Why Does cos(x)^2 Average to 1/3 Over a Sphere's Solid Angle?

[Zamze]

Homework Statement

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

Homework Equations

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

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!

 

[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

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

 

[Zamze]

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

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

Ty very much.

 

[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.