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

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

(page 136 of http://www.hep.phy.cam.ac.uk/~thomson/partIIIparticles/handouts/Handout_4_2011.pdf.)

I am looking for an intuitive answer.

Thank you!

  • $\begingroup$ Would you care to include relevant formulae into the question? $\endgroup$ – Darkseid Oct 2 '17 at 19:48
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    $\begingroup$ @FedorIndutny done! $\endgroup$ – 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.

Let's talk about this precisely.

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...

  • $\begingroup$ Thanks for the answer. But this is what we get mathematically. Also, by saying photon massless, I can't get what I am looking for. Its massless in both the cases. I was looking for an intuitive answer. (I guess it would be more better to say that it is the momentum carried by the photon which accounts for this result, but again this is mathematically) $\endgroup$ – kbg Oct 2 '17 at 20:21
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    $\begingroup$ @kbg : only the combination of the two factors leads to the singular behavior of the cross-section. I'll add some clarifying words to my answer which will probably make my arguments clearer. $\endgroup$ – Name YYY Oct 2 '17 at 20:29
  • $\begingroup$ I am sorry, but I did not understand the last paragraph completely. I am unable to understand : " for massless particles the one-particle contribution to the full propagator in the spectral representation isn't divided from the multi-particle contributions by the gap" $\endgroup$ – kbg Oct 2 '17 at 20:50
  • $\begingroup$ @kbg : I've updated the answer. Please review it. I'll also think how to make the cited statement clearer, and will add the corresponding explanation later. $\endgroup$ – Name YYY Oct 2 '17 at 22:11

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