# Equivalence principle implies uniformly accelerated charge doesn't radiate?

• johne1618
In summary, the conversation discusses whether a uniformly accelerated charged particle radiates electromagnetic waves. The equivalence principle is brought up, stating that a particle in uniform acceleration is equivalent to a particle at rest in a gravitational field. However, the OP argues that this principle does not apply to radiation due to its nonlocal nature. The paper by Richard Feynman is also mentioned, which argues that the usual formula for radiation by an accelerating charge is incorrect. However, the validity of this argument is disputed by other papers, including one that argues the equivalence principle applies locally in curved spacetime. Ultimately, more research and discussion is needed to fully understand this paradox.
johne1618
Does a uniformly accelerated charged particle radiate em waves?

The equivalence principle says that a particle in uniform acceleration is equivalent to a particle at rest in a gravitational field.

A particle at rest in a gravitational field is clearly not going to radiate em waves therefore by the equivalence principle a uniformly accelerated charge won't either.

Is that right?

johne1618 said:
Does a uniformly accelerated charged particle radiate em waves?

Yes.

johne1618 said:
The equivalence principle says that a particle in uniform acceleration is equivalent to a particle at rest in a gravitational field.

False. The equivalence principle explains how to distinguish between a particle in a gravitational field and one in uniform acceleration - that is, a gravitational field will have tides.

johne1618 said:
A particle at rest in a gravitational field is clearly not going to radiate em waves therefore by the equivalence principle a uniformly accelerated charge won't either.

Is that right?

No. You have a history of pushing unconventional viewpoints. I sincerely hope this is not another attempt.

No, an accelerating charge does radiate. The equivalence principle applies only locally. Radiation is inherently nonlocal, so the equivalence principle does not apply.

http://arxiv.org/abs/0806.0464

This paper seems to say that a charge in free-fall in a uniform gravitational field (ie. accelerating) does not radiate:

Ashok K. Singal said:
4. CONCLUSIONS
We have shown that from the strong principle of equivalence the electric field of a freely falling charge in a static, uniform gravitational field would appear to fall along with the charge, remaining everywhere in a radial direction from the instantaneous position of the charge. Accordingly there will be no transverse fields (radiation!) from a freely falling charge in such a gravitational field. It is further shown that in the case of a charge supported in such a gravitational frame, the electric field energy, as measured by freely falling observers instantaneously at rest with respect to the charge, is equal to the Coulomb field energy of a charge permanently stationary in an inertial frame. It follows that in neither of the two cases will there be any electromagnetic radiation.

The OP also seems to have Richard Feynman on his side. This is from a paper "Does a Uniformly Accelerating charge radiate?":

Whether these solutions are realistic or not is an open question, but the predicted absence of radiation for hyperbolic motion has sometimes been cited as a way of reconciling the Equivalence Principle with the fact that a charged particle held stationary in a gravitational field (and therefore undergoing constant proper acceleration) does not radiate. For example, in Feynman's "Lectures on Gravitation" he says "we have inherited a prejudice that an accelerating charge should radiate", and then he goes on to argue that the usual formula giving the power radiated by an accelerating charge as proportional to the square of the acceleration "has led us astray" because it applies only to cyclic or bounded motions. ...​

Thus the radiation reaction force (and therefore the radiated power) is proportional to the third derivative of position, so if the particle is undergoing constant acceleration it does not radiate (according to this formula).​

AM

Andrew Mason said:
This paper seems to say that a charge in free-fall in a uniform gravitational field (ie. accelerating) does not radiate:
The OP also seems to have Richard Feynman on his side. This is from a paper "Does a Uniformly Accelerating charge radiate?":

Whether these solutions are realistic or not is an open question, but the predicted absence of radiation for hyperbolic motion has sometimes been cited as a way of reconciling the Equivalence Principle with the fact that a charged particle held stationary in a gravitational field (and therefore undergoing constant proper acceleration) does not radiate. For example, in Feynman's "Lectures on Gravitation" he says "we have inherited a prejudice that an accelerating charge should radiate", and then he goes on to argue that the usual formula giving the power radiated by an accelerating charge as proportional to the square of the acceleration "has led us astray" because it applies only to cyclic or bounded motions. ...​

Thus the radiation reaction force (and therefore the radiated power) is proportional to the third derivative of position, so if the particle is undergoing constant acceleration it does not radiate (according to this formula).​

AM

Feynman was wrong on this point. (Try references 6 & 7 of http://arxiv.org/abs/0806.0464)

I think those references are not free, maybe try http://arxiv.org/abs/physics/0506049. I think this paper is technically right, but I prefer the resolution that the equivalence principle does not apply due to the nonlocal nature of radiation, rather than that a comoving observer does not measure radiation (but that might be splitting hairs - I agree that in flat spacetime a co-moving observer does not measure radiation.) The reason I prefer the non-locality resolution is that the equivalence principle should apply locally in curved spacetime. In that case, the charge is not "freely fallling" because in addition to the gravitational field, it is acted on by its own field. A similar point of view is expressed in http://arxiv.org/abs/1102.0529 (p144): "In the scalar and electromagnetic cases, the picture of a particle interacting with a free radiation field removes any tension between the nongeodesic motion of the charge and the principle of equivalence.", or http://arxiv.org/abs/0806.0464 (p2): "The principle of equivalence has a local character. The mentioned equivalence is only valid as far as the measurements does not reveal a possible curvature of space. So, if there is a non-local interaction between the charge and the curvature of spacetime, due to the non-local character of its electromagnetic ﬁeld, this may modify the equation of motion of the charge."

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Yes.
False. The equivalence principle explains how to distinguish between a particle in a gravitational field and one in uniform acceleration - that is, a gravitational field will have tides.
No. You have a history of pushing unconventional viewpoints. I sincerely hope this is not another attempt.

I have not followed this OPs other activity, but it sure seems like, for this particular question, he deserves more than a one word answer followed by chiding.This is a very famous paradox, and had been called a "perpetual problem".

Quoting from Alemida (attachment):

radiation emitted by a uniformly accelerated charge has received considerable attention. Eminent figures
such as Pauli, Born, Sommerfeld, Schott, von Laue, and many others have contributed to this debate with different
answers. The relevant questions we consider are: Does a uniformly accelerated charge actually radiate? In a constant
gravitational field should free-falling observers detect any radiation emitted by free-falling charges? Is the equivalence
principle valid for such situations?"I have posted some papers:

Boulware: Annals of Physics 124, 169-188 (1980) - Probably the most well known paper on this subject.
Singal: General Relativity and Gravitation, Vol. 27 No. 9, 1995 - Argues that no radiation occurs (mentioned by Andrew Mason).
Alemida, arXiv:physics/0506049v5, 2005 - fairly recent, includes some history of the debate.
Ch52: http://rickbradford.co.uk/ChoiceCutsCh52.pdf - This is a very good survey of work on this topic, from Born in 1909 through Feynman in 1999.

The consensus right now is that uniformly accelerating charges probably radiate, but not everyone can see the radiation.

The various papers on this topic make it clear that part of the problem is in definition.

For example, what do we mean by "radiate".

According to the Boulware paper, and further in Alemida, a co-moving observer will not see any radiation from a uniformly accelerating charge,
even though the radiation exists, because it is outside of his event horizon and unobservable to him.

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• almeida.pdf
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feynman said that radiation is proportional to(dx/dt)(d3x/dt3) but it can be written as (d2x/dt2)2- d/dt(dx/dt d2x/dt2) and the last term vanishes for periodic motions so it is also written only with first term present which applies only to periodic motion.In fact a very careful treatment is needed which is given in book of schwinger's 'classical electrodynamics'.

john,
this is notoriously difficult question. Although it is easy-to-state, nobody has provided equally easy-to-understand answer. There are these long papers published for decades about it and still no revelation.

I think this has happened because the question addressed is formulated in rather unclear concepts. It uses expressions such as "accelerated charge radiates", "accelerated charge is equivalent to stationary one", which are too vague.

Acceleration is a relative concept - with respect to what should the charge accelerate?

Production of radiation can mean very different things to different people - does it mean that the particle blows energy away? That the flux of the Poynting vector through the surface in large distance distance is positive? That the EM field in the vicinity of the particle is time-dependent? That the EM field is given by the retarded solution of Maxwell's equations ? That the latter contains non-zero 1/r component? That photons are produced?

Also, it is important to keep in mind that the EM field is relative concept as well.

What is meant by the word "particle in sit. A is equivalent to the particle in sit. B"? That the trajectory of the particle is the same? Or that the state of the fields around the particle is the same?

I can see many combinations of choices which lead to many different questions with possibly different answers; unless the question is made clear, there is no hope for useful answer.

I am sure that if you propose clear formulation of the question, it will meet will much better reception and we may even find some good answer.

Perhaps it depends on how the charge is accelerated. If it is accelerated by gravity, there is no interaction of electrical fields. If the charge is accelerated by electrical force, there is an interaction of electrical fields. In the first case there may be no radiation but in the latter there would be.

In other words, radiation is not caused by the charge's acceleration. Rather it is a consequence of an electrical interaction, which necessarily also causes the charge to accelerate. If we could find a particle that possessed charge but no mass, we might be able to determine this experimentally. But it appears that charge and mass cannot be separated.

From what I can gather, the contention that an electron radiates when accelerated is based on a theory that the electron interacts with its own field when accelerated. Feynman appears to have taken different positions on that. He may have ended up believing that it all depends on how you want to look at it, suggesting that there may be more than one 'correct' equivalent explanation. He talks about this in his Nobel lecture, which is well worth reading.

AM

there is no confusion with it.Accelerating charges radiate.In case of discrepancy,stick to larmor law(generalised)

andrien said:
there is no confusion with it.Accelerating charges radiate.In case of discrepancy,stick to larmor law(generalised)

You figured this out yourself?

@atyy
Your references do not address the OPs question. Your references address radiation from a charge that is in orbit around a gravitating body. As Jano L pointed out, there are a multitude of scholarly works addressing exactly the OPs question (I have posted some). Wouldn't it make sense to begin the discussion with those?

the_emi_guy said:
@atyy
Your references do not address the OPs question. Your references address radiation from a charge that is in orbit around a gravitating body. As Jano L pointed out, there are a multitude of scholarly works addressing exactly the OPs question (I have posted some). Wouldn't it make sense to begin the discussion with those?

Did you see my post #5?

I have to agree with the_emi_guy and Janos L. There has been a long-standing debate among eminent physicists about whether a uniformly accelerated charge (ie a charge in gravitational orbit) radiates. I don't think we are going to resolve this debate here.
Without the right experimental data there is going to remain a controversy.

Remarkably, it appears that this is still an unresolved problem in physics. Tying gravity into the rest of physics remains an intractable problem.

AM

Andrew, I disagree. Whether an accelerated charge radiates is settled. This is section 14.3 of Jackson (2nd). An accelerated charge doesn't know that its acceleration is or is not gravitational, so it can't "decide" whether or not to radiate, as you suggest in Post #9.

A full GR treatment agrees, insofar as "radiation" is well-defined. (Observers, especially nearby accelerating ones, may disagree about what is near-field and what is far-field, but that's a quibble.) Where people get confused is when they try and find a shortcut or simplification, like the initial misunderstanding of the equivalence principle.

Andrew, I disagree. Whether an accelerated charge radiates is settled. This is section 14.3 of Jackson (2nd). An accelerated charge doesn't know that its acceleration is or is not gravitational, so it can't "decide" whether or not to radiate, as you suggest in Post #9.
I don't think a charge has to "know" anything. But surely there is a difference between a charge interacting with an electrical field and interaction with a gravitational field. Why should they be the same?

A full GR treatment agrees, insofar as "radiation" is well-defined. (Observers, especially nearby accelerating ones, may disagree about what is near-field and what is far-field, but that's a quibble.) Where people get confused is when they try and find a shortcut or simplification, like the initial misunderstanding of the equivalence principle.
Perhaps you can explain to me, then, how the charge in circular gravitational orbit radiates without losing energy. One of the papers stated: "To explain this puzzle, we need to recognize that the concept of radiation has no absolute meaning, and that it depends both on the radiation field and the state of motion of the observer." While that may well be true, I don't see how radiation can exist without some kind of loss of energy.

It is a very difficult experiment to do, so it may be a while before anyone can experimental data to test the theory. So it may be unresolved for some time.

AM

Don't we have direct evidence of orbiting charges radiating from black hole X-Ray radiation? The metric is the same, so I don't see why it should be much different on Earth's orbit, other than in magnitude, of course.

And surely, for static charge in gravitational field, it cannot be that hard to solve Maxwell's equations with four-current being just j0 using covariant derivatives and Shwarzschild metric.

You figured this out yourself?
you need to see schwinger's book 'classical electrodynamics' for this.He shows there that if a charge is accelerating and you don't care about the starting of motion and other factors,then you will find that accelerating charges don't radiate as opposed to larmor's formula.

K^2 said:
Don't we have direct evidence of orbiting charges radiating from black hole X-Ray radiation?

Could supply a scholarly reference to this radiation that connects it to our classic problem?
K^2 said:
And surely, for static charge in gravitational field, it cannot be that hard to solve Maxwell's equations with four-current being just j0 using covariant derivatives and Shwarzschild metric.

If it is easy then it surely has been done. Can you supply a reference?

andrien said:
you need to see schwinger's book 'classical electrodynamics' for this.He shows there that if a charge is accelerating and you don't care about the starting of motion and other factors,then you will find that accelerating charges don't radiate as opposed to larmor's formula.

You stated in post #10 that accelerating charges do radiate (and supplied no references).
Here you are referring to Schwinger's book to say they do not radiate??

the_emi_guy said:
You stated in post #10 that accelerating charges do radiate (and supplied no references).
Here you are referring to Schwinger's book to say they do not radiate??
I am saying they do radiate but one has to be careful.it is shown there that one must take into account the beginning and ending of acceleration.
we maysay that a uniformly accelerated charge radiates because it is not uniformly accelerated
page 391-395

K^2 said:
Don't we have direct evidence of orbiting charges radiating from black hole X-Ray radiation? The metric is the same, so I don't see why it should be much different on Earth's orbit, other than in magnitude, of course.

I have attached a paper on Hawking-Unruh radiation which includes a discussion on its possible use in resolving our paradox.

To me it shows how non-trivial this problem is that we are dipping into quantum mechanics in order to explain it.

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andrien said:
you need to see schwinger's book 'classical electrodynamics' for this.He shows there that if a charge is accelerating and you don't care about the starting of motion and other factors,then you will find that accelerating charges don't radiate as opposed to larmor's formula.

I haven't seen Schwinger's paper, but Eriksen and Gron in the second of their series of articles say that Schwinger considered constant eternal coordinate acceleration, which is not possible in special relativity because it leads to superluminal motion. Eriksen and Gron do consider constant eternal proper acceleration, and find that the charge does radiate.

Later in their series they reach the same technical conclusion regarding the observability of the radiation as the Almeida and Saa article that the_emi_guy (#6) and I (#5) posted.

My view is that there is no paradox in the first place, since it arises from an abuse of the equivalence principle (which applies locally in curved spacetime). Nonetheless, the question is interesting as a classical special relativistic electrodynamics problem.

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atyy said:
I haven't seen Schwinger's paper, but Eriksen and Gron in the second of their series of articles say that Schwinger considered constant eternal coordinate acceleration, which is not possible in special relativity because it leads to superluminal motion. Eriksen and Gron do consider constant eternal proper acceleration, and find that the charge does radiate.

Later in their series they reach the same technical conclusion regarding the observability of the radiation as the Almeida and Saa article that the_emi_guy (#6) and I (#5) posted.
see post# 20.

andrien said:
I am saying they do radiate but one has to be careful.it is shown there that one must take into account the beginning and ending of acceleration.

page 391-395

This is not the problem that we are discussing. We are not discussing a particle that is at rest and then begins accelerating, this is a change in acceleration (from 0 to some positive value).

the_emi_guy said:
This is not the problem that we are discussing. We are not discussing a particle that is at rest and then begins accelerating, this is a change in acceleration (from 0 to some positive value).
it is always there,and one should carefully include it for not getting non-sense results.

K^2 said:
Don't we have direct evidence of orbiting charges radiating from black hole X-Ray radiation? The metric is the same, so I don't see why it should be much different on Earth's orbit, other than in magnitude, of course.

And surely, for static charge in gravitational field, it cannot be that hard to solve Maxwell's equations with four-current being just j0 using covariant derivatives and Shwarzschild metric.
You need a time dependent electric field or time dependent magnetic field and that seems to depend on what reference frame you are in. If the equivalence principle means anything, there is neither in the reference frame of the charge.

AM

Andrew Mason said:
Perhaps you can explain to me, then, how the charge in circular gravitational orbit radiates without losing energy.

It radiates and loses energy. Its path is not geodesic. It is not in free fall because it is acted on not only by the gravitational field, but also its own electromagnetic field. The equivalence principle does not apply because radiation is non-local. In GR, curved spacetime is only locally flat. If you look at second derivatives of the metric, then you will detect curvature which cannot be transformed away, and the equivalence principle does not apply.

In GR, when we write the equivalence principle, roughly speaking we have to use the form of the laws of physics in which only first derivatives or lower appear. This is because higher derivatives are more "non-local" in the sense that each time you take a derivative, you take a difference of values at different points. Technically, the derivatives all exist at a point after taking the limit, but this is the idea behind higher derivatives being considered "non-local".

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I don't think a charge has to "know" anything. But surely there is a difference between a charge interacting with an electrical field and interaction with a gravitational field. Why should they be the same?

I agree with Andrew, the situations are not the same. However, the circular motion has the additional complication - the radial field cannot be transformed away.

I think better example is the fall of a charged sphere due to gravity in the vicinity of the Earth surface, where the gravitational field is almost uniform. Consider two frames:

1) In the frame of the sphere, there is no gravitational force. The principle of equivalence and experience leads us to believe that both the motion of the sphere is uniform and its fields are given by the standard retarded solution of the Maxwell equations. That is, the sphere is at rest and the EM field in its vicinity is static.

2) In the frame of the Earth, there is gravitational force. The sphere accelerates, but this motion is referred to non-inertial frame of reference. Therefore there is no easy way to find the EM fields of the sphere. The only clue we have is to go back to the situation 1), find the field there and get back to 2) by coordinate transformation.

Now, since the field in 1) is frozen into the sphere, it seems very probable that the "free fall transformation" will make it into EM field which is comoving with the sphere.

Of course, the field will be time-dependent, as the charge moves. But I think we should not call it radiation. This is because in ordinary usage of that word, one has in mind some sort of propagation of phase or signal from the source to infinity, and in this case, there is such propagation. If the fall happens in the z- direction, in the planes xy no propagation happens. In the direction z, there is propagation of the sphere, but this is motion with slow velocity, not an EM wave.

atyy said:
It radiates and loses energy. Its path is not geodesic. It is not in free fall because it is acted on not only by the gravitational field, but also its own electromagnetic field.
So you appear to be saying that there is a real difference between a charge in circular gravitational orbit and a charge at rest in an inertial frame of reference. If that is the case, the same would apply to an electric dipole would it not? How would the principle of equivalence ever apply?

The equivalence principle does not apply because radiation is non-local. In GR, curved spacetime is only locally flat.
Why is radiation non-local? Is there something preventing it from being absorbed locally?

AM

Andrew Mason said:
So you appear to be saying that there is a real difference between a charge in circular gravitational orbit and a charge at rest in an inertial frame of reference. If that is the case, the same would apply to an electric dipole would it not? How would the principle of equivalence ever apply?

Yes, there is a real difference. I'm not sure about the electric dipole specifically, but as a rule of thumb one cannot naively apply the equivalence principle to charged particle trajectories. For example, one might be inclined to think about a "free falling" charge - but generally a charge cannot be free falling because it is acted on by the field it produces. Also, the equivalence principle fails in non-local situations. How much non-locality can the EP tolerate? A rule of thumb is the second derivatives are too non-local for the EP to apply. One can heuristically see this by noting that at a point in curved spacetime, it is possible to make the metric and its first derivatives look the same as in flat spacetime, but if spacetime is curved the second derivatives cannot be transformed away. Thus the EP holds at all points in curved spacetime (if one looks at the metric and first derivatives), and it also fails at all points in curved spacetime (if one looks at second derivatives and higher). Roughly, higher derivatives are considered more non-local than lower derivatives, because they involve more differences of quantities at different spacetime points.

Andrew Mason said:
Why is radiation non-local? Is there something preventing it from being absorbed locally?

Radiation is non-local because it involves second derivatives. Also, the boundary conditions are important in solving such problems - roughly, one cannot solve a problem involving radiation by confining the charge to a small cabin and not looking outside.

So how does one apply the EP to charged particles? In the Lagrangian for all known charged particles, there are at most first derivatives coupled to the metric. So one can apply the EP to the Lagrangian by asserting that the form of the Lagrangian in curved spacetime is the "generally covariant" form of the Lagrangian in flat spacetime. This is known as the "minimal coupling" prescription.

There's a discussion about similar issues in section 24.7 of Blandford and Thorne http://www.pma.caltech.edu/Courses/ph136/yr2006/text.html.

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K^2 said:
Don't we have direct evidence of orbiting charges radiating from black hole X-Ray radiation? The metric is the same, so I don't see why it should be much different on Earth's orbit, other than in magnitude, of course.
This is thermal radiation. Gravity just helps to compress and heat stuff, it does not generate the radiation itself.

atyy said:
Yes, there is a real difference. I'm not sure about the electric dipole specifically, but as a rule of thumb one cannot naively apply the equivalence principle to charged particle trajectories. For example, one might be inclined to think about a "free falling" charge - but generally a charge cannot be free falling because it is acted on by the field it produces.
Does that mean that a charge in free-fall will always have a non-symmetric electric field - right down to the smallest distances? That seems to be something that is testable. It seems also to violate the principle of equivalence because it shrinks the meaning of 'local' to an arbitrarily small region.

I think the dipole is important because one could always create an electric dipole by separating + and - charges in a neutral body. Many molecules (eg. water) have an electric dipole moment. If a dipole in free-fall radiates, then a great deal of matter in freefall must radiate. Mind you, there is likely a quantum mechanical threshold for radiation for an electric dipole ie. quantum harmonic oscillator.

AM

Andrew Mason said:
Does that mean that a charge in free-fall will always have a non-symmetric electric field - right down to the smallest distances? That seems to be something that is testable. It seems also to violate the principle of equivalence because it shrinks the meaning of 'local' to an arbitrarily small region.

Yes, in principle the EP applies only to an arbitrarily small region - a point, and even then only as long as one does not look at spacetime curvature.

Andrew Mason said:
I think the dipole is important because one could always create an electric dipole by separating + and - charges in a neutral body. Many molecules (eg. water) have an electric dipole moment. If a dipole in free-fall radiates, then a great deal of matter in freefall must radiate. Mind you, there is likely a quantum mechanical threshold for radiation for an electric dipole ie. quantum harmonic oscillator.

I don't know the quantitative answer, but here's my guess. Let's first ignore gravity and the equivalence principle, and do classical electrodynamics in flat spacetime. If we put water in a car and accelerate it, we generally do not detect radiation, so water can be treated as effectively neutral and classical at our level of experimental accuracy, ie. we don't worry about water violating classical electrodynamics even though we don't detect radiation when we accelerate it. For this reason, at the same level of accuracy for which water in a car is considered neutral, we also don't expect to detect any violation of the equivalence principle when water is accelerated by a gravitational field.

But at some level, one might think the non-neutrality of water at small scales does come into play. I don't know the answer for what one might expect then. For example, is it a limiting factor in our ability to do extremely precise equivalence principle tests?

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atyy said:
I don't know the quantitative answer, but here's my guess. Let's first ignore gravity and the equivalence principle, and do classical electrodynamics in flat spacetime. If we put water in a car and accelerate it, we generally do not detect radiation, so water can be treated as effectively neutral and classical at our level of experimental accuracy, ie. we don't worry about water violating classical electrodynamics even though we don't detect radiation when we accelerate it. For this reason, at the same level of accuracy for which water in a car is considered neutral, we also don't expect to detect any violation of the equivalence principle when water is accelerated by a gravitational field.
I think you would have to reduce the water temperature to close to absolute 0 as well. Otherwise the thermal radiation would predominate. So it would be a pretty difficult thing to test.

AM

## What is the Equivalence Principle?

The Equivalence Principle is a fundamental concept in physics that states that the effects of gravity are equivalent to the effects of acceleration. This means that an observer in a uniform gravitational field cannot distinguish between being at rest in that field or being in an accelerated frame of reference.

## How does the Equivalence Principle relate to accelerated charge?

The Equivalence Principle also applies to accelerated charge. This means that an observer in a uniform electric field cannot distinguish between being at rest in that field or being in an accelerated frame of reference. In other words, an accelerated charge will not radiate electromagnetic waves in the same way that a stationary charge would.

## Why doesn't a uniformly accelerated charge radiate?

According to the Equivalence Principle, an accelerated charge cannot be distinguished from a stationary charge in a uniform electric field. This means that the charge will not experience any acceleration relative to its own frame of reference, and therefore will not radiate electromagnetic waves.

## Is the Equivalence Principle proven?

The Equivalence Principle has been extensively tested and has been found to hold true in all of the experiments conducted so far. However, it is still considered a fundamental principle and is not yet fully understood. Further research and experiments are still being conducted to better understand this concept.

## How does the Equivalence Principle impact our understanding of gravity?

The Equivalence Principle plays a crucial role in our understanding of gravity. It allows us to treat gravity as a geometric property of spacetime, rather than a force. This has led to the development of Einstein's theory of General Relativity, which has been incredibly successful in explaining the behavior of gravity on a large scale.

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