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By the Newtonian definition, the photon wouldn't count due to its zero mass, but this is a relativistic collision, so you need a relativistic definition of the center of mass. Relativistically, the c.m. frame is the one in which the total momentum four-vector of the system is purely timelike. No, this does not coincide with the electron's frame. What you're ...


11

The reference frame of the center of mass is, by definition, the one where the total $3$-momentum vanishes. It exists almost always also for massless particles as I go to discuss. The total $4$-momentum $P$ of a system of $N$ free particles is the sum of their $4$-momenta of the particles, i.e., $$P = \sum_{a=1}^N P_{(a)}\:,$$ where each $P_{(a)}$ is a ...


2

I wonder if this may be of some help. The optical theorem relates the total cross-section of a scattering with the forward scattering amplitude. For example for a $2\to2$ scattering you get that: $\left\langle n\left|S\right|m\right\rangle \equiv\delta_{mn}+i\left(2\pi\right)^{4}\delta^{4}\left(p_{m}^{\mu}-p_{n}^{\mu}\right)\left\langle ...



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