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Classical electrodynamics states that every accelerating charge emits radiation. Contrary to that, when studying QED most of us have computed scattering processes involving 2 incoming and outgoing charged particles (e-e, e-p,..). In such processes charged particles change their velocities thus necessarily accelerate, however no outgoing photon is taken into account (at least in the textbook examples). In electron-muon scattering this acceleration is prominent - the muon velocity barely changes whereas the electron velocity can change dramatically, even reverse its direction altogether. If these non-radiating processes are indeed inconsistent with classical ED as I expect, I would have expected their QED cross sections to vanish - Yet they don't. What am I missing?

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    $\begingroup$ "In such processes charged particles change their velocities thus necessarily accelerate" It's unclear what this is supposed to mean - this is quantum physics, what does it mean for these particles to "accelerate"? Why do they have to do so? See also e.g. physics.stackexchange.com/q/52008/50583 for why classical intuition about "acceleration" is ill-suited to quantum processes $\endgroup$
    – ACuriousMind ♦
    Commented Sep 8, 2022 at 21:08
  • $\begingroup$ This also probably isn't the best diagram for your point however since this shows electron/positron annihilation and muon/antimuon pair production. Which isn't a case of particles having changed their momenta. $\endgroup$
    – Triatticus
    Commented Sep 8, 2022 at 21:40
  • $\begingroup$ Thank you for the comment, I've replaced the image. Regarding the answer - your comment on classical intuition is in place. However we can ignore the acceleration event and focus just on the change between the initial and final states of the electron, and the question still stands. In the lab frame, the EM field from the electron differs between the initial and final state, hence the EM field changed at a some finite (and short) timescale. Wave equation tells us that this change propagates as a wave-packet, i.e. photon(s). $\endgroup$
    – Amit
    Commented Sep 8, 2022 at 22:17
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    $\begingroup$ This is actually addressed thoroughly in standard textbooks, such as section 6.1 of Peskin. The point is that every real process actually involves the emission of infinitely many low energy photons, which represent the classical radiation. When this is properly taken into account, the diagram you showed doesn’t actually represent the amplitude to emit zero photons, it represents the amplitude to emit zero high-energy photons. $\endgroup$
    – knzhou
    Commented Sep 9, 2022 at 0:15

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Neglecting the emission of photons during the scattering of charged objects is indeed an approximation. It is safe to make this approximation if we are only working at tree level anyway. But at one loop, accounting for the so called Bremsstrahlung process is essential. This is because loop diagrams in a gauge theory, in addition to UV divergences which led to the development of renormalization, also contain IR divergences. Renormalization is not enough to cancel these but adding soft photon emission is. If you're using a good textbook, examples of this will probably come up later. In Peskin and Schroeder e.g. this starts at chapter 6.

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