Justify the following relations:
$$
\begin{aligned}
\int \frac{d^3 p^{\prime}}{2 p_0^{\prime}} \delta^{(4)}\left(p+q-p^{\prime}\right) & =\int d^3 p^{\prime} d p_0^{\prime} \delta^{(4)}\left(p+q-p^{\prime}\right) \theta\left(p_0^{\prime}\right) \delta\left(p^{\prime 2}-M^2\right) \\
& =\frac{1}{2 M} \delta\left(\nu+\frac{q^2}{2 M}\right) \\
& =\frac{1}{2 M A} \delta\left(E^{\prime}-E / A\right)
\end{aligned}
$$
where $A=1+(2 E / M) \sin ^2 \frac{\theta}{2}$, and the step function $\theta(x)$ is 1 if $x>0$ and 0 otherwise.
Inserting (6.43) into (6.46) and using (6.47), we obtain
$$
\frac{d \sigma}{d E^{\prime} d \Omega}=\frac{\left(2 \alpha E^{\prime}\right)^2}{q^4}\left\{\cos ^2 \frac{\theta}{2}-\frac{q^2}{2 M^2} \sin ^2 \frac{\theta}{2}\right\} \delta\left(\nu+\frac{q^2}{2 M}\right) .
$$
Using (6.48), we may perform the $d E^{\prime}$ integration and, replacing $q^2$ by (6.44), we finally arrive at the following formula for the differential cross section for $\mathrm{e}^{-} \mu^{-}$ scattering in the laboratory frame:
$$
\left.\frac{d \sigma}{d \Omega}\right|_{\text {lab. }}=\left(\frac{\alpha^2}{4 E^2 \sin ^4 \frac{\theta}{2}}\right) \frac{E^{\prime}}{E}\left\{\cos ^2 \frac{\theta}{2}-\frac{q^2}{2 M^2} \sin ^2 \frac{\theta}{2}\right\}
$$
A powerful technique for exploring the internal structure of a target is to bombard it with a beam of high-energy electrons and to observe the angular distribution and energy of the scattered electrons. Such experiments have repeatedly led to major advances in our understanding of the structure of matter. Starting in Chapter 8, we describe how this method has revealed the structure of the proton. Equation (6.50) plays a central role in the story.