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If we consider the reaction of a gamma photon with a proton, e.g.,

$$\gamma + p \rightarrow p + \pi^0$$

I wonder what would the linear momentum of the initial proton in the center-of-mass reference frame be. I am confused because this reference frame should be located above the proton, since the photon has no mass. However, by definition, in the center-of-mass reference frame the total momentum should be zero, whereas in this case it is not, since the photon carries a linear momentum $p=\frac{E}{c}$.

Wouldn't therefore the center of mass be placed above the proton, but between it and the photon? What would the linear momentum of the proton be if we assume it at rest in the laboratory frame of reference?

Invenietis
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1 Answers1

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Draw an energy-momentum diagram (adding the timelike 4-momentum of the proton and the lightlike 4-momentum photon (tip to tail) to get the timelike 4-momentum of the system). It'll look like a triangle... in fact like a Doppler-effect problem.

Then, find the component of the proton 4-momentum that is orthogonal to the 4-momentum of the system. (The analogous component of the photon 4-momentum should be opposite this, leading to a total spatial momentum of zero in the center-of-momentum frame.)

Follow the strategy at Lowest kinetic energy of particle for which reaction is possible (invariant mass) applied to that reaction in a recent question What is the minimum energy of a photon for the reaction to occur?.

See also: Momentum diagram for two colliding Particles

robphy
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