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Boltzmann Equation-where'd the 3 come from?

  1. Jul 12, 2005 #1
    Boltzmann Equation--where'd the 3 come from?

    Greetings, I'm a little bit confused about the derivation for the Boltzmann equation for a particle in thermal equilibrium in the Friedman-Robertson-Walker metric. I've been following the exposition in Kolb and Turner, The Early Universe p. 116. I reproduce all the relevant results here.

    In particular:
    We are given (K&T, eq 5.5) that for a phase space distribution [tex]f[/tex] the form of the Liouville operator in the FRW model is given by:

    [tex]\hat{\mathbf{L}}[f(E,t)] = E\frac{\partial f}{\partial t}-\frac{\dot{R}}{R}|\mathbf{p}|^2\frac{\partial f}{\partial E}[/tex]

    Further, the number density [tex]n[/tex] is given by an integral over momenta (K&T eq. 5.6):

    [tex]n(t) = \frac{g}{(2\pi)^3}\int d^3p f(E,t)[/tex]

    where [tex]g[/tex] is the number of internal degrees of freedom.

    The Boltzmann equation, [tex]\hat{\mathbf{L}}[f]= \mathbf{C}[f][/tex], can then be written out by plugging in the above equation for the Liouville operator on the left hand side.

    We can then divide by [tex]E[/tex], multiply by [tex]\frac{g}{(2\pi)^3}[/tex], and perform a momentum space integral to express the Boltzmann equation in terms of [tex]n[/tex].

    Kolb and Turner write the result as:

    [tex]\frac{dn}{dt} + 3\frac{\dot{R}}{R}n = \frac{g}{(2\pi)^3}\int\textbf{C}[f]\frac{d^3p}{E}[/tex]

    I'm confused by the factor of 3 in the seccond term and am not sure how this is resolved. I'm also not sure how to treat the energy in the momentum integral--I assume that since [tex]E^2=\mathbf{p}^2+m^2[/tex], one can rewrite the momentum integral in spherical coordinates where the function [tex]f[/tex] is a function of the radial coordinate alone. I assume some integration by parts is necessary, but this still does not account for the factor of 3.

    Any help would be appreciated,

    Last edited: Jul 13, 2005
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  3. Jul 13, 2005 #2


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  4. Jul 13, 2005 #3


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    I haven't read Kolb and Turner's book, but I think you'll find the factor 3 and the rest of that term comes from:
    [tex]\frac{1}{R^3}\frac{d}{dt}(R^3)n [/tex]

  5. Jun 17, 2011 #4
    Re: Boltzmann Equation--where'd the 3 come from?

    Last edited: Jun 17, 2011
  6. Jun 17, 2011 #5
    Re: Boltzmann Equation--where'd the 3 come from?

    A bit late for the one who asked, but perhaps it helps those who have the same problem

    [tex]E\frac{\partial f}{\partial t}-\frac{\dot{R}}{R}|\mathbf{p}|^2\frac{\partial f}{\partial E} = \hat{\mathbf{C}}[f(E,t)][/tex]
    divide by E
    [tex]\frac{\partial f}{\partial t}-\frac{\dot{R}}{R}|\mathbf{p}|^2\frac{\partial f}{E\partial E} = \hat{\mathbf{C}}[f(E,t)]\frac{1}{E}[/tex]
    [tex]\frac{g}{(2\pi)^3}\int d^3p\frac{\partial f}{\partial t}-\frac{\dot{R}}{R}|\mathbf{p}|^2\frac{\partial f}{E\partial E} = \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}[/tex]
    [tex]n(t)-\frac{\dot{R}}{R}\frac{g}{(2\pi)^3}\int d^3p|\mathbf{p}|^2\frac{\partial f}{E\partial E} = \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}[/tex]
    [tex]E\partial E=p\partial p[/tex]
    [tex]n(t)-\frac{\dot{R}}{R}\frac{g}{(2\pi)^3}\int 4\pi dp|\mathbf{p}|^2|\mathbf{p}|^2\frac{\partial f}{p\partial p} = \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}[/tex]
    [tex]n(t)-\frac{\dot{R}}{R}\frac{g}{(2\pi)^3}\int 4\pi dpp^3\frac{\partial f}{\partial p} = \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}[/tex]
    integration by part:
    [tex]n(t)+\frac{\dot{R}}{R}\frac{g}{(2\pi)^3}\int 4\pi dp3p^2f= \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}[/tex]
    [tex]n(t)+3\frac{\dot{R}}{R}\frac{g}{(2\pi)^3}\int 4\pi dpp^2f= \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}[/tex]
    [tex]n(t)+3\frac{\dot{R}}{R}n(t)= \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}[/tex]

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