A series providing more explanation of how the atmosphere absorbs and emits radiation, including a simple model to provide insight. The model uses fictitious molecules pH2O and pCO2 (which have only a passing resemblance to the real molecules) to demonstrate some key points. Part Six even explains the real equations used.
Part One – a bit of a re-introduction to the subject.
Part Two – introducing a simple model, with molecules pH2O and pCO2 to demonstrate some basic effects in the atmosphere. This part – absorption only.
Part Three – the simple model extended to emission and absorption, showing what a difference an emitting atmosphere makes. Also very easy to see that the “IPCC logarithmic graph” is not at odds with the Beer-Lambert law.
Part Four – the effect of changing lapse rates (atmospheric temperature profile) and of overlapping the pH2O and pCO2 bands. Why surface radiation is not a mirror image of top of atmosphere radiation.
Part Five – a bit of a wrap up so far as well as an explanation of how the stratospheric temperature profile can affect “saturation”
Part Six – The Equations – the equations of radiative transfer including the plane parallel assumption and it’s nothing to do with blackbodies
Part Seven – changing the shape of the pCO2 band to see how it affects “saturation” – the wings of the band pick up the slack, in a manner of speaking
Part Eight – interesting actual absorption values of CO2 in the atmosphere from Grant Petty’s book
Part Nine – calculations of CO2 transmittance vs wavelength in the atmosphere using the 300,000 absorption lines from the HITRAN database
Part Ten – spectral measurements of radiation from the surface looking up, and from 20km up looking down, in a variety of locations, along with explanations of the characteristics
Part Eleven – Heating Rates - the heating and cooling effect of different “greenhouse” gases at different heights in the atmosphere
Part Twelve – The Curve of Growth - how absorptance increases as path length (or mass of molecules in the path) increases, and how much effect is from the “far wings” of the individual CO2 lines compared with the weaker CO2 lines
And Also -
Theory and Experiment – Atmospheric Radiation – real values of total flux and spectra compared with the theory.
[...] [...]
[...] [...]
[...] of radiation (OLR = outgoing longwave radiation) from the climate system into space. See the Atmospheric Radiation and the “Greenhouse” Effect series for detailed explanations of [...]
Scienceofdoom:
I’ve been in a bit of a tussle over at WUWT with a few people. I’ve referred them to my blog. One particular page is this one:
http://johneggert.wordpress.com/2011/09/24/how-well-do-you-understand-radiant-heat-transfer/
Could you comment on the question itself. Is it clear? I thought it was, but I value your input.
Cheers!
JE
John,
It seems clear to me and a very well designed question. I guess you should say infinite plates (to avoid questions about edge effects) and that there is no gravitational field.
Perhaps I can post an article here to ask for answers to your question?
I’m not sure of the answer myself without spending some time on it. First off, I thought that the temperature profile vs distance between the two plates (prior to CO2 being introduced) would be linear, now I am wondering. The heat transfer mechanism for the gas is conduction but as one side is hotter the density must increase, due to the ideal gas law, and this would change the conductivity.
The problem (as CO2 is introduced) is simplified vs the real atmosphere as the pressure is constant, which avoids the complexities of Doppler vs collisional broadening. Still, a great question is one where the problem is clear but you hesitate before trying to work out the answer.
What was the WUWT article where people commented on your blog?
SOD:
Sorry I don’t have a short link.
http://wattsupwiththat.com/2012/11/13/counter-programming-to-al-gores-dirty-weather-report-will-be-on-wuwt-tv-live-starting-wednesday-nov-14-at-8pm-est/
I had a quibble with the cartoon at the top. The statement “There is no observational / empirical evidence for how much extra CO2 raises the global temperature” is counter to what I was taught in my university heat transfer course. My professor showed us how it could be done though the focus of the course was on metallurgical applications. At WUWT, I gave an example of a blast furnace where we calculate the increase in temperature of intervening gases to complete the energy balance. This is pretty clear empirical / observational evidence.
Cheers
JE
By the way, if you use the path length approximation and the emissivity curves in my metallurgical textbook as RTE approximations the plotted curve is approximately logarithmic. If you use the emissivity curves in my heat transfer text book it is more closely approximated by an hyperbola. The tail at high concentrations of CO2 is nearly parrallel to the concentration axis. That is at high concentrations of CO2 the effect of increased CO2 goes nearly to 0. As we have discussed in the past, this may not be relevant to the atmosphere as it is at constant pressure and the emissivity curves stop a 0C, hence in applying to the atmosphere one is going beyond the range of RTE’s used to generate the emissivity curves.
I wrote that post a while back. My wife packed all my school notes and put them into storage, so I don’t have quick access to them. I’ll see if I can find them and send you a link to the answer.
The metallurgical text is below. It may be out of print.
Schumann, Reinhardt, Metallurgical Engineering, Volume 1, Addison-Wesley,
The heat transfer text is:
Bejan, Adrian; Kraus, Allan D. Heat Transfer Handbook. John Wiley & Sons
John Eggert,
You dismissed Frank’s valid criticism on the path length approximation on your site. That led you to make wrong conclusions on the effect of additional CO2. There’s another factor that adds to your error and that’s due to the fact that it’s enough for the warming influence of CO2 that the penetration changes in the upper part of the troposphere.
Then to your question. As you stated the details, there’s no energy transfer mechanism between the plates and the gas with 0% CO2. Thus in that case the heat transfer between the plates proceeds according to the Stephan-Boltzmann law but the gas may have any arbitrary temperature.
With a very small concentration of CO2 the gas gets essentially isothermal with the temperature ((300^4+350^4)/2)^0.25 = 327.8K because radiation from both plates affects equally the whole gap. (1% CO2 is already more than “very small”).
With very high concentration the temperature at each plate is close to the temperature of the plate and changes linearly in mass between these values. Linear in mass is not linear in distance due to the influence of temperature on the density. The heat transfer between the plates is reduced but not very much because the gas is transparent to most wavelengths.
The intermediate cases are – intermediate.
Pekka,
With zero CO2 the gas and the plates interact via conduction as the only heat transfer mechanism. It will take a long time, but in the end there will be an approximately linear temperature profile with distance. I say approximately in case the temperature difference changes the conductivity of the gas.
I don’t disagree on that, but I took literally the total absence of all other energy transfer than radiative even on his point where it’s clearly unphysical.
Having the warmer plate at the top convection does not enter. Thus including only conduction is a valid alternative and that does lead slowly to a similar temperature profile as high concentration of CO2, but a low concentration of CO2 would still result in almost isothermal gas.
I had a stab at this using my atmospheric model just designed. I haven’t checked my butchered up version for your example because I don’t have a reference point (unlike for the atmosphere).
Here’s what I posted on your blog page:
John,
I created a radiative transfer model which you can see at Visualizing Atmospheric Radiation – Part Two and the code at Part Five.
I remembered this example here and quickly butchered up the code to make 2 plates 20 meters apart with fixed temperature (rather than an atmosphere with very different conditions).
I have calculated a few things.
With no CO2 of course the energy supplied to plate 1 (the hotter plate) at 350K is 392 W/m^2 – this doesn’t need an RTE model.
At 1% – energy supplied to plate 1= 364 W/m2
2% – 359
5% – 351
10% – 344
20% – 334
50% – 316
At 1%, the temperatures stablize at (5, 15, 25, 35, 45m): 340 334 327 321 313
50% the temperatures stablize at (5, 15, 25, 35, 45m): 340 334 328 322 315
I have spectra as well for these different conditions.
Does this match up with what you have calculated?
Are the land and the oceans warmed only by the infrared component of sunlight, or does visible light play a significant part too ?
Short answer: Yes. If the light is absorbed, energy is transferred. If it’s reflected it isn’t. Light perpendicular to a clear water surface is mostly absorbed if the water is deep enough. Some short wavelength light is scattered by water more than longer wavelengths, which is what makes the ocean look blue. That’s also true for the sky.
DeWitt,
Just to be clear though, is the warming by visible light significant, ie not just a few percent of the total warming? Would appreciate some references too if you have some to hand. tvm.
Montalbano,
All solar heating systems depend on the heating effect of the radiation. The efficiency of such systems has been calculated using spectral data. The results agree well with standard understanding. The share of visible light is a little less than 50% while all heating is due radiation with wavelengths less than 3µm.
You can find related data from http://rredc.nrel.gov/solar/spectra/am1.5/
Montalbano
Transparency of Pure Water in the Visible Range.
There is a rather nice explanation and graph in this link
http://hyperphysics.phy-astr.gsu.edu/Hbase/chemical/watabs.html
These wavelengths have no physical link to electronic transition levels in pure water.
Any absorption of light is caused by particulates in the water for example sea water phytoplankton, seaweed and inorganic particulates.
The container of the water for example sea and pond bed may also absorb in the visible spectrum.
Bryan,
A source that presents more detailed data both for water vapor and for pure liquid water is
http://www.lsbu.ac.uk/water/vibrat.html
This curve as that you linked is for pure water without any impurities. A liquid can absorb at all wavelengths, it’s not restricted by the set of discrete excited states as a gas is when the pressure is not very high.
Pekka Pirilä
I think that your link (although with much more detail) does not contradict the HyperPhysics link.
You link says;
“Water is almost perfectly transparent to ‘visible’ light, a property which is made good use of by photosynthesis and allowing production of both biomass and oxygen.”
As a general statement the HyperPhysics link is accurate enough for most purposes.
Bryan,
They are not contradictory.
The link I gave is useful as it has a absolute vertical scale and it’s expanded enough to tell some interesting details like:
- blue light has a penetration depth of about 100 m, i.e. about a quarter of blue light penetrates more than 100 m in pure water
- the penetration depth of red light is only about 1 m
- near IR penetrates typically a fraction of one meter
- LW IR penetration depth is around 10 µm. That’s still enough to get some 1000 molecular layers deep into water.
All this is for pure water but the penetration depths are so long that the impurities really matter only for the short wavelength part of the visible spectrum.
The main conclusions are
- Most of the warming of solar radiation happens in the topmost meter or two.
- A significant fraction of warming happens even so at a depth of several meters and up to a few tens of meters. This fraction is significant because it causes warming at those depths and almost all that heat must escape through the surface. I.e. the net energy flux by convection and mixing is upwards and there must be times during which the layers at that depth are warmer than the layers above.
- Practically all interaction of radiation with water occurs deep enough to prevent direct transfer of energy from absorption of radiation to evaporation. I.e., the primary consequence of absorption is an addition to local thermal energy of water. That raises the water temperature and the rest follows as consequence of this increase in the temperature.
Bryan,
You’ve obviously never been scuba diving. The red end of the visible spectrum starts to go away almost immediately. The blue end can penetrate 100m or more. A flash is required even at moderate depths for photography in true colors.
Even the hyperphysics chart doesn’t show water in the visible band as perfectly transparent. It’s orders of magnitude more transparent than at longer and shorter wavelengths, but it’s not perfectly transparent anywhere. What you’re getting is absorption from the tails of the bands at longer and shorter wavelengths. The absorption cross section is small, but not zero.
DeWitt Payne and Pekka
I see nothing wrong with the HyperPhysics link as a general statement.
“narrow window of transparency which includes the visible spectrum.”
Pekka’s link gives an even stronger statement
“Water is almost perfectly transparent to ‘visible’ light, a property which is made good use of by photosynthesis and allowing production of both biomass and oxygen.”
Attenuation is not absorption.
There are strong scattering events particularly for the blue end.
Pekka’s link also gives minority processes such as overtone absorption.
These are third or fourth order events and hardly account for much of the energy of the visible spectrum.
Pekka’s link gives a much more likely destination for a significant fraction of the energy.
“photosynthesis and allowing production of both biomass and oxygen.”
Of course most inorganic particulates and sea beds will thermalise the energy.
Bryan,
That should be ‘ not necessarily absorption’ The general term is extinction. Extinction is due to scattering or absorption. Period. In this case, though, it’s definitely absorption.
If it were scattering instead of absorption, the blue end of the spectrum would penetrate less deeply than the red end. But of course it’s not. Rayleigh scattering is wavelength dependent. Scattering from particles much larger than the wavelength of the scattered light is Mie scattering and is largely wavelength independent. That’s why the sky is blue and clouds are white. The water droplets in clouds are too small to absorb enough of the red end of the visible spectrum to affect the color. But they absorb strongly in the IR.
http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html
If you have access to a swimming pool, take something red that doesn’t float and throw it into the deep end. It won’t look red when it’s on the bottom.
Bryan,
Seriously, read Grant Petty’s book. It’s cheap if bought directly from the publisher. And read it, or read it again if you already have it, with the attitude that he knows more than you, because he does.
Not only is the sky blue, but the sun looks red when it’s low on the horizon. That’s mainly the Tyndall effect from scattering from particles on the same order of magnitude as the light wavelength. Wanna bet that the sun looks red looking up from about 10m down in clear water and gets redder as you go deeper?
DeWitt Payne
You said above
“Some short wavelength light is scattered by water more than longer wavelengths, which is what makes the ocean look blue. That’s also true for the sky.”
Now apparently your not so sure.
Blue light is not absorbed by pure water.
This is standard physics as documented by the two links above.
This fact unfortunately does not fit well with IPCC ‘science’ which assumes 100% thermalisation of solar radiation entering water.
No photosphere or photochemistry for Earths water in their opinion.
Given your chemistry background you should have no difficulty in identifying the electronic transition involved in a photon of blue light being absorbed by a molecule of water.
There are four possibilities for light that has entered water.
The first (and dominating far) possibility is that it gets absorbed and that the energy goes immediately to heat the water.
The second possibility is that it’s absorbed by green plants and used to build organic material. The lifetime of organic material is mostly very short in oceans and most of that energy ends soon up as heat again. A very small part sinks to the bottom as organic material.
The third possibility is that the light penetrates all the way to bottom and warms first the bottom, but that heat ends almost totally in the water again.
The only case where the energy does not stay in the water is that where the light gets reflected or scattered back out of the ocean. The share of that is very small excluding that light that gets reflected already before it has really entered the ocean at all.
Bryan,
Your ability to selectively ignore anything that contradicts you and pull out something that might confirm your point is truly amazing.
No it isn’t. Apparently you are incapable of reading charts and graphs. The absorption coefficient of liquid water NEVER goes to zero either in your hyperphysics link or Pekka’s link. In Pekka’s link there is a table of absorption coefficients at six wavelengths in the visible. The transition at 401nm or 24940 cm-1 is aν1 + bν3; a+b=8. You can look up the rest yourself. Eyeballing the graph of the log of the absorption coefficient vs the log of the wavelength, it looks like the value at 400 nm is about 5E-5 cm-1 or a characteristic depth of 200m.
From Pekka’s link:
[my emphasis]
Does the word ‘almost’ mean anything to you? Or do you just blank out any adjective that interferes with your ability to fool yourself.
Make that a table of transitions rather than absorption coefficients.
Bryan,
You might also try a little light reading here:
http://repository.tamu.edu/bitstream/handle/1969.1/85959/Wang.pdf?…
It’s a Ph.D dissertation titled:
MEASURING OPTICAL ABSORPTION COEFFICIENT
OF PURE WATER IN UV USING THE
INTEGRATING CAVITY ABSORPTION METER
If you don’t want to read the whole thing, look at figure 6.9 on page 87
Obviously, highly pure water does absorb in the near UV and Visible spectrum. It’s been measured by several different people. Well, probably not obviously to you as I have complete faith that you will be able to figure out a way to misinterpret this too and declare yourself right and everyone else wrong.
DeWitt Payne
None of the material you have provided contradicts the two almost identical conclusions above that……..
“Water is almost perfectly transparent to ‘visible’ light, a property which is made good use of by photosynthesis and allowing production of both biomass and oxygen.”
To quantify it further your Ph.D dissertation link has Fig 1.3 showing
Absorption Coefficient UV / Absorption Coefficient blue light = 10^10
Absorption Coefficient IR / Absorption Coefficient blue light = 10^8
I think that this sets the ‘almost’ into context.
Now these results hardly support the IPCC position of 100% thermalisation of Solar radiation interaction with water.
But that is the reality!
How often have we heard a similar generalisation
The atmosphere is almost transparent to short wave Solar radiation.
Now because this is a major part of the greenhouse theory its OK.
No effort is devoted to quantifying the ‘almost’ on this occasion.
So what happens to light energy on entering natural water as found in seas and ponds?
Apparently a large part of it is involved in photosynthesis
http://www.cen.ulaval.ca/warwickvincent/PDFfiles/137.pdf
Why is that important (apart from the obvious, essential for life reason)?
Energy is stored for periods > one day.
This is not part of the energy budget of the IPCC
The first law does not allow the creation of energy, so no 100% thermalisation.
Bryan,
Neither DeWitt Payne nor I has argued against the importance of organic matter in attenuating UV, blue light or green light in natural water. We have only told that the the two links are on pure water and that even in pure water the attenuation is rather strong for all wavelengths. Your later link on the influence of the organic matter is about another issue.
The importance of the photosynthesis in the ocean energy balance can be estimated from the total primary production of oceans. The estimate for that is about 50 billion tons of carbon per year (see the presentation of Behrenfeld from the Ocean Productivity pages
http://www.science.oregonstate.edu/ocean.productivity/presentations.php
The chemical energy content of biomass is roughly 40MJ/kg(carbon). From these numbers we can calculate the average power of biomass production as 63TW. The average solar energy that enters the oceans is approx. 90000TW. Thus photosynthesis takes only 0.07% of the solar energy that enters the oceans. This is far too little to justify your conclusions – and even less so because the lifetime of that organic material is short.
I knew you could do it! And as usual, you modify your position slightly while never admitting that your original contention was wrong and then move the goalposts. You do realize that only a small fraction of the solar energy absorbed by plant life for photosynthesis is converted to chemical energy. The rest ends up as thermal energy. And there really isn’t all that much plant life in the ocean on a concentration basis. If Wikipedia is to be believed, there are only 1-2 gigatons of carbon in the aquatic biosphere compared to 37,400 gigatons of inorganic carbon. Inorganic carbon is mostly in the form of bicarbonate ion which has a concentration in seawater of 145 mg/kg.
But let’s put that 10^8 difference into real perspective. The peak IR absorption depth is less than 1 μm. The minimum of the absorption coefficient curve in the dissertation represents a depth of ~220m at ~420 nm. That’s the same order of magnitude (about 100 m) that Pekka and I have been saying from the beginning and which you originally categorically denied. That’s indeed a difference of >2E08. But the ocean is a lot deeper than 220 m. In a shallow pond, some of the blue end of the spectrum will penetrate all the way to the bottom and be absorbed, unless the bottom of the pond is highly reflective. Even if it were reflected, more would be absorbed on the way back up to the surface. The end result being that nearly all the incoming solar radiation that isn’t initially reflected from the surface is absorbed and converted to thermal energy.
Yet another attempt to deflect the conversation away from your error so you never have to admit you’re wrong about anything. And since irony must always increase, you’re wrong about this too. Total atmospheric absorption of SW radiation is prominently featured in the Trenberth et. al. energy balances. The clear sky spectrum of solar radiation at sea level is well known. What do you think the fuss about the ozone layer was all about?
The SURFRAD database includes measures of downwelling solar, upwelling (reflected) solar, direct normal solar and diffuse solar radiation in W/m² that run 24/7 at seven different sites in the continental US. But I’m sure you will manage to rationalize this as somehow not quantifying the almost.
DeWitt Payne
You seem to be implying that I said there was absolutely no absorption of blue light whatsoever.
What I actually said was….
“Pekka’s link also gives minority processes such as overtone absorption.
These are third or fourth order events and hardly account for much of the energy of the visible spectrum.”
Your link confirms the ratio for UV / Blue light = 10^10.
Most people would regard this as a vanishingly small effect and effectively as near zero as dammit.
“If Wikipedia is to be believed”…….now who is being ironic!
Bryan,
There are areas in the ocean where the water is so pure that the absorption that you state as vanishingly small is the most important form of absorption. In most areas the impurities dominate in the absorption of blue light. In those areas the heating by the solar radiation is restricted to a thinner layer than in the pure water. Otherwise nothing changes significantly in the energy balance.
There’s nothing in this that would somehow contradict main stream understanding or IPCC reports. In particular the 0.07% that goes first to chemical energy makes no significant difference. (It’s important in other ways as discussed by climate scientists in many connections.)
Most people would be wrong because the range of lengths involved are more than covered by the depth of the oceans. The UV absorption characteristic length at the peak near 55 nm is 10 nanometers . 100 meters is 10^10 greater than that. By the same logic, we should be able to ignore gravity because the strong force is more than 10^38 times greater and the electromagnetic force is more than 10^36 times greater.
Bryan,
Wow, short memory too. I’m not implying, I’m stating as a fact. Here’s what you wrote, which I have quoted before above:
November 28, 2012 at 6:48pm
To paraphrase you: Most people would take that as meaning that there was absolutely no absorption of blue light whatsoever. And both those sentences were wrong.
The hyperphysics site while useful, is not a whole lot better than Wikipedia. It’s a place to start but may have been oversimplified.
Pekka and DeWitt
For pure water.
I think we need to balance the less likely processes such as overtone absorption with more likely processes such as Rayleigh scattering.
What is the overall effect?
Do we have a source of unbiased information?
Bryan,
Go back and read what I wrote earlier. It’s really easy to distinguish between mostly scattering and mostly absorption. If it’s scattering from particles the same size or smaller than the wavelength of light, the sun as viewed from below the surface will get redder as you go deeper because the blue end of the spectrum is scattered more efficiently. But it doesn’t. Red goes away first. I’ve been diving off the west coast of Catalina Island where the water is extremely clear and the reefs don’t start until about 100 feet down. Or at least it was that way nearly 50 years ago. It was very blue-green and got bluer as I went deeper. Flash photography at close range with wide angle or macro lenses is pretty much required for decent color even at fairly shallow depths. See the Sunlight section of the Wikipedia article on Underwater photography for example. Sure there’s some scattering, but it’s very small compared to absorption. Look at pictures of the Earth from space. The oceans appear to be a very dark blue.
[...] http://scienceofdoom.com/roadmap/atmospheric-radiation-and-the-greenhouse-effect/ (12-part series) [...]
[...] could be considered as a continuation of the earlier series - Atmospheric Radiation and the “Greenhouse” Effect - but I’ve elected to start a new [...]