Expectation for cos(x)^2

1. Aug 6, 2012

Zamze

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. Aug 6, 2012

clamtrox

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 $\sin x dx$, and the integral you should calculate is

$$E[\cos^2(x)] = \frac{1}{2} \int_0^\pi \cos^2(x) \sin(x) dx$$

3. Aug 6, 2012

Zamze

Ty very much.

Last edited: Aug 6, 2012
4. Aug 6, 2012

andrien

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.