# Why do we get preferential decay of particles in electron-muon scattering

I understood things mathematically but I am not able to convince myself intuitively. The query I am having is regarding the preferential decay of electron in electron-muon scattering. $$\frac{d\sigma}{d\Omega} = \frac{e^4}{8\pi^2 s}\frac{1+\frac{1}{4}(1+\cos\theta)^2}{(1-\cos\theta)^2}$$ (page 151 of http://www.hep.phy.cam.ac.uk/~thomson/partIIIparticles/handouts/Handout_5_2011.pdf).

I can't understand intuitively, why do we get high differential cross section for $cos\theta=1$ (i.e. out going electron(muon) have more probability to keep the same direction and that of incoming electron(muon)? As in the case of electron-positron annihilation, we get a symmetric differential cross section

$$\frac{d\sigma}{d\Omega} =\frac{\alpha^2}{4s}(1+\cos^2\theta)$$

I am looking for an intuitive answer.

Thank you!

• Would you care to include relevant formulae into the question? – Darkseid Oct 2 '17 at 19:48
• @FedorIndutny done! – kbg Oct 2 '17 at 20:01

The reasons for different behavior of the differential cross-sections is that

• these two processes go through different kinematic channels
• the mediator in these processes (which is the photon) is massless

In general, any cross-section has peaks when the squared momentum "carried" by the interaction mediator is near the mass of the mediator (and so the mediator becomes to be "on-shell"). This is the main point of the answer, since for the first process the mediator, which is the photon, can be on-shell, while for the second process it can't be on-shell.

The scattering process $\mu e \to \mu e$ kinematically goes through the $t$-channel, while the annihilation process $e\bar{e} \to \gamma \gamma$ goes through the $s$-channel. Therefore, marking the ingoing particles by the momenta $k_{i}$ and the outgoing ones by $p_{i}$ (the electron has label $1$), we obtain that the electron-muon scattering contains the photon propagator with $$D_{\mu\nu} \sim \frac{g_{\mu\nu}}{(p_{1}-k_{1})^{2}},$$ while the electron-positron annihilation contains the one with $$D_{\mu\nu} \sim \frac{g_{\mu\nu}}{(k_{1}+k_{2})^{2}}$$ The first one contains the singularity in the limit $p_{1} \to k_{1}$, which leads to the mentioned singularity. This is the well-known result in QED. It appears in the t-channel because of masslessness of the photon. In order to demonstrate that these factors - masslessness of the photon and kinematic channel - are the true reasons for the mentioned behavior, note that for the case of non-zero photon mass we'll obtain the denominator $$((p_{1}-k_{1})^{2}-m^{2}) = -(|(p_{1}-k_{1})|^{2}+m^{2}),$$ since there is always $(p_{1}-k_{1})^{2} < 0$, and therefore the singularity will be removed.
Instead, for the annihilation diagram the denominator is now modified to $$(p_{1}+k_{1})^{2} \to (p_{1}+k_{1})^{2} - m^{2}$$ Therefore, for $(p_{1}+k_{1})^{2} = m^{2}$ the cross-section becomes singular, at least for the first sight. Note, however, that this is not the case, since actually the denominator contains also the $i\Gamma m$ piece, where $\Gamma$ is the decay width of our "toy-like" massive photon...