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I'm working on the Eq.$7.92$ in Peskin $($page $252)$

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I have two questions here:

Question $\bf 1$ : That is just what we would expect from the unitarity relation shown in Fig. $7.6~(\text b)$;

The Fig. $7.6~(\text b)$ is

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Obviously, Fig. $7.6~(\text b)$ is for the $t$-channel.

However, in page $233$, he said the $t$-channel didn't contribute.

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In my opinion, this is because when $s>4m^2$, $u<0$ and $t<0$.

But why can the $t$-channel contribute in Fig. $7.6~(\text b)$ ?

Why do we consider Fig.(a) instead of Fig.(b) ?

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Question $\bf 2$ : This dependence on $q^2$ is exactly the same as in Eq. $(5.13)$, the cross section for production of a fermion-antifermion pair.

The Eq.$(5.13)$ is $$\sigma_{\mathrm{total}}=\frac{4 \pi \alpha^{2}}{3 E_{\mathrm{cm}}^{2}} \sqrt{1-\frac{m_{\mu}^{2}}{E^{2}}}\left(1+\frac{1}{2} \frac{m_{\mu}^{2}}{E^{2}}\right)$$ I tried to prove it but I met some difficulties. \begin{align} \operatorname{Im}\left[\widehat{\Pi}_{2}\left(q^{2} \pm i \epsilon\right)\right]&=\mp \frac{\alpha}{3} \sqrt{1-\frac{4 m^{2}}{q^{2}}}\left(1+\frac{2 m^{2}}{q^{2}}\right)\\&=\mp \frac{\alpha}{3} \sqrt{1-\frac{4 m^{2}}{4E^{2}}}\left(1+\frac{2 m^{2}}{4E^{2}}\right)\\&=\mp \frac{\alpha}{3} \sqrt{1-\frac{ m^{2}}{E^{2}}}\left(1+\frac{1 m^{2}}{2E^{2}}\right) \end{align} However, according to Eq.$(7.50)$ $$\operatorname{Im} \mathcal{M}\left(k_{1}, k_{2} \rightarrow k_{1}, k_{2}\right)=2 E_{\mathrm{cm}} p_{\mathrm{cm}} \sigma_{\mathrm{tot}}\left(k_{1}, k_{2} \rightarrow \text { anything }\right)$$ I can't get Eq.$(5.13)$.

How do I prove Eq.$(5.13)$ from Eq.$(7.92)$ by using the optical theorem?

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