Question about entropy of photons

[dextercioby]

I made the calculation explicitely here:

https://www.physicsforums.com/showthread.php?t=58208&page=1

Daniel.

 

[JesseM]

I made the calculation explicitely here:

https://www.physicsforums.com/showthread.php?t=58208&page=1

Daniel.

Your answer was:
$$\langle N\rangle=\frac{2k^{3}\zeta(3)}{\pi^{2}c^{3}\hbar^{ 3}} VT^{3}$$

And my answer was:
$$(16 \pi Zeta[3] k^3 V T^3 ) / (c^3 h^3)$$

So, taking into account that $$\hbar = h / 2\pi$$, it looks like we got the same answer.

 

[kanato]

I tried doing this integral and I didn't get the answer you give above

whoops.. there was supposed to be a proportionality symbol there, showing that N was proportional to VT^3, but apparently I got the wrong TeX command for that.

Well, see the second paragraph of my last post above, I don't think an infinite average fluctuation size necessarily means the average itself is physically meaningless. If it was meaningless, why would your textbook give it at all? Does it make any comments about the significance of the infinite average fluctuation result?

So I was actually basically quoting the textbook. The author shows that formula, then says that it can't really be taken at facevalue as he calls attention to the fact that the fluctuations are infinite. Unfortunately, he doesn't elaborate beyond that.

Also, I was thinking about what it means to say the average size of fluctuations is infinite, and it seems to me it doesn't automatically mean that talking about the average is physically meaningless. For example, imagine a situation where there's a 1/2 chance the fluctuation will deviate from the average by 2, a 1/4 chance it will deviate from the average by 4, a 1/8 chance it will deviate from the average by 8, etc...in this case the average fluctuation would be 2*(1/2) + 4*(1/4) + 8*(1/8) + ... = 1 + 1 + 1 + ... = infinite, but it's still true that in 3/4 of all cases you'll get within 4 of the average. Now, I'm not suggesting the typical fluctuations in photon number would really be so small, but the point is that in principle an infinite average fluctuation size in the value of a quantity doesn't mean that it's meaningless to talk about the average value of that quantity.

hmm.. I see what you are saying.. in your example though, you are considering deviations in only one direction... fluctuations about the average should happen in both positive and negative directions, so the average fluctuation should be zero. The formula I posted was for mean square fluctuation, which in this example would be 4*(1/2) + 16*(1/4) + 64*(1/8) + ... = infinity much faster :) So in 3/4 of the cases this comes out to be 6.. not much different, but also look that the formula is divided by the average number of particles squared.. so the rms deviation is root six or about 2.5 times the number of particles in the system, by this technique, 3/4 of the time. So I think it's not only the fact that the mean square deviation is infinite, but that even divided by N^2 (a very large number) it is still infinite, so it seems to me that indicates it diverges very quickly. But I don't have any analysis to back that up, yet.