I Paradox: Electron Radiates in a Gravitational Field

[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?

 

[Nugatory]

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?

We are considered an object at rest in a gravitational field, such as one sitting on the surface of the earth. Intuitively we expect it not to radiate, but by the equivalence principle it should because this situation is indistinguishable from putting the same object in a spaceship accelerating at 1g in empty space.

 

[Sorcerer]

We are considered an object at rest in a gravitational field, such as one sitting on the surface of the earth. Intuitively we expect it not to radiate, but by the equivalence principle it should because this situation is indistinguishable from putting the same object in a spaceship accelerating at 1g in empty space.

What if the comoving observers (those at rest in the 1g accelerating spacecraft ) have no access to the radiation? I hate to keep harping on this given that I'm an undergrad and this is clearly a graduate level topic, but again, these people seem to have presented a good argument to me. Could someone here dismantle it if it is not valid?

Nevertheless, comoving observers, that is, accelerated observers with respect to whom the charge is at rest, will not detect any radiation because the radiation field is confined to a spacetime region beyond a horizon that they cannot access.

Published in The American Journal of Physics
https://aapt.scitation.org/doi/10.1119/1.2162548

If you don't want a paywall:
https://arxiv.org/pdf/physics/0506049‎
If I understand the set up, tidal forces should have no bearing, and yet their derivation results in a charge accelerating according to an inertial observer radiating, while according to the observer comoving with the charge never having access to the radiation. Again, me = undergrad, topic = grad or upper division at lowest, but why doesn't that solve the issue?

I mean, if the people in the 1 g rocket cannot access the radiation (for whetever reason), the equivalence principle is completely saved: someone in a uniform gravitational field will not see radiation, nor will someone in then 1 g rocket.

 

[Nugatory]

What if the comoving observers (those at rest in the 1g accelerating spacecraft ) have no access to the radiation?

That is indeed one of the possible resolutions to the apparent paradox. Another possible resolution (not generally accepted, but with a respectable pedigree from Pauli to Feynman) is that constant proper acceleration does not produce radiation (that is, the third derivative of position must be non-zero if there is to be radiation).

 

[Sorcerer]

That is indeed one of the possible resolutions to the apparent paradox. Another possible resolution (not generally accepted, but with a respectable pedigree from Pauli to Feynman) is that constant proper acceleration does not produce radiation (that is, the third derivative of position must be non-zero if there is to be radiation).

So... what you're saying is that radiation can be a real jerk?

 

[Sorcerer]

That is indeed one of the possible resolutions to the apparent paradox. Another possible resolution (not generally accepted, but with a respectable pedigree from Pauli to Feynman) is that constant proper acceleration does not produce radiation (that is, the third derivative of position must be non-zero if there is to be radiation).

So... what you're saying is that radiation can be a real jerk?

Of course, if you push him too hard to change, he might snap.

*ba dum tss*

Sorry I can't stop.

 

[Demystifier]

As I pointed out in https://arxiv.org/abs/gr-qc/9909035 , the EM field measured by a local observer depends only on the instantaneous velocity of the observer, not on the acceleration (or higher derivatives). In that sense, the radiation does not depend on acceleration of the observer.

 

[stevendaryl]

As I pointed out in https://arxiv.org/abs/gr-qc/9909035 , the EM field measured by a local observer depends only on the instantaneous velocity of the observer, not on the acceleration (or higher derivatives). In that sense, the radiation does not depend on acceleration of the observer.

I was looking for the definitive answer to the question: Does a Rindler observer see radiation from a "stationary" charge ("stationary" meaning zero spatial coordinate velocity in Rindler coordinates)? Did you give the answer and I missed it?

You seem to be saying, though, that for an accelerating charge the self-force is not isotropic, which I would think would manifest itself as an apparent change to the particle's weight compared to an uncharged particle of the same mass? So, if you have two particles of mass $M$, one has charge $Q$ and one is neutral, then their "weight" on board a Rindler spaceship (the force needed to keep them at constant Rindler spatial location) would be different?

 

[stevendaryl]

What if the comoving observers (those at rest in the 1g accelerating spacecraft ) have no access to the radiation? I hate to keep harping on this given that I'm an undergrad and this is clearly a graduate level topic, but again, these people seem to have presented a good argument to me. Could someone here dismantle it if it is not valid?

Published in The American Journal of Physics
https://aapt.scitation.org/doi/10.1119/1.2162548

If you don't want a paywall:
https://arxiv.org/pdf/physics/0506049‎

There is still a mystery to clear up, in my mind. If a charged particle radiates, according to an inertial observer, then that means that some of the energy used to accelerate the particle goes into radiation, by energy conservation. This means that it requires more energy to accelerate a charged particle than an uncharged particle of the same mass. This in turn means that for a Rindler rocket carrying a charged particle, more fuel is required than if it were carrying an uncharged particle of the same mass. So the charged particle would appear to weigh more.

But then is the same true for a charged particle at rest in a gravitational field?

 

[Sorcerer]

There is still a mystery to clear up, in my mind. If a charged particle radiates, according to an inertial observer, then that means that some of the energy used to accelerate the particle goes into radiation, by energy conservation. This means that it requires more energy to accelerate a charged particle than an uncharged particle of the same mass. This in turn means that for a Rindler rocket carrying a charged particle, more fuel is required than if it were carrying an uncharged particle of the same mass. So the charged particle would appear to weigh more.

But then is the same true for a charged particle at rest in a gravitational field?

Complete shot in the dark here (from a very unqualified person, myself), but: how much energy? And is it comparable to the effect the charged particle would have on the stress-energy tensor by virtue of being charged, and would that be relevant in any way?

EDIT: wait, a charged particle must move to created an EM field, correct? Still, could there be any effect?

Does the presence of a charged particle in some way affect the other particles around it in a way that would create any kind of pressure that could be relevant? Etc, etc.

 

[girts]

Now I do understand that my questions here are more like brakes on a high speed intellectual debate but please bear with me as I too want to understand this topic deeper.I read through the article @Sorcerer posted, and I think I kind of understand the reason why two charged particles traveling at the same speed which is then uniformly increasing due to uniform acceleration (gravitational or electrostatic doesn't matter in this case I assume) can't see each other's radiation but instead see a static field much like two charged particles would see while being at rest with both a third observer and also with one another.
So If I understand correctly the reason for this is that as the particles move along they move with some speed which is less than c, but changes in the EM field travel at c in vacuum, so as the two charges enter every new frame both traveling at the same velocity the changes in their EM field "lines" as perceived by an inertial observer (aka the third observer at rest with respect to the charges) are radiating outwards in all directions with speed "c" but since the charged particle speed itself can never exceed or even come close to c, they themselves (the traveling particles) cannot reach their own created changes in the static field lines.
Nor can they see each other's field lines being distorted because their speed even under acceleration is the same so the field line changes happen simultaneously and appear as not happening at all for the two particles?But still even if this all is so, the question stated by stevendaryl in post#77 and before is valid and interesting, aka where does the energy come from?
I mean gravitational field is static and so is the electric field of a charged particle,so how come two such static fields can produce radiation which is a dynamic property that can radiate (transfer) energy outside these two systems (two fields) without there being any apparent energy input?
Just a quick though experiment along the way, if this is true then in theory we could say somehow have a place here on Earth where the gravitational acceleration constant is higher then nearby and then we could put a bunch of charged particles at that higher gravity and stand next to them and receive radiation, while no energy input would be done, doesn't this somehow violate energy conservation and perpetual motion?

 

[Sorcerer]

Just a quick though experiment along the way, if this is true then in theory we could say somehow have a place here on Earth where the gravitational acceleration constant is higher then nearby and then we could put a bunch of charged particles at that higher gravity and stand next to them and receive radiation, while no energy input would be done, doesn't this somehow violate energy conservation and perpetual motion?

Going to throw more shots in the dark:

I don't think this one would work because the distance for a noticeable gravitational difference is probably too large for the equivalence principle to remain valid, but if it did, presumably there would be different pressure up there, and pressure is part of the stress-energy tensor, right? Maybe the lower pressure can correspond to the "loss in energy?"

 

[Demystifier]

I was looking for the definitive answer to the question: Does a Rindler observer see radiation from a "stationary" charge ("stationary" meaning zero spatial coordinate velocity in Rindler coordinates)? Did you give the answer and I missed it?

My answer is that he sees radiation, provided that one accepts a certain local definition of "radiation". But one does not necessarily need to agree with such a definition. In general there is no "obvious" definition of radiation, so various ad hoc definitions are possible.

 

[stevendaryl]

My answer is that he sees radiation, provided that one accepts a certain local definition of "radiation". But one does not necessarily need to agree with such a definition. In general there is no "obvious" definition of radiation, so various ad hoc definitions are possible.

I thought that others had come to the opposite conclusion (about the same local definition of "radiation").

 

[stevendaryl]

I thought that others had come to the opposite conclusion (about the same local definition of "radiation").

We can operationalize it as: You have a charge Q that is suspended in the middle of an accelerating rocket. Will a hydrogen atom in the wall or ceiling or floor of the rocket be knocked into an excited state? Will a photographic plate in the wall or ceiling or floor be darkened?

 

[jartsa]

We can operationalize it as: You have a charge Q that is suspended in the middle of an accelerating rocket. Will a hydrogen atom in the wall or ceiling or floor of the rocket be knocked into an excited state? Will a photographic plate in the wall or ceiling or floor be darkened?

The rocket is a charged accelerating object, so it must radiate, so its radiation must not be absorbed by itself. I mean the rocket should radiate according to an observer far from the rocket.

If there is also charge -Q somewhere in the rocket ... then the rocket must still radiate - like an accelerating dipole radiates.... I mean if your spaceship is charged the enemies can detect your position from the E-field. If the position velocity changes, information about the new position velocity is transmitted by EM-waves. And Faraday cage or any other wall that moves with the spaceship does not help.

If a charged rocket travels back and forth like a sine wave, what is the radiation like? Not a sine wave, as the frequency of radiation depends on acceleration?? So antennas are (almost) linear devices, while charged rockets aren't?

 

[Demystifier]

I thought that others had come to the opposite conclusion (about the same local definition of "radiation").

Are you sure that it was the same local definition?

 

[stevendaryl]

Are you sure that it was the same local definition?

Parrot, summarizing Singal says:

He shows that the Poynting vector vanishes in the rest frames of certain co-accelerating observers and concludes from this that “in the accelerated frame, there is no energy flux, ... , and no radiation”.

 

[Demystifier]

There is still a mystery to clear up, in my mind. If a charged particle radiates, according to an inertial observer, then that means that some of the energy used to accelerate the particle goes into radiation, by energy conservation. This means that it requires more energy to accelerate a charged particle than an uncharged particle of the same mass. This in turn means that for a Rindler rocket carrying a charged particle, more fuel is required than if it were carrying an uncharged particle of the same mass.

So far I agree.

So the charged particle would appear to weigh more.

Does a light bulb in a rocket weigh more just because it radiates? I wouldnt's say so. The same for the charged particle.

But then is the same true for a charged particle at rest in a gravitational field?

If a charged particle at rest on Earth emits energy, where does it take energy from?

 

[Demystifier]

Parrot, summarizing Singal says:

That doesn't coincide with the definition of radiation I use in the paper.

 

[stevendaryl]

Does a light bulb in a rocket weigh more just because it radiates? I wouldnt's say so.

The "weight" of an object is operationally the force needed to keep it in place. It's just a word. What I meant was: does it require more force to keep a charged particle of mass $M$ at "rest" in a Rindler spaceship than it does to keep an uncharged particle? And is that true in a gravitational field.

If a charged particle at rest on Earth emits energy, where does it take energy from?

You argue in your paper that one can't assume that the self-force of a charged particle in a gravitational field is isotropic. If it's not isotropic, and has a radial component, then that would manifest itself as an additional force needed to keep the particle in place, whether or not it manifested itself as radiation.