7
$\begingroup$

For an accelerated charge to radiate, must an electromagnetic field be the source of the force?

Would it radiate if accelerated by a gravitational field?

$\endgroup$

5 Answers 5

11
$\begingroup$

Your question is somewhat abstruse, but here's what I think you're asking:

Put a charged particle in a uniform external magnetic field. The particle will move in a circular orbit, but since it's accelerating, it will radiate and its orbit will decay.

Now remove the magnetic field. Grab the charge and forcibly swing it around in the same circle as before by some other, unknown means. Does it still radiate in the same way as before?

The answer is yes because Maxwell's equations are linear. Therefore we can analyze any situation in classical electromagnetism by superposition.

$\endgroup$
12
  • 1
    $\begingroup$ So a charge would radiate if accelerated by a gravitational field? $\endgroup$ Mar 16, 2011 at 16:41
  • 1
    $\begingroup$ @John McVirgo Yes $\endgroup$ Mar 16, 2011 at 16:42
  • 1
    $\begingroup$ But then radiation of gravitational waves is also ruled out with the equivalence principle, isn't it? ;-) $\endgroup$ Mar 16, 2011 at 20:48
  • 2
    $\begingroup$ @lurscher: no. Bryce De-Witt wrote the paper on this--falling charges actually should radiate. Also, point-like objects DO radiate gravitational waves, a charge current has a quadrupole moment. The equivalence principle only works inasmuch as you can ignore the gravitational field of test particles. $\endgroup$ Mar 17, 2011 at 22:47
  • 2
    $\begingroup$ @lurscher: The principle of equivalence does not have that conclusion, since radiation is not a local effect. Charge that is accelerating (with respect to some distant observer) will radiate even though locally you can transform to a freely falling frame. In that frame the local electric and magnetic field generated will be that of an inertially moving charge, but that statement only applies locally. Radiation has to do with the behavior of those fields asymptotically, far away from the source, and specifically with how much energy flux (=radiation) they involve. $\endgroup$
    – user566
    Mar 19, 2011 at 21:02
4
$\begingroup$

The radiation, if considered classically, is independent for the reason Mark Eichenlaub gives. But considered quantum mechanically, it is not independent.

In short, photons are bosons. So the presence of radiation of a particular polarization and frequency will increase the probability of the particle radiating that polarization and frequency.

This is a topic I'd not seen before. I'll look around and see if I can find a reference to the effect.


An accelerated electron produces "synchrotron radiation". An example of the electromagnetic field altering the emission of such radiation would be "stimulated synchrotron emission". Stimulated emission was described by Einstein and is the physics behind lasers. An example paper combining these ideas:
Phys. Rev. Lett. 66, 2312–2315 (1991), J. L. Hirshfield and G. S. Park, Electron-beam cooling by stimulated synchrotron emission and absorption
http://prl.aps.org/abstract/PRL/v66/i18/p2312_1

$\endgroup$
1
  • $\begingroup$ Since stimulated emission is by an E&M field, maybe the above should be excluded. However, I think I've answered the question under the usual definition of "accelerated" as opposed to "electromagnetic field". Hmmmm. Maybe the original question is a little loose. $\endgroup$ Mar 16, 2011 at 6:27
4
$\begingroup$

I suspect that two charged objects orbiting one another due to gravitational attraction would radiate, but I can't support that assertion with a citation.

The question of whether or not a charge radiates when it is uniformly accelerated by gravity is an open question; read the link for an excellent discussion of why.

EDIT:
The link I posted isn't inspiring trust, so I searched for peer-reviewed work. I found two relevant papers:
Physical interpretation of the Schott energy of an accelerating point charge and the question of whether a uniformly accelerating charge radiates
The significance of the Schott energy for energy-momentum conservation of a radiating charge obeying the Lorentz–Abraham–Dirac equation

I'm not qualified to comment on the quality of the papers, but both attack the question 'does a falling charge radiate', implying an open question. I didn't find any experimental work on the subject.

This brings up a topic better suited for meta-discussion: I would like to see even more external citations in the answers here. I would also like to see more answers clearly indicate their logical foundation. Is your answer based on...
- Original research?
- Predicted by peer-reviewed theory but unverified?
- Indirectly experimentally verified?
- Directly experimentally verified?
Don't make us guess!

$\endgroup$
11
  • $\begingroup$ I think this is the most germane answer; use gravity to cause the acceleration. Maybe there's a reference in rotating charged black holes. $\endgroup$ Mar 17, 2011 at 0:35
  • $\begingroup$ what about Lurscher's point that a charge in free fall in a gravitational field doesn't radiate? $\endgroup$ Mar 17, 2011 at 21:14
  • $\begingroup$ How could this possibly be an open question? $\endgroup$ Mar 17, 2011 at 21:23
  • 1
    $\begingroup$ @Andrew: indeed, the link seems to be quite strange. It mixes lots of theories together in a weird manner and discusses irrelevant topics like self-interaction in classical theory while for the most part it omits the most relevant theory, QED (or anything quantum for that matter except for incorrect Feynman-Wheeler absorber theory). I would take that reference with a bit more than a grain of salt. $\endgroup$
    – Marek
    Mar 17, 2011 at 23:16
  • 1
    $\begingroup$ Orbiting bodies emit gravitational radiation (there was a Nobel prize given for verifying this). It would be very strange if charged orbiting bodies didn't emit electromagnetic radiation as well. $\endgroup$ Mar 18, 2011 at 11:20
1
$\begingroup$

If you mean an external EMF, the answer is "The radiation is determined with an external filed". The charge acceleration is proportional to the external field, and a single accelerated charge radiates.

If you mean the radiated field influence on the charge motion and subsequent radiation, the answer is "No" because the radiation is only expressed via external field. There is no need to invoke the proper field here.

In QM radiation of hard single photons happens discontinuously in time so QM is somewhat different.

$\endgroup$
-3
$\begingroup$

According to the equivalence principle, a point-like charge in free fall will not radiate EM waves, because its movement its locally equivalent to any other inertial frame.

point-like masses will also not radiate gravitational waves in free-fall by the same reason; gravitational wave emission is a function of the third time derivative of the quadrupole mass moment; this basically means that only extended, non-symmetric masses will give such radiation (being extended means that significant parts of the mass will NOT be in a inertial free-fall frame)

Edit: The only other macroscopic force outside gravity (and the force that is not a force called exclusion principle) is the electromagnetic field. I don't think we have any evidence how accelerated charges radiate in other non-EM forces (i.e: weak and strong) but theoretically acceleration induced on hypotethical macroscopic fields of such should produce the same effect

$\endgroup$
9
  • $\begingroup$ OK, the particle equations of motion do not say whether it radiates or not. How about field equations? What the Maxwell equations say in GR? What is the source of EMW in GR, not acceleration? $\endgroup$ Mar 17, 2011 at 15:28
  • $\begingroup$ classically, i think EM waves are also emitted as the third-derivative of position, but i forgot where i saw that (I think it was a quote from Dirac, but i've never seen the derivation) $\endgroup$
    – lurscher
    Mar 17, 2011 at 15:33
  • $\begingroup$ Let's forget about particle equations of motion and look at the EM field equations. What they say in case of GR? $\endgroup$ Mar 17, 2011 at 15:59
  • $\begingroup$ i don't know, what do they say? $\endgroup$
    – lurscher
    Mar 17, 2011 at 16:00
  • $\begingroup$ In CE the field solutions are explicitly written via particle motion. In case of acceleration there are "propagating" solutions caused with the charge acceleration. The Maxwell equations in GR are modified somehow but they exist anyway. I do not know how their propagating solutions arise in GR either. $\endgroup$ Mar 17, 2011 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.