Average of cos(x)^2 in spherical distribution

  • #1
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I'm particle physics at the moment and I don't get why the average value of cosˆ2 is 1/3.
The section :

Screen_Shot_2017_06_01_at_6_40_52_PM.png


My solution is :
[tex]< cos^{2}(\alpha)> = \frac{1}{\pi - 0} \int_{0}^{\pi } cos^{2}(\alpha)sin(\alpha)d\alpha[/tex]
I substitute
[tex]u = cos(\alpha)[/tex]
so i get
[tex]-\frac{1}{\pi}\int_{1}^{-1}u^{2}du = \frac{2}{3\pi}[/tex]

What im i doing wrong ? :(
 
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  • #2
Ray Vickson
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<< Mentor Note -- Thread moved from the technical forums so no Homework Help Template is shown >>

I'm particle physics at the moment and I don't get why the average value of cosˆ2 is 1/3.
The section :

View attachment 204679

My solution is :
[tex]< cos^{2}(\alpha)> = \frac{1}{\pi - 0} \int_{0}^{\pi } cos^{2}(\alpha)sin(\alpha)d\alpha[/tex]
I substitute
[tex]u = cos(\alpha)[/tex]
so i get
[tex]-\frac{1}{\pi}\int_{1}^{-1}u^{2}du = \frac{2}{3\pi}[/tex]

What im i doing wrong ? :(
$$\langle \cos^2(\alpha) \rangle =
\frac{\int_0^{\pi} \cos^2(\alpha) \sin(\alpha) \, d \alpha}{\int_0^{\pi} 1 \cdot \sin(\alpha) \, d \alpha}. $$
To compute an average like ##\langle f(x) \rangle = \int w(x) f(x) dx,## you need the "weight" function ##w(x) \geq 0## to integrate to 1.
 
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  • #3
$$\langle \cos^2(\alpha) \rangle =
\frac{\int_0^{\pi} \cos^2(\alpha) \sin(\alpha) \, d \alpha}{\int_0^{\pi} 1 \cdot \sin(\alpha) \, d \alpha}. $$
To compute an average like ##\langle f(x) \rangle = \int w(x) f(x) dx,## you need the "weight" function ##w(x) \geq 0## to integrate to 1.
Thank you ! Makes sense!
 

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