Paradox: Electron Radiates in a Gravitational Field

[PeterDonis]

Help me to understand how you can have radiation in one frame of reference but not in another, or how you can detect it in one frame, but not in another.

Talking about "frames" just confuses the issue. The relevant factor is the relative motion of the source and the detector. That should be obvious from the examples @Dale gave.

 

[Sorcerer]

The detectors will have the same results in all frames, regardless of whether or not there is radiation in that frame.

Consider the accelerating charge and an inertial antenna. In the antenna’s inertial frame the accelerating charge emits radiation which is measured by the antenna as a time varying voltage. In the charge’s accelerating frame the charge does not emit any radiation and the time varying voltage is a result of the antenna moving through the static field

Talking about "frames" just confuses the issue. The relevant factor is the relative motion of the source and the detector. That should be obvious from the examples @Dale gave.

Not to messy the waters, but I have a slightly different question: Assuming each observer has their own local detector, and one observer is comoving with the charge, shouldn't that comoving observer see an electrostatic field, because any radiation an inertial observer sees (with respect to the accelerated charge) would be forever inaccessible to the comoving observer due to a kind of event horizon (basically for the observer at rest with respect to the charge, any radiation would be outside of that person's light cone)?

Or am I way off base here?

 

[Gene Naden]

Talking about "frames" just confuses the issue. The relevant factor is the relative motion of the source and the detector. That should be obvious from the examples @Dale gave.

So we are at the point where, if the detector is comoving with the "accelerated" charge (or the charge in a gravitational field) then it doesn't detect radiation because of the speed of light or because of some reason, but if the detector is not comoving with the charge then it can detect radiation? That seems like it would resolve the "paradox" (apologies to those who don't like that word).

 

[Vanadium 50]

Not to messy the waters, but I have a slightly different question

Maybe that better fits its own thread, as the OP misunderstands at least two things, so muddier waters is not what we need.

 

[PeterDonis]

So we are at the point where, if the detector is comoving with the "accelerated" charge (or the charge in a gravitational field) then it doesn't detect radiation because of the speed of light or because of some reason, but if the detector is not comoving with the charge then it can detect radiation?

That's basically what @Dale was saying in post #28.

 

[A.T.]

...shouldn't that comoving observer see an electrostatic field...

This is my understanding as well. If the acceleration is constant, the co-accelerating frame would see a static field as shown in figure b in post #20:

https://www.physicsforums.com/threa...in-a-gravitational-field.950608/#post-6020064

 

[MikeGomez]

This paradox may have come from Feynman's Lectures on Physics, ...

And Feynman says in the lectures that accelerated charges don't radiate. It is the 3rd derivative of position (change in acceleration, not just acceleration alone) that is responsible for the radiation of charged particles.

...But according to the Principle of Equivalence,...

You could in principle construct, instead of an accelerating elevator in flatspace, an elevator which not only accelerates but additionally changes acceleration. Then you could produce radiation from a charged particle, but that would not be an equivalence with the charged particle on Earth which does not radiate.

 

[Vanadium 50]

And Feynman says in the lectures that accelerated charges don't radiate.

Where?

 

[MikeGomez]

Where?

Here (I don’t have the book, but I think it is available online)

https://books.google.com/books?id=G...an accelerating charge should radiate&f=false

Also here is an article by Thayer Watkins of San Jose state university on the subject.

http://www.sjsu.edu/faculty/watkins/quantumEM2.htm

And here is another interesting article. I don’t know who the author is, but the math looks correct.

http://www.mathpages.com/home/kmath528/kmath528.htm

 

[Gene Naden]

And Feynman says in the lectures that accelerated charges don't radiate. It is the 3rd derivative of position (change in acceleration, not just acceleration alone) that is responsible for the radiation of charged particles.

After all that has gone through this thread, I would very much like to see where anyone says that a charge undergoing constant acceleration doesn't radiate. I think I will dig through Jackson on this subject, though I am not really at Jackson's level in my review of physics.

 

[MikeGomez]

After all that has gone through this thread, I would very much like to see where anyone says that a charge undergoing constant acceleration doesn't radiate. I think I will dig through Jackson on this subject, though I am not really at Jackson's level in my review of physics.

Again, Feynman said it.

 

[Gene Naden]

But your first reference says that the radiated power is proportional to $a^2$, not $\frac{d\vec{a}}{dt}$

 

[MikeGomez]

But your first reference says that the radiated power is proportional to $a^2$, not $\frac{d\vec{a}}{dt}$

No. Equation 9.1.1 is the equation that Feynman says has led us astray in our thinking.

 

[Dale]

Note, things like energy and power are frame variant. What is invariant is the outcome of measurements. The measurements we discussed above were simply about constant or time varying voltages, regardless of the power or other frame variant considerations.

 

[MikeGomez]

Note, things like energy and power are frame variant. What is invariant is the outcome of measurements. The measurements we discussed above were simply about constant or time varying voltages, regardless of the power or other frame variant considerations.

Good point.

 

[Sorcerer]

After all that has gone through this thread, I would very much like to see where anyone says that a charge undergoing constant acceleration doesn't radiate. I think I will dig through Jackson on this subject, though I am not really at Jackson's level in my review of physics.

Well, I think there are good arguments for a uniformly accelerated charge not emitting observable radiation according to someone comoving with it.

I don't know how much trust you put in the University of Campinas in Brazil, or FAPESP in Brazil (their wiki page: https://en.wikipedia.org/wiki/São_Paulo_Research_Foundation), or the National Council for Scientific and Technological Deveolopment in terms of peer review (their wiki page: https://en.wikipedia.org/wiki/National_Council_for_Scientific_and_Technological_Development), but they did support the paper I linked to earlier, which, after a lengthy series of derivations and examinations, said this:

This result answers our question. A comoving observer will not detect any radiation from a uniformly accelerated charge. The comoving observer can receive signals only from regions I and IV. The field emitted by the accelerated charge does not reach region IV, and in region I, it is interpreted by the comoving observer as a static field. We note that essentially the same argument was used by Rohrlich to show that in a static homogeneous gravitational field, static observers do not detect any radiation from static charges.

https://arxiv.org/pdf/physics/0506049‎

Now, this is very math heavy, and a bit beyond me. But, as a kind of analogy, not everyone is going to see a magnetic field just because someone else does. One observer moving with respect to an electric field will see a magnetic field. Measurements are what matter. All the measurements have to agree. I believe Dale said this.

Which leads me to believe that if the paper above is correct, then observation of the radiation is frame dependent, but if I understood the paper correctly, the radiation is still happening, it just is impossible to be seen by the comoving observer (due to a what amounts to a "radiation event horizon.")
If any of you post-grad or graduate student people have time, I'd appreciate your insight on the above link, and if I've made any blatant misunderstandings of it. Thanks.

 

[stevendaryl]

There is a discussion of the equivalence principle as applied to a uniformly accelerated charge given here:

https://arxiv.org/pdf/gr-qc/9303025.pdf

Apparently, it's very complicated to untangle all the issues.

 

[Dale]

Apparently, it's very complicated to untangle all the issues.

It is only complicated if you focus on radiation instead of measurements. Hence my insistence from the beginning on defining the measurement procedure.

The EP is very specific. It says that the outcome of an experiment (a measurement) is the same if performed under uniform gravity or under uniform acceleration. It does not say that intermediate quantities, like radiation, are the same.

 

[Sorcerer]

It is only complicated if you focus on radiation instead of measurements. Hence my insistence from the beginning on defining the measurement procedure.

The EP is very specific. It says that the outcome of an experiment (a measurement) is the same if performed under uniform gravity or under uniform acceleration. It does not say that intermediate quantities, like radiation, are the same.

Is this similar to the magnet and conductor issue? I'm referring to the opening paragraph of "On the Electrodynamics of Moving Bodies," where the current is the same but explained differently by two observers. Same measurement, but frame dependent explanations?

 

[Dale]

Is this similar to the magnet and conductor issue? I'm referring to the opening paragraph of "On the Electrodynamics of Moving Bodies," where the current is the same but explained differently by two observers. Same measurement, but frame dependent explanations?

Yes. The EP is about the equivalence of measurements, not explanations.

 

[Vanadium 50]

Here (I don’t have the book, but I think it is available online)

It is not available.

Again, I ask where did Feynman say it? I want to know exactly what he said, because this is a place where context matters.

 

[stevendaryl]

It is only complicated if you focus on radiation instead of measurements. Hence my insistence from the beginning on defining the measurement procedure.

The EP is very specific. It says that the outcome of an experiment (a measurement) is the same if performed under uniform gravity or under uniform acceleration. It does not say that intermediate quantities, like radiation, are the same.

But Parrot is saying that it's not true. You have two different situations:

1. A rocket ship hovering above a massive star.
2. A rocket ship accelerating at constant proper acceleration in flat spacetime.

Parrot is claiming that careful measurements would reveal a difference in these two cases. And not because of tidal effects. He says that in case 1, there is no difference between the rocket power needed to hold up a charged particle of mass $M$ and the power needed to hold up an uncharged particle. In case 2, it requires more force to hold up the charged particle (because some of the energy that goes into accelerating the particle is lost to radiation).

So he's saying that local measurements can in principle tell the difference.

[edit] His conclusion is about measurements. From the abstract:

We argue that purely local experiments can distinguish a stationary charged particle in a static gravitational field from an accelerated particle in (gravity-free) Minkowski space.

 

[stevendaryl]

But Parrot is saying that it's not true. You have two different situations:

1. A rocket ship hovering above a massive star.
2. A rocket ship accelerating at constant proper acceleration in flat spacetime.

Parrot is claiming that careful measurements would reveal a difference in these two cases. And not because of tidal effects. He says that in case 1, there is no difference between the rocket power needed to hold up a charged particle of mass $M$ and the power needed to hold up an uncharged particle. In case 2, it requires more force to hold up the charged particle (because some of the energy that goes into accelerating the particle is lost to radiation).

So he's saying that local measurements can in principle tell the difference.

I'm not saying that Parrot is correct in his conclusions, but only to say that if he's right, it's a violation of the equivalence principle, according to the formulation "local measurements cannot distinguish between blah and blah".

 

[stevendaryl]

It is not available.

Again, I ask where did Feynman say it? I want to know exactly what he said, because this is a place where context matters.

[edit]Kevin Brown discusses Feynman's argument here:

http://www.mathpages.com/home/kmath528/kmath528.htm

 

[Dale]

So he's saying that local measurements can in principle tell the difference.

[edit] His conclusion is about measurements. From the abstract:

Hmm, I don’t think this is an accepted view, but I haven’t finished the article yet

 

[stevendaryl]

John Baez discusses Feynman's argument here:

http://www.mathpages.com/home/kmath528/kmath528.htm

An excerpt from Feynman is found here:

http://www.applet-magic.com/feynmanEM.htm

 

[MikeGomez]

John Baez discusses Feynman's argument here:

http://www.mathpages.com/home/kmath528/kmath528.htm

That link looks oddly familiar. Hmm, just like one that I provided in post #44. Why do you think it is John Baez?
Or did you post the wrong link by accident?

 

[stevendaryl]

That link looks oddly familiar. Hmm, just like one that I provided in post #44. Why do you think it is John Baez?
Or did you post the wrong link by accident?

My mistake. I thought I remembered him writing on that topic, but that particular article doesn't have an obvious author.

[edit] Apparently, it is KEVIN S. Brown.

 

[MikeGomez]

It is not available.

Again, I ask where did Feynman say it? I want to know exactly what he said, because this is a place where context matters.

I actually don't know exactly what he said, and you're right, context matters.

If you wish I can retract my statement that Feynman said that accelerated particles don't radiate.

Edit: I meant accelerated charged particle.

 

[stevendaryl]

I actually don't know exactly what he said, and you're right, context matters.

If you wish I can retract my statement that Feynman said that accelerated particles don't radiate.

Edit: I meant accelerated charged particle.

He definitely did not say that accelerated charged particles never radiate. He was specifically talking about the case of constant proper acceleration.

 

[girts]

can't we use the example of steady DC electrical current, excluding the parts where it is turned on or off otherwise it is constant proper acceleration of electrons which doesn't lead to any radiation as is the reason why DC cannot be used in transformers since the field is static and it can't transfer energy aka radiate.

 

[Nugatory]

can't we use the example of steady DC electrical current, excluding the parts where it is turned on or off otherwise it is constant proper acceleration of electrons which doesn't lead to any radiation as is the reason why DC cannot be used in transformers since the field is static and it can't transfer energy aka radiate.

There is no acceleration of charge in a constant DC current if you exclude the startup and shutdown periods.

 

[girts]

Exactly but that was my point, there is also no acceleration or de-acceleration for a charged particle of constant velocity which was my point. Pardon my misunderstanding but then why we are considering radiation from the examples said previously of a constant accelerated charge? Unless ut somehow changes its speed relative to an observer?

 

[stevendaryl]

can't we use the example of steady DC electrical current, excluding the parts where it is turned on or off otherwise it is constant proper acceleration of electrons which doesn't lead to any radiation as is the reason why DC cannot be used in transformers since the field is static and it can't transfer energy aka radiate.

Are you mixing up constant acceleration and constant velocity?

Constant velocity means zero acceleration. The issue is whether nonzero acceleration necessarily leads to radiation.

 

[girts]

Thanks for clarification , okay so if we say nonzero acceleration then that basically means a charge is accelerating so we are talking about a charge that changes velocity and with respect to an observer at rest with respect to the charge it should see EM radiation , well hasnt this been known for a long time and accepted as the norm or am I still missing something?