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Equation of state of photon gas

  1. Jun 8, 2013 #1
    Dear all,

    I am using stress-energy tensor to derive equation of state of photon gas (assuming it as a perfect fluid).
    I completed all the steps except one:

    average value of [cos(θ)]^2 over unit sphere = 1/3.

    I have no idea how this is so. (θ is polar angle).
    I tried integrating over 0<θ<∏ and 0<φ<2∏, and then divided by the surface area of the sphere, but no luck with that either...

    Please help!
     
  2. jcsd
  3. Jun 9, 2013 #2

    Bill_K

    User Avatar
    Science Advisor

    The element of surface area on a sphere is sin θ dθ dφ. Thus if you integrate this over the entire sphere, ∫∫sin θ dθ dφ you should get the total area.

    Change variables to μ = cos θ and integrate from 0 to 2π on φ and -1 to +1 on μ.

    ∫∫ dμ dφ = (2)(2π) = 4π, the total area.

    Now for your integral,

    ∫∫ μ2 dμ dφ = 2π [1/3 μ3]-11 = 4π/3. Divide this by the total area and you get 1/3.
     
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