# Particle creation (with stationary target particle)

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

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?

Edit:

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?!

• Which is why doing things in the center of mass frame is a good thing. Then translate back to the lab frame. – Jon Custer Oct 12 '16 at 20:13
• 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. – dmckee Oct 12 '16 at 21:00
• @dmckee Do you mean the CoM frame of the initial particles $e^{-}$ and $e^{+}$, such that they have equal and opposite momentum?! – user35305 Oct 12 '16 at 21:03
• Yes, that is the CoM frame for scattering problems. – Jon Custer Oct 12 '16 at 21:05
• 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). – dmckee Oct 12 '16 at 21:06