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.
q = k A dT / s (1)
q = σ T4 A (1)
q = k A dT (1)
Above from engineering tool box.
Here are the basic heat transfer equations. Please note there are no inputs for back radiation. NONE. Your conclusion “The only rational explanation is the back-radiation from the Atmosphere to the Surface.” defies these equations.
There are only two basic requirements for heat transfer 1. path 2. temperature difference. If either ar missing no transfer.
I did not consider shape factor, emissivity differences, etc basics as you wanted to get back to.
Martin Lewitt says:
Okay…Let me explain to you why I think this argument fails…at least from what I understand of the empirical data. Basically, you are saying that the evaporation / condensation increases, transporting more energy from the surface up into the atmosphere where it can be radiated back out into space. However, since the only way that energy can escape to space (in any significant amount) is via radiation and the amount of radiation depends on temperature, what this argument amounts to is the proposition that the upper portion of the atmosphere warms more than the surface.
Now, to the extent that this is indeed expected to be true, it is already incorporated in the models via the lapse rate feedback (a negative feedback)…which says exactly this. The reason that the upper part of the troposphere is expected in the global average to warm more than the surface is that in the tropics is that one expects the lapse rate to closely follow the moist adiabatic lapse rate, which indeed implies more warming at altitude than at the surface. This is the so-called “hot spot” in the tropical troposphere.
So, what you must be proposing is that this amplification of warming at altitude relative to the surface happens to a greater degree than the models predict. However, the data for the tropics currently show, if anything, just the opposite. I.e., people are complaining that the “hot spot” is missing. Whether it is really missing and the degree to which it is missing in fact depend on which analysis of the satellite data and radiosonde data one believes…and there are good reasons to believe there are problems with the data and that the models are basically correct. However, your picture would actually require that the “hot spot” really be more pronounced than the models predict, which seems rather unlikely based on what we do see in the data thus far.
Alternately, I suppose you could argue that the models are correct (or overestimate the hot spot) for the tropics…but that they underestimate the warming at altitude (relative the surface) outside of the tropics by such a degree that the magnitude of the lapse rate feedback is still underestimated globally. However, as far as I know, the empirical satellite and radiosonde data, such as it is, don’t really support that idea either.
So, that is my basic complaint about the argument that a faster water cycle somehow provides a negative feedback: To the extent that this expected to be true, it is incorporated into the models. If you want to argue that the models underestimate the effect, this has testable consequences for the temperature distribution in the atmosphere…And, at this point, the data for the temperature distribution do not seem to support this conclusion, if anything going the other way.
mkelly,
You needed to look a little farther and understand what you are reading. This is from the SAME source you quoted:
Please note there ARE inputs for back radiation (the Tc^4 term).
Ref. mkelly:
May 10, 2011 at 9:25 am
There are no inputs for back radiation because it is taken care of by dT. No back radiation, then dT is simply the temperature of the emitter.
It might be better to consider the back radiation from the atmosphere as a reduction in dT in order to use standard nomenclature. Using dT simplifies the calculation, esp. if emissivity is ignored. However, calling it back radiation clarifies the attribution to greenhouse gases, and allows the separate consideration of the back radiation from various concentrations of different gasses. Emissivity, of course, is a critical quality since it is this factor that makes CO2 and H2O significant contributors to the greenhouse effect, whereas N2 is not regardless of temperature (ignoring conduction, of course).
Bryan says:
Okay, let’s try this one more time:
When you choose what model you use, the model chosen is not independent of the purpose. If I were choosing a model to describe with as much quantitative fidelity as possible the greenhouse effect in the earth’s atmosphere, then the model I would choose would be a state-of-the-art convective-radiative transfer code using the actual composition and empirical absorption / emission lines for the atmospheric constituents. Such models most definitely do not just assume a the Stefan-Boltzmann Equation, as they do the calculation line-by-line (i.e., absorption line by absorption line).
However, our purpose in our reply to G&T was not to model the greenhouse effect with as much quantitative fidelity as possible. Instead, it was just about as far as you could get from that in the other direction: We wanted to have the simplest models possible that illustrate the effect that G&T seemed to imply violated the 2nd Law of Thermodynamics, so simple in fact that nobody could seriously argue about whether or not we had solved them correctly because anybody could solve them on the back-of-an-envelope.
So, that is what we came up with —A few very simple models, such as the one that involves 3 objects: one object A producing thermal energy and radiating energy at a fixed rate, two other objects B and C whose temperature is determined via radiative balance with object A and empty space, with a geometry such that the temperature of object B is higher than that of object C. And, what we wanted to illustrate is that the object C “warms” B in the colloquial sense of the word…i.e., that the presence of object C causes B to be at a higher temperature than if C is absent. By analogy, it then follows that if you choose A to be the sun, B to be the earth’s surface, and C to be the atmosphere, then one can immediately see the fallacy in claiming that “the greenhouse effect violates the 2nd Law because it implies heat flows from the colder atmosphere to the warmer surface” (that’s a paraphrase). In fact, the heat flows in our models are easy to calculate, so it is easy to verify that the 2nd Law is satisfied, but that, nonetheless, the presence of the colder object C causes the temperature of B to be warmer than in its absence.
If you were worried about the issue of whether the T^4 dependence is critical in producing our result, it is not be hard to assume some more general dependence and show that as long as the amount radiated by the object increases with temperature, it doesn’t matter exactly what the temperature dependence is. Such are the advantages of simple models! (Note, by the way, that what is true for a radiating object is that the amount of radiation emitted AT ANY PARTICULAR WAVELENGTH is an increasing function of the temperature, a fact that is not always obvious because people often tend to normalize the emission curves when showing emission curves for different temperatures on the same graph.)
Dave Springer says:
This is not correct. The models…and empirical data…show that clouds have a combination of warming and cooling effects (depending on such details as the altitude and optical thickness of the clouds) but that the net effect of clouds on the earth is currently cooling. (I believe the magnitude quoted is on the order of 20 W/m^2.)
Also, what the models show in terms of what happens with clouds is complicated. In general, to the extent that high clouds increase and low clouds decrease in the models, then the cloud feedback will be positive. You can find somewhere or other an analysis of the various effects due to clouds that are predicted by various models. They usually separate it out as the effect of the cloud feedback on shortwave radiation (i.e., radiation from the sun) and the effect of the cloud feedback on longwave radiation (i.e., radiation from the earth).
A positive cloud feedback does not imply a runaway greenhouse effect. Ray Pierrehumbert would say that the reason why we cannot have a runaway effect on the earth with the sun at its current luminosity is actually quite well-understood. As I noted above, Hansen seems to feel differently, although the details of his reasoning remains obscure.
R Stevenson says:
May 10, 2011 at 1:58 am
I tried to confirm this figure using Plank – Hottel and got 3600m
To get a good figure, you must use the HITRAN data. At 15 cm-1, 99.7% is absorbed in less than 10 meters. At 14.97 cm-1, 99.7% is absorbed in less than 0.8 meters. Both of those measurements are with bandwidths of about 1 cm-1. With a bandwidth of 30 cm-1, 99.7% is absorbed in under 150 meters. To get 3.6 km you must be using a very wide bandwidth.
[I’ll just quickly add what I do know about why Hansen says that a runaway could potentially occur now while it hasn’t in the past: One point he makes is that the sun has slowly gotten brighter, so CO2 levels back several hundred million years ago or more are not directly comparable to levels now with the current luminosity. Another point he makes regards the speed of the change in CO2 levels and that somehow overwhelming negative feedbacks that would otherwise occur if the change in CO2 levels happened over a longer timescale. I am not saying that he has compellingly shown that these two points materially change things, but just wanted to explain why he feels that something that never happened in the past could happen now.]
Dave Springer says:
May 10, 2011 at 8:55 am
CO2 extinction altitude can be seen in the spectrum looking down from 20 kilometers above the arctic ocean. IIRC the spectrum……
Dave Springer,
What is IIRC?
[Reply: IIRC generally means “if I remember/recall correctly,” or similar. This is worth bookmarking. ~dbs, mod.]
Tim Folkerts says:
May 10, 2011 at 9:42 am
mkelly,
You needed to look a little farther and understand what you are reading. This is from the SAME source you quoted:
Mr. Folkerts, you need to read what I wrote regarding the two conditions for heat transfer. Path and temperature difference or gradient. Temperature is not back radiation and trying to name it such does not make it so.
Further please note I said “basic” per Ira’s want to go back to basics. Until you can show back radiation as an input for heat transfer then use the proper fromula and stop renaming things.
Dave Springer you say at May 10, 2011 at 6:49 am
“You can’t heat anything to a higher temperature than the temperature of the photons doing the heating. If you used a parabolic mirror aimed at a warm wall to heat water you could heat it to the temperature of the wall then the heating stops because the object being heated, now at the same temperature as the heat source, emits back as much radiation as it receives. It’s precisely because of back radiation that you can’t heat something hotter than the source of the heat.”
///////////////////////////////////////////
Dave,
What is the temperature of a CO2 re-radiated photon?
What is the (average) temperature of the photons that supposedly downwell about
330 W/m^2 (according to the Trenberth diagram)
“”””” suricat says:
May 9, 2011 at 6:58 pm
Hi Ira.
“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/mm.”
As engineer to engineer. 🙂
A “radiance” is a ‘comparison’ to an accepted ‘standard radiance’ (don’t ask me what frequencies this involves, as that’s defined by ‘the standard’), but an “intensity” is an ‘absolute measure’ of ‘energy’ (don’t ask me how the ‘Planck constant’ is considered there) and seems to be commonly used with a specific spectral wavelength. I hope this helps. “””””
Well units of ” mW/m^2/sr cm^-1 ” are a little bit screwy; and need some brackets to properly set up.
Izzat (mW) / (m^2/sr cm^-1); or is it (mW) /(m^2. sr. cm^-1). I think it is more likely to be (mW/cm^-1) / (m^2. sr). And the proper term is “spectral Radiance”, not “Radiance” because of that , /cm^-1 (per wave number increment) . And it represents Radiant energy in a small frequency (wave number) increment emitted from a small element of surface area (/m^2) into a small elemental solid angle (/sr) to some particular direction in space.
And it is NOT equivalent to “Intensity” which is a property of a point source; so there is no per unit area of the source involved.
We can reasonably talk about “intensity” when dealing with radiation for distant stars; but certainly not when referring to our local star. Well from a computational accuracy point of view, the measure of intensity is accurate to better than 1% so long as you are more than 10 emitting surface diameters away from a non-point source; but the point is there is NO per unit area section in an Intensity specification; but there is in Radiance, or Spectral Radiance, which these graphs properly are, and also in “emittance” which is simply W/m^2 without regard for directional properties, or wavelength or frequency properties, which would intorduce the spectral terms at least.
“Radiance” without the “spectral” carries no information regarding frequency or wave number or wavelength or photon energy or any other such parameter, it simply measures total energy (Watts) emittted into some small solid angle (steradians) at any direction in space, from some small elemental surface area of the radiator. It’s photometric analog, for visible light conditions would be Luminance, or some like to use the term “Sterance” (not me). and “Brightness” would be a common colloquial and somewhat terrifying term, that we should all avoid like the plague; but we won’t, so we should only use it among friends; who can understand what it is we reqally mean. But the lay person will have no idea what exactly we mean.
And the shift in the peak of the BB Spectral Radiance when we change from frequency or wave number over to a wavelength scale, is simply due to that switch from
Robert Clemenzi says:
To get a good figure, you must use the HITRAN data. At 15 cm-1, 99.7% is absorbed in less than 10 meters. At 14.97 cm-1, 99.7% is absorbed in less than 0.8 meters. Both of those measurements are with bandwidths of about 1 cm-1. With a bandwidth of 30 cm-1, 99.7% is absorbed in under 150 meters. To get 3.6 km you must be using a very wide bandwidth.
Robert,
I integrate between spectral band wavelengths of 12.5 to 16.5 microns for CO2.
Joel Shore says:
“…I am not saying that [Hansen] has compellingly shown that these two points materially change things, but just wanted to explain why he feels that something that never happened in the past could happen now.”
If it were not for these endless “what if” scenarios, climate alarmists wouldn’t have much to say.
And the models predicting a tropospheric “hot spot” have been falsified to the point that the backing and filling action now reverts to the stratosphere. But as Prof Richard Feynman makes crystal clear: if the models are contradicted by observations, the models are wrong.
Joel Shore might want to take a few minutes off from writing his lengthy blog comments throughout the work day, and go look out the window. He would see that the climate is normal, and well within its past parameters. The models are wrong, AGW is wildly exaggerated, and the only effect from increased CO2 is on plant growth.
Robert Clemenzi says:
May 10, 2011 at 10:43 am
R Stevenson says:
May 10, 2011 at 1:58 am
I tried to confirm this figure using Plank – Hottel and got 3600m
To get a good figure, you must use the HITRAN data. At 15 cm-1, 99.7% is absorbed in less than 10 meters. At 14.97 cm-1, 99.7% is absorbed in less than 0.8 meters. Both of those measurements are with bandwidths of about 1 cm-1. With a bandwidth of 30 cm-1, 99.7% is absorbed in under 150 meters. To get 3.6 km you must be using a very wide bandwidth.
—–
Excuse me barging in but thanks Robert. I thought that very short mean path length strictly inc CO2’s frequencies was correct. Probably best just converted to tau which used the mean free path as the units maintaining the logrithmic nature of the absorbtion depths. Thanks again, that shores up everything I have been pointing out earlier in these threads in releation to that subject. If the 3600 m had been correct, I would have had to retract much I had claimed earlier, whew!
mkelly,
I’m not sure we are on the same wavelength. I agree with you that “path and temperature difference” are important parameters (especially for conduction). Thermal conductivity is also important, as is geometry. For radiation, emissivity is important, as are the specific temperatures (not just the difference). I think we are in agreement so far.
You said “Until you can show back radiation as an input for heat transfer then use the proper formula and stop renaming things.”
I believe this is indeed the proper formula (give or take a few details like you mention — eg the geometry of the objects and the emissivities of the two different materials.) (And this equation is actually the RATE of energy transfer, which would more commonly be labeled q/t)
q = (ε σ Th^4 A) – ( ε σ Tc^4 A)
or in words
(net energy flow from radiation)
= (“out radiation” energy from the object) – (“back radiation” energy to the object)
I’m not renaming anything. “Back radiation” (which is a function of the temperature of the surroundings) IS an input for finding the heat transfer. It is 1/2 of the equation. I realize you said “basic”, but “basic” cannot involve dropping 1/2 of an equation where both halves are important.
Joel Shore
I said …” A filtered spectrum does not vary with T^4.
The Stephan Boltzmann Law cannot be used for the filtered spectrum of atmospheric gases.”…..
You now agree with me and withdraw this unphysical assumption.
This has implications not just for your absurd paper but for ANY climate model with the T^4 assumption built in!
Its of interest that you do not try to defend the Filtered T^4 atmosphere model that you presented in the article.
Your talk of “simplification” and ” using heat in the colloquial sense of the word”.
Your audience were physicists who must have cringed at the naive mistakes in your article.
I would imagine that virtually all of them knew the thermodynamic meaning of heat.
It is quite clear from your article that none of the Halpern et al team had any clue as to what heat meant.
What was more disturbing is that none of the six of you could read properly.
It is an almost unique example of group delusion!
All of you came to the conclusion that G&T were of the opinion that colder objects could not radiate to hotter objects .
Surely at this point one of you who actually read the paper should have said……
“Hold on, there are a number of diagrams showing two way radiative interaction”.
Another might have said “G&T have an extensive discussion on the Earths two way radiative response to the Sun”.
Perhaps it is your view that your articles comments about HEAT and T^4 should not be taken seriously.
I think your physics audience would take your remarks seriously.
They would form a judgement on Halpern et al based on your comments.
However from now on I will keep in mind that you should not be taken too seriously.
Here is a thought experiment. If we hold everything else equal and double the amount of N2 in the atmosphere, because N2 does not participate in the back radiation, the surface equilibrium temperature should remain “unchanged” according to the radiative transfer model.
However, comparisons of earth and venus suggest that the temperature of the atmosphere will only remain unchanged at 1 atmosphere pressure. As we now have more atmosphere, the pressure at the surface will be increased as compared to present. Under the gravitational model, the surface temperature would be expected to increase as a result of doubling the N2 in the atmosphere.
Which of these two is correct? Adding N2 will not increase temperature, or adding N2 will increase temperature (holding all else equal)?
The problem for me is the lapse rate. Doubling the N2 in the atmosphere will raise the height of the atmosphere. If the surface temperature does not increase, then the lapse rate dictates that since the atmosphere extends higher, that further up, past the present altitude it must get colder than it is at present. That seem nonsensical.
It seems more likely that the top of the atmosphere will remain relatively unchanged, and due to the laspe rate the temperature at 1 atmosphere will be largely as it is now, and deeper down in the atmosphere towards the surface, temperatures will increase. This tells me that the surface temperature is a function of the amount of atmosphere, not (just) the specific gasses, as confirmed by comparison of the atmosphere of verus and earth.
Ira,
You haven’t answered so I’ll give it another try.
Your attribution of the missing 33K from 240W radiation from earths system to “green house effect” assumes an atmosphere that is at 0K AND assumes no reflection (as I understand you, please correct me if I’m wrong). But even without GHGs the atmosphere must be rather warm (the dry adiabatic lapse rate is one lower limit, sensible heat and latent heat is another reason). The atmosphere is also reflective, Trenberth accounts close to 38% of longwave radiation from the ground to be reflected. Third, the atmosphere is heated by sunlight. Again according to Trenberth 78 W/m^2 is absorbed, i.e. 78 W/m^2 of SUNLIGHT is heating the atmosphere.
Let’s assume (for the sake of argument) that the average optical depth is 7 km in the GHG free atmosphere (vs ca 5 km in our). DALR gives us a temperature of roughly 203 K at this point.
If Trenberth is right on 38% reflection, 78 W is heating the atmosphere from above of which half will radiate downwards, and pure adiabatic (dry) lapse rate gives us 203K to begin with, we get 240W*(1-0,38)-78W/2- 5,67E-8*203^4=13,5 W.
So, 13,5 W/m^2 is what is left to be explained by green house gases combined (if the above is correct). Do you disagree? Why so?
Side note. Trenberths “Energy budget” http://mensch.org/5223/other/EarthsGlobalEnergyBudget.pdf has an implicit reflection of the atmosphere of 37% from below but 23% from above. Whenever I fly the clouds are bright white from above, but often dark from below. Should it not be the opposite of what Trenberth claims? Please enlighten me if you know…
Ira, you say in your article above:
“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.”
You are making one assumption there that does not fit in with “Climate Science” at all. It has, ever since the days of Friedrich Wilhelm Herschel and Benjamin Franklin, been known to science that different colors absorb “solar light” at different rates.
Having said all that, I too have been led to conclude that light, and indeed that all kinds of other radiation contains no heat. What light does contain however is “electronic energy”. Or to say it in a different way; light is always “on the move” and is transmitted as a part of the electro-magnetic wave-bands or spectrum, i.e. as radiation. – Only when that radiation is stopped or interrupted by “solids” i.e. atoms and/or molecules is that electronic energy either absorbed (transformed to kinetic energy) or reflected. (We can only see images as reflected light – or it’s source.)
Kinetic energy is used to create molecular movements. – Molecular movements create friction. – Friction creates heat. – Heat can be measured as temperature. – So there it is – as far as I understand it. – Heat is the product of work done by energy provided by an energy source, in this case the Sun’s energy via the curtesy of radiation. Yes, I agree if there is radiation there must also be “back radiation” – Maybe I am not expressing myself very well, but I am hoping to keep my comment as short as possible.
Furthermore, what I find curious with your “workings out” is the same as what I find unexplainable in any other “climate calculations” or models. They all “average” the solar W/m² input over the whole globe, which leads to an unavoidable bias in favor of solar irradiation. In your case, you say:
“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.”
This 0.25 (or ¼) comes about because solar irradiation at the top of the atmosphere (TOA) which you say is 1366 W/m² will only retain it’s full force if it falls perpendicular onto a 2 dimensional disk. (Disk size does not matter as we are talking Watts for each individual, or any one square meter.) The difference between the square area of a disc and that of a sphere is ¼ – so if you divide 1366/4 you get 341.5, but the fact that the Sun is shining on only half of it at any given time, as you correctly state, is not taken into consideration. – According to what I once read about it in “Science of Doom” the thinking behind it was that if radiation out happens from the whole sphere’s surface then “to spread the solar constant” over the whole sphere seems –Fair enough.
And maybe it is, – as the most likely probability is that nobody has a hope in hell of finding the exact average Wattage. But looking at it from the other angle: what if we averaged 0 solar radiation in (which is equally provided by the dark side? Then we would probably freeze to death before waking up in the morning. So of course if we have got sunshine 24/7 then we are going to get warmer! – CO2 or no CO2.
Anyway, my fault is I waffle on, and on, and o – – – – .
Your missing 33K could just be hiding in the old saying that this one is the “Goldilocks’ planet”. Your calculations only encompasses radiation in general and Infrared radiation in particular. It does not take into account anything else – like for example the Earth’s rotation, which is crucial if you wish to find the missing 33K. The Earth never warms “from scratch”, it keeps heat in during the night. In fact you would be hard pressed to register an overnight reduction in temperature in 70% of it (the oceans).
Radiation my dear Ira is moving energy from one place to another – and back again – It may be a cause of “local warming – or cooling”, but is no more a cause of “global warming” than is conduction, convection and advection.
My proof-reader has told me:
Man, your last comment to Ira’s article is too heavy and therefore too dumb.
In that case I shall just say; if you believe in AGW (or may it even just be a bit of warming by CO2), please show me an example where the product has reproduced its maker.
Heat is the product of energy use, so please explain how heat can produce energy.
If you believe CO2 is the producer of heat then please explain how CO2 is produced.
Bryan says:
I would be perfectly happy if you were to just ignore me since you are clearly just interested in peddling your pseudo-science. There are other people here, such as David M Hoffer, who…while we may have strongly different opinions on the importance of AGW…do take me seriously and who I can have serious scientific discussions with. You are off in some make-believe land of your own devising, completely divorced from actual science…and in fact you are actively peddling pseudo-scientific nonsense that I think you are intelligent enough to know is such.
New calcs of radiative GHE shows the GHE is only 8-9 deg. C:
http://activistteacher.blogspot.com/2011/05/radiation-physics-constraints-on-global.html
George E. Smith says: May 10, 2011 at 11:28 am
“””And the shift in the peak of the BB Spectral Radiance when we change from frequency or wave number over to a wavelength scale, is simply due to that switch from”””
That looks to be an unfinished statement George. Would this finish it?
“……, is simply due to that switch from” ‘its inverse form’.
My time is short, so I thank you for your ‘expansion’ for the benefit of ‘Jo public’, but I think Ira understood the gist of my comment.
Best regards, Ray Dart.
JAE says:
May 10, 2011 at 4:03 pm
New calcs of radiative GHE shows the GHE is only 8-9 deg. C:
http://activistteacher.blogspot.com/2011/05/radiation-physics-constraints-on-global.html
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Thanks for the heads up.
The comment by Hans Shruder to that article is particularly interesting. I have repeatedly commented that we will be unable to understand how the atmosphere works as long as we continue to use average figures/average scenarios. This practice really disguises what is really going on. It will therefore be interesting to read the forthcoming paper to which Hans Shrudder refers.