Derivation of Stefan-Boltzmann law from Thermodynamics


by TobyC
Tags: derivation, stefanboltzmann, thermodynamics
TobyC
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#1
Apr29-11, 11:02 AM
P: 87
I've been trying to derive the Stefan-Boltzmann law using thermodynamics, and have resorted to looking up the derivation in the feynman lectures and on wikipedia, and I'm confused by both. I think the wikipedia derivation is the best one to look at, it's here:

http://en.wikipedia.org/wiki/Stefan%...mic_derivation

The line I'm not happy with is this one:

'Because the pressure is proportional to the internal energy density it depends only on the temperature and not on the volume.'

If the pressure doesn't depend on the volume (using temperature and volume as independent variables) then I have no problem in deriving the result, but I'm not sure why the pressure doesn't depend on the volume.

The article claims that it's because the pressure is proportional to the internal energy density, but then isn't that also the case in an ideal gas? And in an ideal gas the pressure definitely does depend on the volume. In fact, the derivation of this law seems to rest on the fact that:

U = 3PV

But in an ideal gas:

U = (3/2)PV

And all the same arguments seem to apply, which would appear to suggest that for an ideal gas at constant volume, the pressure is proportional to T^(5/2), which I know is wrong.

Help on this would be much appreciated.
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Bill_K
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#2
Apr29-11, 04:32 PM
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TobyC, you're absolutely correct! And the derivation in Wikipedia is not. Try this:

dU = T dS - P dV
(∂U/∂V)T = T (∂S/∂V)T - P

We need a Maxwell relation:

A = U - TS
dA = - S dT - P dV
⇒ (∂S/∂V)T = (∂P/∂T)V

Putting that in:
(∂U/∂V)T = T (∂P/∂T)V - P

Let u = U/V be the energy density. Then P = u/3 and ∂U/∂V = u gives you a differential equation

u = T/3 du/dT - u/3

whose solution is u = C T4
TobyC
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Apr29-11, 06:07 PM
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Quote Quote by Bill_K View Post
∂U/∂V = u
Thanks for your reply, I can follow most of your derivation, but I'm still not sure of the above part, why is that true?

Bill_K
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Apr29-11, 06:55 PM
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Derivation of Stefan-Boltzmann law from Thermodynamics


Because by definition u is the energy per unit volume, U = uV.
TobyC
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#5
Apr30-11, 08:37 AM
P: 87
Quote Quote by Bill_K View Post
Because by definition u is the energy per unit volume, U = uV.
But for an ideal gas isn't this statement true:

∂U/∂V = 0

Even though the energy density definitely isn't zero?

Why and how does an ideal gas differ from a photon gas?
enfield
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#6
Dec17-11, 09:30 PM
P: 22
hmm I've been trying to understand this myself recently.

TobyC's derivation is convincing to me! but I don't understand part of wikipedia's thermodynamic derivation still (specifically how the simple differential equation is obtained from equating the second derivatives of entropy). If anyone understands that part and wants to explain I would be very grateful :).

Also, I am unsure why u/3 = P, instead of 2u/3 = P.
Rap
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#7
Dec18-11, 12:58 PM
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The difference is because the relationship between energy and momentum is different for photons than for massive particles. For photons E=pc where E is photon energy, p is momentum, c is speed of light. For massive particles [itex]E=p^2/2m[/itex] where m is the mass of the particle.

The pressure for any kind of particle (assuming ideal gas) is P=nvp/3 where n is particles/volume, v is velocity, p is momentum. For photons, the velocity is c and you get P=u/3, where u is nE, the energy/volume. For massive particles, the average velocity is found from [itex]E=\frac{1}{2}mv^2[/itex], or [itex]v^2=2E/m[/itex], and you get P=2u/3.
enfield
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#8
Dec18-11, 06:32 PM
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thanks for the explanation!

I figured out why by finally being directed here yesterday
http://en.wikipedia.org/wiki/Dispersion_relation.
TobyC
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#9
Dec20-11, 06:25 AM
P: 87
Yeah I am happy with where the U=3PV comes from, but I'm still not sure anyone's answered the question I made this thread for? Although I still think that wikipedia derivation was wrong, I am more happy now with why the pressure and energy density only depend on the temperature. A radiation 'gas' is the state of an electromagnetic field that will be in equilibrium with a set of oscillating radiation emitters at a certain temperature, and an individual oscillator gets into equilibrium with the radiation gas when the energy density reaches a certain value, the size of the box that the oscillator is in shouldn't have anything to do with it.

That's how I'm thinking about it at the moment anyway, although I'd still be interested if someone could address this point properly.
Philip Wood
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#10
Dec24-11, 07:13 AM
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Surely the most significant difference between the photon gas in a constant temperature cylinder, and the gas of molecules in a cylinder is that the number of molecules is fixed, whereas the number of photons is not. Thus as the cylinder containing photons is expanded in volume, you get more photons, in such a way that the number per unit volume stays constant.

To see why the number per unit volume is the same, imagine two cylinders (cavities) at the same temperature, but having different volumes. Suppose they were joined by a very short pipe. If there were net passage of photons from one to the other, we'd violate the second law. So there's no net passage, which will be the case only if the number per unit volume is the same in the large and small container.
Curl
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#11
Dec24-11, 06:10 PM
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This is derived using Bose-Einstein statistics. See Schroeder pages 289-292
TobyC
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Dec27-11, 01:07 PM
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Quote Quote by Philip Wood View Post
Surely the most significant difference between the photon gas in a constant temperature cylinder, and the gas of molecules in a cylinder is that the number of molecules is fixed, whereas the number of photons is not. Thus as the cylinder containing photons is expanded in volume, you get more photons, in such a way that the number per unit volume stays constant.

To see why the number per unit volume is the same, imagine two cylinders (cavities) at the same temperature, but having different volumes. Suppose they were joined by a very short pipe. If there were net passage of photons from one to the other, we'd violate the second law. So there's no net passage, which will be the case only if the number per unit volume is the same in the large and small container.
Yeah the fact that the number of photons is not fixed is probably a crucial difference, although so far I have been trying to think in terms of the classical wave picture.

As for that explanation, I'm not sure that explanation alone is a sufficient reason for why the number per unit volume doesn't depend on the volume. It does show that the density is the same in the large and small container, but they are joined, so the only volume that is relevant is the combined one. For instance, I don't see why your argument, if valid, would not apply to an ideal gas too, but in an ideal gas the density is really a function of volume only.


Quote Quote by Curl View Post
This is derived using Bose-Einstein statistics. See Schroeder pages 289-292
Thanks for that but seeing as black body radiation was investigated over 20 years before Bose-Einstein statistics was developed I'm sure there is a simpler way of explaining it, and also a way of doing it which stays within a classical rather than quantum framework.
Curl
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Dec27-11, 07:21 PM
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Quote Quote by TobyC View Post

Thanks for that but seeing as black body radiation was investigated over 20 years before Bose-Einstein statistics was developed I'm sure there is a simpler way of explaining it, and also a way of doing it which stays within a classical rather than quantum framework.
Earlier does not mean it is simpler. Usually (as is in this case) it is the other way around.
vanhees71
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Dec28-11, 02:57 AM
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The original work by Planck is quite easy to understand. To express it in a modern way, he's counting normal modes of the em. field, i.e., a density of states in a particular way, which (to use modern terminology again) refers to Bose-Einstein statistics. In some sense at this time this has been an ad hoc assumption without a clear foundation from other principles as is nowadays relativistic quantum field theory. Of course no such theory existed at the time, but quantum theory was born.

Planck got the law right, because beforehand he had excellent data, he could fit with the right function, now called Planck's Law. This has been a very well educated guess from a long and deep work to understand the exchange of electromagnetic-field energy with matter.
TobyC
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#15
Dec28-11, 06:18 AM
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Quote Quote by Curl View Post
Earlier does not mean it is simpler. Usually (as is in this case) it is the other way around.
I'm sure the explanation within the framework of Bose-Einstein statistics is simpler, but I don't understand Bose-Einstein statistics yet, and I'd like to feel that I have a complete understanding of where the law comes from, yet doesn't a complete understanding of Bose-Einstein statistics require an understanding of the spin-statistics theorem? Which I think is quite difficult? I could be wrong on this of course not understanding it at the moment.

Quote Quote by vanhees71 View Post
The original work by Planck is quite easy to understand. To express it in a modern way, he's counting normal modes of the em. field, i.e., a density of states in a particular way, which (to use modern terminology again) refers to Bose-Einstein statistics. In some sense at this time this has been an ad hoc assumption without a clear foundation from other principles as is nowadays relativistic quantum field theory. Of course no such theory existed at the time, but quantum theory was born.
But the original derivation of this law was provided by Boltzmann before Planck came along, purely from thermodynamics, without getting into the statistical mechanics, it's his derivation I was seeking to understand when I made the thread.

Also, can I just check, is there general agreement now then that the wikipedia derivation is incomplete?
Curl
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#16
Dec28-11, 04:28 PM
P: 751
if you have a solid understanding of classical statistical mechanics (i.e. Boltzmann statistics) you can pick up the feel of bose-einstein / fermi-dirac statistics fairly quickly, even if you are not too familiar with QM.

Boltzmann derived his classical statistics from thermodynamics, but trying to understand Plank's work without QM is impossible. You can apply thermodynamics to the problem, all you'll get is the ultraviolet catastrophe
TobyC
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#17
Dec29-11, 08:55 AM
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Quote Quote by Curl View Post
if you have a solid understanding of classical statistical mechanics (i.e. Boltzmann statistics) you can pick up the feel of bose-einstein / fermi-dirac statistics fairly quickly, even if you are not too familiar with QM.

Boltzmann derived his classical statistics from thermodynamics, but trying to understand Plank's work without QM is impossible. You can apply thermodynamics to the problem, all you'll get is the ultraviolet catastrophe
Well I'm not trying to understand Planck's work here, I don't want the derivation of the blackbody radiation spectrum, just specifically the Stefan-Boltzmann law, which can and was derived without quantum mechanics. I think I understand the derivation now too. The only bit I was unsure of, that I made the thread for, was how you know that the energy density is a function of temperature only, not volume, but I think I understand this now.
Telemachus
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#18
Jan24-12, 08:01 PM
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Quote Quote by Bill_K View Post
TobyC, you're absolutely correct! And the derivation in Wikipedia is not. Try this:

dU = T dS - P dV
(∂U/∂V)T = T (∂S/∂V)T - P

We need a Maxwell relation:

A = U - TS
dA = - S dT - P dV
⇒ (∂S/∂V)T = (∂P/∂T)V

Putting that in:
(∂U/∂V)T = T (∂P/∂T)V - P

Let u = U/V be the energy density. Then P = u/3 and ∂U/∂V = u gives you a differential equation

u = T/3 du/dT - u/3

whose solution is u = C T4
It's not clear to me this implication ⇒ (∂S/∂V)T = (∂P/∂T)V
I don't know how you get the equality using the first principle and the Helmholtz potential. I'd like to know it, I'm interested on this kind of manipulations to get the Maxwell relations, it seems useful.

Would somebody explain that to me? I'd like to see some intermediate steps to get the equality.


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