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Density of states of photons

  1. May 8, 2012 #1
    1. The problem statement, all variables and given/known data

    density of states of photon gas is proportional to .....

    (a)E^1/2 (b)E (c)E^3/2 (d)E^2

    2. Relevant equations

    i know the relation for density of states of electrons which is proportional to E^1/2. So far i was thinking that electrons and photons shares the same relation for density of states. i found so in some websites too, where they mention about both electrons and photons during derivation procedure. but answer to the above question is E^2.

    my question is, in what way does these two,i.e. density of states of photons and electrons, differs? does the relation for density of states equation for electrons holds good for any another particle? can someone provide me with a link for the derivation of density of states for photons?

    3. The attempt at a solution
     
  2. jcsd
  3. May 8, 2012 #2
    The difference comes from the Hamiltonian. For nonrelativistic particles, [itex] H = \frac{p^2}{2m} [/itex] while for photons, [itex] H = c |p| [/itex].

    The density of states is usually defined as
    [tex] \omega(E) = \sum_{n_i} \delta(E-H(n_i)) [/tex]

    For the nonrelativistic case, you get
    [tex] \omega(E)_{nr} \sim \int_0^\infty dp p^2 \delta(E-\frac{p^2}{2m}) [/tex]
    and substituting [itex]x = p^2/(2m) [/itex],
    [tex] \omega(E)_{nr} \sim \int_0^\infty dx \sqrt{2m^3 x} \delta(E-x) = \sqrt{2m^3 E}, [/tex]
    where I'm missing all prefactors (do it more carefully yourself!)

    For relativistic case, just replace Hamiltonians: the integral is even easier to do in this case.
     
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