Let me provide you all with some context first.
I am studying how to get the differential cross-section formula (in the CoM frame) as explained in Quantum Field Theory's book by Mandl and Shaw (second edition; chapter 8, section 8.1) and I basically got lost in how to compute a certain integral.
The differential cross-section of two particles that, after collision, yield also 2 particles is given by the following formula:
$$d \sigma=f(p'_1, p'_2) \delta^{(4)} (p'_1 + p'_2 – p_1 – p_2)d^3 \mathbf p'_1 d^3 \mathbf p'_2 \ \ \ \ (1)$$
My books says that 'Integrating Eq. (1) wrt $\mathbf p'_2$ yields':
$$d \sigma = f(p'_1, p'_2) \delta (E'_1 + E'_2 – E_1 – E_2) |\mathbf p'_1|^2 d|\mathbf p'_1|^2 d|\mathbf p'_1| d \Omega'_1 \ \ \ \ (2)$$
My issue is that I do not see how to get $(2)$
I suspect that the sifting property of the Dirac Delta function (i.e. $\int f(t) \delta (t-T) dt = f(T)$) has been applied and that's actually why we go from $\delta^{(4)} (p'_1 + p'_2 – p_1 – p_2)$ to $\delta^{(1)} (E'_1 + E'_2 – E_1 – E_2)$. Besides, why $|\mathbf p'_1|^2 d|\mathbf p'_1|^2 d|\mathbf p'_1| d \Omega'_1 = d^3 \mathbf p'_1$? I know, based on what I learned in Calculus, that the solid angle satisfies $d \Omega = \sin \theta d\theta d \phi$ (where $\theta$ is the scattering angle and $\phi$ is the azimuthal angle) but I do not see why it shows up here.
Any help is appreciated.
Thank you.
Best Answer
$\def\d{\delta} \def\s{\sigma} \def\vp{{\bf p}}$The details: \begin{align} d \s &= f(p'_1, p'_2) \d^{(4)} (p'_1 + p'_2 - p_1 - p_2)d^3 \vp'_1 d^3 \vp'_2 \\ &= f(p'_1, p'_2) \d(E'_1 + E'_2 - E_1 - E_2) \d^{(3)}(\vp_1'+\vp_2'-\vp_1-\vp_2) d^3 \vp'_1 d^3 \vp'_2 \\ &\rightarrow \left.f(p'_1, p'_2) \d(E'_1 + E'_2 - E_1 - E_2) d^3 \vp'_1 \right|_{\vp_2'=\vp_1+\vp_2-\vp_1'}\\ &= \left.f(p'_1, p'_2) \d(E'_1 + E'_2 - E_1 - E_2) |\vp_1'|^2 d|\vp_1'| d\Omega_1' \right|_{\vp_2'=\vp_1+\vp_2-\vp_1'} \end{align} The integral over solid angle appears since the author also decides to go to spherical coordinates for the integral over $\vp_1'$, $d^3\vp_1' \rightarrow |\vp_1'|^2 d|\vp_1'| d\Omega_1'$.