I have read these questions:

Can a free electron absorb a virtual photon even though it cannot absorb an ordinary photon?

Where Michael Seifert says:

It is entirely possible for a real electron to emit a virtual photon and remain on its own mass shell; this is exactly what happens in the classic Feynman diagram with two "real" electrons exchanging a virtual photon. The only reason that conservation laws prohibit a "real" electron from emitting a "real" photon is that it is impossible for all three four-momenta (electron before, electron after, and photon) to simultaneously lie on their respective mass shells.

Can an accelerated "free" electron absorb a photon?

Where Anna V says:

Total absorption would mean an incoming photon+ electron , and outgoing only an electron. This cannot happen because the electron has a fixed mass and does not have excited states to absorb all the energy of the photon. If the outgoing (or incoming) photon becomes virtual, connecting with an electric or magnetic field, then the kinematics has to include the originator of the field in energy momentum considerations, and the electron can absorb all the energy of the incoming photon the energy/momentum balance in its rest mass system taken up by the generator of the field that gave the virtual photon.

So one says that for an accelerating free unbound (so not bound to a nucleus) electron it is possible to completely absorb a real photon, if the electron interacts with the magnetic field's virtual photons, and the momentum conservation is kept by the virtual photons of the magnetic field.

The other one says, it is not possible. Because for the four-momenta for a real electron before, after and real photon, they cannot lie simultaneously on the mass shell.


  1. Which one is right? Can a real free unbound (not bound to a nucleus) accelerated electron absorb/emit a real photon?

  2. In this case, does the virtual photon of the magnetic field help keep the momentum conservation?

  • 5
    $\begingroup$ you are misapprehending the question you quote, a free electron is not accelerating, it is in an inertial frame, and acceleration does not belong to an inertia frame, but needs an input dp/dt specific boundary conditions . also the two different answers to the different questions are saying the same thing in different words. $\endgroup$
    – anna v
    Sep 7, 2018 at 17:39
  • $\begingroup$ @annav can you please tell me if an unbound electron, accelerating in a magnetic field, can or cannot emit/absorb a real photon? $\endgroup$ Sep 7, 2018 at 17:42
  • 1
    $\begingroup$ if it is accelerating there exists a third participant in the interaction and the energy momentum balance can happen between the three inputs , the electron the photon and the originator of the magnetic field $\endgroup$
    – anna v
    Sep 7, 2018 at 17:44
  • $\begingroup$ @annav thank you so you are saying that in this case the originator of the magnetic field (I assume another charge) will help keep the momentum conservation? And so the electron will be able to emit/absorb a real photon? $\endgroup$ Sep 7, 2018 at 17:46
  • $\begingroup$ Yes, it is simple if one writes feynman diagrams for the specific situation, like the photon pair production which also needs a field en.wikipedia.org/wiki/Pair_production#Cross_section .i.sstatic.net/kBXX5.jpg , a virtual photon represents the field $\endgroup$
    – anna v
    Sep 7, 2018 at 18:20

2 Answers 2


A free electron, in region of space zero electromagnetic field, can always be analyzed in its rest frame thanks to Lorentz invariance. If the electron were to emit a photon, it would be impossible to conserve the four-momentum of the system in the rest frame — which means it would be impossible to conserve the four-momentum in any reference frame. If the electron were to absorb a real photon, that would be the time-reverse of the process where the electron emitted a real photon, and so you have exactly the same problem.

The virtual photons which are required for a high-precision description of the electron's self-interaction are, in a very real way, part of the electron. You can never turn off the electron's self-interaction. It's important to remember that virtual photons are a computational device, and computing a gyromagnetic ratio or any other real property requires integration over the allowed phase space of virtual photons.

When an electron accelerates, and emits real photons as a result of its acceleration, it is because it has interacted with an electromagnetic field. You can think of the emission of these photons as electron-photon scattering, where the initial state includes the ensemble of virtual photons which make up the field, and the final state includes a real photon.

  • $\begingroup$ Thank you so much! $\endgroup$ Jun 8, 2023 at 5:21
  • $\begingroup$ Instead of comments which say only"thank you," consider using the upvote or accept features. These help the system direct other users to high-quality content in a way that text comments do not. $\endgroup$
    – rob
    Jun 8, 2023 at 13:23
  • $\begingroup$ I did them both now. $\endgroup$ Jun 11, 2023 at 22:41

Thinking and weighing up is necessary to answer this question.

We are talking about an electron moving freely in space, which is accelerated. The change in speed of the electron is supposed to be made by photons.

After acceleration, the electron has a higher kinetic energy. Where does this energy come from? It can only come from photons. Ergo, the electron must have partially or completely absorbed the energy of the photon.

What happens when the electron decelerates? The loss of kinetic energy must be accompanied by the emission of one or more photons.

  • $\begingroup$ Árpád, did you not like my answer? $\endgroup$ May 18, 2022 at 2:49

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.