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Solid angle?

  1. Apr 6, 2009 #1
    So I need to inegrate over a solid angle, in which every possible orientation is considered (we are doing scattering events in which we assume every possible angle is possible), thus I need to solve
    [tex]\int d\Omega_1 d\Omega_2 d\Omega_3[/tex] Eqn (1).

    Now I know

    [tex]\int^{2 \pi}_{0} \int^{\pi}_{0}d\Omega = \int^{2 \pi}_{0} \int^{\pi}_{0} sin(\theta)d\theta d\phi = 2\pi \int^{\pi}_{0} sin(\theta) d\theta = 2\pi (-cos(\theta))^{\pi}_{0} = 4 \pi[/tex]
    so shouldn't that mean that Eqn (1) should integrate to [tex] 64 \pi^3[/tex]?
     
    Last edited: Apr 6, 2009
  2. jcsd
  3. Apr 6, 2009 #2

    tiny-tim

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    Hi physguy09! :smile:

    (have an omega: Ω :wink:)
    I don't understand … how can you have more than one Ω? (more than one solid angle?) :confused:
     
  4. Apr 7, 2009 #3
    the original problem asks us to calculate the density of states
    [tex] d^9 n \propto p_e p_\nu1 p_\nu2 dp_e dp_\nu d\Omega_e d\Omega_1 d\Omega_2 [/tex]
    where each individual particle is allowed to undergo its own possible orientation independent of the other particles (as there is no attraction amongst these, since we have two neutrinos and an e-.
     
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