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I'm having a bit of a slow day, and can't see how to do the following integral in Peskin and Schroeder (page 107 for anyone with the book). We've derived in the centre of mass frame the integral over 2 body phase space

$$\int d\Pi_2 =\int \frac{dp_1p_1^2d\Omega}{(2\pi)^3)2E_12E_2}(2\pi)\delta(E_{cm}-E_1-E_2)$$

where $E_1^2=p_1^2+m_1^2,\ E_2=p_1^2+m_2^2$ and $E_{cm}$ is the total initial energy.

Apparently one can then "integrate over the final delta function" to get

$$\int d\Pi_2=\int d\Omega\frac{p_1^2}{16\pi^2E_1E_2}(\frac{p_1}{E_1}+\frac{p_1}{E_2})^{-1}$$

How is this done? I suppose it's some kind of clever change of variables or trick but I can't see it! Apologies if I'm missing something easy! To be honest I'm just puzzled as to why we're left with any factors of $p_1$ after doing the integral, because it's just an integration variable, isn't it?

A clear argument would be very much appreciated, as would some intuition as to why I'm misunderstanding the notation! Many thanks in advace.

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It is simply a matter of notation. The $p_1$ (and hence $E_1$ and $E_2$) in

$$\int d\Pi_2=\int d\Omega\frac{p_1^2}{16\pi^2E_1E_2}(\frac{p_1}{E_1}+\frac{p_1}{E_2})^{-1}$$

is no longer an integration variable; it has the fixed value that satisfies the delta function $\delta(E_{cm}-E_1-E_2)$ in the previous integral. The factor $(\frac{p_1}{E_1}+\frac{p_1}{E_2})^{-1}$ comes from applying an identity of the delta function: $$\delta(g(x)) = \frac{\delta(x-x_0)}{|g'(x_0)|}.$$

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  • $\begingroup$ Ah of course - I'd forgotten that identity entirely! Thanks a lot! $\endgroup$
    – Edward Hughes
    Commented Jan 4, 2013 at 19:32

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