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Average of cos(x)^2 in spherical distribution

  1. Jun 1, 2017 #1
    << 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 :

    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 ? :(
     
    Last edited by a moderator: Jun 1, 2017
  2. jcsd
  3. Jun 1, 2017 #2

    Ray Vickson

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    $$\langle \cos^2(\alpha) \rangle =
    \frac{\int_0^{\pi} \cos^2(\alpha) \sin(\alpha) \, d \alpha}{\int_0^{\pi} 1 \cdot \sin(\alpha) \, d \alpha}. $$
    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.
     
  4. Jun 1, 2017 #3
    Thank you ! Makes sense!
     
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