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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 $\vec{p_2}$)?

Qmechanic
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CuriousAutomotiveEngineer
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2 Answers2

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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.

Ted Bunn
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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.

Marek
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