Talk:Paradox of radiation of charged particles in a gravitational field

Just starting this page. I have seen a lot of contradictory and confused descriptions on the web. I think this is a pretty good start ManitouLance (talk) 03:53, 7 October 2014 (UTC)[reply]

I think this is still an open question

I like your explanation. It is for sure consistent and I didn't know it. But there is at least a second explanation, at least equally valid, which is based in the Feynman radiation equation.

Classic radiation equation is obtained assuming that a charge does not interact with its own field. Feynman equation does not assume this and obtains a different expression for the radiation [1]

Though maybe it is surprising, there is no experiment that measures directly if an uniformly accelerated charge radiates in an inertial frame. Classic electromagnetic theory says it should, but this has never been confirmed by an experiment, because experiments use circular accelerating charges to verify the classic radiation equation.

For sure, classic equations yield the correct result for circular accelerating charges, but Feynman radiation equation yields the same value in these cases. The big difference is that it yields no radiation for an uniformly accelerating charge seen from an inertial frame.

Should Feynman be right, the charge would not radiate in any frame. This explanation is also consistent, and besides, it preserves energy in any frame without using the Boulware explanation (I thinks it is cleaner). Therefore, I think the article should be extended with at least this alternative explanation.

--Juansempere (talk) 22:13, 3 February 2015 (UTC)[reply]

Explanations might not be exclusive of each other. In general relativity, momentum and energy are not necessarily locally conserved due to the curvature. Stating it otherwise, the covariant derivative has an additional term depending on Christoffel symbols compared to a usual partial derivative. — Preceding unsigned comment added by Klinfran (talk • contribs) 13:13, 17 January 2020 (UTC)[reply]

Free fall - an inertial frame?

The text says: "The Maxwell equations can be applied relative to an observer in free fall, because free-fall is an inertial frame". Since when? This makes no sense. E.g. a person in free fall can feel the accelerations. Within an inertial frame you cannot devise an experiment by which to determine your velocity (in any way shape or form) relative to another inertial frame and an inertial frame is in a state of "constant rectilinear motion". Free-fall would correspond to an accelerated frame of reference.TonyMath (talk) 14:16, 14 October 2015 (UTC)[reply]

I think you are finally grasping the nature of free fall. This is Einstein's core insight; it is called the Equivalence Principle. One way of saying it is that there are no forces in free fall. Gravity vanishes in free fall. You cannot feel the accelerations. Think about the astronauts in the space station. There is plenty of gravity just 100 km above the earth. But they are weightless because they are in free fall. A satellite orbit is free fall.ManitouLance (talk) 03:47, 7 December 2015 (UTC)[reply]

We are not in agreement. Drive a car at constant speed, then press the breaks: you feel the (negative) acceleration and you are no longer in an inertial frame. Jump out of a plane without a parachute and the vertigo you feel and the acceleration you feel will accelerate your heart until your heart stops. You will be dead long before you hit the ground. Accelerations are felt and when they are: you are not in an inertial frame of reference. A space station is in orbit, maintaining a constant radius about the earth because of its centripetal acceleration w.r.t. the earth's center (used as a frame of reference). Energy was imparted to put that space station in orbit. Free-fall cannot be an inertial frame. You know when you are in a state of weightlessness inside that space station. So it cannot be an inertial frame. TonyMath (talk) 19:18, 5 January 2016 (UTC)[reply]

Of course, I am speaking of an inertial frame in Newtonian mechanics (inertial frames are global in Newtonian mechanics and extend to infinity). Inertial frames are very different in General Relativity and are at best local. If you are using the definition from the latter, that should be clearly explained in the article.TonyMath (talk) 05:02, 7 January 2016 (UTC)[reply]

You can not agree, but that doesn't mean you are right. If you feel something when you jump out of a plane, it's because of friction and because you move from an accelerated frame, to an inertial frame (during few seconds before frictions overtake). If there was no air, you would fall as fast as the plane and everything around. Thus feeling no force and no acceleration.Klinfran (talk) 13:10, 17 January 2020 (UTC)[reply]

Improvements needed

This article is not bad insofar as it goes, but it could use some work. For one, the alternative explanation that Juansempere describes would be a great addition. Also, has there been any empirical work on this paradox? Theory is fine and all, but it's not particularly meaningful without an experiment to back it. And this paradox would seem very possibly testable.

Those things being said, it needs to be noted that the given explanation itself needs some serious work. It reads: "So a falling charge will appear to radiate to a supported observer, as expected," but also that the supported charge appears to a free-falling observer to radiate. Now observer-dependent effects have their place in relativity, but they seem out of place here. Either the charge radiates or it does not: there can be no observer-dependence. If it is EM radiation, it carries away energy, so how can "this [falling] charge remains at rest in its free-fall frame, just as a neutral particle would"? There seems to be another paradox here, or at least incompleteness in the explanation. As I said: work needed! JKeck (talk) 04:32, 17 March 2016 (UTC)[reply]

Here's another explanation that is clearer in many ways: http://arxiv.org/abs/physics/9910019. It depends on Rohrlich's 1963 paper, and argues against observer dependence. And here's another paper I found that supports observer dependence: http://xxx.lanl.gov/abs/gr-qc/0006037. JKeck (talk) 04:55, 17 March 2016 (UTC)[reply]

Another paper: http://cds.cern.ch/record/403583/files/9910019.pdf (Harpaz and Soker). And a book: Uniformly Accelerating Charged Particles: A Threat to the Equivalence Principle by Stephen N. Lyle. JKeck (talk) 05:02, 17 March 2016 (UTC)[reply]

Actually that last section on Feynman's proposal is written in poor English. E.g. you do not start a sentence with "And". That last section needs work. Moreover, I am still not satisfied with the physical explanation. Experimental confirmation would be very helpful. TonyMath (talk) 12:12, 21 November 2016 (UTC)[reply]

Authors start sentences with "and" all the time! And there's nothing wrong with that. Betaneptune (talk) 18:55, 16 October 2020 (UTC)[reply]

A free falling charge does not radiate

If the references talk about a charge supported at rest in a gravitational field they do not mean a free falling charge but a charge which is at rest to the planet and therefore accelerated (the free falling charge is not). The latter does not radiate, so I corrected the wrong statements and added some references (click here to compare the changes). --Yukterez (talk) 05:35, 30 September 2018 (UTC)[reply]

@Yukterez: I'm ok with the changes. Try to be more systematic with the sources and cite the official journal articles, you may help yourself with the visual editing feature. Also I am allowing the arXiv citation (Gründler 2015) in the refs, but a better source would be preferable. --MaoGo (talk) 12:02, 2 October 2018 (UTC)[reply]

Minor Edits October 2019

I removed the confusing note from Feynman, and also about Bremstrahlung. I am glad to see that some good references have been added besides Rohrlich, but I think his work is still the definitive treatment. It is striking how many people misunderstand the paradox, so I think by that measure, this article is serving it's purpose. If this article seems in any way incorrect, it probably means you have succumbed to a misconception about gravity or electromagnetism somewhere :-) ManitouLance (talk) 02:12, 28 October 2019 (UTC)[reply]

Falling charge

I also find it very peculiar that nobody tried that experiment. That said I have a question, it seems it isn't stated in the article or I didn't read it correctly. So I understood that a free falling charge didn't radiate in it's frame, but does it in the supported frame? Also, I have a paper where the radiations of a single charge are derived from the acceleration in the accelerated frame. What about the potentials (and the gauges) in different frames, those are bundled in a 4 vector, it should be much more easy to calculate fields from them using coordinate transformations. Klinfran (talk) 13:22, 17 January 2020 (UTC)[reply]

Agreed. But the article says that Lorentz transformations are not applicable here. RadXman (talk) 20:20, 16 July 2023 (UTC)[reply]

The paragraph "Resolution by Rohrlich" doesn't present the resolution by Rohrlich

The mentioned paragraph reproduces the arguments by Rohrlich (references 4 and 6), except for these sentences:

When this coordinate transformation is applied to the electric and magnetic fields of the charge in the rest frame, it is found not to be radiating, as expected for a charge in an inertial frame. and So a falling charge will not radiate, as expected.[9][8].

These sentences completely contradict the findings of Rohrlich. He explicitly says in reference 4 (page 514) "An electron which falls freely in a uniform gravitational field embedded in an inertial frame will radiate, and one which sits at rest on a table in the same field will not radiate; and these two statements do not contradict the principle of equivalence.” Also from his book (reference 6, page 218): "Transformation [...] shows that the observer (an observer in a supported laboratory system on earth) sees this (freely falling) charge radiating".

Besides, the references 9,8 are not even from Rohrlich... Also, reference 8 is not even peer reviewed. I have therefore changed the former of the aforementioned sentences and have removed the latter. I would suggest an additional paragraph (as a sub-paragraph to Rohrlichs solution) entitled "opposing views".

Autobahnvignette (talk) 18:36, 19 January 2020 (UTC)[reply]

Gravity as a fictitious force

Why does the article say gravity is a fictitious force because it can be transformed away? The same is true of EM fields. A pure electric field can be transformed into an electric field and a magnetic field, so a magnetic field is also fictitious? Betaneptune (talk) 19:10, 16 October 2020 (UTC)[reply]

An E field can not be transformed away. It can be (through a Lorentz boost) transformed partly into a B field, but there is still a field left. Gravitational effect (to lowest order) can be completely removed, by a boost. That's the reason gravity is sometimes termed a fictitious force. Mgolden (talk) 22:56, 17 October 2020 (UTC)[reply]
- No, but if you put my scenario in reverse, the magnetic field is eliminated. So then a magnetic field is also a fictitious force. So what is the difference? And gravity isn't fictitious to engineers. — Preceding unsigned comment added by Betaneptune (talk • contribs) 23:12, 17 October 2020 (UTC)[reply]
  E and B fields are not distinct from each other in Minkowski space, they are two different parts of the same field – Electromagnetic tensor. Likewise, both the electric potential and the magnetic (vector) potential together form a single four-vector – Electromagnetic four-potential. When viewed as a single object, electromagnetic fields cannot be transformed away, only the interpretation of what's the electric and what's the magnetic part changes. Slovborg (talk) 21:48, 16 November 2024 (UTC)[reply]