# Approximating high-energy Compton scattering cross section

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 $$s = (k+p)^2$$?

$$$$\frac{d\sigma}{d\cos\theta} = \frac{\alpha^2\pi}{2} \frac{1}{E(E+\omega\cos\theta)}$$$$ Look at the denominator. Start by writing $$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$$. We therefore find \begin{align} E(E+\omega \cos\theta) =& \frac{1}{4} s (1+\cos\theta)+ \frac{1}{2}m^2 \end{align} which gives the desired result.
• I want to share a little more, because I also meet puzzles when I approximated it: $$E\approx\omega\gg m$$, $$s=2p\cdot k=2\omega(E+\omega)+m^2$$ so that $$E\omega \approx \omega^2\approx s/4$$, plugging them into the denominator of cross section, $$E(E+\omega\cos{\theta}) = m^2+\omega^2+E\omega\cos{\theta}\approx m^2+\frac{1}{4}s(1+\cos{\theta})$$ but it's wrong since the formula $$\omega^2\approx s/4$$ discard the term of $$m^2$$ order which we need in the final expression.
• let's first find the transformation between $$(E,\omega)$$ and $$(s,m)$$ $$\begin{cases} s =m^2 + 2\omega(E+\omega)\\ m^2 = E^2 - \omega^2 \end{cases}\Longrightarrow \begin{cases} E = \frac{1}{2} \left( \sqrt{s} + \frac{m^2}{\sqrt{s}} \right)\\ \omega = \frac{1}{2} \left( \sqrt{s} - \frac{m^2}{\sqrt{s}} \right) \end{cases}$$ then the denominator becomes \begin{align} E(E+\omega\cos{\theta}) &= \frac{1}{4}\left(s+2m^2+\frac{m^4}{s}\right)+\frac{1}{4}\left( s-\frac{m^4}{s} \right)\cos{\theta} \\ &= \frac{1}{2}m^2+\frac{1}{4}s(1+\cos{\theta}) + \mathcal{O}\left(\frac{m^4}{s}\right) \end{align} maybe this process would be more clear since we only make approxiamtion in the final step.