physics is what the online Modtran radiation transfer calculator linked by DeWitt uses, our host used in many posts to illustrate the non-intuitive behavior of radiation in simple model atmospheres, and the IPCC relies upon calculate radiative forcing with today’s atmosphere. (AOGCM’s also rely upon these calculations, but they also rely on tunable parameters that are not the product of rigorous science.) Furthermore, the quantitative predictions of the Schwarzschild equation and our database of absorption cross-sections measure in the lab have been validated by numerous experiments in our atmosphere. Most of all, I hope that your dilemma about the doubling of emission associated with doubled CO2 (which I shared) disappeared and you understand what really happens to both absorption and emission when GHGs increase (before anything else changes).

You can make some rough approximations about what the temperature of planets using:

absorbed SWR = outgoing LWR
(S/4)*(1-a) = eoT^4

I’ll let you and DeWitt debate those calculations. (I always like including the sunlit side of non-rotating Mercury, which is much cooler than Venus, in any list of planets I claim to understand.) Remember however, that Planck’s Law and the S-B equation technically shouldn’t be applied to our atmosphere because absorption and emission aren’t in equilibrium at many wavelengths and altitudes. Choosing the correct “effective emissivity” (e) for Earth is a little tricky. If you use surface temperature (288 K), e = 0.61 and if you more accurately assume e = 1 for the atmosphere, you get a temperature of 255 K, about the temperature of the atmosphere where the average photon escaping to space is emitted. The difference between these two temperatures is caused by the lapse rate and a 5 km difference in altitude. The lapse rate in the troposphere is determined by convection, not radiation, which is why you don’t need radiative transfer calculations to explain this temperature difference. These black- and gray-body models do give different values for Planck feedback.

However, you need to recognize that when 2XCO2 slows the rate at which our planet radiatively cools to space (as shown by Schwarzschild’s equation), that is equivalent to reducing the effective emissivity of our planet from 0.615 to 0.605 in a graybody model (a 1.5% change). Or equivalent to raising the altitude from which the average photon escaping to space is emitted by 0.170 km in a blackbody model. And you need to remember that “small” but persistent changes in radiation build up over time to amount to very large amounts of energy. A +1 W/m2 radiative imbalance at the TOA is enough to warm the ocean to a depth of 50 m by 0.2 K/yr. (50 m is the depth warmed and cooled by seasonal changes in irradiation.) And a 1 W/m2 forcing from increased CO2 can persist for decades. I doubt any crude methodology you use for comparing planetary temperatures will be relevant to GHG mediated climate change.

Mark wrote: You don’t need to worry about whether thermal energy is transmitted by conduction, convection and radiation. It is safe to assume that energy in all its forms does its best to even out everywhere.

This is incorrect. When heat is transferred vertically only by convection, you get a stable linear lapse rate that falls altitude at a rate of Cp/g (which needs to be corrected for the heat released by condensing convected water vapor). In a gray model of an atmosphere, when heat is transferred only by radiation, you get a stable linear lapse rate that changes with optical density, which means it rises exponentially (nonlinearly) approaching the surface. That lapse rate is unstable toward bouyancy-driven convection until the tropopause. And conduction alone produces an isothermal atmosphere. So the mechanism of vertical heat transfer does matter. Since convection transfers heat vertically in the troposphere (outside of polar regions) faster than any other process, we observe a linear lapse rate there, but not higher. We also see no fractionation of gases by molecular weight in the troposphere, showing that bulk motion is mixing the troposphere.

Today we have an imbalance at the TOA because heat retained by rising GHGs hasn’t stopped flowing into the ocean. If you insist on applying crude models to our planet, I suggest starting by restating the above equation in terms of a radiative imbalance (I) at the TOA and surface temperature (Ts) and then differentiating:

I = (S/4)*(1-a) – eoTs^4
dI/dTs = -(S/4)*(da/dTs) – 4eoTs^3 – (oTs^4)*(de/dTs)

You’ve got a term for Planck feedback, SWR feedbacks as albedo changes with Ts (as we discussed above), and LWR feedbacks as rising humidity decreases the effective emissivity of our planet by acting as a GHG and by changing the lapse rate in the atmosphere so that the upper troposphere warms faster than the surface. And clouds change with changes in Ts, effecting both albedo and emitted LWR. Of course, this is just the standard formulation used by climate scientists expressed in terms of a graybody model for the planet. Not being physicists, they have given fancy new and misleading names (feedbacks and amplification) to simple derivatives involving albedo and emissivity. (:))

• Frank, thanks again for your explanation yesterday 19/8 at 10.27 pm.

It appears to boil down to this key statement: “more CO2 reduces emissivity from the surface from (say) 0.615 to (say) 0.605”.

This is a bit like Brad’s analogy of shining a torch into a fog (or indeed the more obvious example of being under low lying cloud at night).

Fair enough, that’s either true or it isn’t, and if true, are there other mechanisms that automatically cancel it out (radiation is not thermal energy any more that potential energy or latent heat is thermal energy. Radiation of course converts into thermal energy, in which case it’s not radiation energy any more, is it?).

I shall have to think about that.

• Mark concluded that my comment: “appears to boil down to this key statement: “more CO2 reduces emissivity from the surface from (say) 0.615 to (say) 0.605”.

My comment said that, IF YOU USE A GRAYBODY MODEL, a doubling of CO2 is equivalent to a reduction of emissivity from 0.615 to 0.605. I also said that a graybody model is a flawed model for our climate system, because Planck’s Law and the S-B equation only apply when absorption and emission are in equilibrium. They aren’t at many wavelengths and altitudes in our atmosphere. “Effective emissivity” is merely a fudge factor that allows our planet to emit only 240 W/m2 given a surface temperature of 288 K. In reality, none of the components of our atmosphere has an emissivity this low and the temperature of emitting molecules ranges from 190 to 310 K.

Despite these problems, a graybody model is commonly used and gets some things approximately right. It is also full of bobby traps that can confuse you. I tried to show you how a graybody model can be transformed into something that looks like conventional climate science with “feedbacks” that come from recognizing that da/dTs (the change in albedo with the change in surface temperature) and de/dTs (the change in effective emissivity with surface temperature) are not zero. If rising CO2 is equivalent to a decrease in emissivity, then absolute humidity that rises with Ts makes de/dTs non-zero (along with other LWR feedbacks). When others misuse graybody models, you hopefully have a chance of spotting their problem.

The danger of useful analogies like Brad’s or graybody models is that they all fail at some point if you push them too far. Other analogies – such as saying GHGs act as an insulator, because some people don’t understand that insulation blocks heat transfer by both convection and radiation. My favorite climate cartoon has a picture of a thermometer next to a block of ice and asks if the presence of ice makes it warmer. Yes, a block of ice makes it warmer if empty space a 3 K lies on the other side of the ice!

My take home message is that Schwarschild’s equation contains the physics/chemistry that correctly describes the interaction between radiation and GHGs as it travels through our atmosphere. Everything else is a crude approximation. The terms n, o, T, I, and dI vary with altitude and technically should be written as functions of altitude n(z), o(z), T(z), I(z) and dI(z) (and ds becomes dz using the two stream approximation). Since it is a differential equation that has no analytical solution and must be integrated over all wavelengths to get W/m2, it can be difficult to use if you didn’t understand my intuitive interpretation. If your intuition doesn’t encompass this mathematics or if you need quantitative answers, you need to rely on software that automatic numerical integration of this equation: Modtran, Spectracalc, or the results found in papers that do more sophisticated calculations. And this blog has many excellent posts in the “Understanding Atmospheric Radiation and the Greenhouse Effect” and the “Visualizing Atmospheric Radiation” series that illustrate the surprising phenomena Schwarzschild’s Equation generates. For example: