Guest Post by Ira Glickstein
Solar “light” radiation in = Earth “heat” radiation to Space out! That’s old news to those of us who understand all energy is fungible (may be converted to different forms of energy) and energy/mass is conserved (cannot be created nor destroyed).
My Visualizing series [Physical Analogy, Atmospheric Windows, Emission Spectra, and Molecules/Photons] has garnered almost 2000 comments, mostly positive. I’ve learned a lot from WUWT readers who know more than I do. However, some commenters seem to have been taken in by scientific-sounding objections to the basic science behind the Atmospheric “Greenhouse Effect”. Their objections seemed to add more heat than light to the discussion. This posting is designed to get back to basics and perhaps transform our heated arguments into more enlightened understanding :^)
As I’ve mentioned before, during my long career as a system engineer I’ve worked with many talented mathematical analysts who always provided precise results, mostly correct, but some precisely wrong, usually due to mistaken assumptions. I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on their results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. In those days my analysis was really done using a slide rule and scrap paper, but I now use spreadsheets.
The graphic above is based on an excellent spreadsheet from http://serc.carleton.edu/files/introgeo/models/mathematical/examples/XLPlanck.xls. It uses Planck’s Law to calculate the black body radiation spectrum from the Sun, as observed at the top of the Earth’s Atmosphere. It also may be used to calculate the radiation spectrum from the Earth System (Atmosphere and Surface, see below for explanation) at any assumed temperature. (I will refer to this spreadsheet as “Carleton” in this posting.)
I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).
Since the mean temperature at the surface of the Earth is 288 Kelvins (+15ºC which is 59ºF), that leaves 33 Kelvins (33ºC which is 58ºF) to be accounted for. Guess how we acount for it?
The yellow curve (above left) shows that Solar radiation is in a tall, narrow “shortwave” range, from about 0.1μm (microns, or millionths of a meter) to about 4μm, which we call ultra-violet, visual, and near-infrared. The vertical axis is Intensity of the radiation, measured in Watts/m^2/μm, and the horizontal axis is Wavelength, measured in μm. If you divide the area under the yellow curve into vertical strips, and add up the total area, you get 240 Watts/m^2.
Since we humans sense the visual portion of this radiation as “light”, that is the name we give it, and that has led to the false assumption that it contains no “heat” (or “thermal”) energy.
The violet curve (above right) shows that, assuming a mean temperature of 255 K, Earth System radiation to Space is in a squat, wide “longwave” range, from about 5μm to beyond 40μm, which we call mid- and far-infrared. If you divide the area under the violet curve into vertical strips, and add up the total area, you get the same 240 Watts/m^2 as is under the yellow curve.
DETAILED EXPLANATION
The graph on the left shows the actual observed Solar radiation spectrum (in red) as measured at the top of the Atmosphere. It is superimposed on a black body model (in blue) showing very good correlation. Thus, while the Sun is not exactly a black body, it is OK to assume it is for this type of “sanity check” exercise.
If you calculate the area under the curve you get about 1366 Watts/m^2. That means that a square meter of perfect black body material, held perpendicular to the Sun, would absorb 1366 Watts.
However, the Earth is not a perfect black body, neither is it a flat surface perpendicular to the Sun! So, to plot the yellow curve at the top of this posting, I had to adjust that value accordingly. There are two adjustments:
- The Earth may be approximated as a sphere, with the Sun shining on only half of it at any given time. The adjustment factor for this correction is 0.25.
- The albedo (reflectiveness) of the Earth system, primarily clouds and light-colored areas on the Surface such as ice, causes some of the Solar radiation to be reflected back out to Space without contributing any energy to the Earth System. The adjustment factor for this correction is 0.7.
After applying these adjustments, the net Solar energy absorbed by the Earth System is 240 Watts/m^2.
The graph on the right shows the black body model for an Earth System at a mean temperature of 255 K, a temperature that results in the same 240 Watts/m^2 being emitted out to Space.
Of course, the Earth System is not a perfect black body, as shown by the graph in the upper panel of the illustration below, which plots actual observations from 20 km looking down. (Adapted from Grant Petty, A First Course in Atmospheric Radiation, Figure 8.2, http://www.sundogpublishing.com/AtmosRad/Excerpts/index.html.)
The actual measured radiation is the dark squiggly curve. Note that it jigs and jags up and down between the topmost dashed curve, which is the black body spectrum for a temperature of 270 K and a lower dashed curve which is the black body spectrum for 230 K. This data was taken over the Arctic, most likely during the daytime. The Petty book also has a graph looking down from over the Tropical Pacific which ranges from 300 K down to 210 K. Observations will vary by tens of degrees from day to night, summer to winter, and Tropical to Polar.
However, it is clear that my result, based on matching 240 Watts/m^2, is within a reasonable range of the true mean temperature of the Earth System as viewed from Space.
NOTE ABOUT THE ABOVE ILLUSTRATION
WUWT readers will notice some apparent inconsistencies in the graphs above. The top and bottom panels, from Petty, peak at 15μm to 20μm, while the purple, blue, and black curves in the middle panel, and the Earth System curves from the Carleton spreadsheet I used (see above) peak in the 9μm to 11μm range. Also, the Petty black body curves peak at a “Radiance” around 100 mW/m^2/sr cm^-1 while the black body curves from Carleton peak at an “Intensity” of around 14 W/m^2/μm. Furthermore, if you look closely at the Petty curves, the labels on the black body curves are mirror image! What is going on?
Well, I know some of the reasons, but not all. (I hope commenters who are more fluent in this than I am will confirm my explanations and provide more information about the differences between “Radiance” and “Intensity”.) I have Googled and Wikied the Internet and am still somewhat confused. Here is what I know:
- The horizontal axis in Petty’s plots are what he calls “Wavenumber”, increasing from left to right, which is the number of waves that fit into a cm (centimeter, one hundredth of a meter).
- This is proportional to the frequency of the radiation, and the frequency is the inverse of the wavelength. Thus, his plots are the mirror image of plots based on wavelength increasing from left to right.
- The spreadsheet I used, and my previous experience with visual, and near-, mid-, and far-IR as used in military systems, always uses wavelength increasing from left to right.
- So, when I constructed the above illustration, I reversed Petty’s curves, which explains why the labels on the black body curves are mirror image.
- Fortunately, Petty also included a wavelength legend, which I faithfully reproduced, in non-mirror image, at the top of each plot.
But, that still does not explain why the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet and other graphics on the Internet. I tried to reproduce Petty’s blackbody curves by multiplying the Carleton values by the wavelength (μm) and that did not move the peak to the right enough. So, I multiplied by the wavelength again (μm^2) and, voila, the peaks agreed! (I hope some WUWT reader will explain why the Petty graphs have this perverse effect. advTHANKSance!)
ANSWERING THE OBJECTIONS TO BASIC ATMOSPHERIC “GREENHOUSE EFFECT” SCIENCE
First of all, let me be clear where I am coming from. I’m a Lukewarmer-Skeptic who accepts that H2O, CO2 and other so-called “greenhouse gases” in the Atmosphere do cause the mean temperature of the Earth Surface and Atmosphere to be higher than they would be if everything was the same (Solar radiation, Earth System Albedo, …) but the Atmosphere was pure nitrogen. The main scientific question for me, is how much does the increase in human-caused CO2 and human-caused albedo reduction increase the mean temperature above what it would be with natural cycles and processes? My answer is “not much”, because perhaps 0.1ºC to 0.2ºC of the supposed 0.8ºC increase since 1880 is due to human activities. The rest is due to natural cycles and processes over which we humans have no control. The main public policy question for me, is how much should we (society) do about it? Again, my answer is “not much”, because the effect is small and a limited increase in temperatures and CO2 may turn out to have a net benefit.
So, my motivation for this Visualizing series is not to add to the Alarmist “the sky is falling” panic, but rather to help my fellow Skeptics avoid the natural temptation to fall into an “equal and opposite” falsehood, which some of those on my side, who I call “Disbelievers”, do when they fail to acknowledge the basic facts of the role of H2O and CO2 and other gases in helping to keep temperatures in a livable range.
Objection #1: Visual and near-visual radiation is merely “light” which lacks the “quality” or “oomph” to impart warmth to objects upon which it happens to fall.
Answer #1: A NASA webpage targeted at children is sometimes cited because they say the near-IR beam from a TV remote control is not warm to the touch. Of course, that is not because it is near-visual radiation, but rather because it is very low power. All energy is fungible, and can be changed from one form to another. Thus, the 240 Watts/m^2 of visible and near-visible Solar energy that reaches and is absorbed by the Earth System, has the effect of warming the Earth System exactly as much as an equal number of Watts/m^2 of “thermal” mid- and far-IR radiation.
Objection #2: The Atmosphere, which is cooler than the Earth Surface, cannot warm the Earth Surface.
Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.
Objection #3: The Atmospheric “Greenhouse Effect” is fictional. A glass greenhouse works mainly by preventing or reducing convection and the Atmosphere does not work that way at all.
Answer #3: I always try to put “scare quotes” around the word “greenhouse” unless referring to the glass variety because the term is misleading. Yes, a glass greenhouse works by restricting convection, and the fact that glass passes shortwave radiation and not longwave makes only a minor contribution. Thus, I agree it is unfortunate that the established term for the Atmospheric warming effect is a bit of a misnomer. However, we are stuck with it. But, enough of semantics. Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.
Dave
Regarding your post Dave Springer says:May 13, 2011 at 6:05 am.
My understaning is that theoretical calculations based upon black body calculations for Venus, Mars, Titan and the Moon do not give the correct observed surface temperatures. This may suggest that fundamentally, planets cannot be equated with black bodies. I believe this to be particularly so in the case of a water planet such as the Earth with its variously heated and swirling atmosphere.
Much of the input data is not known with sufficient certainty (eg. the extend of cloudiness which impacts upon how much solar energy is received by the ground, the amount of solar energy absorbed by the atmosphere and the amount of solar energy simply reflected by the ground which impacts upon the amount of solar energy received by the ground, the albedo, the emissivity etc). These uncertainties (some of which may tend ro cancel each other out but which could easily be cummulative) could lead to substantial errors. May be by as much as 10 to 20K. I think your +/- 10% is far to low unless this is applied to the 255K figure.
I also consider that the average temperature of the Earth is not known with sufficient certainty. We are all fanilkier with the arguments as to whether this uncertainty plays a significant role when assessing trends, but this potential uncertainty certainly does have a significant role when seeking to assess absolute (as opposed to a trend) temperature. We have all seen various comments wheresome has set out the IR temperatures around there property (garden, drive, paths, flower beds) which have shown 5 or so different temperatures in a 1000 sqm plot. These differences are everywhere not just in down. If you go for a country walk you can often feel differences as you walk in fields, up hills, in hollows, by hedgerows, near dykes, lakes canal pathways etc.
There have been so many station drop outs I can’t recall the number of measuring stations used, but I would say that if we were to have any prospect of getting a reasonable handle on average globabl temperatures this would have to be increased a billion fold (or by even more than that). It would not surprise me if we were 1 to 3K out on the true and accurate global average temperature which wis within 10% of the assessed 33K figure.
Much of what you say regarding oceans, I agree with or at least in part agree with. I also agree, that once above freezing point, water would become the major player. However, I am far from convinced that even if there was no CO2 (leaving aside issues about life on Earth), all the oceans would be frozen. It may be that oceans around the Equator would not freeze such that there would always be some water vapour.
I am not convinced that there is sufficient evidence of a snowball Earth (although there may well have been). I recall one post on how Earth came out of a snowball, which explanation may be a possibility but then again there may be other explanations (eg., oceanic volcanos splitting in the ice and thereby releasing some water vapour, soot deposits changing albedo, even meteor collision – who knows given the lack of evidence).
Smokey says:
Convection provides a mechanism for moving heat around in the atmosphere. The only way that any significant amount of heat gets transferred out of the earth system (which means, including the atmosphere) is by radiation.
As I discussed in response to Martin Lewitt, some people seem to believe they can vaguely invoke the water cycle to transport heat back out into space. However, because heat can only get out into space via radiation, this proposal has testable consequences: The atmosphere must warm up enough somewhere in order to increase the amount of heat that it emits to space.
JAE says:
And, your evidence of this is, what exactly?
Besides which, like I said, we don’t derive the ~33 K temp difference between the earth’s surface and its effective blackbody radiating temperature by modeling. We get it empirical observation and applying conservation of energy.
mkelly says:
Nobody is calling temperature “back radiation”. What is being called “back radiation” is radiation that the atmosphere emits (back toward the earth) by virtue of having a nonzero temperature.
Bryan says:
Once again, you make a statement that can be interpreted as correct but completely irrelevant to discussions of the atmospheric greenhouse effect. Or, alternately, it can be interpreted as incorrect. You seem to thrive on this ambiguity, just as G&T did, in order to peddle pseudo-scientific nonsense.
Bryan says: May 13, 2011 at 11:02 am
“So by this definition, the Solar Radiation is also “backradiation”!!!”
Sure. Why not? This is simply semantics, not physics.
The sun provides “solar back radiation” or “solar downward radiation” or “solar toward-the-surface short-wavelength EM radiation” to the surface (~ 168 W/m^2). There is no “solar upward radiation” so there is really no need for the directional adjective, but we could use it if we want.
The surface provides “surface forward radiation” or “surface upward radiation” or “surface away-from-the-surface long-wavelength EM radiation” to the atmosphere (350 W/m). It also provides “surface forward radiation” of 40 W/m^2 to outer space.
The atmosphere provides “atmospheric back radiation” or “atmospheric downward radiation” or “atmospheric toward-the-surface long-wavelength EM radiation” to the surface (324 W/m^2).
Personally I’d use a simpler notation like q=radiant energy flow, S = sun, A = atmosphere; G = ground; O = outer space.
Then we would have:
q(S–>G) = 168 W/m^2
q(G–>A) = 350 W/m^2
…
Bryan also says:
“However I think that you should review your opinion that all downwelling Long Wavelength radiation is backradiation.”
Don’t worry about the labels — worry about the physics.
* All q(x –>S) radiation adds energy to the surface
* All q(S –>x) radiation removes energy from the surface
It is as simple as that.
If you want to call some or all of the q(x–>S) “back radiation” or “downwelling radiation” then feel free.
OOPS! I MIXED UP THE ABBREVIATIONS
I said:
Don’t worry about the labels — worry about the physics.
* All q(x –>S) radiation adds energy to the surface
* All q(S –>x) radiation removes energy from the surface
It is as simple as that.
“S” was already taken for “solar”, so “G” was used for ground (which also would include the oceans). That should have read:
Don’t worry about the labels — worry about the physics.
* All q(x –>G) radiation adds energy to the ground (ie adds energy to the surface)
* All q(G –>x) radiation removes energy from the ground
It is as simple as that.
richard verney says:
I don’t understand this description. If you are talking about the earth + atmosphere, it is not treated as a blackbody. There is merely a number given that tells us what its temperature would be if it were emitting as a blackbody. If you are talking about the earth alone, then determining the relation between its temperature and what it radiates using the blackbody formula is a very good approximation because in the IR range of interest, the materials making up the earth’s surface do indeed have emissivities very close to 1.
You have exaggerated the error ranges in the albedo by looking at the separate estimates of land and ocean albedo. For the global albedo, the table that you have quoted from shows four estimates in the range of 31 to 31.3% and two outlier estimates of 28.1% and 33.8%. [Paper here: http://content.imamu.edu.sa/Scholars/it/net/trenbert.pdf ]
At any rate, a better estimate of the “theoretical notional” temperature of the Earth is using the measured emission OLR (outgoing longwave radiation). The table lists four measurements in the range of 233.3 to 237.4 W/m^2 and then two outliers at 245 W/m^2 and 253.9 W/m^2. Even if we consider the most extreme high outlier of 253.9 W/m^2, that only gives an effective blackbody temperature of 258.7 K, which compared to 255 K is far less than a 10 to 20 K error.
The water cycle does not get you around having to obey conservation of energy, which tells us that the earth has a current surface temperature that is roughly 33 K higher than what it would support if its atmosphere were transparent to the radiation that its surface emits (but it otherwise had the same albedo, etc.).
I doubt the higher figure, but this is still way smaller than 33 K.
I still don’t see how all of these words get you around having to satisfy conservation of energy.
I love how the same sort of people who want to tell us about the importance of convection and such in order to get around limits that convection cannot get you around then turn around and make statements like this, ignoring convection and advection when it suits their purpose…and when it is vitally important not to do so. You have presented absolutely no data to support either your contention that there is no correlation nor showing what correlation would be expected once one fully includes other factors.
Again, these arguments are not going to get you around conservation of energy considerations.
As Tim has pointed out, the universe is governed by the laws of physics, not your “commonsence”. If the laws of physics contradict your common sense, I’ll go with the laws of physics. The only reason your argument even seems plausible is that the numbers for DWLIR only exceed the solar energy by a little bit on earth; on Venus, they exceed it by such a huge amount that it would be immediately obvious that your common sense is not even close to correct.
It doesn’t distort what is going on at all. Yes, looking at averages throws out a lot of information, but that just means that the Trenberth budget has its uses and its limitations, like anything does.
This has already been commented upon by others (and myself) upthread.
This is just pernicious nonsense that can only fly with someone who has never actually done a simple radiative balance calculation. It is not just a “delay”; the greenhouse effect reduces the rate of energy loss out into space (for a fixed surface temperature), requiring a higher average surface temperature to restore radiative balance.
Joel Shore says:
“Convection does not occur between the earth and space.”
Unequivocal statement. And unequivocally wrong. Convection does in fact occur between the earth and space, despite the convoluted attempt to explain that mis-statement of fact.
I only mention this because of Joel Shore’s regularly labeling my rejection of CAGW as ‘ideological,’ which is simply projection on his part. I base that rejection on the fact that there is no evidence of global damage from CO2, despite a ≈40% increase. Only the GIGO models predict global harm from that essential trace gas, while real world observations show that CO2 is harmless.
Tim Folkerts
Yes I suppose you could operate like that.
Just as a physics lecturer might decide to do all his lectures standing on his hands upside down.
You would however be the only person on the planet who calls Solar Radiation under the title of “backradiation”.
Others might think you a bit odd,….. but don’t let that put you off.
Bryan,
You are missing my point.
The names do not ultimately matter, only the physics. If people want to use unusual nomenclature, that does not change the physics. There is “backradiation” because people have given that label to “downward directed thermal IR radiation from the atmosphere toward the surface”. It is what I labeled q(A–>G).
There is also q(S–>G). We can call it “sunlight”. Or we could call it “solar backradiation” if we really wanted to emphasize that is heads the same direction as the IR backradiation from the atmosphere. But as long as we both understand the concept, the name doesn’t matter.
Two fundamental physics questions related to climate science are:
1) Does q(A–>G) exist?
2) Is the value around q(A–>G) = 324 W/m^2?
I think the answer to both is “YES!” It seems you think (1) is correct but you question (2).
Some people seem to think that either there is no such thing as q(A–>G), or that it could possibly be larger than q(S–>G). I’d love to hear specific reasoning to support these positions.
THAT should be the starting point for a discussion, not ” ‘backradiation’ is a poor name”!
Joel Shore says:
May 13, 2011 at 11:33 am
“Nobody is calling temperature “back radiation”.
Joel says: “At some point, this comes down to terminology. It is most correct to say that because of back-radiation (in combination with the radiation from the sun), the earth is warmer than it would be if all the radiation from its surface escaped into space.”
Ira says: “The only rational explanation is the back-radiation from the Atmosphere to the Surface.”
You say nobody is calling temperature back radiation. Well it seems as you and Ira are at the very least. You use the word warmer which indicates a temperature change. If Ira wants to say “The only rational explanation is a temperature change in the atmosphere that is causing the surface to warm.” then say that and not use back radiation.
Joel said “Convection does not occur between the earth and space.”
and Smokey questioned this.
Smokey, how about this?
“Energy transfer via convection does not occur between the earth as a whole and space.”
That seems to be very clearly what Joel meant.
I suppose you could warp the sentence to mean:
“Convection does occur within the atmosphere, and the atmosphere is between the earth’s surface and outer space. Thus, convection occurs in the region ‘between the earth and space’ “.
Frankly, I can’t see anyone actually thinking that is what Joel meant!
Or do you have some other meaning for “convection occurring between earth and space” ?
Tim Folkerts,
Thanks for your input. Note that I also commented: “Warmer air both rises and radiates.” But there’s not much wiggle room to parse here:
“Convection does not occur between the earth and space.”
mkelly, back radiation can increase without a temperature increase, for example by just adding CO2. It is very important to separate the temperature effect from atmospheric constituent effect to understand back radiation.
Ira and Joel Shore, thanks for your comments.
Joel, I think that you make the mistake to assume that nothing will be reflected nor absorbed in a GHG free atmosphere. Reflection will be there as long as there are clouds, regardsless of water in vapour phase. (I am assuming here that you do not regard water in liquid or solid state as a GHG.) Particles will also absorb (and to a lesser degree reflect). Whatever the mechanism one has to note that 23% of incoming solar light is absorbed today in the atmosphere – not due to GHGs by definition. And that somewhere between 23% and 38% of outgoing thermal radiation is reflected.
So, if we seek a minimum temperature that earth holds without IR absorbing gases we need to count back radiation of absorbed sunlight, 39 W/sqm, 161 W/sqm absorbed sunlight + 38% reflection of outgoing radiation we get 322 W/sqm or 0 deg celsius at surface. This analysis leaves a lot out (e.g. no spatial analysis of temperature distribution, nor time) but yields us a mere 15°C to be explained by GHGs. It matters as CO2 is at least 12 doublings over the saturation levels of the base band. 15/12 yields a maximum of 1,25 °C per CO2 doubbling. This assumes that no water vapour, methane etc would be present in a CO2 free atmosphere, which is obviously not true and we can safely say that a doubbling of CO2 levels would give less than a degree temperature increase. (All heat flows according to Trenberth, 2009)
Joel and Ira, please comment.
And now, for a beer. Happy weekend to you all.
Tim Folkerts says:
May 13, 2011 at 10:42 am
Well, I’ll be —! Thank you, Tim! By juxtaposing that equation with the K&T diagram, I finally got it through my thick head just what the diagram is trying to show! It now actually makes sense to me (although I still don’t know if I agree with it).
Joel,
I confess that the Trenberth diagram you directed me to leaves me with more questions than answers.
According to the diagram:
– the Earth only absorbs 161 W/m^2 from the sun yet emits 493 W/m^2
– the Atmosphere only absorbs 78 W/m^2 from the sun yet emits 502 W/m^2
Also his numbers don’t add up, though they are out only by a few W/m^2 which perhaps is not important…
Unfortunately this doesn’t make sense to me, if the extra 333 W/m^2 the earth needs to radiate 493 W/m^2 comes from the atmosphere, then where did the atmosphere get the energy from? It only gets 78 W/m^2 from the sun after all…
Also, I have reviewed a couple of books on statistical thermodynamics, and nowhere do I find any evidence that a cold object can heat/transfer energy to a warmer object without the application of external work. As a matter of fact they state that this cannot spntaneously occur. This is not surprising, statistical thermodynamics still have to repsect the first & second laws.
I quote from an excellent book, which you might enjoy:
“In all cases the expressions for the changes in entropy correspond exactly to those deduced from Clausius’s definition, and we can be confident that the classical entropy and the statistical entropy are the same.”
Peter Atkins, Four Laws That Drive the Universe; Oxford University Press 2007
So if, as you claim, the energy from the atmosphere heats the earth, where is the external work that allows this to occur? Unless of course the atmosphere is warmer than the surface of the earth – but then the earth wouldn’t spontaneously heat the atmosphere with the 356 W/m^2 it somehow emits from an input of only 161 W/m^2.
Another point, you Ira and others have stated that it is the NET difference of energy transfered which must go from hot to cold in order to respect the second law.
However the second law states that entropy must stay the same or increase, it can’t decrease as a result of any spontaneous event. So even if a cold object could transfer energy to a hotter object spontaneously, using the numbers in Trenberth’s diagram the net result is a decrease in entropy inasmuch as the 333 W/m^2 emitted from the colder atmosphere represents a greater decrease in entropy than the 356 W/m^2 is an increase in entropy.
So, to me at least, this diagram of Trenberth’s seems very problematic.
While I have enjoyed reading the many threads on adiabatic heating etc… I really can’t get by what appears to me to be a very fundamental problem with this theory of Greenhouse Gas Warming, namely that a cold object can heat a warmer one.
A question that interests me is one that has come up a few times lately, namely:
“So … what specific numbers in the diagram do you think are wrong?”
Below is my version of it. – I know some of you guys have tried to “shoot me down” on this one before, but here I am, once again, as you never did convinced me I was wrong.:
As early as 1859, Gustav Kirchhoff proposed that “At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity” and as far as I can understand, nobody objected and his proposition was accepted as part of “Kirchhoff’s Law”, and, to me, it seems logical and should be unavoidable as it is based on “energy conservation”.
– But even so in 1997 some scientists, Kiehl, Trenberth & al (K&T), published (an IPCC “peer-reviewed”) “Energy Flow Chart” (EFC) which showed equilibrium, in as much as energy in = energy out.
– The surface is, in the EFC, shown to absorb 168 W/m² of incoming Solar radiation, but does not even attempt to conserve any energy as it gets rid of (24+78)= 102 W/m² via thermals and “evapo-transpiration” and then in stead of being contented with radiating away the remaining 66 W/m², it sends out a whopping 390 W/m²
– Yes I have posted questions about this conundrum as to where the surface finds the 324 W/m² that forms the basis for this “Radiation Circuit” (RC) between atmosphere and surface. But none of the answers given have so far convinced me.
– Kirchhoff was kicked into touch and no scientist grumbled.
-Well, I doubt very much that the surface knew when it emitted the 324 extra W/m² of energy from it’s “energy store” that the atmosphere was going to send it straight back and thus conserve that energy on it’s behalf.
Q: “So … what specific numbers in the diagram do you think are wrong?”
A: “ Below is how I would have changed some of the numbers”
To my way of reasoning, if the basics of the EFC is correct, then it would be reasonable to accept that the result of the 168 W/m² absorbed by the surface would be that (24+78) = 102 W/m² are leaving in the form of thermals and “evapo-transpiration” and then the remaining energy 66 W/m², after having done it’s job, would leave via radiation.
– And then, – and only then – as what is lost from the surface is absorbed by the atmosphere (which, by the way has absorbed 67 W/m² directly from the Sun,) would I enter the GHGs and with them “The Atmospheric Window” which gives a free passage to 40 W/m². – In which case there are (66 – 40) = 22 W/m² of energy left to bounce back and forth between surface and GHGs creating a “delay” before those 22 W/m² also left for space, then yes I too may have accepted a bit of warming down here in the Troposphere, but only as a result of the delay in the 22 W/m² leaving the system.
But if the surface is in the habit of radiating away 222 W/m² more than it receives then we must assume that the Moon’s surface will do the same and that the 107 ° C that is reported on as “measured data” on it’s sunny side must be a myth as the Moon gets nothing back from GHGs
Alleyne muses: May 13, 2011 at 3:35 pm
“Unfortunately this doesn’t make sense to me, if the extra 333 W/m^2 the earth needs to radiate 493 W/m^2 comes from the atmosphere, then where did the atmosphere get the energy from? It only gets 78 W/m^2 from the sun after all…”
Here is a different way to think about it that might help. I haven’t seen it explained quite like this before, but it suddenly makes a lot of sense to me to use this approach.
The total thermal energy of a 1 m^2 column of the atmosphere could be roughly estimated as:
U = mcT = 10,000 kg * 1000 J/kg*K * 300 K = 30,000,000 J
There are a variety of little problems with this estimation, but it is clear that a 1 m^2 column has millions of joules of thermal energy. Suddenly the idea of it losing a few hundred of those joules doesn’t seem so noteworthy.
Or think of it this way. At some point in time, an average 1 m^2 column of air will have some amount of thermal energy U (~ 30,000,000 J). During the next second, that column of air will
* give 324 J to the surface via thermal EM radiation
* give 195 J to outer space via thermal EM radiation
* get 350 J from the surface via thermal EM radiation
* get 78 J from the surface via evaporation/condensation
* get 24 J from the surface via convection
* get 67 J from the sun via sunlight
30,000,000 J – 324 J – 195 J+ 350 J + 78 J + 24 J + 67 J = 30,000,00o J
The atmosphere has managed to hold steady, giving up only a TINY fraction of its total energy and absorbing only a TINY fraction of its total energy, with no net gain or loss. (Of course, the number will change from day to night and from winter to summer, but these are “typical” numbers.)
Or just for dramatic effect I could write it as ….
30______ MJ
– 0.000324 MJ
– 0.000195 MJ
+ 0.000350 MJ
+ 0.000078 MJ
+ 0.000024 MJ
+ 0.000067 MJ
———————
= 30 _____ MJ
The atmosphere losing 0.001% of its energy over the course of a second does not sound so dramatic anymore!
Allyne,
Peter Atkins’ handy little book is a really excellent tutorial for anyone interested in understanding the zeroth through third Laws and entropy.
Also, I have bit of a problem with any explanation that goes beyond a simple time delay due to GHGs. There is certainly no empirical evidence that they generate heat. If they do, I want to see testable, replicable and convincing real world evidence. All we have now are models.
And is Trenberth implying that if a warmer object were surrounded by hundreds of slightly cooler objects, all radiating toward the slightly warmer object, that the combined radiation from the slightly cooler objects would warm the single slightly warmer object? Entropy doesn’t allow that, unless work is performed between. What is doing that work?
The observations made by Alleyne at 13, 2011 at 3:35 pm appear particularly pertinent. Trenberth uses the problematic statement “Absorbed by Surface” with respect to the back radiated 333 W/m^2.
Tim Folkerts’ comments at May 13, 2011 at 10:42 am in which he suggests that the surface is (in effect) radiating 66 W/m^2 is a very different take on the diagram and is not simply interpreting the diagram but re writing what Trenberth is saying.
Tim may have been forced into this revision since he has commented several times that radiation can not heat an object to a temperature greater than the temperature of the radiating source. In the case of the the source of the 333 W/m^2 this is back radiated from a height in the atmosphere which is at a temperature lower than the ground surface and therefore this radiation cannot be “Absorbed by the Surface”. Thus Tim’s revised take significantly contradicts the assertion put forward by Trenberth.
As soon as you look at net energy flows, you are looking at the extent to which GHGs merely delay energy departing the Earth’s surface. In which case one needs to properly consider what happens at night (when there is no solar input) and what happens when there are temperature inversions (which must be taking place somewhere on this planet on a reasonably regular basis) which act as a safety valve venting any excess temperature built up (see the article Frostbite Falls by E M Smith on WUWT in which he givces an illustration where local temperatures fell to -43C, ie., 230K which is significantly below the calculated BB temp of the Earth without GHGs)
Alleyne again muses:
“However the second law states that entropy must stay the same or increase, it can’t decrease as a result of any spontaneous event. So even if a cold object could transfer energy to a hotter object spontaneously, using the numbers in Trenberth’s diagram the net result is a decrease in entropy inasmuch as the 333 W/m^2 emitted from the colder atmosphere represents a greater decrease in entropy than the 356 W/m^2 is an increase in entropy.”
I don’t see the problem. Entropy is calculated as
dS = dQ/T
For the sake of argument, lets call the ground 285 K and the atmosphere 250 K (but the values don’t matter as long as the number for the atmosphere is lower than the number for the surface). Let’s look at 1 m^2. And I use the numbers from the original energy balance diagram.
For convection between surface & atmosphere.
The entropy of the ground decreases each second by
dS = – 24 J/ 285 K = – 0.084 J/K
The entropy of the atmosphere increases by
dS = +24 / 250 K = + 0.096 J/K
NET CHANGE from the process is +0.096 J/K – 0.084 J/K
= 0.012 J/K
This is positive, as required.
For Evaporation/Condensation between surface & atmosphere.
The entropy of the ground decreases each second by
dS = – 78 J/ 285 K = – 0.274 J/K
The entropy of the atmosphere increases by
dS = +78 / 250 K = + 0.312 J/K
NET CHANGE from the process is +0.312 J/K – 0.274 J/K
= + 0.038 J/K
This is positive, as required.
For radiation between surface & atmosphere.
Looking specifically at the exchange between surface and atmosphere, the radiation rates are 350 W/m^2 up from surface to atmosphere and 324 W/m^2 down from atmosphere to surface, for a net 26 W/m^2 upward.
For the surface,
dS = – 26 J/ 285 K = – 0.091 J/K
The entropy of the atmosphere increases by
dS = +26 / 250 K = + 0.104 J/K
NET CHANGE from the process is +0.312 J/K – 0.274 J/K
= + 0.0.013 J/K
This is positive, as required.
(You could also use the 350 J & 324 J numbers directly, but that will give the same results).
I’ve actually given numbers to show entropy always increases as required. What specific example of a process with a net entropy decrease were you thinking of?
vindsavfuktare says:
First of all, let me be clear that what I have talked about is an atmosphere that does not absorb terrestrial IR radiation. That is, the total greenhouse effect does include a contribution due to clouds. Note that the clouds do not significantly reflect at these mid- and far-IR wavelengths. They only absorb the radiation or (if not optically-thick) transmit it. So, the way they act is in the same way that the gaseous contributors to the greenhouse effect act (albeit with a broader absorption spectrum).
The rough number for the amount that clouds contribute to the natural greenhouse effect is 25% as discussed here http://pubs.giss.nasa.gov/docs/2010/2010_Schmidt_etal_1.pdf . (These numbers are necessarily rough in that the effects are not completely additive, so you get one number if you start with the current atmospheric composition and take clouds out and another number if you start with an atmosphere with no GHGs and put clouds in.)
I lost you…particularly in the part about how you treat incoming sunlight being absorbed. That doesn’t make sense to me. The point is that the earth’s surface emits as a blackbody at a temperature of ~288K and the earth system as a whole emits an amount of radiation that corresponds to what the earth system absorbs from the sun in shortwave radiation, and corresponds to a blackbody temperature of 255 K. The difference in these temperatures is the greenhouse effect. Of this, about a quarter of it is attributable to clouds, as I noted above.
Woe…There’s a lot of assumptions there (in addition, as I noted to attributing too small an amount to GHGs). First off, an atmosphere without CO2 is indeed expected to lead to the water vapor condensing out, so you do lose its effect. Furthermore, the earth’s surface will become much more ice- and snow-covered leading to a larger albedo and hence a further temperature decrease. So, no, I don’t buy your claim about how much the temperatures will drop when you take CO2 out. See the discussion here: http://www.sciencemag.org/content/330/6002/356.abstract (Technically, that paper looks at removing all of the non-condensable GHGs, so it is an interesting question what the models predict if you just remove CO2 and leave in the others. For removing all of the non-condensable GHGs, they get a temperature drop of about 35 K.)
Second, I am not sure where you get your number for 12 doublings over saturation. It may be right…but I’d at least like to see the reference. And, there are additional questions as to whether the forcing and climate sensitivity would really be constant over that many doublings.
It’s like a hydra ! ! ! :-0
Smokey says:
“There is certainly no empirical evidence that they generate heat. “
Quite true. GHG’s absorb IR energy. They emit IR energy. They are part of heat exchange. But they generate no energy.
Smokey says:
“And is Trenberth implying that if a warmer object were surrounded by hundreds of slightly cooler objects, all radiating toward the slightly warmer object, that the combined radiation from the slightly cooler objects would warm the single slightly warmer object? “
In a word — “NO!”
(If you need more explanation, each cool object, not matter what the size, shape position, or emissivity, will receive more energy from the warm object than it gives to the warm object — energy is conserved and entropy increases.)
richard verney says:
“Tim Folkerts’ comments at May 13, 2011 at 10:42 am in which he suggests that the surface is (in effect) radiating 66 W/m^2 is a very different take on the diagram and is not simply interpreting the diagram but re writing what Trenberth is saying. “
No, I suspect that Trenberth would interpret it just the way I do.
richard verney says:
“Tim may have been forced into this revision since he has commented several times that radiation can not heat an object to a temperature greater than the temperature of the radiating source. In the case of the the source of the 333 W/m^2 this is back radiated from a height in the atmosphere which is at a temperature lower than the ground surface and therefore this radiation cannot be “Absorbed by the Surface”. Thus Tim’s revised take significantly contradicts the assertion put forward by Trenberth.”
No, there is no revision. I have tried to say things several ways, but I always profess that there is “upward thermal IR from the surface” (~ 396 W/m^2, most of which gets absorbed by the atmosphere, but some passes directly thru the atmosphere and out into space) and a “downward thermal IR from the atmosphere” (~ 333 W/m^2, nearly all of which gets absorbed by the surface).
The entropy calculations in my previous post show that there is no problem with entropy for such a process of energy exchange via IR radiation.
Alleyne says:
Well, that’s good because nobody is arguing that point. We all know what the 2nd Law says. And, all models of the greenhouse effect, be they simple shell models or full convective-radiative models have the heat going from the warmer earth to the colder atmosphere.
However, the point I am making is that this does not mean that colder objects don’t radiate toward warmer objects. It only means that the colder object will always absorb more radiant energy from the warmer object than warmer object absorbs from the colder. That is what the real 2nd Law says.
Believing that this means that a cold object won’t radiate toward a warm object (or that the warm object won’t absorb any of this radiation) is believing in what I call “The Magical 2nd Law”, which is not a law of physics and, in fact, is not in accordance with the statistical physics principles on which the 2nd Law is ultimately based. If you don’t believe in “back radiation” or somehow imagine that the final steady-state temperature of the earth doesn’t depend on how much “back radiation” it receives from the atmosphere, then you are a believer in the Magical 2nd Law, not the real 2nd Law.
Furthermore, the equations for radiative transfer that are used in all models of the greenhouse effect will, if correctly applied (e.g., with objects obey Kirchkoff’s Law), will automatically satisfy the 2nd Law.
Look, you are simply confusing terminology here. If you mean by “the atmosphere heats the earth” that the net energy flow (i.e., heat) goes from the atmosphere to the earth, then no, that is not what I am claiming; that violates the 2nd Law.
If you mean by “the atmosphere heats the earth” that the atmosphere causes the earth to be at a higher steady-state temperature than if all of the radiation that the earth emitted went back out into space, then yes, that is what I am claiming; however, it doesn’t violate the 2nd Law because the heat still goes from the earth to the atmosphere. It is just that (for a given earth surface temperature) the heat flow away from the earth is less than it would be if the (IR-absorbing) atmosphere were not present. As a result, the earth’s surface temperature must increase until it reaches a point where the earth system (earth + atmosphere) is radiating back into space as much energy as it receives from the sun.
Read this 20 times until you understand it so that you stop attacking “strawman” arguments and address what we are actually saying.
But…you can’t define your spontaneous event as being just one part of the process. That won’t happen spontaneously. The point of the 2nd Law is that you can’t have the radiation from the colder object toward the warmer occurring without having the warmer object radiating toward the colder object too. Again, you seem to believe in the 2nd Law as some form of magic, rather than understanding the modern basis of it as following from statistical physics.
Alleyne: I see that Tim has interpreted your misunderstanding on the entropy considerations at the end of your post differently than I did. But, hopefully, one or the other or both of our posts will address this issue or issues that are bothering you. Note that for heat transfer between two bodies, the statement that entropy must increase and the statement that heat flows from the hotter body to the colder body are equivalent because of the fact that dS = dQ/T.
Bryan says:
May 13, 2011 at 8:53 am
The RW Wood real greenhsoue experiment (glass vs. rock salt) is bogus. The glass panes heat up when they absorb IR and lose it right away through conduction. The G&T experiment repeats the mistake. A kilometer’s thick layer of greenhouse gases doesn’t conduct very well at all, unlike a thin pane of glass or polyethelene.
G&T need to repeat the experiment with double-paned polyethelene with vacuum separating the two panes to imitate the poor conduction through kilometers of atmosphere. Woods needed to do the same thing with double-paned rock salt.
The better (much more expensive greenhouse panels) are constructed in such a way that they don’t conduct heat very well. IIRC correctly they use some sort of closed cell expansion which makes them sort of like transparent styrofoam. They aren’t clear like glass or solid plastic and don’t transmit visible light quite as well but the increase in R-factor is quite a lot and it’s still far less expensive than double-pane vacuum panels.