Consider the following process: $$e^{-}+e^{+}\rightarrow X+\bar{X}$$ in the case where the positron is the target particle at rest. enter image description here

If the incoming electron has momentum $\mathbf{p}_{e}$, then by momentum conservation, it follows that $$ \mathbf{p}_{e}=\mathbf{p}_{1}+\mathbf{p}_{2}$$ where $\mathbf{p}_{1}$ and $\mathbf{p}_{2}$ are the momenta of the two particles created in this collision.

Why is it the case that $$\lvert\mathbf{p}_{1}+\mathbf{p}_{2}\rvert =2p_{_{X}}\cos\theta\;?$$ Naively, I would've thought it would be $$\lvert\mathbf{p}_{1}+\mathbf{p}_{2}\rvert =\sqrt{\lvert\mathbf{p}_{1}\rvert^{2}+\lvert\mathbf{p}_{2}\rvert^{2}+2\lvert\mathbf{p}_{1}\rvert\lvert\mathbf{p}_{2}\rvert\cos\theta}$$

Is it correct to say that the threshold case is when $\theta =0$? Why is it the case that if the energy is higher than the threshold energy the emitted particles propagate at angles relative to the direction of motion of the incoming particle, whereas in the threshold case they propagate along the same direction?


I think I partially see what I'm missing. If the angles at which the two created particles are emitted, relative to the direction of motion of the positron, are equal, then their 3-momenta are given by $$\mathbf{p}_{1}=\left(p\cos\theta ,p\sin\theta\right)\\ \mathbf{p}_{2}=\left(p'\cos\theta ,-p'\sin\theta\right)$$ where $\theta$ is the angle relative to the direction of motion of the positron.

As such, momentum components along the direction defined by the direction of motion of the original positron, are $p_{1}=p\cos\theta$ and $p_{2}=p'\cos\theta$ and so conservation of (linear) momentum requires that $$p_{e}=p\cos\theta +p'\cos\theta$$ Furthermore, conservation of linear momentum along the direction perpendicular to that of the motion of the positron requires that $$p\sin\theta -p'\sin\theta =0\quad\Rightarrow\quad p=p'$$ and so $$p_{e}=2p\cos\theta$$

I'm still unsure as to why the particles are produced at angles relative to the direction of motion of the positron? Is this just because in general they won't simply be produced such they propagate along the same direction as the initial particle?!

  • $\begingroup$ Which is why doing things in the center of mass frame is a good thing. Then translate back to the lab frame. $\endgroup$ – Jon Custer Oct 12 '16 at 20:13
  • $\begingroup$ As @Jon says you the best way to reason about this is to consider the CoM frame. It makes it quite obvious that you have the right idea. No messing around with nasty radicals. $\endgroup$ – dmckee --- ex-moderator kitten Oct 12 '16 at 21:00
  • $\begingroup$ @dmckee Do you mean the CoM frame of the initial particles $e^{-}$ and $e^{+}$, such that they have equal and opposite momentum?! $\endgroup$ – user35305 Oct 12 '16 at 21:03
  • $\begingroup$ Yes, that is the CoM frame for scattering problems. $\endgroup$ – Jon Custer Oct 12 '16 at 21:05
  • $\begingroup$ The center of momentum frame (and in relativity there may be a distinction between center of mass and center of momentum) is defined as the frame for which the total momentum is zero. So, yeah, equal and opposite momentum (and for electron-positron that means equal and opposite velocities). $\endgroup$ – dmckee --- ex-moderator kitten Oct 12 '16 at 21:06

As discussed in the comments it is always better to start with the center of mass system, and here is a calculation for the reaction with some data shown in the center of mass system taking into account helicities:

angular mumu

So there is an angular distribution for the generation of the pair, and going into the lab the transformation will give a spectrum of angles versus the incoming momentum theta_1 and theta_2 depending on the center of mass angular distribution. It is the sum of the momenta that has to be conserved, so the two muons need not have the same momentum to get the imposed same angle in the question.

A backward going muon will be practcally at rest in the lab and the forward one will take all the momentum, for example. There will be a distribution in the lab too, but calculations and checks with theory are always in the center of mass system.

  • $\begingroup$ Thanks for the link. Why is the case in general the the generated particles will propagate outwards at angles relative to direction of motion of the initial particle (as opposed to propagating along the same direction)?! $\endgroup$ – user35305 Oct 13 '16 at 13:35
  • $\begingroup$ ... Can one heuristically think of it in terms of billiard balls, in the sense that, if a billiard ball strikes another billiard ball (that is initially at rest), then the target ball will experience a force normal to the contact point between the two billiard balls, and therefore, if the collision is not head on, the target ball will move outwards at an angle relative to direction of motion of the incoming ball?! $\endgroup$ – user35305 Oct 13 '16 at 14:11
  • $\begingroup$ But they are moving at an angle relative to incoming electron, each with its own angle which can calculated by transforming from the center of mass system, where the angles are the same except with oposite sign, to the lab where the angles will be different depending on the original cm angle and the lorenz transformation $\endgroup$ – anna v Oct 13 '16 at 14:47
  • $\begingroup$ well it is not billiard balls, it is relativistc mechanics not newtonian $\endgroup$ – anna v Oct 13 '16 at 14:51
  • $\begingroup$ yes, I see, but I was just trying to understand heuristically why the particles produced in the collision move at angles relative to the incoming electron in the first place? What is the physical reason for why this happens? $\endgroup$ – user35305 Oct 13 '16 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.