# Pair production kinematics

In the process $A\rightarrow B+C$ If we know the energy and momentum four vectors of B and C, then how can we find the momentum three vector of A. The particular proces I have in mind is pair production $\gamma\rightarrow e^+ + e^-$

4-vectors for electron is $(E_1,\vec{p_1})$ and for positron it is $(E_2,\vec{p_2})$. Then for the photon all I can deduce is $E = E_1+E_2$ (energy should be conserved) and $|\vec{p}| = |E|/c$ . Where $(E,\vec{p})$ is the four momentum of the photon. Can I also find the three components of the momentum of the photon, given that I have the components of the momenta of electrons and positrons (i.e I know $\vec{p_1}$ and $\vec{p_2}$)?

-
note that a 1->2 process such as this is not possible for a massless particle in the initial state physics.stackexchange.com/questions/13513/… – luksen Aug 20 '11 at 3:55
@luksen we can assume that the photon interacts with some nucleus. In that case, what should be the analysis – samanwita Aug 20 '11 at 4:09
if you assume the nucleus is at rest the photon momentum is simply $\vec{p}_\gamma = \vec{p}_1+\vec{p}_2$ while $E_\gamma+m_nc^2 = E_1+E_2 + m_nc^2$ – luksen Aug 20 '11 at 4:19
@luksen of course meant $m_n c^2$ also in the final term (the nucleus didn't go anywhere and we neglect its kinetic energy after the collision) so that one can indeed deduce $E_{\gamma} = E_1 + E_2$. – Marek Aug 20 '11 at 11:28
yeah of course.. unfortunately can't edit the comment anymore. – luksen Aug 20 '11 at 11:58

## 2 Answers

As luksen and Marek point out, when the incoming particle is massless this reaction can't go, unless there's a third particle around to soak up the momentum. If we allow for the existence of this particle, then I don't think there's enough information to solve.

Let $p_{Xi},p_{Xf}$ be the initial and final 4-momenta of that particle. Then conservation of 4-momentum says $$p_\gamma+p_{Xi}=p_1+p_2+p_{Xf}$$ Assume that the extra particle was initially at rest: $p_{Xi}=(M,0,0,0)$. ($M$ is the mass of $X$, and I'm setting $c=1$.) We don't know anything about any of the components of $p_\gamma$ or $p_{Xf}$, so there are eight unknowns to solve for. The above is 4 equations, so we need 4 additional equations. Unfortunately, we have only two, $$E_\gamma=|\vec p_\gamma|,$$ $$E_{Xf}^2=|\vec p_{Xf}|^2+M^2$$ (Here $E$ means energy and $\vec p$ means 3-momentum in the frame we're working in, of course. These two equations are just $p_\mu p^\mu=m^2$, if you prefer.) So there's not enough information to solve.

The typical assumption is that the extra particle is much more massive than the others. In this case, the energy it absorbs is typically negligible, but the 3-momentum isn't. So if the 3-momentum is what you're interested in, I don't think you can get by with just pretending that particle isn't there.

-
Hmm, I have to look into this, but not tonight. I know that both in bubble chambers and now in detectors, when we see an e+e- pair we add up the four vectors, and the three vector points to the original vertex. Evidently we are working with the assumption that the three momentum transfer to the spectator nucleus is small ( you do not see it in a bubble chamber). – anna v Aug 20 '11 at 18:37
I don't think I know what you mean by "points to the original vertex." Certainly both observed particles will be seen moving directly away from the location of the interaction, but the total 3-momentum of the observed particles won't equal the incoming 3-momentum. Here's a more quantitative statement: The momentum absorbed by the spectator must be at least of order $(E_1+E_2)/c-|\vec p_1+\vec p_2|$ -- that is, the amount by which the sum of the two observed 4-momenta deviates from the zero-mass shell. That'll be small only if both final particles are ultrarelativistic and nearly collinear. – Ted Bunn Aug 20 '11 at 18:44
I mean that we treat the momentum of the pair as if there is no spectator. Then in the event the photon is treated on par with the measured particles and the total information is used in a 4C or 3C fit to get at the exact event composition. In neutrino physics, where the vertex may be invisible, two such manifestations can define the vertex location. Of course we are talking of energies of the electrons over 200Mev and often GeV so they are highly relativistic. Tridents are observed if the interaction is with an electron, which is then also summed. – anna v Aug 21 '11 at 4:16
continued. I have the impression also that the Q2 ( four momentum transfer) dependence of the photon nucleus pair production crossection falls as 1/Q2, which also would suppresses dynamically a measurable change in direction. The links I find by searching are behind paywalls :(. – anna v Aug 21 '11 at 4:29
continued: if you have a 500 MeV photon certainly the pair coming out is practically collinear. We see them as two because of the magnetic field imposed so that one can measure the momentum of charged tracks from their curvuture. In detectors where there is no magnetic field one just sees the overlap and the energy of the photon is measured calorimetrically. – anna v Aug 21 '11 at 5:10

This question is quite elementary but since it's not getting an answer, let me provide one.

We know (hopefully!) that both energy and momentum is conserved, therefore both $E_{\gamma} = E_1 + E_2$ and $\vec p_{\gamma} = \vec p_1 + \vec p_2$. One might get confused by the fact that this process can't occur by itself but this can easily be fixed by adding additional observer particle (by which I mean that it doesn't affect the process in any serious way; beyond, of course, making it possible in the first place) in the initial and final state and then neglect its contribution.

Now, in the four-vector formalism we can express this more succintly as conservation of energy-momentum in the following way $$p_{\gamma} + p_{\text observer} = p_1 + p_2 + p'_{\text observer}$$ and by the definition of observer we have $p_{\text observer} \sim p'_{\text observer}$ and we cancel these observing terms on both sides of the equation.

-
This is confusing to me. You can't assume that the change in 4-momentum of the observer is negligible in this sort of situation. If you could, then you'd have $p_\gamma=p_1+p_2$, but the left side is lightlike and the right side is timelike, so that can't be. In fact, what one typically assumes in this sort of situation is that the "observer" particle carries off 3-momentum but negligible energy. (This works when the observer particle is more massive than the other particles.) In that case, the above procedure will give the correct answer for the energy but not for the 3-momentum. – Ted Bunn Aug 20 '11 at 17:42
@Ted: I thought it is understood that the claimed relation $p_{\gamma} = p_1 + p_2$ (which can never be true by itself) is only approximately correct. I mention neglecting things several times and also write $\sim$ sign explicitly. What is confusing you? – Marek Aug 20 '11 at 17:57
I just don't think that this is ever a valid approximation. – Ted Bunn Aug 20 '11 at 18:10