Blackbody Radiation - Entropy and Internal Energy

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SUMMARY

The discussion focuses on deriving the entropy and internal energy of black body radiation using the total free energy equation, specifically the formula F = (k_b TV/\pi^2) ∫ k^2 ln[1-exp(-\hbar ck/k_b T)]dk. Participants emphasize the importance of using u-substitution to simplify the integral, leading to F = (V/\pi^2)(k_B T)^4/(\hbar c)^3 ∫ u^2 ln(1-e^{-u})du. The conversation highlights the challenges faced in calculating the integral and the necessity of correctly applying limits of integration. Tools like Wolfram Alpha are recommended for solving complex integrals.

PREREQUISITES
  • Understanding of black body radiation concepts
  • Familiarity with thermodynamic equations, particularly free energy and entropy
  • Knowledge of calculus, specifically integration techniques
  • Experience with u-substitution in integral calculus
NEXT STEPS
  • Learn about the derivation of the partition function in statistical mechanics
  • Study advanced integration techniques for solving complex integrals
  • Explore the application of Wolfram Alpha for solving physics-related mathematical problems
  • Investigate the relationship between temperature and entropy in thermodynamic systems
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Students and researchers in physics, particularly those studying thermodynamics and statistical mechanics, as well as anyone interested in the mathematical foundations of black body radiation.

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Homework Statement



Expression for the entropy and internal energy of black body radiation.

Using the below relations:

Homework Equations



Total free energy for black body:
$$ F = (k_b TV/\pi^2) \int k^2 ln[1-exp(-\hbar ck/k_b T)]dk $$
Relationship between partition function and internal energy:
$$ E = -\partial ln(z)/ \partial \beta $$

Where ##\beta## is the inverse temperature given by:
$$ \beta = (1/k_b T) $$
Relationship between the free energy, internal energy and entropy:
$$ F = E - TS $$

The Attempt at a Solution



If I use ## F = E - TS## rearranged to

$$ S = (E-F)/T $$

Then substitute the relations in and calculate.

I make a little progress until I hit the ## F ## part, the integral gives me some problems as I am having trouble calculating it, I tried using Wolfram Alpha as a guide but it won't actually give me an answer which suggested to me that I'm going about it the wrong way.
 
Last edited:
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Use a u-substitution, like

u = \frac{\hbar c k}{k_B T}

then you get

dk = \frac{k_B T}{\hbar c} du

and the integral becomes

F = \left( \frac{V}{\pi^2}\right) \left(\frac{(k_B T)^4}{(\hbar c)^3} \right) \int u^2 \ln{\left(1-e^{-u}\right)}du

which you can easily use Wolfram Alpha to solve
 
Don't forget the limits of integration. The variables are independent of photon's (angular) frequency.
 
kreil said:
Use a u-substitution, like

u = \frac{\hbar c k}{k_B T}

then you get

dk = \frac{k_B T}{\hbar c} du

and the integral becomes

F = \left( \frac{V}{\pi^2}\right) \left(\frac{(k_B T)^4}{(\hbar c)^3} \right) \int u^2 \ln{\left(1-e^{-u}\right)}du

which you can easily use Wolfram Alpha to solve

Oh man, I really should have seen that...

dextercioby said:
Don't forget the limits of integration. The variables are independent of photon's (angular) frequency.

Yeah I have the limits wrote down, I just didn't know how to show them in the post.
 
Code:
\int_{-\infty}^{\infty} x^2 dx

=\int_{-\infty}^{\infty} x^2 dx

You can also right click any TeX equation to see the code that produced it.
 

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