# Earth black body temperature wrong?

#### Andre

The blackbody temperature of Earth is supposed to be -18C or 255K. This can be derived by reworking Stefan Boltzmann law to:

TK = (S*(1-a)/4*rho) ^0,25

http://www.lwr.kth.se/Grundutbildning/1B1292/Compendium_online/ch05s02s01.html [Broken]

in which S is solar influx, we use 1367 w/m2
a is albedo: 0.3
rho is the Stefan Bolzmann constant: 5.667E-8
The factor 1/4 is caused by the difference in influx surface. The cross section of Earth (pi*r2) is intercepting the solar radiation but it has been equally divided over the surface of the Earth (4*pi*r2).

I think the latter reasoning is wrong. You can average out the influx but that doesn not average out the temperatures because the relationship is not lineair.

I may be wrong of course but If I assume the earth to be 180 slices of a sphere and calculate the radiation temperature for every degree of lattitude seperately, then I get an average blackbody temperature of about -21.1C or 251.9K.

Before I show the (simple) calculations, I like to invite everybody to attempt the calculation likewise since independent duplication is the best confirmation.

What would be the consequences on all climate models etc if the blackbody temperature is three degrees lower than always assumed?

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#### billiards

You know, I'm pretty sure that the 'latter reasoning' is correct, and is a simple application of Gauss' Law.

#### sneez

You can average out the influx but that doesn not average out the temperatures because the relationship is not lineair.
Andre nowhere in the derivation it is assumed to be linear and it poses no restriction on average.

I may be wrong of course but If I assume the earth to be 180 slices of a sphere and calculate the radiation temperature for every degree of lattitude seperately, then I get an average blackbody temperature of about -21.1C or 251.9K.
Andre, you must do something wrong. If you use the same values it does not matter if you divide the earth in 180 slices or 1000 slices. When you add them up it should give you the average or 255K since thats how what you are doing anyway you are adding up averages.
[It looks to me as if you divided the earth into 179 slices instead 180. If you do it numerically you should check this. ] Also, for your slice calculations did you use albedo of .3 or .32 ? it will make huge difference.

#### Andre

Sneez, What the original formula is doing is dividing the influx energy over the surface of the sphere before converting it to heat. What I'm doing is converting the incoming radiation to heat first for every degree lattitude and then calculate the average temperature. That's done numerically avoiding cowardly the intergration math.

#### sneez

yes, i undestand and what I said applies. Are you SURE that its 180 and not 179 slices by some mistake?

#### mheslep

Gold Member
The blackbody temperature of Earth is supposed to be -18C or 255K. This can be derived by reworking Stefan Boltzmann law to:

TK = (S*(1-a)/4*rho) ^0,25

See equation under 5.9

in which S is solar influx, we use 1367 w/m2
a is albedo: 0.3
rho is the Stefan Bolzmann constant: 5.667E-8
The factor 1/4 is caused by the difference in influx surface. The cross section of Earth (pi*r2) is intercepting the solar radiation but it has been equally divided over the surface of the Earth (4*pi*r2).

I think the latter reasoning is wrong. You can average out the influx but that doesn not average out the temperatures because the relationship is not lineair.
You know, I'm pretty sure that the 'latter reasoning' is correct, and is a simple application of Gauss' Law.
Interesting. Andre - may I suggest the albedo constant is the place for further inquiry as I believe Billiards is correct w/ respect to Gauss's Law and a black body radiator- the shape of the surface is not important, one simply sums all the energy flux through the surface to derive the BB temp. The http://en.wikipedia.org/wiki/Albedo" [Broken] then its shape w/ respect to the incoming energy distribution does matter. To assume the albedo is constant everywhere (.3) implies the surface & atmosphere must be lambertian everywhere as seen by the incoming solar flux. I'm not familiar w/ the source of the .3 figure, but the actual 'terrestrial albedo' must be a complicated function of surface type and angle of incidence.

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#### Andre

Okay once more to the basics.

Spectral energy hits earth. That energy is turned into heat on the spot of impact according to the basic Stefan Bolzmann law implying the operator 4th square root first. Then we get to the average temperature by calculating the weigted average temperature second.

Now suppose that the hitting energy is 16 and we operate the square root first then we get 4 then we do the area average temp by dividing by 4. The result is one.

Now suppose we do the area averaging first on the hitting energy we get 16/4 = 4 before we do the square rooting then we end up with 2. That's the difference.

The sequence of the operators appears to be wrong in the scholar equation. That's the problem. Temperature first, then averaging, not averaging the energy first before converting it to temperature.

#### billiards

Andre, whatever happens energy must be conserved.

#### sneez

Spectral energy hits earth. That energy is turned into heat on the spot of impact according to the basic Stefan Bolzmann law implying the operator 4th square root first. Then we get to the average temperature by calculating the weigted average temperature second.

Now suppose that the hitting energy is 16 and we operate the square root first then we get 4 then we do the area average temp by dividing by 4. The result is one.

Now suppose we do the area averaging first on the hitting energy we get 16/4 = 4 before we do the square rooting then we end up with 2. That's the difference.

The sequence of the operators appears to be wrong in the scholar equation. That's the problem. Temperature first, then averaging, not averaging the energy first before converting it to temperature.
what are you talking about?

Flux intercepted by circle area = Flux re-emitted from spherical area (conservation of E)
solving for Te we get effective temperature of sphere or what temperature sphere would have if all intercepted energy was re-radiated back to space.

If you play a scientist andre, do it right and post clearly all steps of what you talking about because this is confusing. Either I do not understand or you cannot do simple algebra, i suppose its the first one. If you do honestly want somebody to check your work be decent enough not to waste our time with half baked examples which are not clear.
You are not schooling us Andre even though you like to play that role, that simple Teff derivation is so simple and the laws that its derived from are clear, unlike your counterexample.

#### mheslep

Gold Member
Sneez - pls drop the ad hominem and let him and us work through the problem.

#### Andre

okay I have two handicaps here, I'm neither familiar with the English math terms nor latex. But we'll give it a try.

We take slice of the earth pie between the lattitude of 'a' degrees and 'a+1' degrees. The surface of the cross section of that slice intercepting the solar flux is average length of the top and bottom length: 2*R(cos(a) + cos(a+1))/2 times effective height (cos(a)*(60 nautical miles)). We multiply this surface with the influx and divide it over the effective surface of the slice:
pi*R*(cos(a)+cos(a+1)*absolute height( 60 nautical miles (without the cos(a)!). Now we use the basic stefan boltzman law to calculate the temperature of that particular spheric slice.

We summarize these steps for a=0 to 89, and also sum the temp times surface area divided by sum of surface area (weighted average) to get 251.9 kelvin.

I'm really interested to see what is wrong with that.

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#### sneez

No ad hominem comitted. From a first post im interested with the new idea. However, after 3 posts of clear avoidance of posting the problem, it smells like not science. Since andre is one of few ppl interested in truth Im reminding him of his own medicine. Hope its taken that way.

#### Andre

I was not avoiding btw but just having limited time available. And I have a habit of triggering others to duplicate and scrutinize.

#### Andre

what are you talking about?

Flux intercepted by circle area = Flux re-emitted from spherical area (conservation of E)
solving for Te we get effective temperature of sphere or what temperature sphere would have if all intercepted energy was re-radiated back to space.
I'm sure hat there is no violation of this principle, a uniformly heated earth to 255K is probably emitting the same flux as an Earth with temperatures differentiated with lattitude from ~270K to <100K. But the average of the latter is not the same.

#### siddharth

Homework Helper
Gold Member
I think the latter reasoning is wrong. You can average out the influx but that doesn not average out the temperatures because the relationship is not lineair.
Do you mean that the Temperature isn't uniform? When we model the Earth as a black body, then we assume that the temperature is uniform.

#### Andre

Do you mean that the Temperature isn't uniform? When we model the Earth as a black body, then we assume that the temperature is uniform.
okay but what temperature? The actual average temperature is apparently not the same as the temperature following from an evenly radiating shpere.

Atrtached is a snapshot of the spreadsheet (bottom part) I had to redo it because I worked it out first elsewhere and can't find the original file back. Somehow the rework gives me a new value: 252,5 degrees K. I''ll compare them later.

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#### siddharth

Homework Helper
Gold Member
okay but what temperature? The actual average temperature is apparently not the same as the temperature following from an evenly radiating shpere.

Atrtached is a snapshot of the spreadsheet (bottom part) I had to redo it because I worked it out first elsewhere and can't find the original file back. Somehow the rework gives me a new value: 252,5 degrees K. I''ll compare them later.
But, the point is that if you're going to use Stefan's law, you must assume that the temperature is uniform throughout the sphere. So, what do you mean by an average temperature?

In that case, if $$S$$ is the http://www.britannica.com/eb/article-9068556/solar-constant" (area perpendicular to sun rays), the energy abosrbed by the earth is $$S \pi r^2$$, where r is the radius of the earth. Now, since the earth is assumed as a black body, the energy emitted by the earth, $$\sigma T^4 4 \pi r^2$$ must be the same. From this, solve for T.

I don't know why you are splitting the earth into various parts?

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#### Andre

The question is, what would the average temperature of Earth be as a grey body with an energy absorption of 30%.

For that the current model apears to simple since it assumes interchangeability of the factors emission and temperature. It divides the radiation evenly on the Earth surface before it is converted to temperature.

I'm trying to see what happens if you convert the radiations to temperature first for each degree of lattitude and then see what the average temperaure is. An example of four slices for 0-1 - 29-30 -59-60 - 89-90 degrees lattitude is attached.

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#### Andre

I see I can upload zips

So here is the excel sheet to scrutinize.

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#### sneez

Dont you have to modify albedo as well ? The constant albedo is consistent only with the circular area of intercepted radiation.

Im still working and will be for a longtime , but I will have a crack at this when im off.

#### sneez

Andre, your intercepting area and emitting area are not exact of what it should be, that introduces error. Also as I said above the assumption of constant albedo with sperical earth is very arbitrary one could argue and the temperature difference is not significant as you presume it to be given all that.

And a one technicall, why is the height in emitting area not corrected with the cosine lat? Is that not saying that the distance between 1deg lat at lets say 70-71 is the same as at the equator?

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#### Andre

no, the effective height of the slice intercepting the incoming parallel insolation is corrected for the lattitude with the cosine, whereas the emitting surface area is indeed the complete surface, so the correction is not applied there.

I made an error not showing that in the simple drawing that I attached.

I agree that the one degree slices introduce an error, which diminishes when it is replaced by the integral. I doubt if the error is large enough to explain the difference. Somebody interested to do the math?

Also not sure about the albedo problems. Is anything changing in (bond)albedo between the two methods?

#### sneez

I wrote a program with double precission of those cosine factors and i got 252.71 :D . I was thinking that it could be quite managable to do the integration using calculus technique of rotating curves around y axis. (I will do but when I get to the book, its been a while).

Albedo depends on the angle of incomming radiation. There is whole "science" to it and im not sure what the best approach would be. I do understand that when intercepting earth is presented as a solid circle, the assumption of constant albedo is not that bad. However, once you consider spericity which introduces difference in Temp. the immediate response of a sceptic (my role in this thread) would be the changing albedo.

I have to think what it all means for a while it definitelly very interesting.

#### Andre

meanwhile, assuming that albedo and slice problems will be sorted out, it may be an idea to ponder a bit about the consequences.

BTW. The first time, years ago, that I saw this scholar formula, I wondered if it was that simple, considering the arguments that I gave here earlier. I also realized that reversing the order, dividing radiation equally before transferring to temperature would give a warm bias. So if the reality was colder, then the greenhouse effect was to be stronger >33 degrees. I think I decided then not to pursue the issue further at that time.

But then Skyhunter begged the question in another thread why I could not accept the alleged consensus on global warming in another thread. Obviously, because science is human work in progress and to err is human. I know, I make a lot of mistakes. There is also this issue about honesty in science, regardless if it suits your case or not. So, sceptism and honesty in science is the key to progress and I realized I should not have be silent about my black body dillema, regardless of whose case it would serve. The real Earth temperature appears to be about almost 35 degrees more than the black body temperature instead of 33 degrees. Apparantly there is more greenhouse effect.

There is also this other planet with a blackbody - greenhouse problem. Greenhouse effect does not seem to exist on Mars despite the much higher CO2 concentration. Actually, the several sources declare it's blackbody temperature to be 210K and it's average temperature to be -63C. After a very difficult calculating process we discover that those two are equal. So why conceal that? using kelvin and celsius so it would not show so easily?

So pondering about this, the slice method would get a lower black body temperature. So it seemed we'd finally get greenhouse effect for Mars after all, which is definitely not in the interest of somebody who wants to deny greenhouse effect, would it?

So I googled a fact sheet for Mars:

The vital data: Bond Albedo 0.250; Solar Irradiance: 589.2 W/m2; In the scholar formula this gets us a black body temperature of 210.125K and the sheet says 210.1K. Not bad.

Now we run the numbers in the degree slice model to get: 208.16K. So there you are, at last, about 2 degrees of greenhouse effect for Mars, No?

Look at the fact sheet: Average surface temperature (Celsius): -65C, converted to Kelvin: 208K

Those kinds of results tend to give me the goose bumps.

Editted the BB temp slightly, made another mistake. No impact on the narration

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#### sneez

Andre, im uneasy about the emitting area. The schollary formula implies 4times bigger emitting area. Your formula has 10times biger emitting area (on average). [from 3times bigger to 180time biger at the poles]

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