# The cross section for production of a fermion-antifermion pair in Peskin and Schroeder (Eq.7.92)

I'm working on the Eq.$$7.92$$ in Peskin $$($$page $$252)$$ 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 Obviously, Fig. $$7.6~(\text b)$$ is for the $$t$$-channel.

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