Simple Thermodynamic Proof stumps me

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Davio
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Homework Statement


Show that the average number of photons per unit volume in the cavity for a temperature T is given by

n=( I / Pi^2 . (kT/c(h-bar))^3 )

where I = integral from infinity to 0. dx x^2 ((exp (x)-1))^-1 = 2.404


Homework Equations



You may assume the mean number of photons occupying a state at energy (strange looking e)= hbar omega when the radiation has a temperature T is

<N>=1 over exp (strange looking e/kT)-1

You may also assume that the number of available states in the angular frequency rage omega to omega plus delta omega in a cavity of volume V is given by

g(omega)d-omega = V/c^3 . omega^2 / pi^2 d-omega

The Attempt at a Solution



I know I need to do N/V however I have no idea how available states are useful to me, perhaps I could rearrange for V, but then I dont' know what g(omega) can be used for.
 
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You are given the no. of photons per state, then you're given the number of states in the range [tex]\omega[/tex] to [tex]d\omega[/tex], how would you find the total number of photons?
 
Um... I would times the number of states by number of photons. Is that what g(omega) d omega is referring to? So that means, If I rearrange, and have the above times by c^3 divided by omega squared and times by pi squared.. I should a formula for V...

there's a d omega on both sides, can I just cancel them out or are they important?
 
Not quite...g(w) is not a constant. You'll have to integrate over the whole range of w. Besides, how is the energy of the photon related to w?
 
Would I integrate V/c^3 . omega^2 / Pi^2 with respect to d omega? The limits are from omega to omega + d omega, Would I just integrate it without limits to get a formula :

V. omega^3 = g(omega) omega
-----------
3 C^3 pi^2

and then plug something into E=H(bar) omega.

ps. is there anyway of making these formulas more neat on the forums? I 'm sure I've seen people make actually integral signs come up..