In the previous article in this series, The Earth’s Energy Budget – Part Two we looked at outgoing longwave radiation (OLR) and energy imbalance. At the end of the article I promised that we would look at problems of measuring things and albedo but much time has passed, promises have been forgotten and the fascinating subject of how the earth really radiates energy needs to be looked at.
If you are new to the idea of incoming (absorbed) solar radiation being balanced by OLR, or wonder how the solar “constant” of 1367W/m2 can be balanced by the earth’s OLR of 239W/m2 then take a look at Part One and Part Two.
Introduction
If you’ve read more in depth discussions about energy balance or CO2 “saturation” you might have read statements like:
More absorption by CO2 causes emission of radiation to move to higher, colder layers of the atmosphere
If these kind of comments confuse you, sound plain wrong, or cause you to furrow your brow because “it sounds like it’s probably right but what does it actually mean?” – well, hopefully some enlightenment can be found.
Effective Radiation
The sun’s core temperature is millions of degrees but we see a radiation from the sun that matches 5780K – its surface temperature:
Solar Radiation, top of atmosphere and at earth's surface, Taylor (2005)
In this figure there are two spectra: the top one is how the sun’s radiation looks before it reaches the top of the earth’s atmosphere – contrasted with the dotted line of a “blackbody” – or perfect radiator – at 5780K (5507°C for people new to Kelvin or absolute temperature).
The bottom one – of less interest for this article – is how the sun’s radiation looks at the earth’s surface after the atmosphere has absorbed at various wavelengths.
Why don’t we see a radiation spectrum from the sun that matches millions of degrees?
If we measure the upward longwave radiation from the earth’s surface at 15°C we see an effective “blackbody” radiator of 288K (15°C). But why don’t we see a radiation spectrum of 5000K – the temperature somewhere near the core?
The answer to both questions is that radiation from the hotter inner areas of these bodies gets completely absorbed by outer layers, which in turn heat up and radiate at lower temperatures. In the case of the sun, the radiation spectrum includes hotter areas below the surface that are not absorbed at some wavelengths, as well as the surface itself.
In the case of the earth it’s really the top skin layer that emits longwave radiation.
So when we measure the radiation from the earth with a surface temperature of 15°C (288K) we know we will see a longwave radiation that matches this 288K. This will be a total energy radiated of 390W/m2 with the peak wavelength of 10.1μm. The temperature below the surface is irrelevant.
(Well, it’s not really irrelevant. The hotter layers below warm up the layers above – through conduction and radiation).
This is what the radiation looks like:
Blackbody Radiation at 15'C or 288K
This assumes an emissivity of 1. The emissivity of the surface of the earth varies slightly but is close to 1, typically around 0.98. Watch out for a dull post on emissivity at some stage..
At the top of atmosphere, as many know, the OLR is around 239W/m2. For those confused by how it can be 390W/m2 at the surface and 239W/m2 at the top, the answer is due to absorption and re-radiation of longwave radiation by trace gases – the “greenhouse” effect. See the CO2 – An Insignificant Trace Gas? series, and especially Part Six – Visualization and CO2 Can’t Have that Effect Because.. if you don’t understand or agree with these well-proven ideas.
If the earth’s atmosphere was completely transparent to longwave radiation this spectrum would look exactly the same at the earth’s surface and at the top of atmosphere (TOA).
Here’s what it does look like with some typical blackbody radiation curves overlaid:
Outgoing longwave radiation at TOA, Taylor (2005)
(Note that the spectrum is shown in wavenumber in cm-1. For convenience I added wavelength in μm under the wavenumber axis. Wavelength in μm = 10,000/wavenumber).
For energy balance – if the earth is not warming up or cooling down – we would expect the earth to radiate out the same amount of energy that it absorbs from the sun. That amount is 239W/m2, which equates to an average temperature of 255K (-18°C).
As the text for this graphic shows, when the energy under the curve is integrated this is what it comes to! But as you can see the actual spectrum is not a “blackbody curve” for 255K. So let’s take a closer look.
Everything Gets Through or Nothing Gets Through – a Few Thought Experiments
Imagine a world where the upwards longwave radiation from the earth’s surface didn’t get absorbed by any gases in the atmosphere.
Most people are familiar with that thought experiment – it’s a staple of the most basic radiation model in climate science. The radiation at the top of the atmosphere would look like this (the top graph):
255K radiation, 100% transmittance
This is the blackbody radiation at 255K (-18°C) with 100% “transmittance” through the atmosphere. The area under the curve, if we extend it out to infinity, is 239W/m2.
And of course, because the radiation hadn’t been absorbed or attenuated in any way, the temperature at the earth’s surface would also be 255K. Chilly.
Now let’s think about what would happen if the atmosphere allowed radiation only through the “atmospheric window” and everywhere else the transmittance was zero:
323K radiation through a perfect "atmospheric window", 8-14um
The bottom graph shows how the transmittance of the atmosphere varies with wavelength in this thought experiment.
The top graph in this case is the blackbody radiation from 323K (50°C) only allowed through between 8-14μm. The energy under the curve is 239W/m2. (Note the higher values on the vertical scale compared with the earlier graphs).
So if the atmosphere absorbed all of the surface radiation below 8μm and above 14μm the earth’s surface would heat up until it reached 50°C (323K). Why? Because if the temperature was only 15°C the amount of energy radiated out would only be 141W/m2. More energy coming in than going out = earth heats up. The surface temperature would keep heating up until eventually 239W/m2 made it out through the atmospheric window – which is 50°C.
Closer to The Real World – Illustration of Radiation from Multiple Layers in the Atmosphere
Even in the atmospheric window some radiation is absorbed, i.e. the transmittance is not 1. But let’s assume for sake of argument it is 1. So energy in the 8-14μm band just passes straight through the atmosphere. It’s still a thought experiment.
Lots of gases absorb at lots of wavelengths – which makes thinking about it as a whole very difficult. So we’ll just assume that the rest of the atmosphere outside the atmospheric window all shares the same absorption characteristics – that is, every wavelength is identical in terms of absorption of radiation.
Now let’s try and consider what really happens in the atmosphere. Each “layer” of the atmosphere radiates out energy according to the temperature in that layer. For reference, here is the temperature (and pressure) at different heights:
Atmospheric Temperature & Pressure Profile, Bigg (2005)
The highlighted area at the bottom – the troposphere – is the area of interest. This is where most of the atmosphere (by molecules and mass) actually resides.
In our thought experiment radiation from the surface (outside the atmospheric window) gets completely absorbed by the atmosphere, or at least the amount that gets through is very small. Taken to the extreme we would get the result shown a few graphs earlier where the surface temperature rises up to 50’C.
But just because surface radiation doesn’t get out doesn’t mean that radiation from the atmosphere can’t get out.
Each layer of the atmosphere radiates according to its temperature. Even if the atmosphere’s transmittance is zero when considering the entire thickness of the atmosphere, there will be some layer where radiation starts to get through.
This is partly because there is less atmosphere to absorb the closer we get to the “top”. And also because as we get higher in the atmosphere it gets thinner. Less molecules to absorb radiation. Even if some gas is a fantastically good absorber of energy, there must be a point where radiation is hardly absorbed. For example, at the top of the stratosphere, about 50km, the pressure is around 1mbar – 1000x less than at the surface. At the top of the troposphere (the tropopause) the pressure is around 200mbar – 5x less than at the surface.
The challenge in thinking about the atmosphere radiating is that unlike the surface of the earth where all radiation is emitted from the very surface, instead radiation is emitted from lots of different layers:
- Higher up – less absorption, more radiation makes it through
- Lower down – more absorption, less radiation makes it through
But let’s still keep it simple and think about the surface temperature being our standard 15°C (288K) and the atmospheric window letting through everything between 8-14μm. This means 141W/m2 makes it out through this window.
If we have energy balance, the OLR = 239W/m2 in total = 141 (through the atmospheric window) + 98 (radiated from the atmosphere at some height, wavelengths outside 8-14μm).
What temperature equates to this layer in the atmosphere? Well, assuming no absorption above this radiating layer (not really the case), and only radiation outside 8-14μm, the temperature of the atmosphere would have to be 219K, or -54°C. Take a look back at the temperature profile above – this is pretty much the top of the troposphere, around 11km.
Remember that this isn’t exactly how radiation gets radiated out to space – it doesn’t come from one “skin layer”. We might consider that if the transmittance of the atmosphere is 1 at this height, then maybe at 10km the transmittance is 0.8 and at 9km the transmittance is 0.5, and at 8km the transmittance is 0.1..
So each layer is radiating energy, with higher layers being colder but more of their radiation getting through, and lower layers being warmer – so radiating a higher amount – but less of their radiation getting through.
For many people reading, this is a straightforward concept, why so long.. for others it might still seem tough to grasp..
So here is a sample radiation diagram with illustrative values only (and values a little different from above):
Radiation from different heights in the atmosphere, illustrative values only
What the diagram shows is the radiation outside the 8-14μm band. That’s because in our thought experiment the 8-14μm band doesn’t absorb any radiation (and therefore can’t radiate in this band either).
Take the top layer at 11km. If we calculate the blackbody radiation of 219K (-54°C) and exclude radiation in the 8-14μm band the radiation is 98W/m2. Then the grey block above with “0.6” is the atmosphere above with transmittance of 0.6, so the radiation actually getting through from this layer to the top of atmosphere is 59W/m2. Similarly for the two other layers (with different values).
In total, the energy leaving the top of atmosphere (outside of the atmospheric window) is 98W/m2. (It’s just a coincidence that this is the value of the top layer before any absorption). And inside the atmospheric window was the number we already calculated of 141W/m2, so the total OLR is 239W/m2.
Of course, we all know the real atmosphere is much more complex with lots of different absorption at different wavelengths. But hopefully this “intermediate” example help to explain how the atmosphere radiates out energy.
So finally, onto the real point..
What Happens with More Absorbing Gases?
Remember how this long post started..
If you’ve read more in depth discussions about energy balance or CO2 “saturation” you might have read statements like:
More absorption by CO2 causes emission of radiation to move to higher, colder layers of the atmosphere
Now, maybe this kind of statement will make more sense.
In our model – our thought experiment – above, we had a uniform absorber of radiation outside the atmospheric window. Suppose we increase the amount of this absorber – the skies open and someone pours some more in and stirs it around. Let’s say the amount increases by 10%.
Well, take a look back at the last diagram. See the transmittance values for each layer in the atmosphere – 0.6 at 11km high, 0.25 at 10km high and 0.1 at 9km high.
Regardless of how realistic these actual numbers are, increasing the amount of absorbing gas by 10% will automatically mean that each of the transmittance numbers is reduced by 10%. And so less radiation makes it out to the TOA (top of atmosphere).
Effectively because lower layers are contributing less energy out through TOA the effective radiating height has moved up. It’s not because some directive has been passed down from a higher authority. And it’s not because one layer has stopped and another layer has taken over.
It’s just that lower layers contribute less, so the “average radiating height” is now higher and colder.
(Note: it might look at first sight that the average height is still the same even though the amount of radiation has reduced. This is not really the case, see the note at end).
In our particular example what would happen is that the OLR would reduce from 239W/m2 down to 141+98*0.9=229 W/m2. So the surface would warm up and this would warm up each layer of the atmosphere until eventually a new hotter steady state was reached.
Conclusion
This has been a long post to try and create more of an understanding of how the earth actually radiates energy, and why more of any trace gas increases the “greenhouse” effect.
It does it because more “absorbing gases” reduce the amount of radiation that can make it out from lower layers in the atmosphere. These lower layers are hotter and radiate much more energy. Proportionately more energy will then be radiated from higher layers which are colder, and therefore these radiate less energy.
It’s a not a mystical force that raises the “effective radiating height” in the atmosphere. But the effective radiating height does increase.
Note
In the example above, the three layers together contributed 98W/m2 at TOA. That is an “effective temperature” of 219K – remembering that we are excluding radiation from the 8-14μm window. If we reduce the radiation from these three layers by 10%, we now have 89W/m2 which is about 212K – effectively radiating from a colder level in the atmosphere.
The Science of Doom 
That was really nice. I just read something precisely on the topic, too. You’re as good as having Asimov for a neighbor. I love the integrated presentation – for me it’s eminently digestible.
quibble, sorry, but this one sentence is a bit ambiguous, though it’s all clear in the rest.
“Regardless of how realistic these actual numbers are, increasing the amount of absorbing gas by 10% will automatically mean that each of these numbers is reduced by 10%.”
A serious question, though – why should this be expected to affect surface temperatures very much, if convection and turbulence change the lower layer within which most of the IR is absorbed?
I’m having a bit of difficulty reconciling this idea, too- let me try to explain with another thought experiment – it lets me not have to calculate actual numbers…lol
My window is an infinite heat sink (for this game, as if it were a finned radiator out in space).
My breath condenses on it. The water runs down. The temperature reading is X.
If 2 people are in the room, more breath condenses and more water runs down. The temperature reading is X still because in this game the window is an infinite heat sink.
The warmer the room gets, the more I sweat and the more water condenses on the window. My temperature doesn’t change and the temperature of the window remains X.
Can you have a go at this?
I’m basically seeing the ocean as a great heat sink with fuzzy fins out to space, you see.
What computes is that adding heat capacity to the working fluid would improve the efficiency of the heat flow rather than stifle it.
The red circled area resembles the saturation temperature profile of a refrigerant, btw. Is that pure coincidence?
The critical temperature profile of water is fixed, so I wouldn’t expect a change in blackbody spectrum with additional heat to the system (not talking about heat capacity), but instead, more of it.
Up near the top, there is a typo, I think. You refer to the temperature of the Sun near its center — this should be 15 million K, not 5000 K.
Also, I am a bit puzzled as to why you try to draw such a distinction between the emergent spectrum of our Sun compared to that of our Earth. The physics of the formation of the emergent spectrum is essentially the same. The differences in the two emergent spectra arise from the vastly differing temperatures and opacity sources (because of differences in elemental composition and gas temperatures). For example, it turns out that the Sun does not have a “window” to its central core (lucky us) to allow that radiation to exit from that layer virtually unscathed. But it is still the case that each wavelength of emergent light intensity exits from its own “window” down to some physical depth the Sun’s photosphere — the thin surface layer from which light emerges from the Sun, leaving the matter behind.
Whether it’s the Sun’s or Earth’s emergent spectrum, each wavelength of light that emerges into space does so from its own “photosphere” where the optical depth is ~1. Optical depth, identified by the Greek letter ‘tau’, is a dimensionless measure of the ability of matter to scatter/absorb light of a particular wavelength (lambda), and ranges from 0 to extremely large. For the lower layers in Earth’s atmosphere in which the concept of local thermodynamic equilibrium is an excellent approximation, the emergent spectrum intensity, I(lambda), can be approximated by a sum of weighted blackbody radiators, B_lambda(T), as so:
I(lambda) ~ Integral{B_lambda(T) * exp(-tau_lambda) * d(tau_lambda)},
where I’ve replaced the “sum of” with an “integration of” in taking fine continuous steps, d(tau_lambda), from the surface through layers in the atmosphere, and tau_lambda is the wavelength-dependent optical depth measured from some layer (whose temperature is T) to the top of the atmosphere. The higher the temperature, the more intense the blackbody spectrum is. As you noted above the temperature T varies from Earth’s surface to the TOA, diminishing with height from the surface through the troposphere.
It just so happens that Earth’s atmospheric opacity between 8-14 microns is very small, thereby allowing the generally much warmer (and so brighter) thermal radiation emitted by Earth’s surface (land, oceans, etc) to escape directly. Earth’s atmosphere emits very little light in that window.
[b]In the simplest terms: the intensity of light emerging from Earth at a particular wavelength will be high where the atmospheric opacity is low, because that light is emerging from Earth’s warm surface, and it’s low where the atmospheric opacity is large, because that light is emerging from a generally cooler upper layer in Earth’s troposphere.[/b]
Oops. I just realized that in the post above I should have defined the function “exp(-tau_lambda)” in the equation I gave. This is the exponential function, with ‘e’ as its base. So it’s the same as e[sup]-tau[/sup] (omitting the notation of wavelength dependence of tau), where e’s natural logarithm ln (e) = 1.
And I should have complimented you on what overall was an excellent tutorial. Keep up the good work!
Thanks Science, for another outstanding post.
Dave McK:
Thanks for pointing out the ambiguity, I have fixed up the text.
A very interesting question. Convection is more effective than radiation at moving heat in the “optically thick” lower atmosphere. As we move to “optically thinner” layers of the atmosphere radiation becomes more effective than convection.
So convection determines the temperature profile of the lower atmosphere – the atmospheric lapse rate.
But how does energy make it out of the atmosphere?
Regardless of the profile of the lower atmosphere, if energy doesn’t get radiated out to space then more energy is incoming compared to outgoing. And so the whole atmosphere will heat up.
The lapse rate will determine the “slope of temperature” – i.e. how the energy is distributed in the lower atmosphere.
But the effectiveness of the atmosphere at radiating into space will determine the total amount of energy in the climate system.
And for your further comments this explanation still applies. If all the energy in the climate system was effectively radiated out to space from the surface – it would be a different story.
[…] – Part Three in the series on how the earth radiates energy from its atmosphere and what happens when the amount of […]
Looking at the energy transfer in longwave IR, I hope it is correct to assume that a ‘small’ parcel of gas is effectively a black-body – that is all active molecules are at radiative balance with each other.
My first question is what is the mean distance a longwave IR photon travels in the atmosphere before being absorbed again? That is in lower atmosphere and at a few selected heights above? I also understand that humidity will vary that pathlength – but by how much at different levels?
Next I assume in an isothermal density gradient that as the gas gets less dense the photons travel further before being trapped? And that at the edge of space they simply don’t get trapped at all and keep on going. First appearances to me is that photons / radiant energy would ‘flow’ towards less dense gas and space – at increasingly higher rates of travel – instead of towards denser gas at increasing slower rates of travel?
Finally – I see you said that in the lower atmosphere convection transfers more energy than radiation? How is this possible when radiation is so much faster? Or is the answer to the mean photon pathlength above relevant? If so – how long does it take for energy to radiantly escape from the earth to outerspace?
Jerry:
A blackbody is one that absorbs (and therefore emits) 100% of radiation across all wavelengths under consideration. Generally, no gas radiates as a blockbody. They have wavelengths where they absorb and radiate strongly and other wavelengths where they absorb nothing and radiate nothing.
This doesn’t impact on the slightly confusing: “all active molecules are at radiative balance with each other”.
Generally we can’t say too much about individual molecules. We might be able to generalize about a gas or mix of gases that include a lot of molecules.
I skipped some questions here in case the last answer sheds light on the earlier questions..
If there was no convection the temperature gradient in the troposphere might look something like this (the curve marked “Radiative equilibrium in a gray atmosphere”):
It doesn’t matter how fast the photons travel. The equilibrium temperature profile would be determined by the absorption and re-radiation. Absorption increases as the amount of absorbing gas increases. Radiation from absorbing gases increases as the temperature increases. Photon speed doesn’t come into it.
Check out CO2 – An Insignificant Trace Gas? Part Three
By contrast, convection moves large amounts of air due to the fact that solar radiation warms the bottom of the atmosphere, which therefore expands, and therefore rises.
— Given my answer to this last part, are your earlier questions still valid because I didn’t really understand them, and perhaps the relevance of photon speed is the problem. Anyhow, feel free to ask away.
Nice one, Dr Doom.
A few points which may be of interest:
1. Diagrams – one of the nicest I have seen for outgoing radiation is in John Houghton’s “Global Warming” 4th Edition, Fig 2.5 page 24. I have many cruel opinions of this book, but for once he has a really clear diagram, except that it’s labelled wrong on the upper abscissa. (Any student of your blogs will immediately spot the error, though.)
Anyway this is from a satellite scan over the Mediterranean Sea, with the effects of various gases segregated (as far as possible! ) It is immediately clear that the radiation envelope across the two windows centred at 9 and 12 micrometers corresponds to a radiation temperature of about 12C – as you would expect from the surface of the Med. -, while the radiation from the CO2 band at 15 micrometers corresponds to -53C.
2. Re-radiation – I think it’s worth noting that if a particle absorbs a photon of energy ( if a gas molecule absorbs a quantum of radiation) it can only re-radiate that energy as one or more photons of equal or less energy. Photons of lower energy have (in effect) lower temperatures and longer wavelengths. This is caused by the laws of physics.
Consider a photon which has a wavelength in the CO2 band, leaving the earth’s surface. It will not get far till it’s absorbed by a CO2 molecule. The excited molecule, now travelling a bit faster, will in due course emit another photon, with lower energy and a greater wavelength, perhaps after hitting another gas molecule. A lot of this goes on till eventually, all the energy carried by photons which get absorbed by CO2, will re-appear as photons of greater wavelength. This can take the photons right out of the CO2 absorption band and allow it the radiation to escape into space as photons of much lower energy than left the surface, and of course lower energy implies lower temperature.
For a sort of analogy, think of water running down a gutter and coming to a grating over a drain. The bars of the grating restrict the water from flowing in, so that it there is a big flow, some water may have to traverse the whole grating before it gets away.
This means that the depth of water on the side of the grating where the flow comes from, will increase with the rate of flow, even if all the incoming water does go down the drain. In the same way, the atmospheric temperature increases when the outflow of heat is obstructed, even though all the heat coming in eventually gets out again!
3. The Lapse Rate – Why does the air get colder as you go up?
Is it because greenhouse gases are making it warmer near the surface?
Is it because clouds are keeping the heat in?
Is it because the upper parts of the atmosphere are radiating all the heat away?
Is it because of convection, or the latent heat of water vapour, or what?
Well basically it isn’t. Gases in general cool as they are expanded, even when this is done so slowly and carefully that conduction, convection, and radiation play no part at all, and no energy is lost or gained. This seems paradoxical and it is far from easy to explain. For those who are unafraid of maths, refer to:
P. M. Bellan, A microscopic, mechanical derivation of the adiabatic gas relation, American J. Phys. 72, 679 (2004).
And how it works is described is Wikipedia, “Adiabatic Process”
http://en.wikipedia.org/wiki/Adiabatic_process#Ideal_gas_.28reversible_case.29
The word “ADIABATIC” means a process in which NO HEAT IS TRANSFERRED
The adiabatic law states that
TV^(g-1) = constant
Where:
T is the absolute temperature (in K)
V is the volume of the gas concerned
g (gamma in the original equation, but this blog doesn’t do Greek) = 7/5 for bimolecular gases such as nitrogen and oxygen. (Bellan’s paper explains why gamma has this value.)
Now the higher you go in the air, the less atmosphere there is above you, so the air pressure is less. If the air pressure is less, then the volume is inversely greater. If the volume is greater, then the temperature is less, by the
adiabatic law given above.
In the atmosphere, this change is called the adiabatic lapse rate. If there were no water vapour, it would be about 10C per km., but with water vapour and all the actual things that go on in the atmosphere, such as those mentioned above, the lapse rate is in practice about 6.5C per km.
All this has NOTHING AT ALL TO DO with the greenhouse effect, but you can see that since the two effects exist alongside each other, it is very easy to get them confused.
Thought I would just like to make this quite clear.
(Given the adiabatic law above, it is very easy to calculate roughly the dry adiabatic lapse rate from basic physics without using calculus, but it seems irrelevant to include it here)
At some point in the discussion, I’m hoping things will come around to the solar powered heat pump that is the atmosphere and treat the neglected topic of phase change which has a few major influences quite apart from aerosol effects.
The atmosphere is a dynamic system pumping heat; there is a constant flow. Efficiency of heat transport is a matter of the critical temperature of the working fluid and the molar heat capacity. Increasing the heat capacity of the working fluid should not raise the temperature. It should reduce flow rate.
That’s what happens in a refrigeration unit. Radiative effects are considered, really, only on the radiator end.
Between the heat source and the sink, you have your saturation temperature profile = lapse rate. If your sink stays at a given temperature, no temperatures in the system are affected. Instead, the rate of flow changes.
It slows with an increase in efficiency.
I may have missed this, so apologies in advance:
If CO2 is the culprit in warming, OLR should increase before reaching equilibrium as explained in Part 2, but this has not been measured as far as I am aware.
Could you elaborate, please?
/Mango
MangoChutney:
First, here’s what Part 2 said:
If you add CO2, as an immediate effect less longwave radiation leaves the top of atmosphere (TOA). Therefore, more energy comes in than leaves, therefore, temperatures increase.
Eventually, energy balance is restored when higher temperatures at the surface finally mean that enough longwave radiation is leaving through the top of atmosphere.
Second, the idea in part two defines moving from equilibrium through a warming phase back to a new equilibrium.
In practice this is hard to measure – when were we in equilibrium? Are we now? Are the instruments accurate enough? More on this in the American Thinker article
@James McC (or scienceofdoom):
So is it fair to say, then, that it is this (moist) adiabatic lapse rate that drags the surface temperatures upward as the lower troposphere warms?
I need the calculator again for this one.
By Avogadro, the ideal gas constant has the same value for all gases, so PVT = PVT regardless of the molecule.
1 mole of gas molecules, any kind = 24.45 litres at 298K ( 24.85C, 76.7F) and 101.325 kPa (14.696 psi, 1 atm)
In 1000 liters of air, at STP there are 1000/24.45 = 40.9 moles. As in a previous illustration, let the water component be 1% = 10 liters = 4.09 moles. It doesn’t matter what the other gases are for this.
At the critical temperature of water vapor, this 10 liters condenses and occupies 4.09 * 18g * 1g/cc = 736.20 cc.
So the 1000 liters of air would now be 990.74 liters.
Insofar as PVT = PVT for gases, if the pressure alone changes, it means (using 24.85C and 1 atm)
PVT start is 14.696 psi*1000*(24.85C+273.15K)
final P is
14.696 psi*0.990.74, or 1% pressure drop
To get a 1% pressure drop by changing the temperature alone you need to do from 298 to 292.04 = 5.96 degrees.
——————————-
So- does the atmospheric profile reflect what is expected for a dry gas or is it the profile of a refrigerant, i.e., dominated by phase change saturation profile?
*phew*
Dave McK:
The adiabatic lapse rate of dry air is a nice simple equation and comes out at -9.7K/km
The moist adiabatic lapse rate is a complex equation which depends strongly on the atmospheric temperature. At higher temperatures the saturated lapse rate can fall as low as 0.35 x dry lapse rate.
The environmental lapse rate – what we experience in practice – averages at -6.5K/km (but it varies a lot).
So the temperature profile in the troposphere is strongly affected by latent heat.
Spaceman Spiff:
That’s how some people explain it. When you first asked the question I just saw it as a “handy mental picture”.
But now I’m thinking that – all other things being equal – this is what happens.
All other things won’t be equal, but at least we can say we understand how this bit works in isolation (I hope).
Thank you.
I guess there is also a special case where a thunderstorm creates a sort of pipe straight from surface to stratosphere that virtually bypasses any interaction with co2 or IR input once it’s got going. Is such a system dependent on convection at all? To be sure, convection is not opposing it and would assist it, but latent heat and phase change suffice.
scienceofdoom wrote
The question about effective photon speed was to try and understand the energy flux out of the atmosphere.
I understand in the sun a photon takes thousands of years to reach the surface and be emitted. If the radiative emission / absorbtion processes in the lower earth’s atmosphere require each photon to be absorbed and remitted many many times, each time traveling some miniscule distance than it may take hours or days for that photon to eventually reach upper atmosphere and escape.
I have no idea how long that process takes and I thought it might be relevant to your comment about convection being much more powerful than radiation.
Jerry:
Interesting question.
The real reason convection dominates is not due to the photon absorption time.
I will try and get around to writing a post about convection and radiation and see if that helps.
I’d suggest that photon absorption time may be significant. Simply because air temperature doesn’t plummet once the sun goes down. The transfer of energy from soil to air is far too slow to keep air temperature at a stable level so air must be a very poor radiator.
In fact, it’s the soil that cools fastest at night while the atmosphere stays at pretty constant temperature – hence the development of inversions. There is obviously little radiative transfer between lower atmosphere and the soil, and there appears to be little radiative transfer between lower atmosphere and the ‘sky’ – all these are subject to the caveat of the time scale of around 12 hours of non-sun.
So my question stands rephrased. What is the actual radiative flux in the lower atmosphere based on radiation from gas parcel to gas parcel?
My second question is perhaps related. Why does the soil cool fastest while the atmosphere 10m above cools very slowly if at all? Is it a difference in radiation spectrum or some other effect?
Jerry:
Good questions.
Convection is more effective than radiation, nothing due to photon times.. If we had no convection ever then the temperature profile in the atmosphere would be much “lower”, instead of -6.7K/km it might be -13K/km. If photons were 1000000 times faster it would be the same.
Needs a new post to address these questions, I think.
scienceofdoom said
I thought – though I expect to be corrected – that the wet and dry lapse rates were purely functions of basic gas laws and convection was not a factor. I base this on the fact that all the standard meteroological diagrams ( searches long lost neurons for tephigrams ) do not have any correction for convection but do account for water content.
Jerry:
The wet and dry lapse rates explain convection. Why does a parcel of gas rise? Why does another one sink? Why does another one stay at constant height?
All because of their lapse rate compared with the existing atmospheric lapse rate.
So, convection dominates the lower atmosphere because the basic gas laws determine that this effect is more effective at moving heat around than radiation.
scienceofdoom said
I’m not sure this is a good answer.
In a stable atmosphere the lapse rate is fixed – wet or dry. Agreed – and when you introduce parcels of gas with different temperature and water content then yes, they will move till their buoyancy is equilibriated (yuk – horrid word). Even then they will move a bit more due to momentum factors and gravity effects.
The wet and dry lapse rates set the energy profile of the atmosphere. Convection is a result of introducing differential energy at specific locations. This will then cause motion.
In the example I referred to – where the surface cools but not the atmosphere – you have an instance of convection not dominating and a clear absence of radiation causing equilibrium.
I again ask the question as to the time scale of radiation of energy from gases? If it was quick then there would be no development of an inversion. If it is slow then inversions are quite possible. The only other alternative is a distinct difference in radiation profile between ‘soil’ and air. Given you simply use Boltzman profiles I expect this difference is not considered?
A big subject, I will do a post on it.. watch out for it.
I’ll just add a small amplifier of my point before I go off to make tea.
With inversions it’s obvious that the soil is very much more in radiative equilibrium with the sky than the gas immediately above it even though both are at similar temperatures. It’s almost as though the soil has a transparent window to the sky above it and radiates freely, while the gas a few metres above allows radiation through but at no stage even starts to equilibrate with the sky.
They are both at similar temperatures. They should both radiate in similar ways but they don’t.
Is the difference between the soil radiation spectrum and the atmosphere radiation spectrum significant? Are they essentially independent?
Does the fact that the atmosphere does not quickly equilibriate with the sky mean that the flow of energy at ‘atmospheric frequencies” is severely diminished?
Jerry:
The soil will radiate at close to “blackbody” (perfect) radiation. But the atmosphere can only radiate in frequencies at which it can also absorb, which are limited, and definitely nothing close to a blackbody.
In radiation terms – “somewhat”. The reason for saying “somewhat” is that the atmosphere will absorb radiation from the ground.
Otherwise, in terms of latent (evaporation of water) and sensible (conduction and boundary layer convection) heat, they are not independent, they share energy freely.
scienceofdoom said
Yes but.
First, soil absorbs in frequencies that match its chemical composition and surface texture and then re-radiates temperature shifted. Soil which appears ‘black’ at visible frequencies can be pretty ‘shiny’ and non-radiative at longwave IR frequencies.
Second, I used the term black body in relation to the gas radiation a bit too glibly. I understand the QM absorbtion and emission varies at different frequencies, but the net effect is still the same. Radiation between parcels of gas may be restricted to specific bands but to a close approximation they will be in radiation equilibrium – a poor mans black body.
As I understand it, the earth radiates to some (limited) extent in the long wave IR range. The frequencies that don’t get absorbed well can escape. The others interact with the atmosphere.
From your graphs I would expect the large majority of the earth LW radiation to be captured and only a small portion escape. What I see from practical observation is that the soil radiates very strongly and there appears to be very little interaction with the atmosphere.
The actual numbers – energy vs frequency for terrestrial radiation as compared to CO2 and H20 absorbtion would be very interesting
Just another point relating to my post on another thread. I’m quite interested in the mean free path of a photon and the escape time of a photon from the earth surface to space.
If the CO2/H20 gas is somewhat like an impenetrable fog, the the time for a photon (of the right frequency) from being emitted from the soil or lower atmosphere to finally escape is interesting.
It may be microseconds, or it may be a much longer time. I really have no idea. I guess it depends on the probability of being absorbed by a CO2/H20 molecule in a specific distance, and then the dwell time before being re-emitted.
In the end it will define the energy flux from the surface.
To give a practical example. Assume night, assume clear sky. A LW CO2/H20 peak photon is emitted from the lower atmosphere. It takes microseconds to escape to outer space. In effect the surface and lower atmosphere gases become in thermal equilibrium with the sky – 2-3K. It suddenly gets frikken freezing.
So now take the example where it takes hours or days to escape. At night it might get a bit chilly, but not very much.
So my take is it takes a long time for photons (energy) to escape. My question is how long?
And a final sally re my previous posts.
I’m betting that most – if not all – climate models don’t account for photon transit time. I’d be really pleased to find out the contrary, but I somehow doubt it.
The effect of longer photon transit time is to modify the energy flux and change temperature differentials. Effectively putting a layer of insulation between different climate locii.
Now it may be that the models work at time scales that minimise this effect. It may also be that they don’t and we have some interesting effects to model.
Jerry:
The general equilibrium response of an atmosphere determined by radiative effects is in the order of a few months.
That is, if the atmospheric temperature profile was determined by radiation only, and we changed a factor like solar irradiance, or CO2 concentration – the atmosphere would move to a new equilibrium within a few months.
But the atmospheric profile is determined by convective processes.
scienceofdoom said
I really should go and make tea now, but for the sake of argument.
In a night-time still air situation radiation processes dominate. At night the air at ground level gets cold while the air above stays warm. In this case I know it with certainty as I have done many night-time field trips flying sondes to measure the vertical air profile up to hundreds of metres.
At night within a few hundred metres of ground level radiation effects dominate.
[…] waving” explanations are only ever a second-best “guide”. Also check out The Earth’s Energy Budget – Part Three for explanations about emissions from various levels in the […]
scienceofdoom,
You should probably try to use the meteorologists convention that lapse rate is positive when it decreases with altitude to avoid confusion.
The reason that radiation dominates on a clear, still night is that when the surface cools faster than the air above it, you get what meteorologists call a temperature inversion. i.e. the temperature increases with altitude. A temperature inversion blocks convection because the air below is denser than the air above. The same thing happens at the tropopause. People in Los Angeles know all about temperature inversions. The trapping of air in the LA basin by frequent temperature inversions is why LA had smog before anyone else. You could easily see the demarcation when flying into LA. The top of the smog layer was almost perfectly flat and very sharp, sort of like the thermocline in the ocean.
[…] subject was covered in some detail in The Earth’s Energy Budget – Part Three, but essentially each layer of the atmosphere also radiates energy. If CO2 can absorb radiation at […]
[…] Part Three – which explained how the earth radiated away energy and how more “greenhouse” gases might change that […]
Let mee see if I understand this.
Earth is not a perfect black body absorber, but if it was a perfect back body absorber, it’s temperature would be completely determined by the incoming amount of solar energy and Stefan-Bolzmann’s law. This temperature is the highest temperature that Earth can be at, if incoming solar energy is the only energy source.
Greenhouse gases in the atmosphere make Earth more of a black-body and so Earth’s temperature will rise a bit.
The exact mechanism used to move the energy back to the universe isn’t important for the maximum possible temperature.
Sander van der Wal:
Note quite. You need to define what you mean by “Earth” – strange as it may seem.
The incoming solar radiation is balanced by the outgoing radiation from the earth’s climate system.
This is the important balance for determining the “effective radiation temperature” of the climate system.
We measure this radiation (by satellite) – it is about 240W/m^2. If we convert it to a blackbody temperature it is around 255K.
Think of the climate (earth, ocean, atmosphere) as one system. The whole system is in approximate balance. This by itself doesn’t tell you what the surface temperature will be (or the center of the earth). It just tells you how much radiation is sent out into space.
Not really. The surface temperature is determined by the energy balance between the surface and the top of the atmosphere (TOA). The TOA is in balance with the solar radiation.
More (or less) GHG’s change the relationship between the surface and TOA.
Think of a house..
Suppose you feed 50,000W into the house via solar or internal heating (or a combination). If the whole house is in approximate balance then 50,000W will be leaving the house via radiation and conduction.
But armed with this information you don’t know what the temperature of the lounge will be.
If CO2 forcing is what it’s alleged to be then why does this happen:
http://theresilientearth.com/?q=content/greenlands-ice-armageddon-comes-end
Could other physical factors be a play here or is the heat budget more complicated then theory portends.
Higher surface mass balance of the Greenland ice sheet revealed by high-resolution climate modeling:
Ettema et al. GRL http://www.agu.org/pubs/crossref/2009/2009GL038110.shtml
Empirical data show a different scenario.
I don’t quite understand how satellites can show the average heat budget over a broad diverse albedo topography of the Earth of 30% land and 70% water.
Consider as conveyed by Dave McK in another post:
Gas or Vapor: kJ/kg
Air 0.287
Carbon dioxide 0.189
Water Vapor 0.462
Steam 1 psia.
120 – 600 oF
That’s what it takes to change the temperature 1 degree K.
When CO2 changes from 1 to -1 C, a change of 2 degrees C, it radiates 2(0.189 kJ/kg) = 0.378 kJ/kg.
http://en.wikipedia.org/wiki/Enthalpy_of_vaporization
When water vapor changes from 1 to -1 (and condenses) it radiates 2257 kj/kg + 2(0.462 kJ/kg) = 2257.524 kJ/kg. It does this every single time you see a cloud.
But CO2 has no phase change so it carries no heat: All gases at the same temperature have the same number of molecules per unit volume.
Water, being light, masses 18g/mole and CO2 masses 44 g/mole. Using 1 mole of air, just to make math easy:
We lowball the water in the atmosphere at 1% of the molecules
So, in a mole of atmosphere, we have 0.01 moles of water = 0.18g
Now we highball the CO2 at 500ppm which is 0.0005, or 1/2000 of a mole of CO2. 1/2000 * 44g/mole = 0.000484 moles of CO2 = 0.021296g
So in our mole of air with but 1% H2O and a generous 500ppm CO2- the water condensing radiates 0.18g * 2257.524 kJ/kg = 406.35432 J
while the CO2 radiates 0.021296g * 0.378 kJ/kg = 0.008049888 J
the ratio of 0.008049888/406.35432 = .00001981002195
or that water vapor in the example carries 50,479.50 times more heat than the CO2 does.
And that’s just rain. If it turns to snow- multiply by 5-6.
Conclusion – the CO2 is insignificant retainer of heat in our atmosphere. Water vapor does 50,000 times more work.
(And that was even done with shorting the estimate on water while boosting the estimate on CO2)
I think McK has a point here that water vapor is the big elephant in the greenhouse room. Why else do deserts get hotter and colder than temperate zone? Why do ocean current have a significant effect on nearby land masses? Why has ocean increases been relatively steady for the last 2 centuries?
And could solar influences have more of an effect than previously assumed? Solar energy is not uniform, it has components of ultraviolet, xrays, light, and IR. NASA has indicated that the total quantity has not been adequately picked up by existing satellite sensors.
So where has the forcing gone in the last decade plus?
Scottar:
This post isn’t about attributing anything, it just explains the basic theory of how more radiatively-active gases can change the temperature of the surface “all other things being equal”.
Do you mean the heat budget or the temperature?
If you are talking about temperature – which is what they do measure – different wavelengths are used for different types of measurements.
The same approach (measurement of different wavelengths) is used to measure the temperature of different layers in the atmosphere.
There’s a mishmash of ideas in what you write making it difficult to know where to start.
Water vapor condensing releases lots of heat. This is correct.
However, comparing the specific heat capacity of CO2 and water vapor is not going to help. If that was the one important ratio for how the atmosphere functions then you would be right.
But you can’t just divide two numbers and hope for the best.
How much energy is carried from the earth into the atmosphere by latent heat of water vapor?
How much energy is radiated from the surface of the earth out to space?
How much of this energy is absorbed by the radiatively-absorbing gases, like CO2 and water vapor, in the atmosphere?
If you calculate how much rain falls to the earth you can find out how much latent energy was carried up in the first place (what comes down, must have gone up first).
It turns out that the energy carried by latent heat into the atmosphere is very approximately averaged at 80 W/m^2 (global annual average).
This is substantial but not nearly as much as is radiated from the earth’s surface (396W/m^2) or absorbed in the atmosphere and re-radiated back to the surface (324 W/m^2).
Water vapor has a very strong impact here as well. And in this case it is nothing to do with its latent heat – it is the ability of water vapor to absorb longwave radiation.
Try reading How much work can one molecule do, The Hoover Incident and Sensible Heat, Latent Heat and Radiation
[…] Increasing the concentration of “greenhouse” gases like CO2 has an important effect unmentioned by Chilingar and his colleagues. This effect is explained in The Earth’s Energy Budget – Part Three. […]
[…] See The Earth’s Energy Budget – Part Three for an explanation about why more CO2 means less radiation emitted to space initially. […]
[…] “opacity” of the atmosphere. See The Earth’s Energy Budget – Part Three. Clearly Jelbring doesn’t know about it, otherwise he would have brought it up – and […]
[…] Another way to consider the effect is to think about where the radiation to space comes from in the atmosphere. As the opacity of the atmosphere increases the radiation to space must be from a higher altitude. See also The Earth’s Energy Budget – Part Three. […]
[…] Note 1: Reducing the TOA flux = less heat leaves the planet = the planet warms; all other things being equal. More about this idea in The Earth’s Energy Budget – Part Three. […]
Surely any discussion of the earth’s energy budget should mention the contribution from the earth’s core?
Small amounts of the earth’s core energy are conducted up to the earth’s crust. However large amounts of energy (and gases) are released by volcanoes and equally large amounts of energy are released by earthquakes.
I note that the surface energy released by the 2004 indian ocean earthquake was 1.1×10^17 J which is close to the total energy from the Sun that strikes the face of the Earth each second 1.74×10^17 J (so the earthquake could be considered to have a negligeable effect on the earth’s energy budget). However the total energy released by the 2004 Indian Ocean Earthquake is estimated to be 4×10^22 J compared to the total energy from the Sun that strikes the face of the Earth each day 1.5×10^22J. Thus the total energy contribution of this single event is probably not negligeable for the earth’s energy budget.
There is a tendency to classify of earthquakes in terms of the energy released on the Richter scale however it seems that the work done in displacing the earth’s surface significantly increases the numbers (10^17 -> 10^22 from the example above).
Looking at the yearly occurance of earthquakes and an estimation of their energy release I get a very (very!) rough estimation of 1×10^18 J/year which is tiny compared to the sun’s contribution of 5.5×10^24 J/year However, the total work done by these earthquakes is liable to be significantly larger bearing in mind the Indian ocean earthquake example above.
I think the earthquake/volcano combination potentially involves enough energy to have an influence on the earth energy balance.
Surely any discussion of the earth’s energy budget should mention the contribution from the earth’s core?
Small amounts of the earth’s core energy are conducted up to the earth’s crust. However large amounts of energy (and gases) are released by volcanoes and equally large amounts of energy are released by earthquakes.
I note that the surface energy released by the 2004 indian ocean earthquake was 1.1×10^17 J which is close to the total energy from the Sun that strikes the face of the Earth each second 1.74×10^17 J (so the earthquake could be considered to have a negligeable effect on the earth’s energy budget). However the total energy released by the 2004 Indian Ocean Earthquake is estimated to be 4×10^22 J compared to the total energy from the Sun that strikes the face of the Earth each day 1.5×10^22J. Thus the total energy contribution of this single event is probably not negligeable for the earth’s energy budget.
There is a tendency to classify of earthquakes in terms of the energy released on the Richter scale however it seems that the work done in displacing the earth’s surface significantly increases the numbers (10^17 -> 10^22 from the example above).
Looking at the yearly occurance of earthquakes and an estimation of their energy release I get a very (very!) rough estimation of 1×10^18 J/year which is tiny compared to the sun’s contribution of 5.5×10^24 J/year However, the total work done by these earthquakes is liable to be significantly larger bearing in mind the Indian ocean earthquake example above.
I think the earthquake/volcano combination potentially involves enough energy to have an influence on the earth energy balance.
[…] The greenhouse effect does not saturate with increasing CO2 […]
[…] RealClimate: Lessons from Venus 3.The greenhouse effect does not saturate with increasing CO2 The Earth 4.The CO2 concentration in the atmosphere has risen significantly over the last 200 years 5.This […]
Hi Scienceofdoom,
Thanks for the great blog! I’d be ever so grateful if you could help me please with some pointers for the following questions related to this post.
[1] Held and Soden 2000 has the following comment: “The increase in opacity due to a doubling of CO2 causes Ze to rise by ~150 meters. This results in a reduction in the effective temperature of the emission across the tropopause by ~(6.5K/km) (150 m) ~1 K, which converts to 4W/m2 using the Stefan-Boltzmann law.” I can see how to convert between these different numbers – 150m, 1K and 4W/m2 – but what I can’t quite see is how to get to any one of them in the first place. Is this done using a computer? What is the basis for the calculation?
[2] I noticed in F.W. Taylor’s text book (section 7.6.2), a simple model that gives a value for rise in Ze of 3 km! Why the big discrepancy compared with the value given in HS?
[3] You’ve got a nice diagram containing the tropospheric temperature profile in June at 45N. How does this profile vary with latitude and season? What does the temperature profile look like when there are lots of tall cumulus clouds or during a thunderstorm?
Apologies if you’ve already answered these elsewhere and I’ve missed them. If it’s easiest, please send me off to do some more reading — but I only have access to publicly available journal articles.
Sorry, one more question occurs to me….
[4] Probably not practical, but has anyone attempted to estimate Ze experimentally perhaps by combining measurements of temperature profiles and albedo?
Philip R:
The first number (150m) is calculated using the radiative transfer equations (RTE) – there isn’t a simple relationship that can be applied. These equations can only be solved numerically on a computer.
If you take a look at the Atmospheric Radiation and the “Greenhouse” Effect series, especially Part Six – The Equations and also Part Nine you will see the basis for the RTE.
The 150m converts into ~1K simply from the lapse rate (0.15 * 6.5). Note that this is a “no feedback” calculation, which is why the lapse rate is assumed constant.
The 4 W/m2 can be seen very approximately from E=σT4. But in practice the real calculation of radiative forcing is done using the radiative transfer equations – these are not linear so whereas the peak of the CO2 band is radiating from up in the stratosphere, other parts of the CO2 band are radiating from lower in the troposphere, and so on.
Armed with a temperature profile and the concentration of radiatively-active gases, from before and after, the RTE produce a “change in upwards emission”.
Because he is demonstrating a very simple model.
Good question and a reasonable sized topic. Something I will have to cover at a later date, because it requires pulling information together from various sources. In general the lapse rate is a higher value at higher latitudes and higher altitudes – due to less water vapor being present. The dry adiabatic lapse rate is around 10 K/km, but the most moist adiabatic lapse rate is under 4 K/km.
Scienceofdoom,
Thank you very much for your kind reply. I will educate myself some more regarding the RTE.
This whole site is pretty extensive.. and I have not read it all. However I got interested in deriving the temperature profile for an atmosphere uniformly semi-opaque to IR and transparent to solar energy, for my own benefit. I have been able to get the temperature at the top of the atmosphere, at ground , the discontinuity at ground and the profile with altitude for emissivities independent for density (or for a constant density fluid).. and then I got stuck, I need to read more.
1) Are there good links to information online on the development of a gray atmospheric model?.
I looked up some of the references (Goody etc.) and these books are $90 plus, the research papers are $25 to download. Its unfortunate but I am not in the position to spend near $1000 just for an intellectual hobby. The next step I am looking for is how to include the density into the emissivity and then integrate it closed form. I know It can be done numerically, the excellent rte programs by SOD (which I am still learning about) help do that.. but I am really curious about the close form results others have come up with and how they went about it.
2) Is there a better post or subforum in here to talk about the gray atmosphere model? if so I would post there.
3) Is there a good number for the blackbody temperature of the earth (and what is the proper name for it?).
What I am looking for is shown looking at the radiated spectrum like figure 2.9 by Taylor in the original post, it looks like the black body temperature is slightly less than 275K (the peaks of the spectrum should cap out following the blackbody emitted spectrum). Then the ratio of blackbody to Effective temp would be (275/255)^1/4
4) Trying to compare my results to those listed by Miskolzci in ‘The greenhouse effect and the spectral decomposition of the clear-sky terrestrial radiation’ I have come up independently with equations 1), 2), 3), 4) but for a vertical atmosphere (or a flat earth), so I dont understand yet the factor of 3/2 on tauA = 3/2tau, nor do I know what the characteristic optical thickness tauc is. I wanted to follow up deeper and figure out where the e^-tauA term comes from in equation 5. Is there any place where this is explained? (My understanding is these are general equations in the literature.. Milskoczi didnt come up with them)
What throws me for a loop and want to understand better is that Miskolzci indicates that equations 1,2,3,4 are for an unbounded atmosphere and proceeds to develop other equations for a bounded atmosphere. But my equations where derived for a bounded atmosphere with a finite height, and a finite temperature at the top layer, so I don’t follow his comment that they are for a semi-infinite atmosphere. I guess my own derivation is more relevant to a tall glass of fluid like a copper sulfate solution at constant density.
Sorry for the long post. Thanks for an excellent forum.
Have you read the Miskolczi threads here ( https://scienceofdoom.com/roadmap/miskolczi/ )? There’s a lot of detail in the comments.
The problem with gray atmosphere is that energy is redistributed over the entire spectral range at each altitude. The real atmosphere doesn’t behave that way. The molecular lines do change width so the atmosphere is grayer near the surface, but having a spectral range where there is little or no absorption (the window) does make a difference. For one thing, it helps remove the temperature discontinuity at the surface. I think there’s a link to a free copy of a Ramanathan(?) paper that talks about that buried in there somewhere.
A. Vanags:
The article considering the “semi-gray atmosphere” is The Mystery of Tau – Miskolczi – Part Five – Equation Soufflé.
The article does include the mathematical derivation in the comments.
Unfortunately from your perspective the article is written around the confusion of Miskolczi rather than from the “introduction to grey atmospheres” approach. You might find some of it useful. However, trying to understand the Miskolczi approach is not recommended except for people who like to invest uncountable hours on stuff that is wrong.
As DeWitt Payne said, there is a link to a free Ramanathan paper. This is an excellent paper – Deductions from a simple climate model: factors governing surface temperature and atmospheric thermal structure, Weaver & Ramanathan, JGR (1995).
You could say that what they are trying to do is extend the models (which have analytical solutions) as far as possible – the semi-grey approximation is very limited – but even with their improvements, which provide good insight, their model is still extremely limited. Of course, they don’t claim to have overturned anything in atmospheric physics, only extended the range of a teaching model.
A. Vanags:
You might be thinking of the “effective radiating temperature” of the earth. This is simply the actual radiated flux (outgoing longwave radiation) converted into a blackbody temperature.
However, I recommend avoiding this altogether and just using the radiated flux. Armies of people who have learnt their climate science in blogs are convinced that the shorthand convention called “effective radiating temperature” which is widely used, actually means that climate science assumes everything radiates as a blackbody.
The radiated flux globally annually averaged = 239 W/m2 or thereabouts. If we made the heinous mistake of converting it into the convenient shorthand of “effective radiating temperature” it comes out to 255K.
So I’m not sure exactly what you are looking for.
Do you want to compare the radiated energy from the surface of the earth with the top of atmosphere value?
I’m in discussion with someone who doubts AGW and he is saying that satellites such as ERBE are showing that the amount of LWR radiation escaping the TOA has increased over the past 30 years. I’ve looked at the satellite data and they appear to show this increase, which seems to contradict this post. How is this so? Am I missing something or failing to interpret the data correctly?
I’m doing my best to understand all the science but I’ve less science background than most on a site like this. Can anyone give me a reasonably quick answer.
Without looking at the numbers myself, the first question you should ask is if the increase is significant considering the precision and accuracy of the measurements. You have different satellites measuring over relatively short time spans separated by ~15 years. Looking at TOA flux alone could be misleading as there are other factors that contribute like albedo variability. That’s not to mention that the conversion of the radiance measurements to TOA flux is still a work in progress.
Can you supply a link to the data you’re looking at?
Your right, it conflicts and this site is off base.
Wayne:
What data are you looking at? (Do you have a reference or a link?)
It’s important to understand the key principles at work as well.
Let’s say that solar radiation absorbed by the climate system hasn’t changed over 30 years. For now, it’s an assumption to illustrate a principle..
So if the OLR (outgoing longwave radiation = radiation emitted by the climate system) has increased then the earth has been cooling. And as a first approximation there are three ways the OLR could increase.
First, the opacity of the atmosphere has reduced – in simple terms, allowing surface radiation to escape more easily.
Second, the surface temperature is higher, therefore emitting more radiation
Third, the lapse rate (the temperature profile of the atmosphere) has changed, which has made the upper atmosphere hotter, therefore emitting more radiation to space
You can look at the second and third ideas as perhaps redistributing total heat within the climate system to allow it to be more effectively radiated to space.
Hopefully the above points make sense.
If the earth is warming we would either expect the cause to be a higher absorption of solar radiation or a reduction in OLR (or both). Of course, if we just considered the surface warming it might be a redistribution of heat from the oceans. But as best as we can tell, the ocean heat has also increased over the past 30 years.
And finally, it is also important to remember that even if the earth was not heating up or cooling down the earth would still not actually be in a radiative equilibrium – this is more a useful approximation. Over one year or one decade there will be a small imbalance in one direction or the other due to the “random” fluctuations of climate. So with surface temperature fluctuations from month to month and year to year we don’t expect to see a perfect flat line – or a monotonic trend.
Thanks for the quick and easy to understand reply.
“What data are you looking at? (Do you have a reference or a link?)”
The link I looked at was at the NOAA web site. You can create a NOAA Interpolated OLR GrADS image for different time periods. If I make one for the period of June 1974 to June 1980 it shows a map with areas divided by contour lines of OLR values. The minimum is 114.111 and the maximum is 290.586 w/sqm. If I do the same for June 2004 and June 2010 ( another 6 year period) the map has a max 302.207 and a minimum of 114.117. Upon closer inspection the various contour there doesn’t seem to be much difference in the values but it’s obviously a gross . What I’m wondering about is that maximum number.
The link to the site is: http://www.esrl.noaa.gov/psd/cgi-bin/DataAccess.pl?DB_dataset=NOAA+Interpolated+OLR&DB_variable=Outgoing+Longwave+Radiation&DB_statistic=Mean&DB_tid=30972&DB_did=24&DB_vid=905
I recognize that what I’ve done is likely a ridiculously simplified examination of the data, but I don’t have the expertise (or time) to learn or go through it all. It just seems if I create visualizations for different six year periods (no idea why I picked 6) as the years advance the maximum gets higher.
Wayne:
The NOAA website data may be “reanalysis data”.
For explanation of reanalysis, see the subheading Reanalysis – or Filling in the Blanks in Water Vapor Trends.
I’m not certain this data is reanalysis.. but if it is – reanalysis data, while very useful, can be problematic for assessing long term trends, especially over multiple instruments. Detecting long term small trends isn’t the main use or the strength of reanalysis.
Comments from other commenters on this dataset?
E.g. PSD Climate and Weather Data linked by the original page from Wayne.
There are probably some papers on long term OLR trends and this might be an interesting subject for an article.
I did start writing one a while ago on TOA energy balance but got bored with it, because a lot of the substance needs to be about error analysis. Hard to stay awake.
Here is the data (OLR) plotted from the CERES instrument. This covers the period from 2000 and has a much higher accuracy than the ERBE instrument:
You can plot data yourself by visiting the CERES data page.
From that webpage I could export the monthly data, so calculated the mean annual OLR from 2001 – 2010 (2000 excluded because data only available from Mar-Dec for that year).
As you can see, variability year to year, but no discernable trend over this 10 year period and definitely not an increasing trend.
This is about accuracy – considering that ERBE was followed by CERES..
From Toward Optimal Closure of the Earth’s Top-of-Atmosphere Radiation Budget, Loeb et al, Journal of Climate (2009)
Here is a comparison of various measurements between different measuring systems. One of the important points being that the absolute accuracy is not accurate enough to be used for trend analysis between ERBE and CERES:
Click for a larger image
And reviewing other papers, like Clouds and Earth Radiant Energy System (CERES), a review: Past, present and future, Smith et al, Advances in Space Research (2011), it seems that the two major satellites didn’t have coincident periods so that they can be compared for absolute accuracy.
SOD and DeWitt: thanks for the responses.
I think I broke through where I was stuck. I seems that if the transmissibility has the form e^-ko*rho*dL then if I define layers of the same mass (equipartition of the total mass) then I get the same equation form I had before, and the same answers for the temperature at the top of the atmosphere, at the bottom of the atmosphere and at the ground. The difference from a constant density atmosphere is that the altitude is different function and consequently the lapse rate is different.
Its mind-boggling. This implies that (at least for a grey atmosphere) it doesn’t matter what the distribution of the density is, or the temperature profile or the pressure profile. As long as the total mass is the same, the temperature at the ground will be the same since ko is constant. The temperature at the ground for a given Solar input is solely due to the total optical thickness not the temp or pressure profiles.
Thanks for the link to the Ramathan paper. So far its over my head and talks about improvements on the grey model. It provides some answers I can check mine against, it will take a (long) while for me to digest it. In retrospect, equipartitioning the mass like I did is the same as equipartitioning the total emissivity geometrically or equipartitioning tau, that’s probably why Ramathan integrates versus tau and not versus height.
“Do you want to compare the radiated energy from the surface of the earth with the top of atmosphere value?” Yes.
I was wondering if it could be roughly estimated from a spectrum like figure 2.9.
Yes, I know a grey model is only a crude approximation, its just I was impressed with Miskolczi, Goody and others that can represent a complicated thing like the atmosphere with a few global equations. I just thought it was cool. I have long ways to go, and way down the line I would like to try the effects of H2O and CO2 increasing the optical thickness.
I will move now to the ‘equation suffle’ posts as you suggested.
Understanding the interaction of radiation with the atmosphere is a difficult topic.
Like any new topic my recommendation would be to try and work through more than one textbook which explains the subject.
For myself I usually find one textbook leaves me confused on some points. I’m the kind of person that can’t quite make progress until all aspects are clear. A blessing or a curse? I’m not sure which. But multiple sources usually help to clear up the mysteries.
If you have the benefit of a university library I recommend seeing the cost of joining to be able to borrow books and then in their climate section find a book introducing atmospheric physics.
Then definitely recommended and worth its price many times over is A First Course in Atmospheric Radiation, by Grant Petty which you can find reviewed in Find Stuff Out and Book Reviews.
It can. This spectrum can be numerically integrated to calculate the TOA flux. The value is already known from the average of all of the satellite measurements of total flux, and is about 239 W/m2.
I have a question regarding emissivity. Given a climate sensitivity lambda (as defined in IPCC 2007) and the incoming radiant flux (say 236 W m^-2) , one can calculate the inferred effective emissivity from the Stephan-Boltzmann radiant balance law. Using lambda=.986 K m^2/W per the cenral value in IPCC 2007 and confirmed in many sources, the effective emissivity comes out to be only .564. I’ve seen some sources that seem to indicate this is correct (or at least it might be as low as .64) and some that say it should be close to 1. Can you shed any light on this subject? Thanks.
I can’t understand the proposed relationships between these numbers.
How are “effective emissivity”, λ, and the incoming absorbed flux related?
The Stephan-Boltzmann law defines a transfer function between flux and temperature. It’s derivative (dTe/dF evaluated at the equilibrium temp) then, defines its small-signal “gain”, equivalent in function to Hansen’s “no feedback sensitivity k, but superior in that it preserves the inverse relationship between k and Ts (or F), The derivative has e as a dependent variable so given k, (or equivalently λ and the total feedback factor b) and F, one can calculate e. Hansen and others have estimated k= .313 (also used in IPCC 2007) which implies e=.564
I used the term effect e because its value comes from the so called zero dimensional model. In a multi-layer atmosphere there is an e for each layer. But it strikes me as odd that the effective emissivity would come out so low, just 10% or so above what we’d expect from a planet perpetually shrouded in cloud cover.
Another odd thing is that if you use e=.564 to calculate the surface temp given Trenbeths energy budget, you get a very hot planet! e=1 gives the correct result. I’m trying to understand this apparent inconsistency.
Small problem, emissivity isn’t a variable. When you calculate Teff, ε ≡ 1 If it were a variable, then you have to use the chain rule to differentiate. But then you have an additional unknown, dε/dF, because you don’t actually know how ε depends on F. Even if your calculation were correct, you still can’t use that ’emissivity’ to calculate the surface temperature from the S-B equation.
ΔF=γS – εσTc4 …[1]
where ΔF = change in net flux at TOA
γ = proportion of solar radiation absorbed (=1-albedo)
S=solar
ε = “effective emissivity” of the climate system
σ = 5.67×10-8
Tc = temperature of the “effective climate system” – not as clear a value as people might like, due to the fact that some part of the climate system radiates from the surface, most from varying parts of the troposphere and a little from the stratosphere
And, defining climate sensitivity, λ – and assuming, for now, it is a constant:
λ = ΔF / ΔTs ….[2]
where Ts = temperature of the surface
How do we link together the variables you identified?
We can suppose there is a linear relationship between Ts and Tc if that helps. We should assume ε and γ are functions of temperature.
If we make the false assumption that Ts = Tc, then we can write:
ΔF=γS – εσTs4 …[3]
and differentiating with respect to Ts:
∂ΔF/∂Ts = S ∂γ/∂Ts – σTs4∂ε/∂Ts – 4εσTs3 …[4]
if we assume that ∂γ/∂Ts = 0, ∂ε/∂Ts = 0, and writing ΔF / ΔTs ≈ ∂ΔF/∂Ts we can equate [2] = [4] and so:
λ = 4εσTs3 and so:
ε = λ / 4σTs3 …[5]
Which is where I think you got your relationship from?
If we try to use the correct temperature in the equations.. Tc = a + bTs (ie a linear relationship) and say that b is not far off 1.. then Tc4 ≈ b4Ts4 then we are going to end up with eqn 5 becoming:
ε = λ / 4σb4Ts3 …[6]
[Note for readers who want to write equations – lots of tags as shown in Comments and Moderation – so after you post the comment and find the mistakes just add a comment and I will fix it up]
That kind of tihinkng shows you’re an expert
Actually the I used the zero dimensional model and solved for the boundary condition at the surface but I get almost the same answer using your approach which I like even better because it better models the atmosphere.
Zero net flux at the TOA requires:
εσTc4 + (1- ε)σTs4 = y S / 4 …. [1]
At the surface, assuming the atmosphere radiates equally up and down we have
y S / 4 + εσTc4 = σTs4 …… [2]
rearranging terms and adding to cancel the So terms:
2 ε σ Tc4 = ε σ Ts4 ==> Tc=Ts/(4th root of 2) ~ .84 Ts …[3]
That’s the linear relationship you were looking for (but I think you are missing a factor of 4 on the lhs). Substitute for Tc in [1] using [3] and solve for ε:
ε = 2- [y S/ (2 Ts4 σ) ] …….. [4]
Once we have ε we can calculate the theoretical sensitivity. First get the small signal equilibrium gain from differentiating d[Ts = ( F/(εσ))1/4]/dF and eliminating F by substituting F= εσTs4. partial derivatives not required, we’re assuming no feedback (but calculate K at the operating point it sees – critically important). This gives:
K = 1/(4εσ Ts3 ) …….[5]
then add the feedback to get the “closed loop” sensitivity:
λ = K/(1- b K) where b is the feedback “gain”. IPCC estimates b=2.16.
Alternately if you know λ and b you can solve for K and from K solve for ε.
Sorry, I tried to follow the directions for entering equations but it didn’t work. The numbers after the Ts should be exponents so Ts3= Ts^3 etc
[Pls check my changes in case I messed anything up – SoD]
I forgot to post the conclusion, Using your TOA approach I get ε=.774 and λ=.492. While setting K=.313 (λ =.966) and solving gives ε=.590).
Jeff Patterson,
Your equation 1 is correct assuming we change notation slightly from my earlier comment, including ε = emissivity of the atmosphere (rather than the whole climate system).
Equation 2 appears to be:
[Solar energy absorbed by the atmosphere and earth] + [Energy radiated downwards by the atmosphere] = Energy radiated upwards by the surface.
What is the basis for this relationship? Conservation of energy? If so, it is not correct.
The conservation of energy for the surface would be:
[Upwards radiation from the surface] + [Sensible and latent heat net from the surface to the atmosphere] = [Solar radiation absorbed by the surface] + [Downward radiation from the atmosphere absorbed by the surface].
In the case of my equations I allowed an assumption of a linear relationship between the surface and the atmospheric temperature simply to see what happens. I don’t believe there is such a relationship on theoretical grounds but it is probably one of the simplifications that should be made in a simple model – to see what we can learn from such a simplification.
You can see the results of some very simple assumptions in Simple Atmospheric Models – Part Two and in the preceding article to that.
SoD our
Thank you for your reply. Before I respond let me say that I’ve been perusing your site and it is by far the best on climate science I’ve seen and I’ve seen them all. So many want to grind their axes on the heads of those who simple have questions or who like to work things out for themselves. You answer all with patience and clarity. Well done and much appreciated.
Equation 2 comes from some old undergrad Thermo notes I dug up. I poked around last night and found a similar treatment in an online reference here which appears to be a college text book used at UT. See section 2.3.2. Note he takes e as absorption. Replacing e with 1-ε and rearranging a bit I can get both eqs 2 and 3 above.
Math aside, I’m trying to understand how Hansen and others arrive at what they call the no feedback sensitivity k (BTW you seem to have these terms inverted in your blog on climate sensitivity. λ and k should both have units K m^2/W). I think (but am not sure) that the missing terms you point to can be treated as feedback and so can be ignored for deriving k. In any case lets ignore them for now. The main thing I want to verify is that they derived k in situ. In non-linear dynamics it is critically important that when ascertaining the feed forward, small-signal gain, you do so at the operating point that gain will see at equilibrium. This is usually very tricky but if one simply breaks the feedback loop (heuristically or experimentally) as one might in a linear system analysis, you’ll be lead badly astray because by definition, the forward gain is a function of its input value in a non-linear system. In any case, at this point I’m just trying to build a simple mental model and checking my results against reality as I add complexity. I ran into this unexpectedly low value for ε and realized something’s off. Does the low effective ε surprise you? If we were a space observer calculating earth’s temp from its TOA near blackbody temp, I doubt one would estimate that you’d need to use ε=.54 to get the surface temp based on observation of our blue ball.
You can get those equations if you start with different assumptions.
The text you pointed me to uses energy balance and some very simplified assumptions:
– an atmosphere that is totally transparent to shortwave
– isothermal atmosphere
– no convection
(You can see similar derivations in the links I provided earlier). This model is used for illustration and conceptual understanding.
For the more real-world model we are discussing equation 2 is not correct.
The atmosphere absorbs some solar radiation. The atmosphere is not isothermal. Convection takes heat from the surface into the atmosphere.
For an approximation it turns out to work quite well.
We know that for a gray body:
Ts = (y S/(εσ))1/4 ………[1]
Assume y S is constant i.e. y S = Fo so that the temperature depends only on ε.
ε can thus be inferred from Ts:
ε = Fo/(σ Ts4) ……….. [2]
The no-feedback sensitivity for a grey body is
Kg = d Ts / d F = 1 / (4 σ ε Ts3) ….[3]
Eliminate ε by substitution from [2]
Kg = Ts/(4 Fo) …. [4]
Wow. That came out better than expected. For Ts=288 and Fo=239 Kg=0.3013, very close to the accepted value.
Note that the positive correlation of Kg with Ts does not imply runaway. [4] just gives the no feedback gain for small perturbations around a particular Ts. d(Kg)/dTs = 3/(Fo Ts) which shows the sensitivity decreases with Ts as expected.
Thanks for helping me noodle through this. The take away is that emissivity from a simple model does not translate well into the real world and indeep the planet has an effective emissivity of around .54
I did not read every word of this long discussion, but nowhere in what I did read did anyone deal with the heat sources within the Earth. Treatment of the Earth and its atmosphere as a “closed system” that seeks a balance temperature at which incomiing solar equals outgoing energy needs to account for the internal source.
It’s usually not discussed, because it’s both very small and essentially constant. The heat from the interior of the Earth is roughly 1/4000 of the incoming solar radiation. A change of this magnitude to the energy balance would not be insignificant, but as a constant contribution the effect is much smaller than other uncertainties in the constant parts of the energy balance.
I was discussing the global warming with someone that was skeptical of the science, and they directed me to the following post:
http://claesjohnson.blogspot.com/2015/03/a-basic-model-of-greenhouse-effect-with.html
to illustrate why CO2 in the atmosphere can never make much difference.
I read it over a few times, but I don’t follow what the author is trying to say with the following:
Radiative emission of 240 W/m2 from ground directly to outer space through “fully open atmospheric window” with emissivity 0.7 at 280 K = 7 C according to Stefan-Boltzmann 240=0.7×σ×2804 with σ=5.67×10−8Wm−2K−4.
We compare with present observation of ground temperature of 15 C with emission of 40 W/m2 directly from the ground to outer space, with thus “1/6-open atmospheric window”.
Extrapolation to fully closed window, then predicts a ground temperature of 15+ 8/5 C, say 17 C.
I found this article from the always great Science of Doom, which is much clearer and comes to a very different conclusion. Does anyone have any insight as to what is wrong with the conclusions above, if it is even decipherable.
Joe: It is worth remembering that Klaus is a mathematician, not a scientist. Mathematicians postulate a set of axioms and then make a series of logical deductions/theorems from them. Testing those theorems to see if they are consistent with observations is what differentiates science and the scientific method from mathematics. One can postulate that zero, one or many lines can be drawn through a point not on a line and parallel to that line. All three possibilities are valid fields of study and none was “invalidated” by experiment. General relativity came along much later demonstrate that non-Euclidean geometry had some utility – that it made useful predictions about how things behave. Given the century of increasingly sophisticated experiments that have been performed since the QM and relativity revolutions, there is little chance that Klaus’ mathematical postulates about radiation will be at the heart of another major revolution in physics. In my experience, Klaus never discusses situations where his postulates and established theories make different predictions and what experiments might be performed to decide which is correct. Challenge you friend to show Klaus applying the scientific method to determine whether conventional theory and his alternatives are correct.
Klaus has POSTULATED a model for earth without GHGs with the following properties and then makes some mathematical deductions based on these postulates:
1) Radiative absorption of 240 W/m2 out of incoming 340 W/m2 from the Sun with absorptivity 0.7 by Earth+atmosphere system.
2) Thermodynamic transfer of 240 W/m2 from Earth+atmosphere to Earth ground surface (without loss).
3) Radiative emission of 240 W/m2 from ground directly to outer space through “fully open atmospheric window” with emissivity 0.7 at 280 K = 7 C according to Stefan-Boltzmann 240=0.7×σ×2804 with σ=5.67×10−8Wm−2K−4.
Unfortunately, none of these postulates agree with observations: 1) SOD has pointed out that the emissivity of the earth’s surface is not 0.7. With the right assumptions, one can come up with an emissivity* for the earth/atmosphere system WITH GHGs of 0.7, but the emissivity of the surface with no interference from GHGs is well above 0.9. 2) 280 degK is simply the temperature needed to emit 240 W/m2 from a surface with an emissivity of 0.7, not a figure based on observations. 3) WIthout GHGs, we wouldn’t have clouds and the earth’s albedo (1-absorptivity) would be much lower. 4) No mechanism is given for step 2).
To be fair, those who claim that the GHE is 33 degK also make some dubious postulates when modeling an earth without GHG’s. Ignore dubious models. The surface emits an average of about 390 W/m2. About 240 W/m2 leaves the top of the atmosphere. That difference is caused by GHGs (colder than surface temperature) absorbing and emitting photons. If the difference were smaller, the surface of the earth would be cooler.
*The concept of emissivity can be tricky when applied to gases or transparent materials. When you are dealing with a homogeneous isothermal object and a single wavelength, emissivity is simply the ratio of observed emission to blackbody emission. Usually we integrate the output over all wavelengths and calculate emissivity in terms of the total power emitted compared with a blackbody at the same temperature. This definition is sometimes applied to the earth-atmosphere as a whole, ignoring the fact that it isn’t homogeneous and has temperatures ranging from 210 to 310 degK, nearly a 5X difference when you raise temperature to the fourth power.
Blackbody radiation is emitted by an isothermal object when absorption and emission of photons have come into a temperature-dependent equilibrium before leaving the object. Emissivity less than one is caused by reflection/scattering as radiation crosses the interface between two media, and the symmetry of this process results in Kirchhoff’s Law; absorptivity = emissivity at a given wavelength. Depending on wavelength, absorption cross-sections and density of GHG, radiation leaving our atmosphere may or may not reach equilibrium before leaving for space or the surface. When equilibrium isn’t reached, then the calculated emissivity of the gas depends on how much gas is present. This is also true for thin films of solids and liquids. Then simple ideas arising from blackbody concepts must be abandoned and changes in radiative flux are calculated using the Schwarzschild eqn (aka radiation transfer equation)
dI/ds = n*o*B(lamba,T) – n*o*I_0
Joe Yangtree,
The emissivity of the earth’s surface varies but is not 0.7. I assume that is what the statement you cited is saying.
The emissivity of the ocean, covering over 70% of the surface of the earth is about 0.96. See Emissivity of the Ocean.
So even if the emissivity of the land area was 0.0, which is isn’t, you would be almost at an average emissivity of 0.7.
The emissivity of other land surface areas varies – here is Hartmann (1994):
Various blogs are confused about the greenhouse effect.
Once these blogs finally figure out that the 240 W/m2 globally annually averaged radiation at top of atmosphere doesn’t match the surface temperature they invent surface emissivities to try and reconcile the observations with “no greenhouse effect”. I can point you to some hilarious blog articles if you like..
Claes “marketing genius” Johnson is in a little bit of a different category. He doesn’t believe in the photon. Radiation theory – as it has been understood for the last 100 years or so – is not correct. So he claims.
His marketing genius is to not explain this to his many admirers, who believe he has found a flaw specifically in climate science, rather than a flaw in basic radiation theory. I’m not particularly interested in debating whether basic physics is correct, as outlined at the end of the Etiquette.
If we accept into evidence that basic physics textbooks are correct, then the measurements of radiation with pyranometers and pyrgeometers are correct, the Stefan Boltzmann equation is correct, and measurements of emissivities are therefore meaningful, verifiable and can be demonstrated to be correct.
If you go back to Claes’ blog and bring up emissivities, he will then tell you that pyrgeometers have problems, and .. long story short.. eventually you will realize that you are debating a subject with someone who doesn’t believe a surface emits radiation and also absorbs radiation. There are no photons.. But please, the “no photon” thing.. we don’t want to discuss it here..
This statement is not quite right.
“This is the blackbody radiation at 255K (-18°C) with 100% “transmittance” through the atmosphere.”
If you have 100% transmittance you can have no clouds and clouds are responsible for a great deal of the albedo.
The real albedo would be 10 to 20%, meaning more absorbed power from the Sun, and the temperature would be 267K or -6 celcius.
Besides that i have a question of how clouds influences the effect of GHG’s.
Clouds have their own effect on the radiation balance, but they also shadows for the outgoing radiation (all else equal).
Svend,
There wouldn’t be clouds because, for the atmosphere to be transparent, you can’t have water vapor. No water vapor means no clouds. But you’re right, the albedo would change too. However, by convention, we assume that the albedo is still 0.30 for this sort of thing.
The effect of clouds is controversial. It’s quite clear that the distribution of clouds that exists now cause the surface to be cooler. They reflect more incoming radiation than they reduce radiation emitted. Note that’s reduced. Cloud tops emit LW radiation with an emissivity very close to 1. If you use MODTRAN, it shows that, for example, the tropical atmosphere with a clear sky emits 289.225 W/m² to space. Add Cumulus clouds and the emission only drops to 261.97 W/m². Cumulus clouds reflect a lot more than 10% of incoming solar radiation.
But some types of clouds, particularly high cirrus clouds, may increase surface temperature. Clouds of ice crystals can reflect LW IR. Using MODTRAN again with the tropical atmosphere but at the surface looking up, for clear sky, the surface sees 347.912 W/m² incoming LW IR. Add the NOAA cirrus model clouds and that increases to 350.11W/m².
Most climate models have clouds as a positive feedback long term. I don’t think that’s correct, but that’s just my opinion. The problem with models is compounded because models don’t do clouds very well at all. The grid resolution is nowhere near fine enough.
Thanks DeWitt Payne
I know that clouds are a big joker in the climate and global temperature, but my question was how clouds may change the outgoing LW radiation by simply shadowing it like clouds shadow the incomming sunshine.
In the so called window they reduce the amount of IR that goes out, like they shadow for the Sun.
The radiating effect of clouds itself must depend on the base temperature and top temperature, but that is independent of its shadowing effects for both sunshine and IR.
“MODTRAN again with the tropical atmosphere but at the surface looking up, for clear sky, the surface sees 347.912 W/m² incoming LW IR. Add the NOAA cirrus model clouds and that increases to 350.11W/m².”
Then i ask how much is the incomming sunshine reduced by cirrus clouds?
It is a question of dividing the effects to some parts where you can say something with confidence.
This link has some data on radiation seen from ground:
http://www.esrl.noaa.gov/gmd/grad/surfrad/
With the sky covered by clouds, there is very little radiative energy transfer from the ground to the clouds or vice versa. The energy transfer is by convection only. Tropical atmosphere cumulus clouds, surface looking up: 418.248 W/m², surface looking down: 417.306 W/m².
I’m not sure of the reflectivity of cirrus clouds for solar radiation or what percentage of the sky is covered in the different cirrus cloud models. But there is a setting in MODTRAN for sub-visual cirrus clouds that still increases DLR by 0.3 W/m².
Sorry for very late reply, but i had to think about it.
You say: “With the sky covered by clouds, there is very little radiative energy transfer from the ground to the clouds or vice versa. The energy transfer is by convection only.”
What then about the energy transfer in the atmospheric window?
It is anyway 141W/m2.
It is absorbed by the cloud and might heat the cloud, and based on the cloud base temperature it is radiated down as IR in a broader band than the 8 to 14um that was received.The same holds for the top. It radiates up based on the cloud top temperature, but in a broader band so that more are lost to the atmosphere.
The effect is that the cloud now sends some radiation back that else would have made it to TOA, and the radiation it sends up to TOA is lower because more is lost outside the 8 to 14um band.
Svend,
There is little radiative transfer in the thermal IR from surface to cloud because the temperature of the cloud base is very close to the temperature of the surface. That means the emission from the cloud base is the same as from the surface. Up equals down so the net radiative energy transfer is very close to zero. There is also less energy being absorbed by the surface from solar radiation because that is reduced by reflection of radiation in the visible wavelengths from the cloud top and absorption of near IR solar radiation by the clouds.
There is no window when there is no clear sky.
Could anyone tell me what the reference is for the “Outgoing Long Wavelength Radiation” plot by Taylor, 2005? (Fig 2.9 above) In particular I would like to know how high the satellite/satellites are. I have looked and looked for this.
Douglas,
A major source of recent data of this type is the CERES experiment on the Terra satellite orbiting at about 710 km. See
https://en.wikipedia.org/wiki/Terra_(satellite)
for orbital details.
https://en.wikipedia.org/wiki/Earth_Radiation_Budget_Satellite
An earlier source was the Earth Radiation Budget Experiment at 572-599 km:
https://en.wikipedia.org/wiki/Earth_Radiation_Budget_Satellite
I think the trick is knowing what the satellites are called.
Douglas (and Mike),
The source is a book, Elementary Climate Physics by Prof FW Taylor. I don’t own the book, I borrowed it from the university library, so I don’t have his source.
The CERES instrument doesn’t give a spectral graph like this – it has shortwave (0.3um – 5um), total radiation (0.3um – 100um) and the “atmospheric window” (8-12um). The shortwave will give you reflected solar radiation and the total minus shortwave will give you emitted longwave radiation.
AIRS is another modern instrument (on the same satellites as CERES) that does give a spectral breakdown – with the infrared spectrum in 2378 individual frequencies.
But the source for Taylor’s graph is likely one of the many experiments which has measured the spectra of terrestrial radiation, for example this one which you can find discussed on this site in Theory and Experiment – Atmospheric Radiation:
The Infrared Interferometer Experiment on Nimbus 3, Conrath et al (1970) is the source.
The paper is sadly behind a paywall, the abstract starts:
I don’t know much about the Nimbus satellites – wikipedia article.
You can see a number of measurements of OLR in Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Ten.
I can dig out more sources if you like.
.. but I can email the paper if you are interested. My email is scienceofdoom you know what goes here gmail.com. However, it is a lot about the technical aspects of the interferometer that allows for the measurements across the spectrum and overall very technical.
“The highlighted area at the bottom – the troposphere – is the area of interest. This is where most of the atmosphere (by molecules and mass) actually resides.”
It is very interesting Indeed. It shows how the surface only heats the air that is directly above it to the same temperature. The rest of the troposphere gets so little heat that the temperature drops by ~70 degrees in 10km. At the tropopause we see that the temperature is that of dry ice, co2, emitting according to a temperature of 210-220K, which is the temperature kept into the stratosphere where it starts to rise. What I see is how co2 is a “cold lock” for the troposphere. It looks as it is what keeps the temperature at that low level at high altitudes.
You can count all the photons you want, the spectrums and the gradient clearly show how co2 cools the atmosphere and keep it cool. Dry ice absorb heat like a boss. How did you come to the conclusion that adding a powerful heat absorber in a situation where thermal energy is constant and limited, will cause increasing temperatures? More cold molecules absorbing the same amount of heat means less heat per molecule, and that is the definition of lower temperature.
Reading this column brought into focus three apparently contradictory physics principles.
1. From this column, “Each layer of the atmosphere radiates according to its temperature.”
2. From one of your other columns (hoping, here, that my memory is accurate enough to skip the effort of tracking down the exact quote), a statement that a specific portion of the atmosphere can only radiate according to the extent that it contains GHGs, and then only according to the absorption lines of those GHGs.
3. From Raymond T. Pierrehumbert, regarding CO2, a statement that for the “transitions that are most significant in the thermal IR, the lifetimes tend to range from a few milliseconds to a few tenths of a second.”
and (expanding out here the scientific notation because I cannot do superscripts), “In contrast, the typical time between collisions for, say, a nitrogen-dominated atmosphere at a pressure of 10000 Pa and temperature of 250 K is well under 0.0000001 seconds. Therefore, the energy of the photon will almost always be assimilated by collisions into the general energy pool of the matter and establish a new Maxwell–Boltzmann distribution at a slightly higher temperature. That is how radiation heats matter in the LTE limit.”
Note: LTE is local thermodynamic equilibrium (LTE).
(Source: “Infrared radiation and planetary temperature”, Raymond T. Pierrehumbert, January 2011 Physics Today, American Institute of Physics.
Link– https://t.co/9rWJBi0MZ8 )
Barring a typo, it looks like Pierrehumbert’s statement applies up at around the Tropopause. Separately, in a lecture, William Happer has made a comparable statement for near surface conditions.
I realize that the CO2 situation might not apply to other GHGs, but CO2 is so significant to GHE theory that it deserves it’s own analysis.
So here is the dilemma:
IR is emitted at the surface. Some escapes to space, some is absorbed by CO2, some… etc. The IR energy absorbed by the CO2 serves to warm the surrounding packet of air by collisions between the energized CO2 and other air molecules, but the CO2 itself never has a chance to re-radiate a photon, at least anywhere near the surface.
So if CO2 were, in this thought experiment, the only GHG present, and if CO2 can never radiate for itself by emitting an IR photon, how is it possible for the layer of atmosphere containing that CO2 to radiate according to the temperature of that layer?
I do not doubt that radiation from that layer exists, BTW. The question is: How is it being generated, if not from CO2?
RationalClimate,
Assuming I understand your dilemma correctly..
The molecule of CO2 that aborbs radiation at some wavelength (lets say 15.00um) doesn’t re-emit that radiation immediately. Instead, the energy absorbed is transferred by collision to surrounding molecules, so increases the local temperature.
Emission of radiation at 15.00um is from CO2 (it could be from water vapor, but CO2 overwhelms at that wavelength).
Just not from the specific CO2 molecule that absorbed the radiation ( it’s possible, just a vanishingly small probability).
Hopefully that helps?
Yes, thanks. I can pin down my dilemma more precisely.
If it takes at least one millisecond for a CO2 molecule to “unwind”, so to speak, its mechanical stress in such a way as to release a photon, and during that millisecond a number of other molecules collide with it, does it ever actually complete that process, or, does each new collision amount to being the logical equivalent of pressing “restart” on a PC?
To use a baseball analogy: the pitcher is winding up to deliver a pitch to home plate when the opposing team starts throwing baseballs at the pitcher from their dugout. Does the pitcher ever get the pitch off?
Dewitt’s response following looks to address this question from a slightly different perspective so I will think about his response as well.
RC,
Collisional excitation. At LTE, the fraction of excited CO2 molecules depends only on the local temperature, not the incoming or outgoing radiation at that wavelength. Or, more properly, over the wavelength range of the excited molecule populationo, as there is a range of vibrational energy levels that can be excited to and emitted from, not to mention the broadening of the individual lines by the Doppler effect and pressure broadening. The intensity of the emitted radiation depends only on the number (fraction x number density) of excited molecules, not the lifetime of the excited state for an individual molecule. Btw, LTE applies at altitudes well above the tropopause, i.e. to about 100km.
If you really want to learn about atmospheric radiation transfer, then you should obtain a copy of A First Course in Atmospheric Radiation by Grant W. Petty. ( https://sundogpublishingstore.myshopify.com/products/a-first-course-in-atmospheric-radiation-g-w-petty ). It’s currently on sale for $36, which is only $2 higher than when I bought my copy many years ago.
Thanks much both for the explanation and for what looks to be an excellent reference. If I need to get too much deeper into this question I may well track it down.
Your explanation points to a different way for me to think about this, and also suggests several different conceptualizations.
The one that makes the most intuitive sense (but of course I would need to try to do a reasonableness check on the numbers of molecules) is that there are so many excited CO2 molecules in so many different energy states any one instant, that even though the probability of any *one* of them making it through the complete release “process” (as outlined in my above reply to SoD) is vanishingly small, there are so many of them that they simply swamp the small probability of individual success by collective overwhelming brute force, so to speak. I dunno.
But I like your temperature-based explanation (regardless of whatever the detailed emissions process underlying it turns out to be) because it does resolve how downwelling radiation can persist so well even at night.
FYI. I thought I had inadvertently deleted this post so I resubmitted it. That is why there is a duplicate further down. Apologies for that.
RationalClimate and friends: Exactly what do we mean when we say a group of molecules are in Local Thermodynamic Equilibrium?
Petty (p126) tells us that LTE exists in a group of molecules when they are exchanging kinetic energy with each other via molecular collisions much faster than they are exchanging energy with the radiation field or other source of energy. I – perhaps incorrectly – therefore conclude that LTE exists if and only if molecular collisions have produce a Boltzmann distribution of excited and ground states in a group of colliding molecules. Then I look at the “kinetic definition” of temperature – proportional to the mean kinetic energy of a group of rapidly colliding gas molecules – more precisely a group of molecules with an average kinetic energy of (3/2)kT, where k is the Boltzmann constant. And I go a step further and conclude that a system where LTE doesn’t exist, doesn’t have a “temperature” according to this definition. (There is a second definition of temperature that involves entropy.)
The semi-classical derivation of Planck’s Law postulates the existence of a Boltzmann distribution of energy between ground and excited states. The exp(-E/kT) term from the Boltzmann distribution appears in the form of exp(-hv/kT) or exp(-hc/lambda*kT) in Planck’s Law.
Petty tells us that LTE doesn’t exist in LED lights, lasers, and fluorescent lights. These devices manage to create excited states without relying on molecular collisions so they can emit visible light without being above 1000 K. These are systems where it isn’t clear that heat always flows from hot to cold – unless you assert they don’t have a properly-defined temperature. And LTE doesn’t exist above the stratosphere, a location where some scientists refer to a “Boltzmann temperature” (based on kinetic energy) and a Planck temperature (based on the rate of photon emission).
When Planck derived his law, he went one step further and postulated thermodynamic equilibrium between quantized oscillations (today molecules) and radiation. In other words, the rate of absorption of photons equals the temperature-dependent rate of emission of photons (and that temperature dependence arises from a Boltzmann distribution of excited states). The atmosphere doesn’t emit radiation with a blackbody spectrum, because temperature changes with altitude and radiation doesn’t always travel far enough or interact strongly enough with gas molecules for absorption and emission to reach equilibrium. This is why applying a black- or gray-body model to the atmosphere can be problematic. Most liquids and solids are much denser than the GHGs in the atmosphere, so absorption and emission are more likely to be in equilibrium at the local temperature, and these materials tend to emit radiation of near blackbody intensity.
When absorption and emission are not in equilibrium, Planck’s Law doesn’t apply and we must fall back on Schwarzschild’s equation for radiation transfer, which is what most climate scientists use to calculate radiative forcing and transfer. o is the cross-section for absorption of a photon and o*B(lambda,T) is the cross-section for emission of a photon – when LTE exists.
dI = n*o*B(lambda,T)*ds – n*o*I*ds
We are lucky that the only portion of our atmosphere that absorbs and emits a significant number of photons (the troposphere and some of the stratosphere) is in LTE, so radiative transfer calculations are relatively simple. And the radiation field in the atmosphere is weak enough that we can ignore stimulated emission. We are unlucky because our education system usually stops with Planck’s law, which provides an inadequate explanation for the behavior of radiation in the atmosphere.
That’s really helpful. The literature I run into often seems fuzzy or contradictory regarding what is happening in the atmosphere. Your explanation was clear, and helps highlight things to keep an eye out for when reading.
Reading this column brought into focus three apparently contradictory physics principles.
1. From this column, “Each layer of the atmosphere radiates according to its temperature.”
2. From one of your other columns (hoping, here, that my memory is accurate enough to skip the effort of tracking down the exact quote), a statement that a specific portion of the atmosphere can only radiate according to the extent that it contains GHGs, and then only according to the absorption lines of those GHGs.
3. From Raymond T. Pierrehumbert, regarding CO2, a statement that for the “transitions that are most significant in the thermal IR, the lifetimes tend to range from a few milliseconds to a few tenths of a second.”
and (expanding out, here, the scientific notation because I cannot do superscripts), “In contrast, the typical time between collisions for, say, a nitrogen-dominated atmosphere at a pressure of 10000 Pa and temperature of 250 K is well under 0.0000001 seconds. Therefore, the energy of the photon will almost always be assimilated by collisions into the general energy pool of the matter and establish a new Maxwell–Boltzmann distribution at a slightly higher temperature. That is how radiation heats matter in the LTE limit.”
Note: LTE is local thermodynamic equilibrium (LTE).
(Source: “Infrared radiation and planetary temperature”, Raymond T. Pierrehumbert, January 2011 Physics Today, American Institute of Physics.
Link– https://t.co/9rWJBi0MZ8 )
Barring a typo, it looks like Pierrehumbert’s statement applies to the atmosphere up around the Tropopause. Separately, in a lecture, William Happer made a comparable statement for near surface conditions.
Analysis
I realize that the CO2 situation might not apply to other GHGs, but CO2 is so significant to GHE theory that it deserves it’s own analysis.
So here is the dilemma presented by 1, 2, and 3 above:
IR is emitted at the surface. Some escapes to space, some is absorbed by CO2, some… etc. The IR energy absorbed by the CO2 serves to warm the surrounding packet of air by collisions between the energized CO2 and other air molecules, but the CO2 itself never has a chance to re-radiate a photon, at least anywhere near the surface.
So if CO2 were, in this thought experiment, the only GHG present, and if CO2 can never radiate for itself by emitting an IR photon, how is it possible for the layer of atmosphere containing that CO2 to radiate according to the temperature of that layer?
I do not doubt that radiation from that layer exists. The question is: How is it being generated, if not from CO2?
RationalClimate wrote: “IR is emitted at the surface. Some escapes to space, some is absorbed by CO2, some… etc. The IR energy absorbed by the CO2 serves to warm the surrounding packet of air by collisions between the energized CO2 and other air molecules, but the CO2 itself never has a chance to re-radiate a photon, at least anywhere near the surface.”
Since the surface behaves approximately as a blackbody, it emits all wavelengths of thermal IR according to the Planck’s Law and the SB equation. HOWEVER, GHGs in the atmosphere emit (and absorb) at MOST of the same thermal IR wavelengths as the surface. For every 10 photon emitted by the surface into the atmosphere, about 1 photon escape directly to space, perhaps 100 more photons are emitted by GHGs in the atmosphere, about 5 of these photons escape to space, 9 are absorbed by the surface (DLR) and 95 of photons from both sources are absorbed by the atmosphere. So the average outward flux at the surface is reduced from about 390 W/m2 near the surface to 240 W/m2 at the edge of space (0.6 photons for every one emitted by the surface). This 10:100 ratio may be inaccurate, but the principle that most photons in the atmosphere were emitted by the atmosphere and will be absorbed by the atmosphere is correct.
Although absorption of thermal IR photons locally heats the air where absorption occurs and emission of thermal IR photons locally cools the air emitting photons, the temperature in the troposphere is mostly controlled by upward convection of latent heat from the surface. In general, the troposphere loses more heat by emission than it gains by absorption. The difference is made up by convection of latent heat from below and release of that heat by condensation. Our atmosphere is too opaque to thermal IR for radiation alone to carry away all of the energy delivered to the surface by SWR. Convection carries the remaining heat away from the surface and through the troposphere – leaving behind a lapse rate that must average about 6.5 K/kilometer in much of the troposphere – regardless of absorption and emission. If you noticed above, I said that the atmosphere emits 100 photons and only absorbs only 95. The energy lost by the other 5 photons is provided by convection.
By the tropopause, the atmosphere has thinned and become more transparent to thermal IR. Above the tropopause, radiative cooling to space balances incoming SWR and temperature is determined by absorption and emission thermal IR (and the absorption of solar UV by ozone).
RC wrote: “So if CO2 were, in this thought experiment, the only GHG present, and if CO2 can never radiate for itself by emitting an IR photon, how is it possible for the layer of atmosphere containing that CO2 to radiate according to the temperature of that layer?”
You are forgetting that 99.9+% of the GHGs in the atmosphere that are in an excited state (and capable of emitting a thermal IR photon) were excited by MOLECULAR COLLISIONS. And 99.9+% of those excited states are relaxed by molecular collisions before they can emit a thermal IR photon too. The FRACTION of GHGs that are in an excited state at any time depends on the kinetic energy delivered by collisions (temperature), not the negligible amount of excitation and relaxation provided by absorption and emission. This temperature-dependent FRACTION of excited GHG molecules, the total number of GHG molecules and the average time it takes for an excited state to emit a photon determine the overall rate of emission. That fraction is determined a Boltzmann distribution of energy among ground and excited states. When we say LTE exists, we mean the fraction of excited states deltaE above ground state is determined by a Boltzmann exp(-deltaE/kT) term, not by absorption and emission. Planck’s Law contains an exp(-hv/kT) term because it was derived by assuming a Boltzmann distribution of excited and ground states.
This is why RC faced this dilemma: “So if CO2 were, in this thought experiment, the only GHG present, and if CO2 can never radiate for itself by emitting an IR photon, how is it possible for the layer of atmosphere containing that CO2 to radiate according to the temperature of that layer?”
CO2 emits thermal IR photons because CO2 molecules are excited by collisions. Occasionally an excited state emits a photon. Occasionally a ground state absorbs a photon. But the vast majority of the transitions between ground and excited states of CO2 involve collisions, not photons. The phrase “local thermodynamic equilibrium” means a Boltzmann distribution of ground and excited states exists, a distribution that is not significantly perturbed by absorption or emission of radiation.
RC also wrote: “Each layer of the atmosphere radiates according to its temperature.”
It helps to add some information to this statement. The radiative flux TRAVELING THROUGH a layer of atmosphere depends on how much radiation enters the layer, how much is lost by absorption, and now much it is added by emission. Emission depends only on the temperature of the layer (not how many photons it absorbs).
RC wrote: “From one of your other columns (hoping, here, that my memory is accurate enough to skip the effort of tracking down the exact quote), a statement that a specific portion of the atmosphere can only radiate according to the extent that it contains GHGs, and then only according to the absorption lines of those GHGs.”
Both absorption and emission at a particular wavelength depend on the density of GHG molecules present (n) and the “propensity” of those GHG molecules to interact with photons of that particular wavelength. “Propensity” is quantified in terms of a “cross-section” (o, area/molecule) that applies to BOTH absorption and emission. The wavelengths of thermal IR that are most strongly absorbed by a GHG are exactly the same wavelengths most often emitted by the GHG. The product n*o is important to both absorption and emission. The only real difference between absorption and emission is that amount of radiation absorbed by a layer of atmosphere is proportional to the intensity (power) of the radiation entering the layer and emission from the layer is proportional to the Planck function B(lambda,T), roughly T^4.
Frank,
IMO, Rational Climate is not posting here to actually learn something. He’s a tr011 like a person who was banned here whose name started with Br. I understand that you feel you should correct his errors. For example, he has already been told at least once about collisional excitation. Doing it again and again won’t do any good. Don’t feed him.
DeWitt is completely incorrect and his ccomment is insulting.
Anyone reading my posts, and I’m relatively new to Science of Doom so there aren’t many yet, can clearly see that I am personally struggling to come to a good understanding of climate factors, and at the same time to try to think them through and point out any inconsistencies I find.
Also, my posts are technical in nature. I analyzed several articles that had been suggested within these comments and pointed out a definitive omission in each that undermined the rationale being presented in them.
SoD — Unless I missed it, DeWitt never posted any objections to the technical points I had raised concerning those articles, which are completely on solid ground as far as I know, and yet he posts something like this?? This kind of demeaning personal comment should be unacceptable and out of bounds on a high quality blog like this one.
RC: DeWitt is usually accurate in his assessment of trolls (who have wasted lots of our time here at SOD). And he is more accurate and generous with his scientific expertise. Years ago, he performed the radiative transfer calculations that you liked showing the large number of photons emitted and absorbed by the atmosphere compared with the number entering and leaving the atmosphere. And if I made any serious mistakes in my comments (which has certainly happened in the past), he has generously pointed them out.
Frank,
Thanks for your comments. In this case I can assure you that he has made a mistake, at least as I understand the meaning of that term:
“In Internet slang, a troll is a person who starts quarrels or upsets people on the Internet to distract and sow discord by posting inflammatory and digressive, extraneous, or off-topic messages in an online community with the intent of provoking readers into displaying emotional responses and normalizing tangential discussion, whether for the troll’s amusement or a specific gain. But Wikipedia.”
I have run into people like that. I don’t like them. I would never consciously do that. Where I have questions they are sincere, and on topic. If I return to a topic a second time because I still have a question, I always try to reflect what I have learned up to that point and scope down the question to a narrower sub-topic. If I return to a topic because I either don’t understand a point or because I see an unconvincing aspect to the argument being advanced, I always try to explain more specifically what the issue is so the dialogue can cycle in toward a point of closure rather devolve into an endless circular argument. I believe every post I have made adheres to these general principles of good practice.
The trolls I have run into, and have been bothered by, don’t do that. They tend to repeat the same thing over and over again, while ignoring any explanations given as to why their point cannot be completely correct.
DeWitt’s technical comments to me have been good, and I have taken them seriously. I have either folded them into my thinking or pointed out, in response, any aspects that seemed unconvincing, with an explanation as to why.
I have only been discussing climate on the internet for about six months, after some years of independent reading. One thing I ran into during that time was Nikolov and Zeller’s work, which impressed me. I don’t know if it is right, but it seemed to me that there was enough substance to it that it would merit a serious discussion. I naively assumed that I would find that within the internet discussions. And there is some of that, make no mistake. But mainly what I found was a concerted effort to ridicule N&Z’s work, and to similarly ridicule anyone who wanted to have the science behind it given some serious consideration. Such individuals are often beset by actual trolls, and part of what those real trolls have been doing is to accuse people who are not fully on board with the current greenhouse effect theory of being … trolls.
This is an insidious problem. It is absolutely hampering an honest scientific discussion of whether or not there is any validity either to Nikolov and Zeller’s work or to several other alternative theories that have been advanced. I strongly value having open and constructive scientific discussions concerning the multiple serious alternative theories for explaining climate change that now exist.
I hope this explanation helps clarify where I have been coming from intellectually; why I find highly valuable, and take very seriously, discussion forums like this one; and why I would never knowingly abuse them.
What an outstanding explanation. Wow. Thank you.
You explained things that I instinctively believed must somehow be true, but for which I did not understand exactly *how* they would work out to be true. I also learned some things I would not previously have known to even ask about. And lastly, you provided some actual numbers regarding the distribution of photons. I know they were only meant to be an approximation, but that was very helpful for understanding your description.
It will take me some days to digest this and work it into the rest of what I think I know. The convection explanation in particular puts a lot of other things into perspective, as does the final sentence.
I *greatly* appreciate your efforts here.
RC: Many people have difficulty moving from the macroscopic world to the microscopic world of molecules and photons. Mastering the microscopic world is rarely necessary. Grant Petty, the author of a useful and inexpensive text , “A First Course in Atmospheric Radiation”, spends 30 pages reviewing the physics of light as electromagnetic waves and 2 pages discussing light as photons. He concludes: “The quantized nature of radiation comes to the forefront when considering absorption and emission by INDIVIDUAL molecules. Finally, for calculations of large scale transport of radiation in the atmosphere, the effects of both types of interactions [waves or photons] will have been deeply buried in some generic extinction and scattering coefficients, and can be conveniently put out of your mind altogether.” In other words, you can simply rely on Schwarzschild’s equation for radiation transfer (see 2/27/2020 comment), which I first learned about here and recently wrote about for Wikipedia. The product of density and absorption cross-section (n*o) discussed above is an absorption coefficient, n*o*B(lambda,T) is an emission coefficient, and n*o*s is optical depth (aka optical density and similar to absorbance in chemistry. s is distance radiation travels). Unfortunately, most people simply refer to “radiative transfer calculations” and never mention what equations are used and how they are done. For example, this online resource for performing such calculations (Modtran) is certainly using some variety of Schwarzschild’s equation.
http://climatemodels.uchicago.edu/modtran/
The biggest problem occurs when skeptics claim that the Second Law of Thermodynamics (2LoT) prevents thermal IR photons emitted by GHGs in the cooler atmosphere from being absorbed by the warmer surface. Individual molecules and photons are not constrained by 2LoT; they obey the laws of quantum mechanics! When applied to large groups of molecules and photons, the laws of quantum mechanics cause the NET FLUX of radiation between two large groups of colliding molecules to flow from the warmer to cooler group. Individual molecules have a kinetic energy (which is changed by collision about 10^9 times per second), but they don’t have a temperature (a stable property that is proportional to the mean kinetic energy of the molecules in a group).
SOD has some really good posts about lapse rate and the “radiative-convective equilibrium” that controls temperature in the troposphere.
https://scienceofdoom.com/2010/02/10/co2-–-an-insignificant-trace-gas-part-five/ especially Figure 2.9.
https://scienceofdoom.com/2012/02/09/density-stability-and-motion-in-fluids/
Below hopefully is a classic figure showing that the surface of our planet would be about 340K without convection. Where convection controls temperature, the temperature varies roughly linearly with altitude. Where radiation controls temperature, temperature of a homogenous atmosphere would vary linearly with pressure. (There is a constant lapse rate of about 10 K/km on Venus, so we can deduce convection is controlling temperature there.)
The more I re-read your last two atmospheric explanations the better I like them. The concept of an energized atmosphere has already begun influencing how I interpret what I read. The different elements of your descriptions fit together hand-in-glove, from the micro level to the macro level. To put it simply: those explanations have the ring of truth to them. Thanks again.
Just FYI: After reading the first of these two explanations I came across a column that presented a very similar view of atmospheric mechanics, although from a more macro level. It also referenced MODTRAN within its explanation.
I think there are some small differences, which I am working to understand, and maybe some that I have missed. Here is the link:
Dr. Joachim Dengler:
Physics of the green house effect
http://klima-fakten.net/?page_id=1245〈=e
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