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I've had two special relativity courses so far but none really gave me a clear description of the process.

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  • $\begingroup$ Should your question then read "how" instead? $\endgroup$ – Ignacio Vazquez-Abrams Jun 10 '12 at 19:55
  • $\begingroup$ Perhaps a quick look at bremsstrahlung might be useful $\endgroup$ – twistor59 Jun 10 '12 at 20:00
  • $\begingroup$ That doesn't explain why it happens. And with "why" I don't mean stuff like "why is there an electric charge" but why does an accelerating charge produce photons? I see that that particle gains/loses energy, but aren't there other ways to do that? (unless my question, in fact, really is too fundamental/philosophical) $\endgroup$ – Jan M. Jun 10 '12 at 20:08
  • $\begingroup$ I've added a quantum electrodynamics tag since this question can be interpreted as a variant of the issue of how an electron recoils (decelerates in this case) when it emits a photon. I'll attempt a short answer along those lines, but I'm not sure if it will do much to answer the more classical intent of the question. $\endgroup$ – Terry Bollinger Jun 10 '12 at 22:23
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This answer is in part an attempt to demonstrate how you often can apply very different perspectives to the same physics question.

In his PhD thesis, Richard Feynman asked this question: "Why does an electron recoil when it emits a photon?" He asked this question from a rather peculiar perspective, in which he assumed that fields do not exist and particles only interact directly."

Your question simply reverses the order of causation: "why does an electron emit a photon when it recoils (decelerates)?"

So, the answer from Feynman's thesis applies nicely to your question. The details of Feynman's answer were incorrect on one point, where Feynman assumed that an electron could not interact with itself. @RonMaimon pointed this out here in Physics.SE, but also why it does not undermine Feynman's main answer.

So what would Feynman's answer be to your question likely would have been this: An electron emits an ordinary photon in response to being struck by a reverse-time ("advanced") photon that has traveled backward in time from some point in the future. "Some point in the future" could range from femtoseconds (or less) away up to billions of years in the future. The real photon that is emitted in response to the recoil-inducing, backwards-in-time photon travels along the classical time path as a normal or "retarded" photon (no I did not make that up), and eventually strikes the very same target that emitted the advanced photon sometime in the future.

The advanced photon solutions, incidentally, were a second set of solutions to Maxwell's equations that traditionally were just ignored as irrelevant, for obvious reasons. That changed when John Archibald Wheeler, Feynman's thesis adviser, suggested them as possible solutions to Feynman's seemingly hopeless quest to create a self-consistent theory of how particles might interact "directly" (whatever that really means), without using any intermediate classical fields.

While I realize this is likely not the type of answer you were expecting, the success of the QED mathematical framework shows that as an answer, it is both self-consistent and correctly predictive. It just approaches the question more form a quantum theory perspective than from a special relativity perspective.

So, to recapitulate: (I always wanted to use that word once [only] in my lifetime!)

Electrons emit ordinary photons when they recoil in response to being struck by an "advanced" photon that has traveled backwards in time. Once emitted, the ordinary or "retarded" photon travels along with the rest of us through ordinary classical time until it completes the circuit -- possibly quickly, possibly very slowly indeed -- by striking the same electron that was the source in the future of the advanced photon.

Addendum 2012-06-13.20:40 EDT - In response to a good comment about causality:

Causality, remarkably, is preserved throughout all of this, as proven in a series of papers by Feynman and Wheeler. And what that says is mostly that our concept of "now" is really quite a bit more complicated than we usually think.

More specifically, quantum particles embedded in ordinary thermodynamic matter require longer integration lengths into what we think of as the past and future -- and those extensions can be very long indeed. I like to think of this fluctuating "toothiness" of quantum particles embedded in the ordinary thermodynamic matter as the "ragged edge of now", but that's just my personal visualization... one that a book author and some Hollywood folks must have had many years before I did. But rows of temporal teeth aside, all I mean by this amusing visualization is that the time distances over which path integrals must be integrated increases as the particles remain quantum for longer periods of time.

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    $\begingroup$ Let's also be clear that "backward in time" should not be taken literally, it is only an interpretation based on the mathematical definitions and equations... A particle travelling backward in time can be recast into its antiparticle travelling forward in time. $\endgroup$ – Chris Gerig Jun 11 '12 at 2:59
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    $\begingroup$ Chris, thanks, I should have emphasized that. Feynman and Wheeler went to a lot of trouble and wrote many papers to show that their model did not result in violations of causality. I think the fact that it worked surprised them as much as anyone. Such "atemporal" views apply only within the context of QED style path integral representations, before such representations have been "tested" by a classical observer. Classical time flow does not and cannot work the same way within an event that is still quantum. $\endgroup$ – Terry Bollinger Jun 11 '12 at 4:26
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This is a hand waving answer.

You ask in the comments: I see that that particle gains/loses energy, but aren't there other ways to do that?

Down in the particle world everything is quantum mechanical and the only things that exist are the standard model particles, which may sometimes act as waves according to strict QM rules.

What are the possible interactions of an electron? Weak and electromagnetic. Weak is orders of magnitude weaker than the electromagnetic ( hence the name) and can be ignored.

Thus any measurable interaction an electron can have has to be electromagnetic. In the microcosm dimension any gaining or losing energy has to go through photons .

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  • $\begingroup$ All good answers, I can't choose :D $\endgroup$ – Jan M. Jun 12 '12 at 21:55
  • $\begingroup$ thats fine :) no problem $\endgroup$ – anna v Jun 13 '12 at 3:30
  • $\begingroup$ Well if it helps a bit I'm giving @annav a point myself! I like her analysis of the limited selection between the major forces, and I hadn't even thought about that aspect. $\endgroup$ – Terry Bollinger Jun 15 '12 at 21:40
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Charged particles are permanently coupled to the electromagnetic field, it's an experimental fact and the very essential feature of charges. As any coupled (compound) system, the system (electron + EMF) has its center of mass variables ans the "relative motion" (or "internal") variables. Generally, when you act on one of a constituents of a compound system, you transfer energy to its center of mass and to internal variables, both energies being additive. Just like hitting a ball - you push it as a whole and you make it vibrate "internally" (you excite the "shape dynamics", too).

I think when you act on the electron, the previous state of "relative motion" of the compound system (electron + EMF) gets perturbed and one observes relaxation of this perturbation as electromagnetic waves.

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Ahhh... let's see!

Any electron that changes speeds or direction emits blue light. When an electron moves to difference valence levels emits blue and white light.

For an electron to change speed or direction it must have interaction with other particles or other EMFs. If the electron is spinning around an atom it will behave according to black body radiation theory. If it is a free electron then it can retain broader bands of energy or absorb energy differently than if in a valence level restricted model, and if free, will never uniformly absorb and emit because angle of particle interaction and external field strengths encountered and the starting speed of every electron will vary. Valence electrons require specific photons to jump valence levels. So a free electron can be hotter, faster when absorbing, cooler and slower when re emitting. So the electron is not creating the photons itself. It is absorbing, deflecting, transferring energies that have already been given to it from external source.

At rest, and cold an electron doesn't emit photons because it isn't absorbing energy and changing its energy state. For an electron to move something has to hit it.

The source of light when changing direction is related to the emf electrons carry. This means then any photon coming from an electron is not the electron's photon, it was given addition energies through heating, absorption, and acceleration of speeds.

For anything to change speed or direction it must give up its momentum and since very little mass in electrons most of the energy is created by temperature and velocity, any of which should change, will cause emission of light.

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  • $\begingroup$ Your first two sentences are not correct. Radio antennas, with accelerated charges, should not be emitting blue light, but radio waves. And transitions between different atomic/solid state levels can create all kinds of colors (like a red LED). The rest is, well, not particularly correct either. $\endgroup$ – Jon Custer Oct 5 '16 at 2:09

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