When an electron loses energy and emits a photon, what determines the direction that this newly-created photon will travel? The electron, being a point entity, has no internal physical structure, so I'm assuming the photon's motion is inherited from the motion of the electron at the instant when the photon was emitted?
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6$\begingroup$ (1) A free electron does not emit photons. (2) An electron attached to an atom is not a point entity. $\endgroup$– Jon CusterCommented Sep 14, 2018 at 14:08
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3$\begingroup$ @ManudeHanoi - sure, when they hit something. A freely propagating electron does not emit photons. $\endgroup$– Jon CusterCommented Sep 14, 2018 at 14:43
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2$\begingroup$ @ManudeHanoi that's just it: the particles are accelerated by the magnetic field: they are effectively absorbing photons and are very unfree indeed. $\endgroup$– Cosmas ZachosCommented Sep 14, 2018 at 15:04
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2$\begingroup$ @DavidWhite - and that requires shaking the electrons around - they are only propagating 'freely' between the undulators, and in the undulators they are forced to, well, undulate... $\endgroup$– Jon CusterCommented Sep 14, 2018 at 17:11
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2$\begingroup$ @ManudeHanoi that is fine. You never specified in your above comments this is what you meant by "free". All of this would have been avoided if you had just said what you meant by free. No one here is wrong, just talking about two different things. Typically in physics, free means no interactions, and the electrons you speak of would be not bound to any atoms. But if you want free to mean not bound to atoms, then specify this rather than getting upset with people who are just using different terminology than you. $\endgroup$– BioPhysicistCommented Sep 15, 2018 at 15:05
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3 Answers
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Electrons and photons are quantum mechanical entities and obey solutions of the appropriate quantum mechanical equations.
When an electron loses energy and emits a photon,
The "loses energy" already describes an interaction: it is called bremsstrahlung
what determines the direction that this newly-created photon will travel?
The calculation of these Feynman diagrams
will give the probability distribution that the photons will have to obey. Remember in quantum mechanics it is the probability distributions that are strictly determined. Individual scatters/events conform to that distribution.>
The electron, being a point entity, has no internal physical structure, so I'm assuming the photon's motion is inherited from the motion of the electron at the instant when the photon was emitted?
No , it is random BUT obeying energy momentum conservation AND the accumulation of these events has to follow the calculable probability distribution. If the probability distribution peaks in the direction of the incoming electron a sampling of one scatter will probably fall in that region, but there is a probability that it may have a larger angle. Thus it will depend on the conditions assumed . The "nucleus" in the diagram may just be an electric or magnetic field .
In this paper the calculations are done for high energy electrons and positrons in storms , to estimate the gamma (high energy photons) emission.
So the probability distribution for angular effects has a bias towards the incoming electron , but there is an appreciable probability with larger angles, in the solution of this specific problem.
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Perhaps one needs to provide some context. The OP does not specify a free electorn or bound electron. If the electron is bound in an atom, then the spontaneous radiation of a photon would occur when the electron moves from an excited state to the ground state, for instance. In such a case the direction of photon emission is completely random. It simply occurs because the excited state is unstable and there is no way to predict to direction.
Another scenario is compton scattering. Here a photon is first absorbed by the electron and then another photon is radiated by the electron. If one would ignore the absorption of the first photon, one would think that the electron just spontaneously emitted a photon. However, neither of these two event can be considered in isolation, because neither of them can in general conserve energy-momentum. So one would need to consider the whole process. In that case, the direction of the emission is governed by the condition of the initial photon and electron prior to the absorption of the first photon. One can compute the propability distribution for all the possible directions that the last photon can be radiated.
There are many other scenarios, however, they can all be considered in a similar way as these two scenarios.
answered Sep 15, 2018 at 11:43
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$\begingroup$ I was thinking an electron bound to an atom that takes part in a reflective surface. Who is it that considers the photon's angle of incidence relative to the surface, to decide in which direction to emit the photon? $\endgroup$– uKERCommented Feb 12, 2019 at 10:04
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$\begingroup$ For just one such atom the direction of emission is random. However, in a reflective surface there would be a superposition of photons emitted from multiple atom. The resulting interference would then determine the direction in accordance with the laws of reflection. $\endgroup$ Commented Feb 12, 2019 at 10:36
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you'd better think in terms of 3d waves, the electron has an electrical field, when it moves that creates a special 3d shaped wave related to the direction of oscillation(shape here). That 3d wave is light/EM radiation, and you'll get a photon where light is caught.
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$\begingroup$ i'm surprised at the amount of downvotes for a question no one has provided another answer yet. $\endgroup$ Commented Sep 15, 2018 at 6:30
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2$\begingroup$ Yes, I don't think your answer is so bad that it deserves so many down votes. Perhaps people can leave a comment when to give a down vote. $\endgroup$ Commented Sep 15, 2018 at 11:28