1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    Science Advisor
    Homework Helper

    $$\langle \cos^2(\alpha) \rangle =
    \frac{\int_0^{\pi} \cos^2(\alpha) \sin(\alpha) \, d \alpha}{\int_0^{\pi} 1 \cdot \sin(\alpha) \, d \alpha}. $$
    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!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted