Simple Thermodynamic Proof stumps me

In summary, we are asked to show that the average number of photons per unit volume in a cavity at temperature T is given by the formula n = (I/Pi^2)(kT/c(h-bar))^3, where I = 2.404 and is obtained from the integral from infinity to 0 of x^2(exp(x)-1)^-1. We are given the mean number of photons occupying a state at energy h-bar*omega at temperature T and also the number of available states in the angular frequency range omega to omega plus delta omega in a cavity of volume V. We need to find the total number of photons by integrating the function V/c^3 * omega^2 / Pi^2 with respect to d omega
  • #1
Davio
65
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.
 
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  • #2
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?
 
  • #3
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?
 
  • #4
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?
 
  • #5
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..
 
  • #6

FAQ: Simple Thermodynamic Proof stumps me

What is a simple thermodynamic proof?

A simple thermodynamic proof is a logical and mathematical demonstration of a thermodynamic principle or equation. It involves using basic principles and equations to derive an equation or relationship between variables in a system.

Why is it important to understand simple thermodynamic proofs?

Understanding simple thermodynamic proofs allows scientists to have a deeper understanding of the fundamental principles and equations that govern the behavior of matter and energy in a system. It also allows for the application of these principles to real-world problems and the development of new technologies.

What are the key components of a simple thermodynamic proof?

The key components of a simple thermodynamic proof include the initial assumptions and conditions of the system, the application of fundamental thermodynamic principles, and the use of mathematical equations and logic to derive a conclusion.

How do scientists use simple thermodynamic proofs in their research?

Scientists use simple thermodynamic proofs in their research to validate theoretical models, develop new equations or relationships, and gain a deeper understanding of the behavior of matter and energy in a system. These proofs also serve as a foundation for more complex thermodynamic analyses and experiments.

What are some common challenges in understanding and using simple thermodynamic proofs?

Some common challenges in understanding and using simple thermodynamic proofs include the complexity of mathematical equations, the need for a strong understanding of fundamental thermodynamic principles, and the ability to apply these principles to real-world situations. Additionally, different thermodynamic systems may require different approaches to proof, making it a versatile but sometimes challenging subject to grasp.

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