Simple Thermodynamic Proof stumps me

Click For Summary

Homework Help Overview

The discussion revolves around deriving the average number of photons per unit volume in a cavity at a given temperature T, using a specific formula involving integrals and constants related to thermodynamics and quantum mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the number of photons and the available states in a cavity, questioning how to utilize the provided equations and definitions effectively. There is discussion about the integration of states and the implications of the limits of integration.

Discussion Status

Participants are actively engaging with the problem, raising questions about the integration process and the relationship between energy and frequency. Some guidance has been offered regarding the need to integrate over all frequencies, but no consensus has been reached on the specific steps to take.

Contextual Notes

There is mention of constraints regarding the limits of integration and the need to consider all possible frequencies, as well as a reference to formatting equations within the forum.

Davio
Messages
65
Reaction score
0

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.
 
Physics news on Phys.org
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..
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
25
Views
4K