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

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Homework Help Overview

The discussion revolves around calculating the average value of cos(x)^2 over the solid angle of a sphere, specifically addressing why this average equals 1/3. The original poster references a context involving Einstein A and B coefficients and mentions differing approaches from other courses regarding unpolarized, isotropic radiation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use an average integral equation but consistently arrives at 1/2 instead of the expected 1/3. They explore integrating from 0 to 2π and using trigonometric identities.
  • Some participants suggest that the original poster is not correctly weighting points on the sphere, emphasizing the importance of the uniform probability measure involving sin(x) in the integral.
  • Another participant introduces the concept of solid angle and spherical coordinates, proposing a specific integral setup for calculating the average.

Discussion Status

The discussion is ongoing, with participants providing insights into the correct approach for integrating over the solid angle. There are multiple interpretations being explored regarding the weighting of points and the setup of the integral, but no explicit consensus has been reached yet.

Contextual Notes

Participants are addressing potential misunderstandings related to the integration limits and the correct measure for averaging over a sphere. The original poster's attempts indicate a struggle with the mathematical setup required for the problem.

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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!
 
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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
 
clamtrox said:
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.
 
Last edited:
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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