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Given a scattering event where a photon and electron go in and a photon and electron come out, what is the center of mass frame?

I'd say, since the photon has no mass, it's the rest frame of the incoming electron.

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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 non-vanishing, future-oriented, causal $4$-vector (i.e., it is either timelike or lightlike).

By standard properties of sums of causal vectors for $N\geq 2$, one realizes that $P$ is causal as well, and it cannot be lightlike unless all $4$-momenta of the system $P_{(a)}$ are parallel and lightlike. If at least one particle is not massless, like in the case you consider, $P$ turns out to be timelike and thus the center of mass is well defined.

Indeed, when the total $4$ momentum $P$ is timelike and future directed (up to $4$-translations and spatial rotations which do not change kinematic) there is a future oriented Minkowskian reference frame (*) whose temporal axis $\partial_t$ verifies $P = k \partial_t$ for some $k>0$.

In that reference frame, $P$ has no components perpendicular to $\partial_t$. In other words, the $3$-momentum vanishes. That is the center of mass reference frame of the physical system.

The fact that even a system of massless particles has a center of mass (excluding the case pointed out above) should not be surprising because the (invariant) mass of the particles does not matter here. Differently form the classical case, it is the energy which matters and it is strictly positive also for photons. Let us illustrate this fact. In the classical picture, one has to look for a reference frame where

$$\vec{0} =\sum_{a=1}^N m_a \vec{v}_a\:.$$

Instead, in the relativistic realm, instead, the relevant equation is

$$\vec{0} = \sum_{a=1}^N \vec{P}_a = \sum_{a=1}^N E_a\vec{v}_a $$

where the $4$-momentum is decomposed into its energy (temporal) component and its $3$-momentum (spatial) components as
$$P_a = (E_{(a)}, \vec{P}_{(a)})$$ (I assume $c=1$) and consequently the $3$-velocity is $$\vec{v}_{(a)}= \frac{\vec{P}_{(a)}}{E_{(a)}}\:.$$

You see that, also if some of the particles are photons, we have $E_{(a)}>0$, because $E_{(a)}^2 - \vec{P}^2_{(a)}=0$ and $P_{(a)}\neq 0$ in that case.


(*) This is consequence of the fact that, up to dilations, the orthochronous Lorentz group $O(3,1)_+$ acts transitively in the interior of the future lightcone.

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  • $\begingroup$ One massless particle never has a CoM frame. But you can consider several massless particles travelling in non-parallel directions and find the center-of-mass frame of a “system” of these particles. $\endgroup$ – Incnis Mrsi Oct 23 '14 at 15:36
  • $\begingroup$ Sure! I make more precise my answer, I was referring to $N\geq 2$ particles... $\endgroup$ – Valter Moretti Oct 23 '14 at 15:52
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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 describing is Compton scattering. In the c.m. frame, both the electron and the photon are scattered at 180 degrees with an energy the same as that with which they came in.

To determine the c.m. frame, add the momentum four-vectors and normalize the result. This is the velocity four-vector of the c.m. observer.

There are cases where this definition fails. For example, if the system consists of a single ray of light, then the momentum four-vector is lightlike, so you can't normalize it; there is no frame of reference in which it's at rest. But, e.g., if your system consists of two rays of light moving in different directions, then there is a c.m. frame.

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    $\begingroup$ And this is why various authors say "center-of-momentum frame" instead, where conveniently/confusingly it has the same acronym. $\endgroup$ – user10851 Oct 23 '14 at 17:59
  • $\begingroup$ Wrong. By the Newtonian definition, the photon would count if only because of non-zero momentum. $\endgroup$ – Incnis Mrsi Oct 26 '14 at 13:23
  • $\begingroup$ @IncnisMrsi: I suppose it depends on what you take as the definition of the Newtonian center of mass and Newtonian center of mass frame. Newtonian mechanics doesn't actually apply to light, so it's not surprising that the definitions don't apply cleanly. $\endgroup$ – Ben Crowell Oct 26 '14 at 16:14
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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 mass of a system of several non-interacting particles is greater or equal than the sum of mass of constituents, and the “equal” case happens only when all 4-momenta are parallel (centers of mass of constituents have zero velocity in some reference frame for massive constituents, travelling in parallel direction for massless ones, and never happens when there are both massive and massless constituents). Yes, a massive particle and an emitted photon always have together greater mass than the particle alone. Not relevant to current case, just for the record: for interacting (namely, attracting) bodies this mass defect may have either sign.

Another approach: you do not need to speculate how photon’s mass is small (if any) while its momentum is definitely non-zero.

So, throw the “photon can be ignored” thought away and calculate the momenta.

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