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The Bremsstrahlung effect happens when an electron is decelerated by changing its direction typically around a nucleus and then a photon beam is released.

We know that when a scattering happens, let's say Compton scattering, the particle nature of the photon presents itself and the photon hit a free (or valance) electron like a billiard ball then they change courses of motion with different energies and angles.

Now the change in the direction of motion is a change in the direction of the velocity of the electron, which results in the acceleration (or let's say deceleration of a moving electron) of that electron. So this looks like a very quick, sudden and short Bremsstrahlung as in a quick change in motion of an electron when the scattering happens.

If you say that accelerating an electron does not cause an electron to create Bremsstrahlung radiation, you can think this Compton scattering as one of the decelerating types: Incident electron hits a moving electron with a constant velocity and decelerates it or much better changes its direction of motion (acceleration again caused by the change in direction of motion).

Does this cause Bremsstrahlung then because of the same logical assessment?

I mean, is this way of thinking correct? Does Bremsstrahlung happen when any type of scattering happens?

EDIT: Wikipedia : "Broadly speaking, bremsstrahlung or braking radiation is any radiation produced due to the deceleration (negative acceleration) of a charged particle, which includes synchrotron radiation (i.e., photon emission by a relativistic particle), cyclotron radiation (i.e. photon emission by a non-relativistic particle), and the emission of electrons and positrons during beta decay. However, the term is frequently used in the more narrow sense of radiation from electrons (from whatever source) slowing in matter." en.wikipedia.org/wiki/Bremsstrahlung

This link says that acceleration of a charged particle causes Bremsstrahlung radiation also: astro.utu.fi/~cflynn/astroII/l3.html

medical physics
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  • A single electron cannot produce a beam, see the Feynman diagram for brems strahlung here https://physics.stackexchange.com/questions/249057/how-does-bremsstrahlung-occur-in-a-vacuumized-particle-accelerator – anna v Jun 07 '23 at 11:24
  • @annav Would you explain what you mean by "a single electron" please? My OP does not talk about a single electron. The explanation in the link you provided is not clear to me, and I do not think it answers my OP. – medical physics Jun 07 '23 at 11:27
  • This link says that acceleration of a charged particle causes Bremsstrahlung radiation also: https://www.astro.utu.fi/~cflynn/astroII/l3.html – medical physics Jun 07 '23 at 12:47
  • It is complicated, There is classical electrodynamics where a charged particle is composed of a large number of charges and can give off brems radiation. A single electron has to be treated as the quantum mechanical particle it is, with quantum electrodynamics, which is explained with feynman diagrams and interactions as in the picture in the link I gave. Photons from a great number of electrons add up to the classical brem radiation. – anna v Jun 07 '23 at 12:58

3 Answers3

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Bremsstrahlung means braking radiation, and involves at least two charged bodies - the body that is braking (e.g. an electron), and the body that causes the braking (e.g. a nucleus, or a group of them).

Various EM radiation scattering scenarios do not have to involve two charged bodies - the radiation would be scattered already by a single charged body (e.g. an electron).

Both bremsstrahlung and EM wave scattering involve at least one accelerated charged body.

EM wave scattering off one charged body is not bremsstrahlung, and does not involve bremsstrahlung.

Bremsstrahlung is not a scattering of an existing radiation, but creation of new radiation.

Accelerated charged particle radiates changes in EM field, and EM field can be ascribed some energy. Changed EM field means changed energy of EM field.

In classical EM theory, scattering of an existing radiation means that electron changes its velocity (in quantum EM theory, electron field changes its state) and produces new (secondary) radiation. Superposition of the primary and secondary radiation is total radiation that can be observed. In the simplified billiard model, the scattered photon is a new photon, and the old photon disappears.

Ján Lalinský
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  • Although, I specifically say "Bremsstrahlung", I actually mixed up things. I was actually asking that as we know that accelerated charged particles radiate energy, does scattering itself cause an electron to radiate a new EM wave as it changes its direction or velocity by the hit of an incident photon, thus accelerating? What would you say to that? – medical physics Jun 07 '23 at 13:32
  • Wikipedia : "Broadly speaking, bremsstrahlung or braking radiation is any radiation produced due to the deceleration (negative acceleration) of a charged particle, which includes synchrotron radiation (i.e., photon emission by a relativistic particle), cyclotron radiation (i.e. photon emission by a non-relativistic particle), and the emission of electrons and positrons during beta decay. However, the term is frequently used in the more narrow sense of radiation from electrons (from whatever source) slowing in matter." https://en.wikipedia.org/wiki/Bremsstrahlung – medical physics Jun 07 '23 at 13:46
  • I've edited my answer. – Ján Lalinský Jun 07 '23 at 14:28
  • Why would old photon disappear? – medical physics Jun 07 '23 at 14:32
  • In the billiard view of scattering, one photon comes in, one comes out. If two came out, radiation energy would increase and it would not be called scattering, but emission. – Ján Lalinský Jun 07 '23 at 14:34
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It's all kind of arbitrary. Thomson's classical scattering calculation was based on the acceleration of an electron by electromagnetic waves. Then, in the Compton effect is the outgoing photon the "same" as the incoming one? It's not something that an experiment can decide, so it's not a proper physical question.

Nature doesn't care about how we classify phenomena.

John Doty
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Buzz Jun 08 '23 at 18:16
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There is a lot of misleading and confusing information out there regarding Bremsstrahlung. This is because there really are two definitions - a broad one and a narrow one, and they are all too often conflated. Allow me to elucidate on the broad one.

Bremsstrahlung is created by the acceleration of any charged body. Classically, this can be understood from creating a time-dependence in the electric field emitted by that body. As a time-dependent electric field creates a time-dependent magnetic field, and vice versa, this results in the propogation of radiation emitted from the charge. This happens regardless of the cause of acceleration.

Yes, quantum mechanics provides its own description, but it ultimately results in some spectral variance, not a total change in whether the emission occurs at all (bound states being a special case, that I'm not going to get into).

This means that bremsstrahlung is the result of any interaction that results in an acceleration on a charge. This includes all scattering events, Lorentz force interactions, gravitational interactions, and so on.

Logan J. Fisher
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  • Scattering of EM wave off a charged particle involves acceleration of this particle and production of secondary emission, but this is not a case of bremsstrahlung. – Ján Lalinský Jun 07 '23 at 21:00
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    As I said, there isn't a single consistent definition of bremsstrahlung. By a broader definition of simply being radiative emission caused by the acceleration of a charge, this does qualify. – Logan J. Fisher Jun 07 '23 at 21:04
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    There is - braking radiation when matter stops the particles. It is true the term is used for other scenarios as well, such as for cyclotron/synchrotron radiation. But it is almost never used for scattering. – Ján Lalinský Jun 07 '23 at 21:06
  • My intent was to give as broad of an answer as possible. Nature doesn't care how we categorize things - all that really matter is the underlying operations resulting in emission. I would certainly not include any sources of radiation that don't strictly depend on the acceleration of a charge. – Logan J. Fisher Jun 07 '23 at 21:09
  • Every radiation emission is caused by acceleration of a charge. – Ján Lalinský Jun 07 '23 at 22:28
  • The primary radiation emitted by particle-antiparticle annihilation does not. That's not to say there's no secondary radiation emitted from them being mutually attracted and accelerating toward each other, but that's secondary. – Logan J. Fisher Jun 07 '23 at 22:30

  • "The primary radiation emitted by particle-antiparticle annihilation does not." Source? How could a two particle state annihilate without current density changing in time?

     – Ján Lalinský Jun 07 '23 at 22:39
  • Sure, and that causes secondary radiation. The annihilation itself isn't responsible for that though - it's responsible for the primary radiation, which isn't inherently due to that. Realistically, they're inseparable processes, but I'm strictly talking about primary emissions from any given process. – Logan J. Fisher Jun 07 '23 at 22:47

  • "Realistically, they're inseparable processes" That's my view as well, so we agree. Source on "primary radiation" being different from the "secondary radiation" during annihilation?

     – Ján Lalinský Jun 07 '23 at 22:51
  • That's just by definition. There's the annihilation event itself and the necessary conditions (e.g. the particle and anti-particle approaching) for that event. Radiation caused by the annihilation event are primary, and radiation caused by the necessary conditions are secondary. – Logan J. Fisher Jun 07 '23 at 22:53
  • Source of that definition where the terms primary and secondary are used in that way? – Ján Lalinský Jun 07 '23 at 22:54
  • Specifically, "necessary conditions" or "consequences of". Anything that isn't inherent to the action itself. Your own first comment used it in this fashion. Secondary in this context doesn't mean "after", it just means "not primary". I'm sorry, but I'm not going to hunt down a source for this since it's simply an application of basic English. – Logan J. Fisher Jun 07 '23 at 23:00
  • You can't rely only on basic English knowledge to make an argument in theoretical physics. Some terms have established meaning. Specifically in scattering, primary radiation means incoming radiation of an external source, and secondary radiation means radiation produced by the charged particle due to it being accelerated by the primary radiation. Maybe in annihilation these terms have different meaning - that's why I'm asking about some source where that meaning is used. But it seems you just used basic English to describe your own understanding of the annihilation. – Ján Lalinský Jun 07 '23 at 23:31
  • Even in the context of scattering, I would argue that these are best understood as "essential" and "consequential". That is, radiation that is an essential part of the process, and radiation that is simply emergent from that interaction combined with other physics. Otherwise, it would have always made more sense to simply refer to radiation as inbound and outbound. – Logan J. Fisher Jun 07 '23 at 23:45
  • That's not an established terminology in scattering. And it does not help your argument that annihilation radiation has a component that is not connected to accelerated charge. – Ján Lalinský Jun 08 '23 at 13:34
  • I meant that "primary" is to be understood as meaning "essential" and "secondary" is to be understood as meaning "consequential", not that "essential" and "consequential" are the used terminology. – Logan J. Fisher Jun 08 '23 at 14:31
  • Ok that's what you meant, but that does not support any of your claims about bremsstrahlung or annihilation. – Ján Lalinský Jun 08 '23 at 22:12