# How does the coefficient in the Boltzmann equation comes from

1. Aug 10, 2011

In the boltzmann equation,
${\bf{L}}\left[ f \right] = {\bf{C}}\left[ f \right]$, the right which is the collision term and in general it is${\bf{C}}[f]=\sum\limits_{p,p_1} {|Amplitiude|^2}{f(p)}$. and explicitly, the collision term for decaying process is
$\begin{eqnarray} {\bf{C}}\left[ f \right] &=& \frac{1}{{E\left( p \right)}}\int {\frac{{{d^3}p}}{{{{\left( {2\pi } \right)}^3}2E\left( p \right)}}\frac{{{d^3}{p_2}}}{{{{\left( {2\pi } \right)}^3}2E\left( {{p_2}} \right)}}}\nonumber\\ &\times& {\left| {{\cal M}\left( {p \to {p_1} + {p_2}} \right)} \right|^2}f\left( p \right)\nonumber\\ &\times& {\left( {2\pi } \right)^4}{\delta ^4}(p - {p_1} + {p_2}) \end{eqnarray}$
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 $E(p)$comes from

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

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

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

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