Register to reply 
Derivation of StefanBoltzmann law from Thermodynamics 
Share this thread: 
#1
Apr2911, 11:02 AM

P: 87

I've been trying to derive the StefanBoltzmann 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. 


#2
Apr2911, 04:32 PM

Sci Advisor
Thanks
P: 4,160

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 T^{4} 


#3
Apr2911, 06:07 PM

P: 87




#4
Apr2911, 06:55 PM

Sci Advisor
Thanks
P: 4,160

Derivation of StefanBoltzmann law from Thermodynamics
Because by definition u is the energy per unit volume, U = uV.



#5
Apr3011, 08:37 AM

P: 87

∂U/∂V = 0 Even though the energy density definitely isn't zero? Why and how does an ideal gas differ from a photon gas? 


#6
Dec1711, 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. 


#7
Dec1811, 12:58 PM

P: 789

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. 


#8
Dec1811, 06:32 PM

P: 22

thanks for the explanation!
I figured out why by finally being directed here yesterday http://en.wikipedia.org/wiki/Dispersion_relation. 


#9
Dec2011, 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. 


#10
Dec2411, 07:13 AM

PF Gold
P: 962

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. 


#11
Dec2411, 06:10 PM

P: 757

This is derived using BoseEinstein statistics. See Schroeder pages 289292



#12
Dec2711, 01:07 PM

P: 87

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. 


#13
Dec2711, 07:21 PM

P: 757




#14
Dec2811, 02:57 AM

Sci Advisor
Thanks
P: 2,537

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 BoseEinstein 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 electromagneticfield energy with matter. 


#15
Dec2811, 06:18 AM

P: 87

Also, can I just check, is there general agreement now then that the wikipedia derivation is incomplete? 


#16
Dec2811, 04:28 PM

P: 757

if you have a solid understanding of classical statistical mechanics (i.e. Boltzmann statistics) you can pick up the feel of boseeinstein / fermidirac 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 


#17
Dec2911, 08:55 AM

P: 87




#18
Jan2412, 08:01 PM

P: 552

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. 


Register to reply 
Related Discussions  
Integral in StefanBoltzmann law  Classical Physics  2  
StefanBoltzmann lab  Introductory Physics Homework  1  
Irradiance contrast , stefan boltzmann  General Physics  0  
Planck,StefanBoltzmann law  Advanced Physics Homework  3  
StefanBoltzmann Law  General Physics  6 