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How does the coefficient in the Boltzmann equation comes from

  1. Aug 10, 2011 #1
    In the boltzmann equation,
    [itex]{\bf{L}}\left[ f \right] = {\bf{C}}\left[ f \right][/itex], the right which is the collision term and in general it is[itex]{\bf{C}}[f]=\sum\limits_{p,p_1} {|Amplitiude|^2}{f(p)}[/itex]. and explicitly, the collision term for decaying process is
    [itex]
    \begin{eqnarray}
    {\bf{C}}\left[ f \right] &=& \frac{1}{{E\left( p \right)}}\int {\frac{{{d^3}p}}{{{{\left( {2\pi } \right)}^3}2E\left( p \right)}}\frac{{{d^3}{p_2}}}{{{{\left( {2\pi } \right)}^3}2E\left( {{p_2}} \right)}}}\nonumber\\
    &\times& {\left| {{\cal M}\left( {p \to {p_1} + {p_2}} \right)} \right|^2}f\left( p \right)\nonumber\\
    &\times& {\left( {2\pi } \right)^4}{\delta ^4}(p - {p_1} + {p_2})
    \end{eqnarray}
    [/itex]
    I want to know exactly how does the coefficient in the equation comes from,for it is very important for the result.
    My questions are:
    1) how does the [itex]E(p)[/itex]comes from

    2) [itex]\begin{equation}
    \int {\frac{{{d^4}p}}{{{{\left( {2\pi } \right)}^4}}}\left( {2\pi } \right)\delta ({E^2} - {m^2} - {p^2})} = \int {\frac{{{d^3}p}}{{{{\left( {2\pi } \right)}^3}}}\frac{{\delta (E - \sqrt {{m^2} + {p^2}} )}}{{2E}}}
    \end{equation}[/itex],the[itex](2\pi)^4[/itex]in the left of the equation is the space element with[itex]\hbar=1[/itex],but how does the [itex]2\pi[/itex] comes from?

    3) in the equation above, why cannot we write the [itex]\delta[/itex]function is the left with[itex]\delta (E - \sqrt {{m^2} + {p^2}} )[/itex] directly, if so, then [itex]2E[/itex]does not exist in denominator of the right equation.

    4) why there is [itex](2\pi)^4[/itex]in the equation[itex]{\left( {2\pi } \right)^4}{\delta ^4}(p - {p_1} + {p_2})[/itex].

    can anyone give give explanations or recommend some books to which I can refer?
     
  2. jcsd
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