Daniel M. Sweger: Response to Nikolov and Zeller’s Unified Theory of Climate

Posted: March 30, 2012 by tchannon in Astrophysics, atmosphere, climate, Gravity

Response to “Unified Theory of Climate”

Dr. Daniel M. Sweger National College

Introduction

During the last two years that I have been teaching Environmental Science and attempting to explain global warming to my students, I have been frustrated at the lack of solid theoretical understanding of the primary processes coupled with actual data. Speaking as a research physicist, I have been saying to others, “Wait until the physicists get into action. They have an entirely different approach to solving physical problems”.

Sure enough, two such physicists have published their analysis of their fundamental understanding of the climate. In October 2011 Nikolov and Zeller presented their analysis in a poster entitled “Unified Theory of Climate” at the World Climate Research Program in Denver. Their analysis was indeed from fundamental physical and mathematical principles. A month or so later their poster was converted into a document and appeared on Tallbloke’s Talkshop. From there it spread to WUWT and many other blogs, and it has generated a tremendous amount of discussion.

I would like to add my two-cents to that discussion.

Calculating the Average Temperature of the Earth

It can be something of a difficult problem to define what is meant by the average temperature of the earth. Some have argued that there is no such thing as the “average” temperature; that temperature is a localized measure. However, temperature is a measure of the energy in a system, and energy can be averaged.

Virtually all the energy received on the surface of the earth comes from solar radiation. The sun acts like a black-body with a temperature of 5780 K. According the Stefan-Boltzmann law, the energy it emits is

Here j 0 is the irradiance measured in watts/m2, T is the temperature in Kelvins, and σ is the Stefan-Boltzmann constant. Of the 63 million watts/m2 given off by the sun, approximately 1366 watts/m2 is received at the edge of our atmosphere.

However, not all of the solar radiation from the sun has a direct effect on the surface temperature of the earth. Energy that is reflected does not provide any warming effect. That fraction of the energy reflected is called the albedo, which for the earth is about 30%. The amount of the energy absorbed is emissivity, denoted by the symbol ε. Thus,

Equation 1 is thus modified as:

Knowing the radiance, then, the temperature at any given point is:

Thus one only needs to know the average irradiance to calculate the average temperature.

The IPCC Approach

The earth, however, is a rotating sphere, and the problem of calculating the average irradiance is not apparent at first sight. The IPCC made two assumptions to simplify the calculation:

1. At any point the half of the earth is illuminated and half is in darkness. Therefore, the IPCC assumed a hemisphere illuminated by half of the irradiance and:
2. Since the surface area of a hemisphere is exactly twice the surface area of a disc with the same radius, the effective irradiance is halved again:

This process is illustrated in Figure 1.

Using this approximation for the irradiance, j, and applying an albedo of 0.30 the IPCC concluded the average temperature of the earth was about 255K. Since the average temperature of the earth is considered to be 288K then the difference between the calculated and actual average temperatures, which is -33oC, represents the effect of the atmosphere. That is, the presence of earth’s atmosphere raises the average temperature by 33oC. Moreover, the only way to account for this is via the “greenhouse effect”.

The Approach of Nikolov and Zeller (slightly altered):

Nikolov and Zeller took another approach towards calculating the average irradiance, and thus the average temperature. They said the approximations applied by the IPCC were incorrect, and that the only valid method of finding the average irradiance is to perform a “proper spherical integration of the planetary temperature field.” They further specified that they wanted to calculate the average surface temperature of the earth without the presence of an atmosphere at all. They called this “planet” a “gray-body”.

It is at this point that I think many people get confused, since they are not accustomed to differential calculus. So I am going to present a slight altered approach.

At any given point on the sunlit portion of the moon the irradiance is given by

and

or

Here the longitude is given by the angle θ¸ relative to the horizon and ranges from 0 to Π. The latitude is given by the angle φ relative to the equator and ranges from (-Π/2) at the south pole to zero at the equator to (Π/2) at the north pole. Consider the first assumption, that is, the one that accounts for the earth’s rotation, made by the IPCC. If the irradiance as a function of longitude on the surface was a linear function of the sun’s angle then an arithmetic average would be sufficient. However, the angle of the sun is non-linear, so rather than performing an arithmetic average, we need to perform the proper way to average a function over a surface. This is done by integrating the function over the range of the given variable from the minimum value of the variable, a , to the maximum value, b . The average of a function therefore has the general form,

The two angles, Θ and φ, are orthogonal, and are thus independent of one another. That means we can perform the integration of the two sequentially. That is, we can integrate with respect to one angle while holding the other a constant, then, using the results of that integral as a constant, integrate with respect to the other angle. Consider the following scenario:

1. At the equator, the angle φ=0 and cos φ=1, which is a constant. There is no solar energy until the sun crests the horizon in the east. At this point the angle of the sun is very low, and θ is zero. During the course of the day the sun rises to its apex at noon and then declines until it sets, and θ = Π. The surface temperature at any time during the day is therefore a function of the angle θ and is given as: Since the albedo for a gray-body, such as the moon, is approximately 0.11, from equation 2 above the emissivity is thus 0.89.
2. In order to find the average daytime temperature one must now integrate this function with respect to θ over the range of angles for 0 to Π as per equation 10. Everything within the radical is a constant except for sin (θ) and can be factored outside the integral. Thus the integral becomes: As far as I can tell, the above integral does not have a closed form and must be evaluated numerically (Vanovschi). When this is done the value of the integral is :

This yields the average daytime temperature: Taking j 0 to be 1366 w/m2, the average daytime temperature would thus be:

3. Assuming that the nighttime temperature is approximately a constant and lasts the same length of time as the daytime, the average temperature at the equator during one day is

Determining T night is not as simple as setting the irradiance to zero. There are three factors that affect the nighttime temperature:

• The background deep-space radiation of 2.72K, which is the smallest affect.
• The irradiance from the earth. Assuming the irradiance of the earth is equal to 1366 w/m2, the maximum irradiance on the lunar surface is about 0.37 w/m2. This much irradiance would produce a maximum nighttime temperature of approximately 50K.
• The effect of the heat capacity of the moon’s surface. The stored energy from the sun during the day would result in a lag in reaching the minimum temperature at night. According to the published results of the Divinerproject, the minimum nighttime temperature at the equator would be about 90-95K.

Using the third value of 93K yields an average equatorial temperature of about 211K. This compares quite favorably with Diviner measurements.

The second assumption by the IPCC, that the effective surface area is half of the area of a hemisphere, suffers from the same fault. Again, one must integrate the function over the latitude of the sphere.

Having determined the average daytime temperature at the lunar equator from equation 16, we can now average this over the latitudes from pole to pole. At any given latitude

The value of the integral of (cos φ)1/4 in equation 18 is the same as that in equation 14 above. Thus,

We can now average with the nighttime temperature, which is about the same at all latitudes, to get an overall gray-body temperature:

Notice that the results are significantly different than that of Nikolov and Zeller, which was about 155K.1 While 33K higher, it is still significantly lower than the results of the Stefan-Boltzmann approach of the IPCC. Following the approach of the IPCC, the average calculated temperature of the moon would be:

Given the results so far from the Diviner project the IPCC approach is clearly wrong.

Almost all the predictions from the IPCC and others are predicated on the results of models, as if those model results were equivalent to actual, real-live data. These models, however, were all predicated on assumptions concerning the nature of the effect of the atmosphere on temperature.

Nikolov and Zeller have done us a great service by pointing out the errors of the most fundamental of assumptions: the role of the greenhouse effect. Whether you agree with their method or not, their method begins with firm foundational principles, and their results are outstanding. We must come to the place where we follow where the data leads, not our pre-conceived ideas.

Works Cited

Vanovschi, Vitalii. The Number Empire–Definite Integral Calculator.
www.numberempire.com/definiteintegralcalculator.php.

Nikolov & Zeller. Unified Theory of Climate.
tallbloke.files.wordpress.com/2011/12/unified_theory_of_climate_poster_nikolov_zeller.pdf .

Dr. Sweger did his undergraduate work at Duke University, graduating in 1965, and earned his Ph.D. in Solid State Physics at American University in 1973. He is currently adjunct faculty at National College, where he teaches Environmental Science, among other subjects. He previously has posted a paper entitled Earth’s Climate Engine which can be found at
http://junksciencearchive.com/Greenhouse/Earth-s_Climate_Engine.pdf. From the Executive Summary:

“While models can be useful, the results must be compared to actual measurements, i.e. data. Data is the language of science, but little has been done in that regard with the climate change models. It is the premise of the author that water vapor is the dominant influence in determining and understanding global climate. Water vapor is much more abundant in the atmosphere than carbon dioxide, and its physical properties make it more important as well. During daylight hours it moderates the sun’s energy, at night it acts like a blanket to slow the loss of heat, and it carries energy from the warm parts of the earth to the cold. Compared to that, if carbon dioxide has any effect it must be negligible. Thus, the purpose of this paper is to explore the effect of water vapor on climate. Detailed calculations and analysis of data from several locations clearly demonstrate that the effect of water vapor on temperature dominates any proposed effect of carbon dioxide. Furthermore, it is clear from the data presented that water vapor acts with a negative feedback on temperature, not a positive one. That is, the data demonstrate that increasing the level of water vapor in the atmosphere results in a decrease of temperature, not an increase as predicted by the climate models. In essence, atmospheric water vapor acts as a thermostat.”

1 According to the posting on Tallbloke today [probably 2012-03-29], which gives a Diviner-calculated mean lunar temperature of from 192-197K, this is very good agreement.

Document prepared for web, with some necessary format changes, by Tim Channon, co-moderator. There are probably errors in the hand editing of HTML symbols.

[updated PDF Response to Unified Theory of Climate-update-1]

1. Stephen Wilde says:

It is very satisfying to see that others are getting the point. Thank you Dr. Sweger.

Applying the implications verbally for lay readers the implications would appear to be as follows:

If the atmosphere gets warmer due to a composition change then it will expand to reduce density at the surface and weaken the Atmospheric Thermal Enhancement proposed by Nikolov and Zeller and the consequent cooling at the surface will maintain system equilibrium by offsetting the warming of the atmosphere.

That doesn’t reduce or increase the rate of cooling of the surface. It simply gives the surface a proportionately reduced influence on the whole energy budget.

The thicker the atmosphere the more it detaches the surface from the energy budget. One could conceive of an atmosphere thick enough to prevent any insolation reaching it at all but the surface will still be very hot because of pressure and density plus downward conduction.

What we have then is a sun/atmosphere thermal interaction which is initially dependent on conduction from the surface but the thicker the atmosphere, the more GHGs or the more aerosols the more that interaction becomes taken over by the atmosphere whilst the influence of the surface declines.

A completely transparent atmosphere would remain reliant on conduction from the surface but even pure Nitrogen has some absorption capability so even a pure Nitrogen atmosphere could theoretically become thick enough to detach the surface from any direct thermal interaction with the sun. Probably, that would occur when pressure at the surface causes the Nitrogen to turn into a liquid or a solid.

The only difference that GHGs make in such a scenario is to reduce the solar interaction with the surface to that applicable at a lower atmospheric density than would be the case without them.

The logical outturn is that GHGs warm the atmosphere but cool the surface for a zero net effect because they serve only to replace a portion of the sun/surface interaction with a sun/atmosphere interaction.

2. David Appell says:

This is interesting, but I think it goes astray with equation 8. The S-B Law only holds for a flat blackbody. Otherwise you have to consider the curvature of the surface before you integrate over all solid angles, not after, and the area element dA will depend on the longitude.

3. wayne says:

Thank you Dr. Sweger for looking into this matter in detail, many need to thoroughly understand why the IPCC is so wrong in their assumptions. Your integrations are very close to the numerical integrations I performed to check N&Z’s results back in late December.

However, how do I put this, I think you have made a mistake that causes you to end up with a different answer than Drs. Nikolov and Zeller. You have set the emissivity to (1-albedo) renaming that variable, which is fine with me, I can just adjust the nomenclature. But, you left out the true emissivity that is included in N&K’s calculations. They state the relationship as αs ε(renamed)=(1-α)/ε, not just ε=(1-α). I think you will find that is where your results diverge. My first mistake was to adjust the cosine weighting in both orthogonal directions but I was missing the third cosine weight for the decrease in latitudinal area. I was also coming up with a higher figure till I realized my mistake.

Just don’t want too many further mistakes to end up clouding this topic. Let’s get it perfectly correct and stick with it. If your is correct and N&Z is wrong I do want to know exactly why, mathematically.

4. Brian H says:

The date is different (06/29). Maybe you should put up some redirects to the correct link.

[ Ouch! Yes the date is wrong. Is this a bug in WordPress… I spent many hours on the 29th playing ping-pong with the server, the date it was uploaded. It wasn’t published until the 30th, the date on the emails. Don’t think there is a workaround: managed servers. Consider Jabber/XMPP as a better way to get notified.
http://en.support.wordpress.com/jabber/ –Tim]

5. Bryan says:

Ive just finished reading Daniel M. Swegers last paper!
Its interesting to find out how his interesting ideas meld with other of your recent features.
The climate jigsaw will emerge from contributions like those highlighted here.
Michele Casati might even be persuaded to post a feature on his climate ideas!

6. Brian H says:

Since you can’t average the energy in a bucket of water at 300K and an equal mass of steel at 300K, because they have different specific heats, why should you be able to average the energy contents of packets of air, even controlling for density, at widely different RH levels?

7. Daniel Sweger says:

@wayne: Dan Sweger says, thanks for your comment. 2 points: 1: In N-Z’s “Reply to Comments, Part 1”, page 3, top, they state that the emissivity of rock, which in the denominator, is about 0.95. When you take the fourth root of that, the result is 0.99, they rounded to 1. 2: The reason for the difference is how we did the integration. N-Z used a substitution method (eq. 5, page 5), which is not correct for the integral of (sin x)^.25 or (cos x)^.25. I was reluctant to make that explicit point in the paper. Perhaps I should have.

8. wayne says:

Dan, thank you for your comment back. You seem to be correct on that point, I have since literally integrated (boole) those equations, and guess what, 2.6999…, I stand corrected there and that is exactly what we need, correctness to settle this down a bit! Thanks leaving this where everyone should be able to understand exactly why, numerically.

On the emissivity, I know it’s a small correction, after fourth root, just hate to leave it out for then many think it actually doesn’t exist at all and end up argue on that point… it just wastes time and space in already lengthy blog threads. I’d almost wish it was left in even if it is actually one, but really it never is. Maybe I’m just too picky.

9. tallbloke says:

Note David Appell’s recently approved comment:

“This is interesting, but I think it goes astray with equation 8. The S-B Law only holds for a flat blackbody. Otherwise you have to consider the curvature of the surface before you integrate over all solid angles, not after, and the area element dA will depend on the longitude.”

10. Stephen Wilde says:

As regards Dr Sweger’s earlier paper about the significance of water vapour on climate the most important need is to rebut the theory promulgated by Realclimate that more energy in the air from GHGs slows down the rate of energy loss from oceans to space thereby warming the oceans.

I went into that in some detail back in 2009 and came to much the same conclusions as Dr, Sweger about the significance of the water cycle as a thermostat.

http://climaterealists.com/index.php?id=4245

“Greenhouse Gases Can Cause Cooling”

11. wayne says:

“The reason for the difference is how we did the integration. N-Z used a substitution method (eq. 5, page 5), which is not correct for the integral of (sin x)^.25 or (cos x)^.25. I was reluctant to make that explicit point in the paper. Perhaps I should have.”

Daniel, I do have one more question. I’m know wondering how I could have possibly wrong in my numerical integrations that, in fact, exactly matches N-Z’s formula. That program is basically a Monte Carlo of evenly distributed, though not evenly spaced, points on a hemisphere and merely first calculating the decrease in radiation from the zenith angle (simple cosine), converting to an effective temperature, standard equation… ((So+Cs)(1-α)/(ε σ))**1/4, at that point, and doing it a million times. The temps are all accumulated and averaged then divided by two. I get 154.3K as they did. How can that possibly not be matching your figure of 188K? Have I made some mistake in the point-by-point integration? Or, have you yourself actually numerically integrated the entire question?

Heck, I just took the time to write a third version with a more concentric target type geometry instead of latitude and longitude (an over the equator at the prime meridian) version. All say 154.4 K.

using System;
namespace _
{
class Verify_NZ_3
{
static Random rand = new Random();
static void Main()
{
// k is the constant core of Equation 2 - std physics
double k = (1364 + 13.25e-05) * (1 - 0.12) / (0.955 * 5.6704e-08);
double accumTemp = 0;
double accumWgts = 0;
for( int i = 0; i < 1000000; i++ )
{
// theta(lon) irrelevant here, due to circular symmetry
//double theta = rand.NextDouble() * Math.PI * 2; // never used
double phi = rand.NextDouble() * Math.PI / 2;
double cosPhi = Math.Cos(phi);
double sinPhi = Math.Sin(phi);
double temp = Math.Pow(cosPhi * k, 0.25);
accumTemp += temp * sinPhi;
accumWgts += sinPhi;
}
double avgT = accumTemp / accumWgts / 2;
Console.WriteLine("avgTemp: {0:f1} K", avgT);
}
}
}


There’s my code and I, once again, can find nothing wrong, but the answer you will get when you run it is 154.4 K. Please help show where ‘I’ am wrong too. Are you absolutely sure your math is right?

I can be very stubborn when it comes to details in physics.😉

12. Tenuk says:

Thanks Daniel Sweger, for a nice clear post, which broadly confirms the N&K findings. however
<b<Calculating the Average Temperature of the Earth
“It can be something of a difficult problem to define what is meant by the average temperature of the earth. Some have argued that there is no such thing as the “average” temperature; that temperature is a localized measure. However, temperature is a measure of the energy in a system, and energy can be averaged.”

Thermodynamic variables come in two varieties: extensive and intensive.

Extensive variables:
These variables relate to the size of the system and can be added. Examples In this category are mass, volume, particle number, energy, entropy e.t.c. For any one of these extensive variables it is perfectly valid to calculate the mean and a meaningful answer will result. So if we have 5 potatoes of different sizes and weigh them to get their total mass, we can easily calculate the average mass of each potato by dividing by 5, the number of components.

Intensive variables:
These variables are independent of system size and represent a quality of the system, such as temperature, pressure, chemical potential e.t.c. Combining two systems will not yield an overall intensive quantity equal to the sum of its components. For example two identical subsystems do not have a total temperature or pressure twice those of its components. A sum over intensive variables carries no physical meaning. Dividing meaningless totals by the number of components cannot reverse this outcome. A good example of a meaningless average is the one you get by finding the arithmetic mean of all the telephone numbers in the phone book. The answer is mathematically correct, but the result has no usefulness.

In some systems averaging might approximate the equilibrium temperature after mixing, but this is irrelevant to the analysis for a highly dynamic system, like the Earth’s climate, which displays spatio-temporal chaos.

13. Roger Longstaff says:

Should not emissivity be on the other side of equation 3?

Wayne – I have been following your work with interest. For what it is worth I think you are on exactly the right track to add value to the work of N&Z and Huffman. Please keep up the good work and publish when you are ready, or at least offer a post here.

14. mkelly says:

” While 33K

higher, it is still significantly lower than the results of the Stefan-Boltzmann approach of the IPCC. Following the approach

of the IPCC, the average calculated temperature of the moon would be:”

There is that pesky number 33 again. There is something in doing the maths that brings out the number. Find why this number pops up and then things can be reconciled.

15. Daniel Sweger says:

Clarification: Concerning the problem of performing the integral of (sin x)^.25 and (cos x)^.25. N-Z’s method of substitution was to let u=sin x and du=dx, and then [(sin x)^.25]dx=u^.25 du. The error in that substitution is that, if u=sin x, then du = cos x dx. In order for that substitution to work, the integrand would have to be [(sin x)^.25] (cos x) dx=u^.25 du. That clearly is not the case. There is a nice description of the substitution method at http://www.math.wisc.edu/~park/Fall2011/integration/Integration%20substitution.pdf.

Wayne: I think the problem in your code is in the next to the last line: accuTemp +=temp * sinPhi. The last term, sinPhi, falls under the radical, so it should be taken as the fourth root.

Roger: my use of epsilon is slightly different than N-Z. It is the same as their (1-a) in “Reply to Comments”, equation 4.

16. tallbloke says:

Stephen says:
“even pure Nitrogen has some absorption capability so even a pure Nitrogen atmosphere could theoretically become thick enough to detach the surface from any direct thermal interaction with the sun.”

ferd berple says:
January 9, 2012 at 1:23 am

Willis Eschenbach says:
January 9, 2012 at 12:55 am
ferd, the N2 is the most unlike the others because the line strength is many, many orders of magnitude weaker than that of the others.

Perhaps you misread the reference? From what I see, N2 line strength is 10-28, CO2 is 10-23, which is 5 orders of magnitude. However, N2 has 10 times wider spectrum (600 cm-1 versus 50 cm-1). In addition, there are 4 orders of magnitude more N2 in the atmosphere than CO2. So, on this basis it is hard to see that N2 absorbs/radiates significantly less than CO2.

In contrast to CO2, H2O line strength is 10-19 which if 4 orders of magnitude stronger than CO2. As well it has a much, much wider spectrum than CO2. The absorption strength and spectra of water so overwhelms CO2 as to make it CO2 a joke when you consider the amount of H2O in the atmosphere as compared to CO2.

http://www.coe.ou.edu/sserg/web/Results/Spectrum/n2.pdf

17. tallbloke says:

Joel shore has this to say:

“Daniel ( https://tallbloke.wordpress.com/2012/03/30/daniel-m-sweger-response-to-nikolov-and-zellers-unified-theory-of-climate/#comment-21618 ): It is you, not N&Z, who get the mathematics wrong. N&Z are doing an integration over a sphere where the polar coordinate theta is defined with the sun being directly overhead. For such an integration, the solar irradiance is proportional to cos(theta). However,[because] the differential area element on the sphere is sin(theta)*d(theta)*d(phi). Since, sin(theta)*d(theta) = -d[cos(theta)], they can do the integration analytically and have it done it correctly.”

Actually he has quite a lot more to say too, but it’s provocative and obfuscatory, so I won’t bother the open forum with it.

18. Thanks for this.

Inch by inch I’m being dragged back into maths

My gut instinct is that Nikolov and Zeller have got the integral correct. But Sweger has helped me understand better the nature of their double integral. Of course, the delta point has both latitude and longitude, orthogonal to therefore independent of each other. Of course, it’s integration from point thru line to surface, has to be a double integral. And I expect it is a standard integral too, (hemi)surface of sphere.

But ah, I like the inclusion of earthshine! However………. recalculating your figures from N&Z formula, I only get a maximum nighttime temperature of 40K … and… Holders Inequality sees Earthshine effects shrinking into insignificance.

Watch the effects for a hemisphere, using N&Z formulae:

Sunshine, no absorption J=1362 W/m^2, T=309K
Sunshine, with absorption factored in J=1356 W/m^2, T=309K
Night, absorption only emitted J=6.5 W/m^2, T=81K
Night, absorption emitted + max earthshine J=6.9 W/m^2, T=82K
Average temperature Tavg = (309+81)/2 = 195K

It takes less than half a percent average absorption of sunlight (6.5 W/m^2 from 1362 W/m^2) to raise the lunar average temperature from 155K (pure radiative effects) to 195K (as measured).

Game set & match I suspect.

19. wayne says:

Daniel, I better drop out of this discussion, not because I don’t realize what I am doing, but the math vocabulary, that is my lack of math vocabulary, is going to get in the way.

The sine you are saying is under the radical… it is not and should not be, that sine is the fix to correct for the fact that you will end up with an overweighting of points at the pole. The cosine is under the radical as you see. In this numeric integration the zero is directly above, at the pole. You are using a sine with zero on the horizon. By the way, I love your papers so far and are going to read them for reference and I’m not trying to pick without a reason. Reading in your papers so far, it seems you can make that mental jump from a sine centric spherical view to a cosine view.

NED, HELP! I need someone who can talk calculus! I can do it, I just can’t speak it.

Best I can say is you appear to be integrating to get the mean fourth power of all sines and calculating the mean temperature once. I have always viewed N&Z taking the cosine (zenith being zero) at each to calculate a per-point temperature and then averaging as long as points are evenly distributed. To me, that is the reason N&Z made the substitution, not an integration of the cosine itself. In math speak, I probably can’t explain it any better than that. I just can’t figure out why N&Z and all three of my numeric integrations in different geometries are all wrong at the same time if you are correct on that point.

This does temporarily seem to put a hold on some work that critically relies on this very critical topic and equations. I want them right.

20. mkelly says:

There is that pesky number 33 again. There is something in doing the maths that brings out the number. Find why this number pops up and then things can be reconciled.

Talk to the freemasons. It’s clearly a conspiracy

21. David Appell says:

Daniel, to make my comment above (and Joel Shore’s) clearer: The SB Law does not hold pointwise, as you assume in equation 8. Notice that equation predicts T=0 at certain points on the sphere — a clear impossibility and a violation of the 3rd law of thermodynamics.

22. Roger Longstaff says:

Dr Sweger,

Thank you for your clarification. But may I respectfully suggest that you stick to standard nomenclature in physics (eg the SB equation), otherwise other physicists can not, or will not have the inclination to, follow your thesis. (I gave up after the “mistake” in equation 3).

I am sorry to be such a pedant but may I suggest that you reformulate your analysis, using standard nomenclature?

23. Wayne is correct and Dr Sweger is not.

Not only does Dr Sweger unaccountably fail to understand that emissivity is not the same as “one minus albedo”, but also he gets his re-working of N&Z’s formula wrong.

Like Wayne, I have developed an Excel VBA program to perform an independent test, namely a finite numerical integration of the N&Z equation, and I can absolutely confirm that N&Z got their spherical integration completely correct.

With an iteration count of 10,000,000 (that is dividing up the spherical planetary body into 10,000,0000 spherical segments of equal width (and therefore equal zonal area), my calculated value for the mean surface temperature is the same as that obtained from the N&Z formula to seven decimal places.

The rationale behind the N&Z formula is physical not mathematical. Anybody who cares to spend the time applying the same physical rationale to developing a computer program will, like Wayne and me, come to exactly the same conclusion. If they make a mistake in their programming they will not get a coincidence of results correct to seven decimal places, so this is a genuinely independent check.

Of course it is always possible to challenge the underlying physical assumptions behind the formula. I do not do so and, as far as I can tell, neither does Dr Sweger. But it is not remotely possible that N&Z got their spherical integration wrong.

Back to the drawing board, Dr S!

24. davidmhoffer says:

If I read correctly, Joel Shore confirmed the accuracy of this part of N&Z.

According to trusted sources, the news of this caused Willis’ head to spin completely round three times on two different axis simultaneously.

This, to me, is one of the most important aspects of N&Z and it is unfortunate in the extreme that the major players in the blogosphere have gotten themselves into a tiff over the paper, with positions becoming increasingly entrenched. People please! There are real human beings out there who are really losing their jobs or really starving to death because of the obscenity that has become CAGW. Can we PLEASE put aside the vitriol over the state of the bathwater and remember that there is a baby in it?

I can’t say for certain if N&Z, Dr S, Wayne, David S, Lucy or anyone else is right or wrong. The point is that ALL their approaches arrive at a value about 100 degrees less than the IPCC! While we split hairs over the “right” way to do it, we should be singing the same song at the top of our lungs from every megaphone we have.

THE IPCC DIDN’T JUST GET IT WRONG, THEY GOT IT MASSIVELY WRONG!!!

Anthony, a large portion of this rant is aimed at you. The largest megaphone that the skeptic side has available is yours. This isn’t about proving N&Z right. This is taking the parts of N&Z that there is agreement on using it to show that the IPCC is WRONG.

25. mkelly says:

David Socrates says:

March 31, 2012 at 5:26 pm

Wayne is correct and Dr Sweger is not.

Not only does Dr Sweger unaccountably fail to understand that emissivity is not the same as “one minus albedo”,

Correct albedo is reflection divided by incident.

26. David Appell says:

Someone may have pointed this out before, but by the same reasoning as mine and Joel Shore’s above, N&Z make a similar mistake in the way they use the SB law.

Equation 4 is incorrect in their paper at

If you examine the derivation of the SB Law from the Planck formula for the blackbody radiation spectrum, you find that it holds only for flat surfaces, and only after integrating over all solid angles from the surface.

But N&Z are applying the SB Law to a nonflat surface — specifically, to the nonflat infinitesimal area elements at each point on the sphere.

Thus, they are missing part of the emitted radiation — the parts at angles greater than 90 degrees from the normal.

If this isn’t clear, imagine you’re standing on a sphere. Hold out your right arm at a 90 degree angle. There will be space between your arm and the surface as it curves away. That’s the piece N&Z are missing, and that’s why their calculated temperature is too low.

You really have to be careful about including the entire solid angle from an emitting surface, planar or not.

27. tallbloke says:

David: See Lucy’s comment above. N&Z’s temperature is now in good agreement with DIVINER’s empirical data. You’re ‘missing heat’ is nowhere to be found.

Their integration is correct. It is the flat disk approximation which is wrong. Using Moon albedo of 0.11 gives 270K. Way high. Using Earths albedo for the moon gives ~255K, which is a fudge.

A hemisphere can be considered as a collection of flat surfaces anyway. See wiki.

28. Harriet Harridan says:

I can’t say whether Dr Sweger’s maths is correct, or Dr’s N&Z’s is correct, but I do know that Dr Sweger has done a great job of helping the process along. The numbers are different, but as noted above by Mr Hoffer they are both a lot closer than the IPCC, and perhaps the key problem of their approach has been identified which is the important result. The maths will iron itself out given open minds and a bit of time and braincells, and there seems to be a lot of braincells about here!

Thanks Dr Sweger.

29. David Appell says:

> A hemisphere can be considered as a collection of flat surfaces anyway. See wiki.

Not when you’re integrating over it. Which is what is done to derive the SB Law — an integration of the radiation over all solid angles coming off the surface. This clearly includes, for a point on a sphere, angles at 90 degrees greater than the normal. It is those angles missing in N&Z’s analysis — thus their temperature is too small.

30. tallbloke says:

David: N&Z have resolved the issue of the nightside emission and are now getting a theoretical result which matches the DIVINER empirical result. Their temperature is correct. The game has changed. Since the radiation is from the surface, there can be none emitted at more than 90o which would not be re-absorbed immediately.

31. David Appell says:

> David: See Lucy’s comment above. N&Z’s temperature is now in good
> agreement with DIVINER’s empirical data.

I have tried, but can’t follow Lucy’s numbers because I don’t know where she’s pulling them from. I hope she will put in some links to the data.

Also, without a very careful consideration of the position of the Earth in its orbit, and the phase of the moon, I’m quite dubious that one can look at lunar data for a particular day, read off its “average temperature,” and compare that to a simple blackbody formula. There are a lot of astronomical subtleties to consider.

[Reply] Wait for the N&Z update, details are currently embargoed. The DIVINER data is integrated from empirical measurement taken over a long period. These people are not stupid, and have factored the Moon’s various perturbations and perigrinations into their calculations. At last we have properly done science giving us real results we can rely on.

32. David Appell says:

Since the radiation is from the surface, there can be none emitted at more than 90o which would not be re-absorbed immediately.

Of course there can be, because the surface is curved.

Imagine the surface was concave instead of convex, such as inside a sphere. By N&Z’s reasoning, and yours, the same SB equation would still hold. But for a concave surface, the parts of the surface away from whatever particular point we’re considering would radiate toward that point. Clearly, curvature matters.

[Reply] It matters to the extent of the coarseness of the integration. David Cosserat’s multi-million point iterating model gets exactly (to seven decimal places) the same result as N&Z. Your argument is spurious. The ‘falloff’ due to curvature is covered by adjacent integration points and is negligible.

33. wayne says:

Dr. Sweger, Daniel, I so respect your approach I fought with myself before even posted above. I hate to be the harbinger of bad news. However, it’s not that bad, just off 34 K, you will find it make it a harder problem that at first sight. I kind of wish the extra 34 K was there for there then would be no question that the Earth is receiving enough energy to account for the surface temperature totally without any consideration to atmosphere at all. But do continue on your track for I like the rest of your approach.

I owe thanks to TB, Joel Shore, mkelley, and anyone else I might have overlooked for the clarity, I can now continue on my trek. That equation does have some aspects not very intuitive. I too was at first trying to integrate the cosine to the fourth root back in January, really, that is why I went all of the way down to by hand program the integration where I could actually see the integration step-by-step.

34. wayne says:

> A hemisphere can be considered as a collection of flat surfaces anyway. See wiki.
David Appell: Not when you’re integrating over it.

So what David do you think integration is then? Is it not the sum of infinitesimal flat surfaces? And at ‘over 90 degrees’ there is no incoming radiation. I guess I just don’t see the point you are trying hard to make. Are you looking at the ‘any direction’ outbound radiation instead of the parallel ray inbound radiation? I don’t see where N&Z are ‘missing’ anything. They know in their words that the equations are in a pure form and ignore any thermal inertia from the surface mass and diffusion.

35. Daniel Sweger says:

Folks, thank you for pointing out the error in terminology. I meant to use the term absorptivity instead of emissivity; mea culpa. Also, I chose to use emissivity of 1, since the value for the moon is about 0.95. However, that does not change the process of finding the average temperatures.
What is being overlooked by some of the commenters are the results.On 3/28 Tallbloke posted some updated empirical results of the Diviner project: a) the average temperature at the equator is 213K; and b) the overall average temperature for the moon is between 192-197K.
If I now include the emissivity of 0.95, the result for eq. 16 (average daytime temperature at the equator) is 333K. Using a slightly higher nighttime temperature of 100K (which is reasonable based on the figure in that post) in the calculations the result for the average temperature at the equator in eq. 17 is 216K. This compares favorably to the empiriical result (213K). Extending those calculations to the overall average temperature, the result is 286K for eq. 19 and 193K for eq. 20. This is also a favorable comparison to the empirical result (between 192-197K).
This confirms in my mind that the integrations were done correctly.

36. I am not going to critique the maths, just the typing
2. Since the surface area of a hemisphere is exactly twice the surface area of a disc with the same radius, the effective irradiance is halved again:

[ twack! –Tim]
Thanks

37. wayne says:

Exactly Dr. Sweger! Your right, that is the more important point of your whole paper and you have some good science within. I’m going to enjoy referring back to it, for your overall logic is right on the spot!

38. tchannon says:

MODERATOR WRITES: An updated PDF has been added to the article.

39. Phil says:

Problem could be that LW has no vector at the surface, whatsoever..it is just an amount. SW from the Sun has a vector which needs to be accounted for to get the LW production value at the surface.

Both sides of the debate feel hell bent to combine these values in direct inter-relation in SPEED and RATE which is not true, S-B himself states this cannot be done.

I don’t see how the IPCC can be so clueless, it’s as if they know they’re wrong but won’t fess up.

40. John Andrews says:

Willis Eisenbach has pointed out elsewhere that the albedo of the earth is not a constant, but may be changed by the thunderstorm cloud process. (The Thermostat Hypothesis, WUOT, Posted on June 14, 2009 by Anthony Watts, Guest Essay by Willis Eschenbach)

41. David Appell says:

I left a comment here many hours ago showing how the standard SB law easily explains the lunar temperature — was it received?

[Reply] Patience please David, I’m a one man operation in a different time zone to you, and I do sleep occasionally.

Summarizing Pierrehumbert pp 152-153: The moon has little atmosphere, and low heat conductance of its surface material, so at the point directly facing the Sun, where the temperature will be its maximum, the SB law says

sigma*T^4 = S*(1-alpha)

where the solar constant S is effectively that of the Earth’s. With an albedo=0.11 you find

T=382 K

which agrees exactly with the NASA data by Vasavada (since we’re not really sure *exactly* what values to take for the solar constant or the albedo).

As the moon rotates this point will move and the incident radiation on it will vary like

S*cos(2*pi*t/T)

where t is time and T is the period of rotation of the moon. So you get a factor of

cos(angle)^(1/4)

which is exactly the curve the lunar temperature follows in Vasavada’s data:
https://tallbloke.wordpress.com/2012/03/28/empirical-results-from-diviner-confirm-s-b-law-was-misapplied-to-moon/

Then the moon’s rotation takes it around and out of the sunlight after 1/4th of a lunar day, or 6 lunar hours. At that point the temperature is determined by whatever radiation is incident — the cosmic microwave background, galactic cosmic rays, etc — plus the small but nonzero heat conductance of the lunar regolith provides whatever heat it does, essentially a constant — after all, the dark side of the moon does not have T=0.

So you get the rest of the rest of Vasavada’s curve, though the value of about 95 K isn’t solely calculable by radiative considerations — the surface material properties matter too.

So the standard SB Law explains the lunar temperature quite well.

As I explained earlier, both N&Z and Daniel Sweger do their averaging wrong. This shows it.

It’s astonishing — and sad, really — that commenters here like “Phil” believe the IPCC knows they are wrong about something this basic but is hiding that or repressing it or…something. That’s absurd, and not how science — or scientists — operate.

[Reply] ‘Standard SB Law’ produces a Moon average of ~255K (if you use Earth’s albedo – a ludicrous fudge), this is what we have been fed for years to support the ‘grey body’ Earth temp. It gets around 270K (also ridiculously high) if you use the Moon’s albedo . N&Z do the job correctly, with properly derived equation, on a rotating sphere not a flat plate, and get the right answer with their new term for regolith heat retention factored in. There will be an official update to their ‘Reply to comments Part 1’ soon. We won’t allow apologists like David to re-write hitory and we are building a dossier of the official documents stating incorrect figures which proves him wrong in this case.

42. Roger Longstaff says:

Dr Sweger – thank you for correcting the terminology in the PDF, your analysis is easier to follow now.

Having followed the debate on and off for about a year now I am very excited by the work of N&Z, Huffman and those here – the physics is starting to make sense, but I think it is still incomplete. Albedo, for me, is the elephant in the room (and perhaps emissivity). So here is my dumb question – has anybody tried to determine a relationship between albedo and emissivity using empirical planetary data, and could this then be translated into a continuous function that could be included in the calculus?

43. David Springer says:

Problems:

1. Albedo is generally a measure of reflectivity of visible light. Emissivity on the other hand makes no artificial distinction. Only about half of the radiation from the sun is visible light. How is “albedo” defined in your missive and how is it measured?

2. Nothing except more power or less emissivity can raise something to a temperature above the S-B temperature. Greenhouse gases in effect lower emissivity. There is therefore a limit that the greenhouse effect can acheive. This limit is ostensibly the S-B temperature for 342W/m2 with emissivity of zero. That temperature happens to be 279K.

3. The calculated “average” temperature of the earth is a ballpark estimate. Because the temperature/power relationship is not linear and the S-B equation is for an evenly illuminated body in equilibrium, the real number might be significantly higher or lower if effective albedo is not evenly distributed from low to high latitudes.

4. The greenhouse effect is due to the atmosphere. The difference between a greenhouse gas and a non greenhouse gas is the former is transparent to shortwave energy from the sun and opaque to longwave thermal radiation from insolated surfaces. Liquid water is transparent to shortwave energy from the sun and quite opaque to longwave thermal radiation. Given the mass of the ocean heated by the sun is many times the mass of the atmosphere above it and pound for pound water has twice the heat capacity of air then we can easily see that the greenhouse effect comes primarily from the ocean not the atmosphere.

5. The Diviner data illustrates the problem with trying to calculate an S-B average temperature for a body that is never evenly illuminated. The temperature swing on the moon at the equator is ~100K to ~400K. At 100K the surface is emitting 6W/m2 while at 400K it is emitting about 1500W/m2. Taking the mean power and using that to calculate mean temperature is an exercise in futility for a surface which undergoes such extreme temperature swings. This calculation would only be valid if a surface at 100K emitted 25% of the energy of a surface at 400K. The reality is that a surface at 100K emits less than 1% the energy of a surface at 400K.

The calculation in #5 works much better for the earth because the earth is over 70% covered by a a deep ocean which practically nullifies diurnal temperature variation. In this case the mean power input correlates very well with mean temperature.

44. David Springer says:

Stephen Wilde says:
March 30, 2012 at 11:31 pm

“The thicker the atmosphere the more it detaches the surface from the energy budget. One could conceive of an atmosphere thick enough to prevent any insolation reaching it at all but the surface will still be very hot because of pressure and density plus downward conduction.”

We usually discount internal heat of formation and radioactive decay because the internal leakage at the surface is tiny compared to insolation from the sun. However, if the atmosphere is so thick that there is little insolation from it reaching the surface then the internal leakage becomes a large factor in surface temperature. In other words the thick atmosphere alters the thermal gradient from molten core to crustal surface. This is what makes Venus so hot at the surface. Venus’ atmosphere traps internal heat and where on the earth it quickly escapes once it reaches the surface making its effect neglible, on Venus it cannot easily escape once it reaches the top of the crust so it builds up until the temperature is so high that it can escape via conduction instead of radiation from rocks and/or convection in the ocean on the earth.

One must be very very careful comparing other bodies in the solar system with the earth. No others are covered by liquid water and water is radically different than rock in its physical properties. Engineers appreciate the different physical properties of materials better than anyone else. Lives and treasures depend on it. Nobody dies or goes bankrupt when an academic fails to take all the facts into account in a paper. Worse, if they’re tenured they can’t even be fired for incompetence.

[Reply]Greenhouse gases don’t ‘trap heat’ because convection rules. They are certainly not going to raise surface temperature over a long timescale as internal heats leaks out. The ‘coincidence’ that Venus’ surface temp above grey body temp matches the DALR x the height from tropopause to surface just as Earth’s does, is a strong clue that atmospheric mass rather than solely composition is the key to understanding ‘the greenhouse effect’. GHG’s are necessary for cooling to space at altitude, their role in warming the surface is minor to non-existant IMO.

45. Roger Longstaff says:

David, “How is “albedo” defined in your missive and how is it measured?”

Albedo in the suggested form of analysis would have to include reflectivity at all frequencies, measured by spacecraft and as a function of incidence. I think ESA and NASA have the data, at least for some planets – do you think it is worth trying to find out?

A further problem is that any general expression must apply to planets with atmospheres and those which do not – perhaps requiring an expression for condensation that collapses to zero for a grey body?

Like I said, probably a dumb question, but if it could derive a term that could be included in N&Z the whole thing could make a lot more sense (at least to me).

I agree with your point about mean power and mean temperature (an exercise in futility), the calculation has to include a proper integration.

Finally, you use the term “greenhouse effect”. I never use that term any more as it means different things to different people.

Cheers, Roger

46. David Springer says:

Tallbloke, is there some particular reason that my comments are getting routed into the spam queue instead of the regular moderation queue? When I was running a WordPress blog I’d put certain email addresses into the blacklist so they’d get special scrutiny before approval. 😉

[Reply] Yes. You don’t get to be unresponsive without consequences. It’s bad etiquette. The presumption that you can pontificate about here without the need to respond to others is part of your arrogance which leads you to refer to us as “science cranks” and we don’t have to put up with it.

47. David Springer says:

Roger Longstaff says:
April 1, 2012 at 10:50 am

David, “How is “albedo” defined in your missive and how is it measured?”

Albedo in the suggested form of analysis would have to include reflectivity at all frquencies, measured by spacecraft and as a function of incidence. I think ESA and NASA have the data, at least for some planets – do you think it is worth trying to find out?

It’s probably a pretty accurate average for the moon across all frequencies (but it’s still worth knowing for sure how it was obtained) but it’s a nearly intractable problem for the earth getting an accurate average albedo. Depending on who you ask the value is given anywhere from 30% to 40%. Several attempts have been made to obtain an accurate measure year over year but there is no satisfactory agreement between them. Google “earthshine” for a good synopsis of the situation. Earthshine was one attempt that ran for about 5 years IIRC (2000-2005) measuring the illumination of the moon lit only by the earth.

The lack of ability to accurately measure and track the earth’s albedo to better than +-5% makes the whole notion of being able to determine what the earth’s temperature should be a poor joke as a 1% diffence in albedo is greater than all the anthropogenic forcings combined.

48. David Appell says:

> N&Z do the job correctly, with properly derived equation

Incredible. Even shown how standard theory provides the correct numbers — not just the average, but across the entire dataset, for all 24 lunar hours — you still cling to your illusions.

Didn’t you say that the empirical data should be the deciding factor?

[Reply] Even before the regolith heat retention factor is included, N&Z’s original estimate is about twice as close to the empirical value compared to the BB temperature given in the NASA Moon factsheet.

49. David Appell says:

Roger Longstaff says:
David, “How is “albedo” defined in your missive and how is it measured?”

It’s just the standard Bond albedo of the moon:
http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html

> A further problem is that any general expression must apply to
> planets with atmospheres and those which do not

Why? Who says there must be a expression that covers both cases? The same physics covers both cases, but a planet without an atmosphere behaves very differently from one with — its surface is not anywhere close to thermal equilibrium.

[Reply] Earth’s surface isn’t anywhere close to equilibrium either. Compare annual average temperature at the poles to that at the equator.

50. David Appell says:

[Reply] Earth’s surface isn’t anywhere close to equilibrium either. Compare annual average temperature at the poles to that at the equator.

Sure it is. If you perturb the Earth system, say by increasing the solar irradiance by some amount, you will see a corresponding increase in the temperature across all regions, to first order — which is all the standard application of the SB law gives you anyway.

On the other hand, such a solar irradiance increase will only affect lunar surface temperatures directly where the radiation impacts, and not on, for example, the dark side of the moon. It’s nowhere near equilibrium.

[Reply] You are contradicting your earlier argument. Holder’s Inequality comes in to play here, and your argument that the BB temperature is good for Earth because it is near equilibrium fails.

51. David Appell says:

[Reply] Even before the regolith heat retention factor is included, N&Z’s original estimate is about twice as close to the empirical value compared to the BB temperature given in the NASA Moon factsheet.

NASA has the correct lunar BB temperature. What you have incorrect is thinking that that one BB temperature applies to the entire surface. It does not, since the surface is not in equilibrium (since there is essentially no atmosphere).

[Reply] Not I, but all the warmist webpages I’ve found assuring us that the Moon’s average surface T is around 255K are the ones that have it wrong. I know they’re wrong. And so do you. I’ll snip further spurious argument from you now so the thread doesn’t get derailed by your deluge of deliberate misinterpretation of other’s words.
.

52. David Appell says:

[Reply] Not I, but all the warmist webpages I’ve found assuring us that the Moon’s average surface T is around 255K are the ones that have it wrong. I know they’re wrong. And so do you.

So they’re wrong. So what? There’s loads of incorrect science on loads of blogs. This is the beginning of correcting them — and it would help if you corrected your earlier headline “Empirical results from DIVINER confirm S-B Law was misapplied to Moon” and attached a note up top saying that standard theory does provide the correct numbers.

[Reply] Heh, not so fast. I’ll wait until you give me your alpha and sigma quantities and the link to data plus derivation thanks.😉

53. David Appell says:

It does not, because as Joel Shore and I have been pointing out, N&Z apply the SB law incorrectly and do not get their solid angle integration right. And the DIVINER data shows it — their method gives the wrong answer for the average temperature, while the standard theory gives exactly the right answer for both the average and for the curve.

[Reply] Are you sure that’s what Joel Shore has been pointing out? Here’s what he said:

“N&Z are doing an integration over a sphere where the polar coordinate theta is defined with the sun being directly overhead. For such an integration, the solar irradiance is proportional to cos(theta). However, the differential area element on the sphere is sin(theta)*d(theta)*d(phi). Since, sin(theta)*d(theta) = -d[cos(theta)], they can do the integration analytically and have it done it correctly.”

Doesn’t look like Joel and you agree to me. So which one of you two is Wrong David? Hmm?

54. tallbloke says:

“One moment in annhiliation’s waste
One moment at the well of life to taste
Come, for the Moon is setting, and the caravan
Starts for the dawn of nothing, Oh make haste!”

– Omar Khayyam – ‘ Rubayat’

55. David Appell says:

[Reply] Heh, not so fast. I’ll wait until you give me your alpha and sigma quantities and the link to data plus derivation thanks.

I’ve given them: sigma is the SB constant, and alpha=0.11, according to NASA:
http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html

56. David Appell says:

[Reply] Are you sure that’s what Joel Shore has been pointing out? Here’s what he said:
…Since, sin(theta)*d(theta) = -d[cos(theta)], they can do the integration analytically and have it done it correctly.”

It’s unclear what Shore means, due to the mistype at the end of the sentence. Does he mean they have done it correctly, or they can do it correctly with the correct solid angle element?

If they have done it correctly, why didn’t their result agree with the DIVINER measurements? Standard theory agrees with all of it.

[Reply] Give it up David, their spherical integration is correct and your ‘beyond 90 degrees’ argument is spurious. Let it go. Their result didn’t agree with the latest DIVINER estimate because they didn’t include a term for heat retention in the regolith. Update will be sometime this week. Relax and let people do some science, instead of trying to score petty points.

57. David Appell says:

Doesn’t look like Joel and you agree to me. So which one of you two is Wrong David? Hmm?

What do you mean? Should the calculation agree with the data or not? Standard theory agrees with the data — ALL OF IT.

What more can possibly be necessary?

[Reply] You still haven’t shown how ‘standard theory’ gets the value for the night side of the Moon.

58. wayne says:

Dr. Sweger, take a look at what I have narrowed down to be the difference between your calculations and N&K’s. I’m not going to call either right or wrong, I simply don’t know but I can now see the very subtle difference in the way the calculus is performed. If you notice my derivation of N&K’s figures does not match the way they have explained it at all but the end answer ends up the same. The double square roots are just for brevity, and you will notice some reductions I left in so you can see where they came from!

case.A = 1/2 * √√(2.0/m.pi)² * √√(1362*(1-0.11)/(0.955*SB)) = 154.3 K
N&K

case.B = 1/2 * √√(2.7/m.pi)² * √√(1366*(1-0.00)/(0.880*SB)) = 188 K
(yours)

Here is a detailed breakdown of what I have been working on, partly to further my understanding of the calculus itself (I’m self taught).

Starting with this difference that N&K are basically integrating WITHIN the radical of cosine, not the fourth root the cosine (or equivalent sine with different domains). And I sure hope all of this latex formats correct!

$\large T_{eff. \; mean} = \sqrt[4]{\frac{S_0\;(1-\alpha)\;\left[ \frac{1}{\pi/2-(-\pi/2))}\int_{-\pi /2}^{\pi /2}cos(\theta)\right]\;\left[ \frac{1}{\pi/2-(-\pi/2))}\int_{-\pi /2}^{\pi /2}cos(\phi)\right]}{\epsilon\; \sigma}}$

Simplify a bit:

$\large T_{eff. \; mean} = \sqrt[4]{\frac{S_0\;(1-\alpha)\;\left [ \frac{1}{\pi}\int_{-\pi /2}^{\pi /2}cos(\theta)\right ]\; \left [ \frac{1}{\pi}\int_{-\pi /2}^{\pi /2}cos(\phi)\right ]}{\epsilon\,\, \sigma}}$

The integral of cosine is exactly two so:

$\large T_{eff. \; mean} = \sqrt[4]{\frac{S_0\;(1-\alpha)\;\left [ \frac{1}{\pi}\int_{-\pi /2}^{\pi /2}cos(\theta)\right ]\; \left [ \frac{1}{\pi}\int_{-\pi /2}^{\pi /2}cos(\phi)\right ]}{\epsilon\; \sigma}}$

Regroup to a final form:

$\large T_{eff. \; mean} = \sqrt[4]{\frac{S_0(1-\alpha) \frac{2}{\pi} \frac{2}{\pi}}{\epsilon\,\, \sigma}}$

In the top text equations I have moved the 4/pi² out of the radical and that’s what I now see causes the difference between the two, a subtle different way to see the cosine weighting across the sphere. Maybe yourself or Ned might expound on which one is correct and why. I am now a bit confused once again on the finesse in the math.

59. wayne says:

Dr. Sweger, one point not said is that the T.eff.mean is of the lit hemisphere only. I always seem to leave a few things out… darn. Just fill in other places if necessary.

60. David Appell says:

(Resubmitting) I was discussing this with a colleague, and he made an obvious point that easily shows N&Z and Sweger can’t possibly be right. It has nothing to do with their integration, but with their assumptions.

Specifically, they assume the Earth is not in equilibrium, when in fact it is close to equilibrium.

Look at Sweger’s equation 9 above. It says that T=0 at the poles and at two points along the equator. That is obviously wrong.

Or consider points only along the equator (phi=0). Take the derivative of T with respect to theta. That equations predicts a far, far larger rate of temperature change along the equator than anything observed on Earth.

In short, N&Z and Sweger assume the temperature at any point on Earth depends only on the sunlight it receives. Actually it also depends strongly on the fact that there *is* an atmosphere and oceans that readily spread heat around.

The standard method assumes the Earth *is* in equilibrium — an approximation, but a far, far better one. Just look at the data of the variation of temperature of the Earth — it’s nothing like that seen on the moon.

N&Z and Sweger are wrong in their basic assumptions.

[Reply] You are missing the point big time. N&Z use this technique to determine the true grey body temperature, not the temperature with oceans and atmosphere. Then they go on to show that pressure, coupled with insolation, not greenhouse gases, are responsible for the rise to the real surface temperature of Earth in further equations.

61. David Appell says:

[Reply] You still haven’t shown how ‘standard theory’ gets the value for the night side of the Moon.

Radiative considerations alone are insufficient to calculate the nightside lunar temperature, because that depends on heat conductance through the regolith. This is discussed in Vasavada et al (1999), where they do attempt to model it:

[Reply] Exactly. So you borowed the figure from this blog. I won’t allow you to dominate this thread with more spurious argument, so be prewarned.

62. David Appell says:

[Reply] Exactly. So you borowed the figure from this blog. I won’t allow you to dominate this thread with more spurious argument, so be prewarned.

Now I can’t discuss or use data if it’s on this blog? Others have, in particular tchannon, whose circuit model is tuned to the measured nightside temperature.

These aren’t spurious arguments. You seem very biased against the explanation based on standard physics, and annoyed that I keep insisting (and showing that) standard science can easily explain the observed lunar temperature. Why?

[Reply] Tim states the source of his data in his opening paragraph. We have been told *for years* by ‘physicists’ such as Joel Shore that the standard form of the S-B equation which makes use of the fact that the area of a hemisphere is equivalent to a disc of twice the radius is sufficient for computing grey-body temperature. Now N&Z show that the properly derived integration method they have developed gives a more accurate result, even when heat retention isn’t included.

Then you want to claim that’s how ‘standard physics’ did it all along, and use Sweger’s integration, and Ned’s properly integrated calculation using DIVINER data of the night side temperature without attribution to ‘prove it’. The level of hubris and arrogance is astounding. Learn some humility and give credit where it is due. Acknowledge them and their achievements.

Your attempts to belittle them and negatively criticise their achievements are sour grapes which will be tossed in the vinegar barrel. The fine wine will be used to toast the success of Ned and Karl, and those who contribute positively and productively, like resident co-mod and author Tim C and Dr Sweger have. Write up a convincing article using previous techniques and data and submit it for consideration if you wish, I’ve had enough of your unproductive thread bombing and spurious argument.

This blog is where the science, which is being advanced by the people who have been mocked and abused by Joel Shore, Willis Eschenbach and others on WUWT is being first published. Ned and Karl know that I will defend a space for them to disseminate their findings from, and maintain a vibrant, uncluttered and productive forum for discussion and development of their work and our knowledge.

63. Daniel Sweger says:

David, just to answer you. Nowhere have I done any calculations concerning the temperature of the earth. Nor did I ever say the temperature of the MOON was zero at the poles or the horizons. What the equations show is the radiance from the sun, which does go to zero on the dark side, wherever the sun is not above the horizon. That does not mean the temperature is zero. There are many other factors which determine the temperature, such as the heat capacity and thermal conductivity of the regolith, i.e. the surface dust/rock. Using the S-B equation give us a fairly good approximation of the surface temperature of the moon, since the moon, unlike the earth, has no atmosphere. I suggest you get your facts straight, particularly when you are talking to others.