I’m trying to obtain the approximation (5.94) on page 164 of Peskin and Schroeder’s “Introduction to QFT”.
Let an electron with momentum $p = (E,-\omega\hat z)$ scatter off a photon with momentum $k = (\omega, \omega \hat z)$ in the center of mass frame. Then, for $E\gg m$ and $\theta\approx \pi$, the differential cross section is approximately:
$$\begin{aligned}\frac{d\sigma}{d\cos\theta} &\approx \frac 12\cdot\frac 1{2E}\cdot \frac 1{2\omega} \frac {\omega}{(2\pi)4(E + \omega)}\cdot\frac{2 e^4 (E+\omega)}{E+\omega\cos\theta}.\end{aligned}\tag{5.94a}$$
The RHS equals
$$\frac{1}{16(2\pi)E}\cdot\frac{4\pi}{4\pi}\cdot\frac{e^4}{E+\omega\cos\theta}=\frac{4\pi}{8E}\cdot\frac{e^4}{(4\pi)^2}\cdot\frac{1}{E+\omega\cos\theta}=\frac{2\pi\alpha^2}{4E(E+\cos\theta)}.$$
How can this be further approximated to
$$\frac{2\pi\alpha^2}{2m^2+s(1+\cos\theta)},\tag{5.94b}$$
where $s = (k+p)^2$?
Best Answer
\begin{equation} \frac{d\sigma}{d\cos\theta} = \frac{\alpha^2\pi}{2} \frac{1}{E(E+\omega\cos\theta)} \end{equation} Look at the denominator. Start by writing $E(E+\omega \cos\theta) = E\omega(E/\omega + \cos\theta)$. Next, from $m^2=E^2-\omega^2$ find $E/\omega =\sqrt{1+m^2/\omega^2} \approx 1+m^2/2\omega^2$. Therefore \begin{align} E(E+\omega \cos\theta) =& E\omega \left(1+\frac{m^2}{2\omega^2} +\cos\theta\right) = E\omega(1+\cos\theta) +\frac{Em^2}{2\omega} \end{align} Now $s=(p+k)^2 = 2p\cdot k = 2 E(E+\omega)$ and in the high energy regime $0\approx m^2 = E^2-\omega^2$ and therefore $E\approx \omega$ and so $s/2= E(E+\omega) = 2E\omega$ and $E/\omega \approx 1$. We therefore find \begin{align} E(E+\omega \cos\theta) =& \frac{1}{4} s (1+\cos\theta)+ \frac{1}{2}m^2 \end{align} which gives the desired result.