This is inspired by a recent post on why a free electron can't absorb a photon, though my question below is about something considerably more general.

The argument in the accepted answer goes (in essence) like this:

Taking $c=1$, and working in the rest frame of the electron, let $p\neq 0$ be the momentum (and hence the energy) of the photon and let $m\neq 0$ be the mass (and hence the energy) of the electron. Then the combined particle must have momentum $p$ and energy $m+p$, which means that its mass $\hat{m}$ must satisfy $\hat{m}^2=(p+m)^2-p^2=m^2+2mp$.

Now if we suppose that the combined particle has the mass of the original electron, we get $2mp=0$, which is a contradiction.

This shows that the combined particle cannot be an electron. It does not, by itself, show that the combined particle cannot exist. This leads me to wonder how the argument can be completed, and to the more general question of how one shows that a given particle cannot exist. (I expect this is a very naive question.)

It's been suggested (in comments on the referenced post) that the argument for non-existence can be completed by invoking additional conservation laws, such as conservation of lepton number. But I don't see how this can possibly suffice, as we can always suppose that the lepton number of the combined particle is the sum of the lepton numbers of the electron and the photon, and likewise for any other quantity that needs to be conserved.

In the end, then, we have a particle with a prescribed mass and prescribed values for a bunch of other conserved quantities.

My question is: What tells us that this particle can't exist?

My guess is that this comes down to some exercise in representation theory, where the particle would have to correspond to some (provably non-existent) representation of the Poincare group.
Is this guess right, or is there either more or less to it than that?

  • 2
    $\begingroup$ The answer doesn't argue that such a particle doesn't exist. The question is about an electron absorbing a photon, and absorbing is commonly understood to mean that the thing that absorbs something is still there afterward. One cannot exclude a reaction $e+\gamma \to ???$ with a conservation argument, only the $e+\gamma\to e$. $\endgroup$
    – ACuriousMind
    Dec 25, 2015 at 15:30
  • $\begingroup$ @ACuriousMind: Thank you for that clarification. (I hadn't been aware of the conventional meaning for "absorb".) My question about $e+\gamma\rightarrow ???$ does, however, remain. $\endgroup$
    – WillO
    Dec 25, 2015 at 15:31
  • $\begingroup$ Just inspect the standard model. The only vertex it has with an electron and a photon is the 3-vertex with two electrons/positrons and the photon, you just don't get any other particle. Of course, you can let the electron interact after that with $W$ or $Z$ bosons and try to change it into a muon or something, but this then has nothing to do with the photon. $\endgroup$
    – ACuriousMind
    Dec 25, 2015 at 15:33
  • $\begingroup$ @ACuriousMind: Thanks for that too. I'm not sufficiently familiar with the standard model to know whether the existence/non-existence of various vertices is due entirely to the existence/non-existence of certain group representations, so I'm still not sure I know the answer to my question. (I hope it's clear that this is not meant as a complaint, and that I'm very grateful for your comments.) $\endgroup$
    – WillO
    Dec 25, 2015 at 15:37
  • $\begingroup$ I wonder if the photon wave bounces or if it is emited back by the electron $\endgroup$
    – user46925
    Dec 25, 2015 at 19:38

1 Answer 1


There are only two stable particles in the Universe, the electron and proton, excluding EM waves. All other particles are excited electron and proton states or from their interactions, like neutrinos. They, and the Up and Down quarks constituting protons, are precise arithmetic or exponential reciprocal multiples of Sommerfeld's alpha = 0.007297 fine structure constant correlating light speed to hydrogen ground state potential energy root. This energy density root factor is coincident in the nuclear, atomic and gravitational energy domains, relating them all to the hc = 1.986445684 x 10^-25 Jm electromagnetic impedance of space. Thus the reciprocal of the proton's radius and light year distance correlate closely since the light speed events within the particle and gravitational domains both relate to c. These correlations are enumerated on page 21 of A Nuclear-Gravitational Electrodynamic Framework at QuantumEnergySystems.com. Because of this energy density framework electrons can only absorb photon energies corresponding to their muon, pion and tau energy states. The quarks in the proton however offer a much wider range of photon absorption states.

  • 3
    $\begingroup$ This is wrong in all possible respects. To start with, there's a full collection of stable particles other than the electron and the proton (in particular all allowed combinations of elementary constituents with decent lifetimes are stable). Plus, the relations among all those constants don't really mean all the above (rather they are just mere multiples of each other). Moreover, absorption and emission processes are regulated by other factors as well (as all the conservation laws) rather than just energy densities. $\endgroup$
    – gented
    Dec 25, 2015 at 17:43
  • $\begingroup$ It's correct. Lifetimes mean unstable, they decay. All particles except electrons, protons, photons and neutrinos, are positive energy quantum states. The flaw in Standard Model particle interpretation is that Einstein did not write E = mc^2, he wrote mass = E/c^2, in Boltzmann's P = e^(S/k) probability principle format. Energy E has two available entropic degrees of freedom, space's permeability-permittivity impedance and 4-D space-time, a light speed resonance between particle and field forms. If the forms equate they are stable, if not they are unstable and decay. Read the paper. $\endgroup$ Dec 26, 2015 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.