Equation of state of photon gas

[4everphysics]

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!

 

[Bill_K]

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.