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Maxwell-boltzman distrubution

  1. Mar 17, 2010 #1
    If I have two particles that follows the maxwell velocity distribution:
    [tex]\phi(v_i)dv_i=4 \pi v_i^2 \left ( \frac{m_iv_i}{2\pi kT} \right ) ^{3/2}e^{\frac{-m_iv_i^2}{2kT}}dv_i[/tex]
    Why is their combined distribution:
    [tex]\phi(v)dv=4 \pi v^2 \left ( \frac{\mu v}{2\pi kT} \right ) ^{3/2}e^{\frac{-\mu v^2}{2kT}}dv[/tex]
    where mu is the reduced mass and v=v2-v1
    I have these questions because I don't quite follow these derivations.
    http://dissertations.ub.rug.nl/FILES/faculties/science/2007/a.matic/c2.pdf [Broken]
    Namely, I not sure why the following holds:
    [tex]\int_0^\infty \int_0^\infty \phi(v_1) \phi(v_2) v_1 v_2 \sigma dv_1 dv_2 = 4\pi \left ( \frac{\mu v}{2\pi kT} \right ) ^{3/2} \int_0^\infty v^3 \sigma e^{\frac{-\mu v^2}{2kT}}dv[/tex]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 18, 2010 #2


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    You have v=v1-v2. Let u=v1+v2. The new integration is straightforward as long as the integral limits are -∞ to ∞. You then need |v1v2| instead of v1v2 in the integrand. The u integration should leave you with what you want.
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