Simple Thermodynamic Proof stumps me

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 \omega to d\omega, 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...

Theres 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..
 
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