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Bose-Einstein Statistics

  1. Dec 13, 2006 #1
    1. The problem statement, all variables and given/known data
    (a) Find the average energy per photon for photons in thermal equilibrium with a cavity at temperature T.
    (b) Calculate the average photon energy in electron volts at T = 6000K.

    2. Relevant equations
    [tex] u(E)dE = \frac{8 \pi}{(hc)^3} \frac{E^3 dE}{e^{E/k_B T} - 1}[/tex]

    3. The attempt at a solution

    Integrate both sides of the equation.
    [tex] \frac{E}{V} = \int_0^\infty u(E)dE = \int_0^\infty \frac{8 \pi}{(hc)^3} \frac{E^3 dE}{e^{E/k_B T} - 1}[/tex]

    Use the fact that
    [tex]\frac{z^3 dz}{e^z - 1} = \frac{\pi^4}{15}[/tex]

    and that the equation can be rewritten as
    [tex] \frac{8 \pi (k_{B}T)^3}{(hc)^3} \int_0^\infty \frac{ (\frac{E}{k_B T})^3 dE}{e^{E/k_B}-1} [/tex]

    which finally gives
    [tex] \frac{E}{V} = \frac{8 \pi (k_{B}T)^3}{(hc)^3} \frac{\pi^4}{15}[/tex]

    Did I do this right?

    Part b will be easy, just plug in the value for T if (a) is right.
    Last edited: Dec 13, 2006
  2. jcsd
  3. Dec 13, 2006 #2

    Physics Monkey

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    Ok, your approach so far is good, but you've missed three factors of kT (check your units!). Now you're asked to find the energy per photon, and you've found the total energy per volume, so you must now find the number of photons per volume and divide.
  4. Dec 13, 2006 #3
    Whoops, must have missed translating the cubed into typing because it's on my paper.

    Okay, so I should find N/V and divide. So for N/V
    [tex]\frac{N}{V} = \int_0^\infty n(E)dE = \int_0^\infty g(E)f_BE(E)dE[/tex]
    and after some algebra
    [tex] \frac{N}{V} = \frac{8 \pi}{(hc)^3}(k_B)^2 \int_0^\infty \frac{z^2}{e^z - 1}[/tex]
    [tex]z= \frac{E}{k_B T}[/tex]
    and since
    [tex]\int_0^\infty \frac{z^2}{e^z - 1} = 2.41[/tex]
    the number per volume becomes
    [tex]\frac{N}{V} = \frac{8 \pi}{(hc)^3}(k_B)^2 (2.41)[/tex]
    after taking the energy per volume and dividing it by the number per volume
    [tex] \frac{E}{N} = k_B T \frac{\pi^4}{15(2.41)}[/tex]
  5. Dec 13, 2006 #4

    Physics Monkey

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    Which is of course just what you would expect: kT times some number of order one. This is because the only energy scale in the problem is kT.
  6. Dec 14, 2006 #5


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    THe average energy per photon is simply

    [tex]\langle E\rangle =kT\frac{\zeta(4)\Gamma(4)}{\zeta(3)\Gamma(3)} [/tex]

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