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During an electron-electron-scattering in any case the electrons change their directions and by this undergo accelerations.

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Shouldn’t the electrons emit photons in this time, loosing a part of its kinetic energy and slowing down a bit?

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    $\begingroup$ aren't you describing "radiative corrections"?--which are such a pain for the experimentalist. $\endgroup$ – JEB Dec 21 '18 at 16:21
  • $\begingroup$ As always a single Feynman diagram represents one term in a series expansion. The full physics description involves all the terms (and thus all the diagrams) that agree with the physics you are computing. But for QED you generally get away with the approximation represented by only the first few orders. $\endgroup$ – dmckee Dec 22 '18 at 0:31
  • $\begingroup$ @Jeb, :-) is the reaction to your comment. Seriously, the question occurs after this answer physics.stackexchange.com/questions/449431/… $\endgroup$ – HolgerFiedler Dec 22 '18 at 16:21
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Shouldn’t the electrons emit photons in this time...?

Yes. They do. Just as in classical physics, an accelerating (or scattering) electron in QFT emits EM radiation. If desired, this radiation can be described in terms of photons, although (as usual) that's not necessarily the most natural description.

To see this, consider the same Feynman diagram but with one or more external photon lines emanating from one or more of the electron lines. These amplitudes are non-zero, so this process does occur. In fact, when we consider higher-order terms in the small-coupling expansion (and these diagrams with extra photon lines are of higher order), we must include this effect in order to get a meaningful result.

Weinberg's book (The Quantum Theory of Fields, Volume 1) has an entire chapter devoted to this subject (Chapter 13: "Infrared Effects"). Page 544 emphasizes that we must account for this effect even if we don't care about or don't detect the EM radiation:

...it is not really possible to measure the rate... for a reaction... involving definite numbers of photons and charged particles, because photons of very low energy can always escape undetected. What can be measured is the rate... for such a reaction to take place with no unobserved photon having an energy greater than some small quantity..., and with not more than some small total energy... going into any number of unobserved photons.

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  • $\begingroup$ Dan, in my imagination the electron is like a sponge for photons. Hitting an electron, some amount stays at the electron and during a scattering process of the two electrons the sponges get squeezed and photons (like water droplets) splash out, mostly in the direction away from the direction of collision. See please my answer here physics.stackexchange.com/questions/449431/… $\endgroup$ – HolgerFiedler Dec 22 '18 at 7:45

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