Express the ep $\rightarrow$ eX differential cross section (8.34) in terms of $\sigma_{T, L}$. That is, show that
$$
\left.\frac{d \sigma}{d E^{\prime} d \Omega}\right|_{\mathrm{lab}}=\Gamma\left(\sigma_T+\varepsilon \sigma_L\right),
$$
where
$$
\begin{aligned}
\Gamma & =\frac{\alpha K}{2 \pi^2\left|q^2\right|} \frac{E^{\prime}}{E} \frac{1}{1-\varepsilon}, \\
\varepsilon & =\left(1-2 \frac{\nu^2-q^2}{q^2} \tan ^2 \frac{\theta}{2}\right)^{-1} .
\end{aligned}
$$
EXERCISE 8.17 The formalism has been set up in such a way that, when $q^2 \rightarrow 0$,
$$
\begin{aligned}
& \sigma_T \rightarrow \sigma^{\text {tot }}(\gamma \mathrm{p}) \\
& \sigma_L \rightarrow 0,
\end{aligned}
$$
where $\gamma$ is a real photon and $\sigma^{\text {tot }}(\gamma \mathrm{p})$ is given by $(8.45)$.
Despite its appearance, convince yourself that $W_{\mu \nu}$ must not be singular at $q^2=0$. Hence, show that
$$
W_2 \rightarrow 0 \quad \text { and } \quad\left(W_1+\frac{\nu^2}{q^2} W_2\right) \rightarrow 0
$$
as $q^2 \rightarrow 0$, and so establish that $\sigma_L$ vanishes.