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Expectation for cos(x)^2

  1. Aug 6, 2012 #1
    1. The problem statement, all variables and given/known data
    This question comes from calculating the Einstein A and B coefficients. I am supposed to find the average value of cos(x)^2 over the solid angle of a sphere which is 1/3. And I need to show this.
    A similar course in a different uni just says that For unpolarized, isotropic radiation, the expectation of cos(x)^2=1/3



    2. Relevant equations

    cos(2x)=2cos(x)^2-1



    3. The attempt at a solution

    I tried using the average integral equation however i always end up with 1/2. I've tried
    1/pi *∫cos(X)^2dx and just use the trig equation that I have given. However the answer comes out as 1/2 and I do not know how to get 1/3. I also tried integrating from 0 to 2pi etc.

    Thankful for any help!
     
  2. jcsd
  3. Aug 6, 2012 #2
    You are getting wrong result because you don't weight the points on the sphere correctly. There are less points corresponding to each value of x near the poles. The correct uniform probability measure is [itex] \sin x dx [/itex], and the integral you should calculate is

    [tex] E[\cos^2(x)] = \frac{1}{2} \int_0^\pi \cos^2(x) \sin(x) dx [/tex]
     
  4. Aug 6, 2012 #3
    Ty very much.
     
    Last edited: Aug 6, 2012
  5. Aug 6, 2012 #4
    In fact it is taken over solid angle dΩ and it is easy to write in spherical coordinates where
    dΩ=d(cosθ)dψ and your average will be

    =(∫(cos2θ) d(cosθ)dψ)/4∏,where 4∏ is ∫d(cosθ)dψ,limits are from o to ∏ for θ and 0 to 2∏ for ψ.
     
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