# Question about relativistic collision between particles

1. Oct 1, 2015

### Frank Einstein

1. The problem statement, all variables and given/known data
Good morning-afternoon. I have been tasked with solving the next exercice; but I am unable to spot the error. Please can anyone point me in the right direction?

A)Calculate the thresold energy of a photon to disintegrate in an electron and a positron in the presence of an electron.

B)Show that you can't have the former disintegration in the vacuum

C)If the disintegration happens in the presencie of an atomic core, show that it's momentum is miminal.

Consider c (light speed =1)
2. Relevant equations
Conservation of energy: Ei=Ef
Conservation of momentum: Pi=Pf; since the photon is a maseless particle, Pγ=Eγ

3. The attempt at a solution

A) Ei=me+Eγ. Ef=2me+(me^2+p^2)^0.5 Then, I apply conservation of energy (I eliminate me at both sides)
Eγ=me+(me^2+p^2)^0.5 Eγ-me=(me^2+p^2)^0.5 Eγ^2+me^2-2Eγme=me^2+Eγ
So I arrive to 2Eγme=0. That has to be wrong.

B) I apply conservation of momentum. Pi=Eγ. If we go th the centre of mass, Pf=0; so the conservation of momentum is violated, so the desintegration doesn't happen.

C) Same as A), but substituting the mass of the original electron with the mass of a core.

2. Oct 1, 2015

### fzero

The final state has three particles, not two. There is the original electron plus the new electron-positron pair.

3. Oct 1, 2015

### Frank Einstein

I can't see my mistake; the three particles are aknowledged; Ef=2me (both of the particles formed from the photon)+(me^2+p^2)^0.5, which is the electron in whose presence the disintegration happens.

4. Oct 1, 2015

### fzero

OK, I see now that you're trying to work in a frame where the original electron carries all of the momentum that the photon had. It doesn't appear that there's any such frame. To check this, give the new electron and positron a small component of momentum $p$ in the direction of the photon ($p$ can be $<0$), so that
$$E_f = 2 \sqrt{ m^2 + p^2} + \sqrt{ m^2 + (E-2p)^2}.$$
In the limit that $|p|\ll m$, you will find that $E \approx m^2 /(2|p|)$, so we cannot take $p\rightarrow 0$.

It is easy to get bogged down in algebra in this problem because the final momenta are not very restricted. One estimate for the threshold would be under the assumption that each of the final state particles, being of the same mass, carry the same amount of momentum $E/3$. Another sloppy estimate is to say that $E\ge 2 m$, but this neglects the fact that there is less energy available in the center of mass frame. I'm not quite sure what they're looking for, since there are more complicated ways to compute a threshold here as well that would take a lot of algebra.

5. Oct 1, 2015

### vela

Staff Emeritus
Wouldn't the threshold be where the two electrons and positron are at rest in the center-of-momentum frame?

6. Oct 2, 2015

### Frank Einstein

One of the electrons has to have more energy than the other two particles; it comes from the conservation of momentum. The original electron, at rest first, "inherits" the momentum of the photon. Since the photon is a maseless particle, it's momentum can be described as Eγ/c; since I am using natural units, c=1, so the mometum of the electron is equal to the energy of the photon.

7. Oct 2, 2015

### fzero

Ah yes, this actually gives the same answer as the equimomentum case.

I think if you solve the problem fully, you find that the produced pair particles almost always have nonzero momentum, but that the probability is peaked around having the components of momentum transverse to the photon momentum be equal and opposite. That's beyond the scope of the problem given here though.

I would suggest that you follow vela's suggestion, but you will have to convert the initial data into the center-of-mass frame first.

8. Oct 2, 2015

### vela

Staff Emeritus
In the lab frame, it's true that at least one of the particles has to be moving to conserve momentum, but your assumption that the initial electron absorbs all of the photon's momentum isn't correct. All you can say is that the total momentum of the three resulting particles has to equal the momentum of the photon.

In the center-of-mass frame, the photon and electron approach each other with equal and opposite momenta, so the total momentum of the system is 0.

9. Oct 2, 2015

### Frank Einstein

Allright; I'll try to solve the system giving some energy to the new particles.
Thak you all for your help.