I Paradox: Electron Radiates in a Gravitational Field

[Gene Naden]

This paradox may have come from Feynman's Lectures on Physics, or I may have dreamed it up myself. I am not sure. It has been around for a while and if you have already seen it, I apologize. I am not aware of any resolution.

An electron is at rest in a gravitational field. We know from experience that it doesn't radiate. But according to the Principle of Equivalence, it is equivalent to an electron undergoing acceleration. So it should radiate. Does it radiate or not?

 

[Dale]

We know from experience that it doesn't radiate.

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?

 

[Gene Naden]

If an electron at rest were to radiate, the energy would have to come from somewhere. Assume that the particle is supported, so that it doesn't fall. Since the rest mass of an electron is invariant, it is not clear where the energy would come from. Also, would it merely go on radiating forever?

However, the radiation of accelerated electrons is well-known, as I am sure you are aware, for example synchroton radiation and bremsstrahlung,

 

[PeterDonis]

Assume that the particle is supported, so that it doesn't fall.

How do you propose to support it? Ordinary objects are electrically neutral; they are supported by forces between the electrons in their atomic orbitals, but those are not single electrons and their charges are canceled by the charges of the atomic nuclei. You can't use the same method to support a single electron.

 

[Gene Naden]

How about a magnetic bottle? Or what about everyday electrons in solids? They are supported by the lattice in which they are embedded. Their charge is separate from the charges of the atomic nuclei.

Electrons in particle accelerators do not radiate until they are deflected by electromagnetic fields, apparently. Perhaps further experimentation is in order.

 

[PeterDonis]

How about a magnetic bottle?

I don't think you can hold an electron motionless in one, although you could contain it in a closed orbit. But of course it would be accelerated and would radiate.

Or what about everyday electrons in solids?

I already covered that case in my previous post. The only thing to add would be to understand that electrons in atoms are in stationary states and, assuming the atoms are in their ground states, the electrons cannot radiate because there is no lower energy state for them to go to. This cannot be understood purely in terms of classical EM; it requires quantum mechanics.

Electrons in particle accelerators do not radiate until they are deflected by electromagnetic fields

More generally, the only way to accelerate an electron is to subject it to an EM field. That was the point of my question about how you would you propose to support the electron. Any method of doing so will involve EM fields and so the electron will be accelerated by them and will radiate because of that. (See, for example, my response to the magnetic bottle example above.)

 

[Gene Naden]

Well, if the magnetic bottle were good enough the orbit could be quite small, hence little radiation. As for the quantization of electron states, here I think we have a contradiction between the predictions of general relativity and those of quantum mechanics. GR says the electron should radiate. QM says it cannot. This gets back to my earlier point that the rest mass of the electron is invariant. We might regard zero momentum as the ground state of an electron.

I do not understand the objection that in a solid the electronic charges cancel the charges of the atomic nuclei. They are separate.

 

[PeterDonis]

if the magnetic bottle were good enough the orbit could be quite small, hence little radiation

Making the orbit smaller would increase the electron's acceleration and make the radiation stronger, not weaker.

As for the quantization of electron states, here I think we have a contradiction between the predictions of general relativity and those of quantum mechanics.

No, we have a situation that GR was never intended to model in the first place.

GR says the electron should radiate.

No, GR makes no prediction about whether the electron should radiate, since GR treats matter as continuous and does not even attempt to model its quantum properties.

We might regard zero momentum as the ground state of an electron.

This won't work since "zero momentum" is frame dependent, but changing your frame of reference doesn't change whether the electron is in its ground state or not.

Free momentum states can't, in general, be ground states (except for the trivial edge case of the vacuum, with zero particles present).

I do not understand the objection that in a solid the electronic charges cancel the charges of the atomic nuclei. They are separate.

From the GR perspective (where, as I said above, we treat matter as continuous), they cancel. From a quantum perspective, yes, you can model the electron using the potential well of the nucleus, but then, as I said above, you're not using classical EM anyway.

 

[Dale]

If an electron at rest were to radiate, the energy would have to come from somewhere. Assume that the particle is supported, so that it doesn't fall. Since the rest mass of an electron is invariant, it is not clear where the energy would come from. Also, would it merely go on radiating forever?

However, the radiation of accelerated electrons is well-known, as I am sure you are aware, for example synchroton radiation and bremsstrahlung,

Please actually answer the specific questions I asked to you.

Btw, we can avoid all of the QM problems by considering a classical charged conductive sphere supported by an insulated stick. Something like a Van De Graf generator. The QM issues go away, but not the measurement issues I am trying to get you to address.

 

[Gene Naden]

What specific measurement supports that claim that it doesn’t radiate in a gravitational field

I can only say that in accelerators, they do not radiate until they are deflected by electromagnetic fields. However, no one has reported electrons radiating from a position of rest, as far as I know.

what exact measurement is the equivalent one for an accelerating charge?

Synchrotron radiation and brehmstrahlung.

Is there actually a different prediction for the two?

Yes there is a different prediction; The principle of equivalence says that a particle in a gravitational field should radiate. The evidence, as I just mentioned, is that it doesn't..

How do you propose to support it?

A magnetic bottle, or simply embedded in the lattice of a solid. If the bottle is large enough then there should be little acceleration (other than the gravitational field)

 

[Dale]

A good description of a measurement involves identifying clearly the entire experimental setup including both the thing to be measured and also the measuring device.

I can only say that in accelerators, they do not radiate until they are deflected by electromagnetic fields.

A particle accelerator (thing to be measured) with a scintillator located at a bend (measuring device) would produce scintillations both in a uniform gravitational field and also if uniformly accelerated in space. The equivalence principle works for this case. It works for any case you can come up with if you actually try what I suggested in the beginning.

Yes there is a different prediction;

No, there isn’t. There is only a failure to correctly specify the setup such that the equivalence principle actually applies.

 

[Vanadium 50]

e. But according to the Principle of Equivalence, it is equivalent to an electron undergoing acceleration. So it should radiate.

Not so. Do the calculation and show me where. If you want to get around the objection of Peter, make it a charged sphere. But do the calculation.

And why is it always "It's a paradox! It's a paradox!" and never "here is something I don't understand!" And if you can't do the calculation (not saying you can't, but you haven't shown it), isn't that a good reason to be more along the lines of "don't understand" and less along the lines of "paradox!"

 

[Gene Naden]

Does it require a calculation to show that an accelerated charge radiates?

Sorry if you don't like the word "paradox." For sure, I don't have the resolution to this, so if you prefer the phrase "I don't undrestand" then fine, I don't understand.

 

[Demystifier]

This paradox may have come from Feynman's Lectures on Physics, or I may have dreamed it up myself. I am not sure. It has been around for a while and if you have already seen it, I apologize. I am not aware of any resolution.

An electron is at rest in a gravitational field. We know from experience that it doesn't radiate. But according to the Principle of Equivalence, it is equivalent to an electron undergoing acceleration. So it should radiate. Does it radiate or not?

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.

 

[stevendaryl]

It might help to look at a heuristic argument for why accelerating charges radiate. Here's a picture of a scenario:

A charged particle is initially at rest at point $A$. Naturally, the electric field points radially away from that point, as is shown by the black lines. At some point, the charge quickly accelerates and moves to point $B$ and then again comes to rest. Afterward, the electric field lines once again point radially away from the location of the charge, as is shown by the red lines.

But because information travels at lightspeed, it takes time for the rest of the universe to learn that the charge has moved. The information spreads out in a spherical shell centered on the point where the state of motion changed. So at a later time, there will be an inner region where the electric field lines point to the new position of the particle, and an outer region where the electric field lines point to the old position, and a transition region where the field lines rapidly change from one to the other. This transition region is a spherical shell of rapidly changing electromagnetic fields that spreads out at the speed of light. That's a pulse of radiation.

So heuristically, electromagnetic radiation in the case of an accelerating charge can be thought of as a "correction" to the static electromagnetic field.

According to this heuristic, if there is a frame in which the situation is static (unchanging with time) then there will be no radiation (at least, none that is observable in that frame). Conversely, if there is no frame in which things are static, then there will be radiation observable in every frame.

The heuristic would tell you two things. (I don't know enough to say whether the heuristic holds up in these cases, but if it's wrong, I think understanding why it's wrong would be illustrative. So I'm covering my bases here: I'm either right, or it's a "teaching moment")

1. In a static gravitational field, a particle at rest will not radiate.
2. A particle undergoing "Rindler motion" (constant proper acceleration) will not radiate.

In both cases, even though the particle is accelerating, the long-range fields (both gravitational and electromagnetic) are unchanging, so there is no radiation.

A big caveat is the possibility that radiation could be present in some frames and not others. In the Rindler case, this is possible if the radiation only appears in the region of spacetime that is inaccessible to accelerated (Rindler) observers. (Discussed here: https://pdfs.semanticscholar.org/2963/958637b374307b495b94b96d7afe1b0e1372.pdf)

Imagine a charge at rest

 

[Vanadium 50]

Does it require a calculation to show that an accelerated charge radiates?

An object at rest is not accelerating.

 

[Demystifier]

In fact, the expression "charge radiates" is rather misleading. This expression suggests a picture in which a charge emits a bit of radiation, after which this radiation propagates freely, independently of the charge. But that picture is wrong. In reality, a charge creates an EM field (not yet a radiation), but then this EM field back-reacts on the charge, after which the charge creates a new field that gets superimposed with the previous one, etc. The full electrodynamics is a non-linear theory. Hence the radiation is not created only be the charge. It is created by a spatially extended non-linear system consisting of the charge and the EM field around it. Since it is a spatially extended system, the additional effect of gravity cannot be fully described by the local equivalence principle.

For details see e.g. http://lanl.arxiv.org/abs/gr-qc/9610053 .

 

[Demystifier]

An object at rest is not accelerating.

How about an object at rest (e.g. on Earth) in a static gravitational field? It accelerates, in the sense that it does not move along a geodesic. The 0-component of proper acceleration $a^{\mu}=u^{\nu}\nabla_{\nu}u^{\mu}$ does not vanish.

 

[Dale]

Does it require a calculation to show that an accelerated charge radiates?

It requires at least a clear description of the experiments being compared in order to determine if/how the equivalence principle even applies.

The equivalence principle is quite specific on how it works. It says that apples accelerating uniformly in the absence of gravity are equivalent to apples at rest in uniform gravity. It does not say that apples in gravity are equivalent to oranges accelerating uniformly. And that is what you are trying to claim.

 

[A.T.]

It might help to look at a heuristic argument for why accelerating charges radiate. Here's a picture of a scenario:
View attachment 227493

1. In a static gravitational field, a particle at rest will not radiate.
2. A particle undergoing "Rindler motion" (constant proper acceleration) will not radiate.

In both cases, even though the particle is accelerating, the long-range fields (both gravitational and electromagnetic) are unchanging, so there is no radiation.

In that context it might be helpful to use images where the acceleration is constant (b), not changing (a):

From: https://arxiv.org/abs/1503.01150

 

[Gene Naden]

Some folks have asked what I am comparing. I am comparing a charged particle sitting in the Earth frame, apparently not radiating, with a charged particle undergoing a change in velocity, which does radiate. What is ambiguous about these pictures? The "experimental setup" for the first is simply any particle whose momentum and energy in the Earth frame does not change. The "experimental setup" for the second is any particle undergoing acceleration, such as Bremstrahlung or synchroton radiation.

Some folks have asked that I "do the calculation." Would that be a calculation from GR or from classical E&M? The calculation from GR is beyond me. The calculations from classical E&M are in Jackson's textbook I am sure.

 

[A.T.]

...a charged particle undergoing a change in velocity, which does radiate...

Does it radiate, if its proper acceleration is constant? If you observe it from its rest frame, you will see a static E-field, with curved field lines (see figure b above). Same as with a charged particle hovering in an uniform gravitational field, in its rest frame.

 

[Dale]

I am comparing a charged particle sitting in the Earth frame, apparently not radiating

What measuring device are you using to determine that it is not radiating? How is that measuring device positioned relative to the charge?

a charged particle undergoing a change in velocity, which does radiate.

What measuring device are you using to determine that it is radiating? How is that measuring device positioned relative to the charge?

I can't believe that it is this difficult for you to answer this question. This is now the 23rd post in the thread, and this information was asked of you in the 2nd post!

 

[Gene Naden]

Well for a collection of charged particles, you could use an antenna to pick up the radiation, connected to an amplifier. For a single particle, you would use a photomultiplier tube. Both would be positioned anywhere around the charged particle. It doesn't matter what angle, perhaps perpendicular to the direction of acceleration or the direction of the gravitational field.

 

[Dale]

Well for a collection of charged particles, you could use an antenna to pick up the radiation, connected to an amplifier.

OK, good, let's go with that so that we can avoid the quantum mechanics issues raised above.

It doesn't matter what angle, perhaps perpendicular to the direction of acceleration or the direction of the gravitational field.

Excellent. In the case of gravity is the antenna in free fall or is it also at rest? In the case of the accelerating charge is it inertial or is it also accelerating with the charge?

 

[Gene Naden]

For gravity, it is at rest relative to the Earth. For an accelerating charge, it is also at rest relative to the Earth. I think there is an assumption here that if radiation is seen in any reference frame then it is seen in all reference frames. Is this reasonable?

 

[A.T.]

For gravity, it is at rest relative to the Earth. For an accelerating charge, it is also at rest relative to the Earth.

If the antenna is at rest relative to the charge in one case, then it also must be at rest relative to the charge in the other case, to have equivalence.

 

[Dale]

For an accelerating charge, it is also at rest relative to the Earth.

The equivalence principle relates behavior at rest in gravity to behavior accelerating in the absence of gravity. So there cannot be any Earth in the accelerating case. That is why the motion must be specified relative to the charge.

I think there is an assumption here that if radiation is seen in any reference frame then it is seen in all reference frames. Is this reasonable?

Not without specifying how the radiation is seen. It matters substantially.

In the gravitational case, if the antenna is at rest it detects no radiation, but if the antenna is in free fall it detects radiation. In the accelerating case, if the antenna is co-accelerating then it detects no radiation, but if the antenna is inertial it detects radiation.

 

[Gene Naden]

In the gravitational case, if the antenna is at rest it detects no radiation, but if the antenna is in free fall it detects radiation.

What happens as time goes on? If the antenna is at rest then there is no problem - the situation can go on forever. If the antenna is in free fall and it detects radiation, can this situation go on forever? Wont the charged matter run out of energy after a while?

In the accelerating case, if the antenna is co-accelerating then it detects no radiation, but if the antenna is inertial it detects radiation.

Same question, almost. If the antenna is co-accelerating, then the situation can go on forever. If the antenna is inertial, can this situation go on forever? Won't the charged matter run out of energy after a while?

I think there is an assumption here that if radiation is seen in any reference frame then it is seen in all reference frames. Is this reasonable?

I asked if this was reasonable. I now claim that it is in fact reasonable.

 

[A.T.]

If the antenna is inertial, can this situation go on forever? Won't the charged matter run out of energy after a while?

Since the charged matter is accelerating in the inertial frame of the antenna, something must be doing work on the charged matter in that frame.

 

[Dale]

What happens as time goes on? If the antenna is at rest then there is no problem - the situation can go on forever. If the antenna is in free fall and it detects radiation, can this situation go on forever? Wont the charged matter run out of energy after a while?

The equivalence principle does not apply forever. It only applies for a small region of space and time where the spacetime curvature is negligible. The curvature is not negligible forever.

If the antenna is co-accelerating, then the situation can go on forever. If the antenna is inertial, can this situation go on forever? Won't the charged matter run out of energy after a while?

No, if the charge is being accelerated forever then it is being given an infinite amount of energy. However, the amount of energy captured by the antenna is not that large, although that is a calculation that I will leave to the interested reader.

I asked if this was reasonable. I now claim that it is in fact reasonable.

I am not sure how you came to the sudden conclusion that a false claim is reasonable: https://en.wikipedia.org/wiki/Unruh_effect

But the important question in terms of the equivalence principle is not whether or not radiation exists in the frame, but whether or not the detector detects it.

 

[Gene Naden]

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.

 

[Dale]

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.

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

 

[Sorcerer]

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.

This may be relevant

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

 

[Gene Naden]

Yeah it looks relevant

 

[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.