5

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?

uKER
  • 153
  • 6
    (1) A free electron does not emit photons. (2) An electron attached to an atom is not a point entity. – Jon Custer Sep 14 '18 at 14:08
  • 1
    actually, the photon direction is random in the spontaneous emission. and there is no rule! – Persian_Gulf Sep 14 '18 at 14:24
  • @JonCuster I believe electrons in a cathode ray tube can emit photons (x ray) – Manu de Hanoi Sep 14 '18 at 14:42
  • 3
    @ManudeHanoi - sure, when they hit something. A freely propagating electron does not emit photons. – Jon Custer Sep 14 '18 at 14:43
  • @JonCuster what about cyclotron radiation? https://en.wikipedia.org/wiki/Cyclotron_radiation – Manu de Hanoi Sep 14 '18 at 14:58
  • 2
    @ManudeHanoi that's just it: the particles are accelerated by the magnetic field: they are effectively absorbing photons and are very unfree indeed. – Cosmas Zachos Sep 14 '18 at 15:04
  • @CosmasZachos from wikipedia "Cyclotron radiation is emitted by all charged particles travelling through magnetic fields, not just those in cyclotrons." – Manu de Hanoi Sep 14 '18 at 15:14
  • I dont know why u claim that free electrons cant emit photons. To emit photons u need a quick change in the electric field, a moving electron can do that – Manu de Hanoi Sep 14 '18 at 15:17
  • @CosmasZachos yeah yeah, then if all it takes is for a force to apply to an electron to make it not free, then there is not a single free electron in the universe. Again from wiki: "Electromagnetic waves are emitted by electrically charged particles undergoing acceleration." – Manu de Hanoi Sep 14 '18 at 15:20
  • 1
    @ManudeHanoi Absolutely. Acceleration is the result of an interaction; "free" is an idealization where such interactions are ignorable. – Cosmas Zachos Sep 14 '18 at 15:22
  • @CosmasZachos there is no free electron in the universe according to your definition. Therefore your definition of "free electron" is useless. You should know that's not what we mean by free electron – Manu de Hanoi Sep 14 '18 at 15:24
  • Behold. – Cosmas Zachos Sep 14 '18 at 15:38
  • @JonCuster, there is, of course, the free-electron laser. See https://en.wikipedia.org/wiki/Free-electron_laser – David White Sep 14 '18 at 15:46
  • 2
    @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... – Jon Custer Sep 14 '18 at 17:11
  • It is better to think of a photon being absorbed or emitted by the whole atom rather than by the electron. But even so, for spontaneous emission by a free atom nothing determines the emission direction, all directions have equal probability – PM 2Ring Sep 14 '18 at 17:11
  • Are you familiar with the computation of QFT propagators? – Gabriel Golfetti Sep 15 '18 at 12:06
  • @ManudeHanoi I have to agree with JonCuster here. Acceleration is needed for radiation emission. By definition, "free" means no other interactions. So free electrons can't accelerate and emit photons. Of course truly free electrons can't exist, but that doesn't mean the idea is useless. It's like when we assume frictionless surfaces or neglect air resistance. The assumptions aren't true, but they are close enough to reality that they are still useful with the bonus of being easier to work with. – BioPhysicist Sep 15 '18 at 13:46
  • @AaronStevens Ive used and been taught in high school that metals conduct electricity because they have free electrons. Free meaning not being bound to a particular atom. – Manu de Hanoi Sep 15 '18 at 14:50
  • 2
    @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. – BioPhysicist Sep 15 '18 at 15:05

3 Answers3

3

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

bremss

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.

bremss

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.

anna v
  • 233,453
1

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.

flippiefanus
  • 14,614
  • 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? – uKER Feb 12 '19 at 10:04
  • 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. – flippiefanus Feb 12 '19 at 10:36
-2

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.

Manu de Hanoi
  • 1,566
  • i'm surprised at the amount of downvotes for a question no one has provided another answer yet. – Manu de Hanoi Sep 15 '18 at 06:30
  • 1
    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. – flippiefanus Sep 15 '18 at 11:28