Answer: Casimir trick
In a typical collider experiment the momentum vectors are
$$
p_1=\begin{pmatrix}E\\0\\0\\p\end{pmatrix}\qquad
p_2=\begin{pmatrix}E\\0\\0\\-p\end{pmatrix}\qquad
p_3=\begin{pmatrix}
E\\
\rho\sin\theta\cos\phi\\
\rho\sin\theta\sin\phi\\
\rho\cos\theta
\end{pmatrix}
\qquad
p_4=\begin{pmatrix}
E\\
-\rho\sin\theta\cos\phi\\
-\rho\sin\theta\sin\phi\\
-\rho\cos\theta
\end{pmatrix}
$$
where $E$ is beam energy, $p=\sqrt{E^2-m^2}$, $\rho=\sqrt{E^2-M^2}$,
$m$ is electron mass $0.51\,\text{MeV}$,
and $M$ is muon mass $106\,\text{MeV}$.
The spinors are
\begin{gather*}
u_{11}=\begin{pmatrix}E+m\\0\\p\\0\end{pmatrix}\quad
v_{21}=\begin{pmatrix}-p\\0\\E+m\\0\end{pmatrix}\quad
u_{31}=\begin{pmatrix}E+M\\0\\p_3^z\\p_3^x+ip_3^y\end{pmatrix}\quad
v_{41}=\begin{pmatrix}p_4^z\\p_4^x+ip_4^y\\E+M\\0\end{pmatrix}
\\
u_{12}=\begin{pmatrix}0\\E+m\\0\\-p\end{pmatrix}\quad
v_{22}=\begin{pmatrix}0\\p\\0\\E+m\end{pmatrix}\quad
u_{32}=\begin{pmatrix}0\\E+M\\p_3^x-ip_3^y\\-p_3^z\end{pmatrix}\quad
v_{42}=\begin{pmatrix}p_4^x-ip_4^y\\-p_4^z\\0\\E+M\end{pmatrix}
\end{gather*}
The last digit in a spinor subscript is 1 for spin up and 2 for spin down.
Note that the spinors are not individually normalized.
Instead, a combined spinor normalization constant $N=(E+m)^2(E+M)^2$
will be used where needed.
This is the probability density for muon production.
Symbol $s=(p_1+p_2)^2=4E^2$,
symbol $s_j$ selects the spin of spinor $j$,
and $e$ is electron charge.
\begin{equation*}
|\mathcal{M}(s_1,s_2,s_3,s_4)|^2
=\frac{e^4}{s^2}\frac{1}{N}\left|(\bar{u}_3\gamma_\mu v_4)(\bar{v}_2\gamma^\mu u_1)\right|^2
\end{equation*}
The expected probability density $\langle|\mathcal{M}|^2\rangle$
is computed by summing $|\mathcal{M}|^2$ over all spin states
and dividing by the number of inbound states.
There are four inbound states.
\begin{align*}
\langle|\mathcal{M}|^2\rangle
&=\frac{1}{4}\sum_{s_1=1}^2\sum_{s_2=1}^2\sum_{s_3=1}^2\sum_{s_4=1}^2|\mathcal{M}(s_1,s_2,s_3,s_4)|^2
\\
&=\frac{e^4}{4s^2}\sum_{s_1=1}^2\sum_{s_2=1}^2\sum_{s_3=1}^2\sum_{s_4=1}^2
\frac{1}{N}\left|(\bar{u}_3\gamma_\mu v_4)(\bar{v}_2\gamma^\mu u_1)\right|^2
\end{align*}
Another way to compute $\langle|\mathcal{M}|^2\rangle$ is to use the Casimir trick.
\begin{equation*}
\langle|\mathcal{M}|^2\rangle
=\frac{e^4}{4s^2}
\mathop{\rm Tr}\left((\not p_3+M)\gamma^\mu(\not p_4-M)\gamma^\nu\right)
\mathop{\rm Tr}\left((\not p_2-m)\gamma_\mu(\not p_1+m)\gamma_\nu\right)
\end{equation*}
Here is a third way to compute $\langle|\mathcal{M}|^2\rangle$.
\begin{equation*}
\langle|\mathcal{M}|^2\rangle
=\frac{e^4}{4s^2}
\left(
32 (p_1\cdot p_3) (p_2\cdot p_4) +
32 (p_1\cdot p_4) (p_2\cdot p_3) +
32 m^2 (p_3\cdot p_4) +
32 M^2 (p_1\cdot p_2) +
64 m^2 M^2
\right)
\end{equation*}
For the momentum vectors given above the result is
\begin{equation*}
\langle|\mathcal{M}|^2\rangle
=e^4\left(1+\cos^2\theta+\frac{m^2+M^2}{E^2}\sin^2\theta+\frac{m^2M^2}{E^4}\cos^2\theta\right)
\end{equation*}
The Stanford Linear Collider
had a collision energy of $2E=91$ GeV.
For beam energies such as SLC where $E\gg M$ the above equation can be approximated as
$$
\langle|\mathcal{M}|^2\rangle=e^4(1+\cos^2\theta)
$$
The differential cross section is
$$
\frac{d\sigma}{d\Omega}
=\frac{\langle|\mathcal{M}|^2\rangle}{64\pi^2s}
=\frac{e^4}{256\pi^2E^2}(1+\cos^2\theta)
$$
Recall that $e^2=4\pi\alpha$ hence
\begin{equation*}
\frac{d\sigma}{d\Omega}=\frac{\alpha^2}{16E^2}(1+\cos^2\theta)
\end{equation*}
The total cross section calculation requires the following definite integral.
$$
\int_\Omega(1+\cos^2\theta)\,d\Omega
=\int_0^{2\pi}\int_0^\pi(1+\cos^2\theta)\sin\theta\,d\theta\,d\phi
=\frac{8}{3}\int_0^{2\pi}d\phi
=\frac{16\pi}{3}
$$
Hence the total cross section is
$$
\sigma
=\int_\Omega d\sigma
=\int_\Omega\frac{\alpha^2}{16E^2}(1+\cos^2\theta)\,d\Omega
=\frac{\alpha^2}{16E^2}\frac{16\pi}{3}
=\frac{\pi\alpha^2}{3E^2}
$$
