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The “photon has no mass” argument is inherently flawed for at least two reasons. First, exact formulation is “photon’s invariant mass equals to 0”. It does not mean that the quantity is undefined. It is defined, but has zero value. Second, invariant mass in relativity does not sum (is not an extensive quantity) even for non-interacting particles. Invariant ...

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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 ...

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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 ...

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Actually the adiabatic hypothesis is usually not required in potential scattering. In case the potential has sufficient decay in coordinate space one can show the existence of the Moller wave operators from which the scattering operator is then obtained. With some modifications this even works for the slowly decaying Coulomb potential. Thus no switching on ...

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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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