Paradox: Electron Radiates in a Gravitational Field

[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.

 

[Tom Kunich]

You have to be careful here. What specific measurement supports that claim that it doesn’t radiate in a gravitational field, and what exact measurement is the equivalent one for an accelerating charge? Is there actually a different prediction for the two?

Where are you suggesting an electron is getting the energy to radiate?

 

[Dale]

Where are you suggesting an electron is getting the energy to radiate?

I’m not. Please read the many subsequent posts for clarification.

 

[PeterDonis]

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?

The proper acceleration required to be "at rest" in a particular Rindler spaceship is fixed by the worldline of the spaceship; it's the path curvature of the worldline. Saying that it takes more force to keep a charged particle of mass $M$ at rest in the same spaceship as an uncharged particle of mass $M$ seems like saying that two objects following the same worldline can have different proper accelerations, which is impossible.

 

[Tom Kunich]

So far I agree.Does a light bulb in a rocket weigh more just because it radiates? I wouldnt's say so. The same for the charged particle.If a charged particle at rest on Earth emits energy, where does it take energy from?

I don't think we need to think of acceleration in a rocket. This entire universe is accelerating one way or the other. Therefore any motion other than that it is presently in is an acceleration of one sort or another.

 

[stevendaryl]

The proper acceleration required to be "at rest" in a particular Rindler spaceship is fixed by the worldline of the spaceship; it's the path curvature of the worldline. Saying that it takes more force to keep a charged particle of mass $M$ at rest in the same spaceship as an uncharged particle of mass $M$ seems like saying that two objects following the same worldline can have different proper accelerations, which is impossible.

[edit]

The total force must be the same for charged and uncharged particles, but in the case of the uncharged particle, the only force acting is the upward force of the rocket, while for the charged particle, there is also an electromagnetic force.

To accelerate a charged particle you need: $F^\mu_{rocket} + F^\mu_{field} = M A^\mu$, while for an uncharged particle, you only need $F^\mu_{rocket} = M A^\mu$

 

[Ian J Miller]

At the risk of being seen as missing something, has anybody ever observed radiation emitted by an electron, or any charged particle, falling in a gravitational field? By that, I mean there must be no change of electromagnetic potential during the charge falling. Also excluded must be heat generated radiation, e..g. radiation from mass falling into an accreting star.

 

[eachus]

At the risk of being seen as missing something, has anybody ever observed radiation emitted by an electron, or any charged particle, falling in a gravitational field? By that, I mean there must be no change of electromagnetic potential during the charge falling. Also excluded must be heat generated radiation, e..g. radiation from mass falling into an accreting star.

This is easy. Use a van de Graaff generator to charge a ball, let's say you cover a tennis ball with aluminum foil. Now drop it through a coil of wire. You will see a pulse of current. If you want to get rid of the effect of the coil, use two coils connected together and wound in opposite directions. (Or wind two strands in a coil and connect the far ends together.) Don't connect the other ends of the wires to anything. Now measure the voltage from one (free) end of a coil to where the coils are joined. Drop the ball and you will still see a spike.

 

[romsofia]

This topic has been explored in the 60s by DeWitt, and is not a paradox. Here is the paper: http://inspirehep.net/record/44837/files/vol1p3-20_001.pdf

 

[jartsa]

This is easy. Use a van de Graaff generator to charge a ball, let's say you cover a tennis ball with aluminum foil. Now drop it through a coil of wire. You will see a pulse of current. If you want to get rid of the effect of the coil, use two coils connected together and wound in opposite directions. (Or wind two strands in a coil and connect the far ends together.) Don't connect the other ends of the wires to anything. Now measure the voltage from one (free) end of a coil to where the coils are joined. Drop the ball and you will still see a spike.

The ball should not collide with anything except gravity field. Otherwise we are studying collision of charge with something else than gravity field. So:

Use a van de Graaff generator to charge a ball, let's say you cover a tennis ball with aluminum foil. Bore a hole through the earth. Now drop the ball into the hole, where it will move back and forth. Measure the generated radio-wave.Oh yes, no detectors near the hole. They would measure "near field effects". In other words the ball would collide into the electric fields of the devices.

https://en.wikipedia.org/wiki/Near_and_far_field

 

[Demystifier]

To accelerate a charged particle you need: $F^\mu_{rocket} + F^\mu_{field} = M A^\mu$, while for an uncharged particle, you only need $F^\mu_{rocket} = M A^\mu$

I think you are doing a category mistake here. You cannot add together forces that belong to different categories. The force of electric field belongs to the category of fundamental microscopic forces, together with gravitational, weak and strong force. The force of rocket belongs to the category of forces caused by various macroscopic objects, such as rocket, Earth, hammer, etc. The force of rocket on the charge is of the electromagnetic origin, so you are doing a kind of double counting. (Which reminds me of the "proof" that horse has 8 legs; 2 left legs, 2 right legs, 2 forward legs and 2 backward legs. )

 

[Demystifier]

I don't think we need to think of acceleration in a rocket. This entire universe is accelerating one way or the other. Therefore any motion other than that it is presently in is an acceleration of one sort or another.

If you mean accelerated expansion of the Universe caused by dark energy, that's "acceleration" in a different sense which is not relevant here.

 

[MikeGomez]

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.

What Demystifier said in post #88 and 100, and PeterDonis in post #93.

I think the problem is easier if it is assumed that the rocket engine thrust is continuously adjusted for constant acceleration, regardless the effects of the lab experiments, and regardless of the variable mass due to fuel consumption, etc. Consider the chamber of the rocket as separate from the engine that is accelerating it (similarly with the rope accelerating the chamber in the case of an accelerated elevator).

 

[MikeGomez]

The principle of equivalence is valid only locally. The concept of radiation, on the other hand, is a global concept. The claim that an accelerating charge in Minkowski spacetime "radiates" means that the flux of Poynting vector through a sphere far from the charge does not vanish. Hence you cannot apply the principle of equivalence to determine whether the static charge in gravitational field radiates. The equivalence principle is only helpful to study EM field in a vicinity of the charge, but this short-range effects do not tell much about radiation.

Can we achieve a valid EP region by making the region much larger, perhaps such as on a planet of far greater radius than the earth? Or what if we put a filter on the apparatus such that a parallel beam of radiation is emitted? Or what if we consider high energy radiation with very small wavelength?

 

[stevendaryl]

I think you are doing a category mistake here. You cannot add together forces that belong to different categories. The force of electric field belongs to the category of fundamental microscopic forces, together with gravitational, weak and strong force. The force of rocket belongs to the category of forces caused by various macroscopic objects, such as rocket, Earth, hammer, etc. The force of rocket on the charge is of the electromagnetic origin, so you are doing a kind of double counting. (Which reminds me of the "proof" that horse has 8 legs; 2 left legs, 2 right legs, 2 forward legs and 2 backward legs. )

This is not actually clarifying anything. Let me try to explain again.

Let's pick an inertial coordinate system. Suppose that at time $t_0$, you have a situation where there is a region of space that can be described in this coordinate system by:

• There is a charged particle with mass $M$ and charge $Q$.
• The particle have 3-velocity $\vec{v}$.
• There is an electromagnetic field $F^{\mu \nu}$.

I think everyone would agree that in accounting for energy conservation, you can't just use $E = \gamma M c^2$. The total energy must also take into account electromagnetic energy.

So if you replace the charged particle by an uncharged particle, and keep the 3-velocity $\vec{v}$ the same and keep the mass $M$ the same, the total energy will not be the same, because the electromagnetic energy will be different. Does anybody disagree with this?

If you agree, then it follows that you can't just use $E = \gamma M c^2$ to calculate how much energy was needed to accelerate the mass $M$ to achieve velocity $\vec{v}$. The charge $Q$ and the electromagnetic field $F^{\mu \nu}$ must be taken into account, as well.

If it turns out that the energy required to accelerate a particle from rest to velocity $\vec{v}$ is independent of whether it's charged, that's a pretty remarkable thing. I'm pretty sure it's not true, in general.

Now, instead of looking at the general case, let's look at the special case where $F^{\mu \nu}$ is the electromagnetic field due to the charge $Q$. In that case, maybe you can say that some of the electromagnetic energy is already accounted for in the mass $M$. That is, the mass $M$ already takes into account the electromagnetic self-energy. Does it then follow that the amount of energy required to accelerate a charged particle is the same as for an uncharged particle of the same mass? Maybe that's true, but it certainly isn't true by definition. It requires an argument.

 

[stevendaryl]

What Demystifier said in post #88 and 100, and PeterDonis in post #93.

Those three posts don't answer the question, at all. There is no reason, a priori, to believe that it takes the same amount of energy to accelerate a charged particle in an electromagnetic field as it does to accelerate an uncharged particle of the same mass. You can't compute the energy by simply knowing the mass and the final velocity.