If a charged particle is free falling in the gravitational field-

[magnetar]

If a "charged particle" is free falling in the gravitational field----------

If a "charged particle" is free falling in the earth`s gravitational field, it will emit electromagnetic waves?

(1)If a person stands on the ground, he will say:Yes ! because,the charged particle is accelerated by the gravitational field!

(2)If another person is free falling along with that particle,he will say:No! because according to the "equivalent principle" he and the particle have the same amount of acceleration, from his point of view,the particle has no acceleration,it can not emit electromagnetic waves!

(3)Paradox! I want to ask: who is right?

. Thank you

 

[Demystifier]

Acceleration is not relative, it is absolute. A free fall is actually a motion along a geodesic, which is locally the same as motion with uniform velocity without presence of the gravitational field. An analogous question is: Will a charge at rest (without a gravitational force) radiate from the point of view of an accelerated observer? The answer is - no.

 

[pervect]

This problem has created a surprising amount of controversy. One paper that I rather like on the topic is http://xxx.lanl.gov/abs/gr-qc/0006037. As the authors note

The problem of radiation of a uniformly accelerated charged particle has an interesting and controversial history

and this one paper shouldn't necessarily be regarded as definitive. A key issue seems to be what one means when one says "emits electromagnetic waves" or "radiates".

 

[Haelfix]

Yea this is a complicated question. The resolution is still controversial, but people generally agree that the paradox dissappears.

The first thing that helps is to forget ones intuition about what radiation is and instead focus on say, its effects. Namely does an accelerating particle heat up say, a glass of water? When properly formulated, a stationary observer and say someone in an elevator passing by, will both agree that this happens.

The root of the nastiness and the cause of the paradox is that radiation is defined with respect to moving a test particle away to infinity. So there is implicitly a preferred inertial frame where 'classical radiation' is defined with respect to a far field. This definition is slightly problematic b/c you can see its somewhat clashy with relativity. Making it mesh, is the technical problem people clash about.

 

[Stingray]

Regardless of how you define radiation, a freely-falling charge does not necessarily move on a geodesic (even in the absence of other charges). It violates the equivalence principle by virtue of being inseparable from its electromagnetic field. This makes the particle act as though it had a large radius no matter how small it actually is. The equivalence principle only holds (as far as it goes) for experiments which can only sample an infinitesimally small region of space.

 

[masudr]

Remember: a charged particle in an EM field is not freely falling.

 

[Garth]

Does a supported electron sitting stationary on a laboratory bench radiate?

According to the equivalence principle it is accelerating at g upwards and therefore according to Maxwell's equations it should radiate, however, if so, where is the power for this radiation coming from?

Garth

 

[robphy]

This question has been bothering me for while... especially when I have received convincing-at-the-time answers from knowledgeable professors which I now cannot reconstruct [or even recall whether it's yes or no, let alone any of the caveats].

As pervect's reply suggests, one must first find a clear and precise mathematical formulation of the problem [intrepreting the physics mathematically], prove a resulting mathematical theorem, then interpret the mathematics physically. Assuming the mathematics is done correctly, there's nothing there to argue over. The key issues must lie in the translation to or from the mathematics... e.g.,
a particle is said to "radiate" when... <insert a mathematical condition regarding spacetime curves and tensor fields>,
the equivalence principle states... <insert a mathematical statement>.

Is there such a formulation?
(The paper referenced by pervect looks like a good starting point to review the literature... time-permitting, of course.)

Underlying my reply is the question:
Is the question being posed even well-defined?
Or does it need to be better formulated before a resolution can be obtained?
Does the question capture the underlying issues involved?

 

[MeJennifer]

An analogous question is: Will a charge at rest (without a gravitational force) radiate from the point of view of an accelerated observer? The answer is - no.

Hmm, perhaps I miss some essential element here, but, for instance an electron radiating, simply means that one or more photons are emitted, right?

So, to me that seems like a clear space-time event. So how do you conclude that that event does not exist for an accelerated observer?

To make it even more specific, what if the electron emits a photon in the direction of the observer, and assuming the case that there is no Rindler horizon here, are you saying it won't hit him since he is accelerating?

 

[fireball3004]

I'm not qualified to answer the question but it's to interesting a paradox not to. It seems to me that the very reason why it wouldn't be visible is the reason why it can exist. The during laspse in time before the radiation or effects of radiation reach you will have moved.

 

[pervect]

Hmm, perhaps I miss some essential element here, but, for instance an electron radiating, simply means that one or more photons are emitted, right?

So, to me that seems like a clear space-time event. So how do you conclude that that event does not exist for an accelerated observer?

To make it even more specific, what if the electron emits a photon in the direction of the observer, and assuming the case that there is no Rindler horizon here, are you saying it won't hit him since he is accelerating?

In the past, I've argued that particle number is not conserved under acceleration, pointing to the Unruh effect as an example. "Rindler" particles are not the same as "Minkowski" particles, in particular it turns out that Rindler particles correspond to Minkowski particles with a negative frequency. (This much is mentioned in Wald).

Thus, by this explanation, there is not any reason to expect the number of photons to agree in the accelerated and non-accelerated frame.

I see that at least one paper on the topic, by Demystifier aka Hrojve Nikolic, http://arxiv.org/abs/gr-qc/9909035, disagrees with this explanation, however.

(This might be another good paper to read with the one I first suggested (ideally, along with all of the referenced papers in both articles)

 

[magnetar]

So far ,is there any relevant "experiment test" have conducted??

 

[Demystifier]

Hmm, perhaps I miss some essential element here, but, for instance an electron radiating, simply means that one or more photons are emitted, right?

So, to me that seems like a clear space-time event. So how do you conclude that that event does not exist for an accelerated observer?

To make it even more specific, what if the electron emits a photon in the direction of the observer, and assuming the case that there is no Rindler horizon here, are you saying it won't hit him since he is accelerating?

Yes, you miss an essential element. More precisely, you do not read carefully. I was talking about a charge at rest, which certainly does not radiate from the point of view of an inertial observer. Thus it should not be surprising that it does not radiate from the point of view of an accelerated observer as well.

 

[Demystifier]

I see that at least one paper on the topic, by Demystifier aka Hrojve Nikolic, http://arxiv.org/abs/gr-qc/9909035, disagrees with this explanation, however.

(This might be another good paper to read with the one I first suggested (ideally, along with all of the referenced papers in both articles)

Thanks for mentioning it. For a more explicit critique of the Rindler particles see also
http://arxiv.org/abs/gr-qc/0103108

 

[MeJennifer]

Yes, you miss an essential element. More precisely, you do not read carefully. I was talking about a charge at rest, which certainly does not radiate from the point of view of an inertial observer. Thus it should not be surprising that it does not radiate from the point of view of an accelerated observer as well.

Now you still got me interested here.

So you say a charge at rest does not radiate, right?
But you also say that an observer who accelerates away from such a charge does not measure radiation, right?

So then I would like to know what you consider a charge at rest. At rest relative to what?

 

[Demystifier]

So then I would like to know what you consider a charge at rest. At rest relative to what?

Now you are right, I was not sufficiently precise.
At rest with respect to an inertial frame in flat spacetime. OK?

 

[Garth]

Now you are right, I was not sufficiently precise.
At rest with respect to an inertial frame in flat spacetime. OK?

And what happens with a supported charge at rest relative to the Centre of Mass in a gravitational field with 'curved' space-time?

Garth

 

[pervect]

Thanks for mentioning it. For a more explicit critique of the Rindler particles see also
http://arxiv.org/abs/gr-qc/0103108

I have a few questions:

Would you agree with Wigner's field-based defintion of a particle in flat space-time as an irreducible representation of the Poincare group?

Is your conclusion basically that we have no good defintion of a particle at all in curved space-times, or am I misreading your conclusion? (I.e. to use Wigner's defintion, no Poincare group -> no particles).

 

[pervect]

Some more comments:

As I understand it, an inertial observer will see an accelerated charge as emitting particles. A co-accelerating observer will see the same charge not as emitting particles, but as absorbing them.

I think this implies that particle number is not a tensor quantity, but I probably need to think about this more.

 

[Haelfix]

Again the key thing is that both observers will agree on the measurement of any experiment, its just the interpretation that will change (for instance one person sees 'radiation' the other comoving observer sees inductance)

Particle number is most definitely not a conserved quantity in curved space, as that only makes sense when the relevant isometry group is Poincare. Only the Smatrix is conserved, given suitably nice asymptotics and time like killing vectors

 

[Demystifier]

And what happens with a supported charge at rest relative to the Centre of Mass in a gravitational field with 'curved' space-time?

In this case, the charge does not move along a geodesic, so it radiates. The energy for radiation comes from the supporting force.

 

[Demystifier]

Some more comments:

As I understand it, an inertial observer will see an accelerated charge as emitting particles. A co-accelerating observer will see the same charge not as emitting particles, but as absorbing them.

I think this implies that particle number is not a tensor quantity, but I probably need to think about this more.

This is why I believe that the usual definition(s) of particles, including the Wigner definition, is not a physically correct definition of particles. For the alternative(s) I propose see
http://arxiv.org/abs/gr-qc/0111029
http://arxiv.org/abs/hep-th/0202204
http://arxiv.org/abs/hep-th/0205022
See also the introduction in
http://arxiv.org/abs/hep-th/0601027
and Sec. II.D of
http://arxiv.org/abs/hep-th/0702060
For an additional argument why particles could be more fundamental than fields see also
http://arxiv.org/abs/gr-qc/0611037

 

[Demystifier]

1. Would you agree with Wigner's field-based defintion of a particle in flat space-time as an irreducible representation of the Poincare group?

2. Is your conclusion basically that we have no good defintion of a particle at all in curved space-times, or am I misreading your conclusion? (I.e. to use Wigner's defintion, no Poincare group -> no particles).

1. As I said above, the answer is - no.

2. As I argue in the papers mentioned above, particles exist in curved spacetime, but their formal definition in terms of fields requires a preferred foliation of spacetime. In the last two papers above, I argue that it is actually a problem for fields, not for particles. This is why in
https://www.physicsforums.com/showthread.php?t=144746
I vote as I vote.

 

[Garth]

In this case, the charge does not move along a geodesic, so it radiates. The energy for radiation comes from the supporting force.

But if it is sitting on a laboratory bench, stationary in the Earth's gravitational field, what work is the supporting force from the bench doing? And where is the energy of radiation coming from?

Garth

 

[Demystifier]

But if it is sitting on a laboratory bench, stationary in the Earth's gravitational field, what work is the supporting force from the bench doing? And where is the energy of radiation coming from?

The work is force times the traveled path. The traveled path is the one perceived by a freely falling observer. Note, however, that everything is static from the point of view of the stationary observer. In particular, the electromagnetic field produced by the charge does not change with time, just as properties of bench do not change with time. The question is: Then what does it mean that the charge "radiates"? The answer is: the field does not fall as 1/r^2 as in the Coulomb law, but as 1/r as in the case of ordinary radiation.

 

[Garth]

The work is force times the traveled path. The traveled path is the one perceived by a freely falling observer.

So a 'laboratory bench observer' can extract this 'work', by absorbing the radiation? A free lunch?

Note, however, that everything is static from the point of view of the stationary observer. In particular, the electromagnetic field produced by the charge does not change with time, just as properties of bench do not change with time.

In what frame? Surely the field does change with time in the freely falling frame as per the Equivalence Principle?

The question is: Then what does it mean that the charge "radiates"? The answer is: the field does not fall as 1/r^2 as in the Coulomb law, but as 1/r as in the case of ordinary radiation.

But are you saying it should still radiate?

Garth

 

[Demystifier]

1. So a 'laboratory bench observer' can extract this 'work', by absorbing the radiation? A free lunch?

2. In what frame?

3. Surely the field does change with time in the freely falling frame as per the Equivalence Principle?

4. But are you saying it should still radiate?

1. Recall that energy and energy-conservation law are actually not well defined in curved spacetime, except locally.

2. In the frame in which metric is time independent.

3. Of course.

4. Depends on the definition of radiation. With the invariant definition I use (see above), it radiates.

Perhaps the main lesson is that the question if charge radiates or not - is the wrong question. One should ask questions only in terms of covariantly defined quantities such as the local electromagnetic tensor. If this tensor is large at some spacetime point far from the charge (where "large" roughly means falling as 1/r with a suitably defined r), then it is large for any observer. Notions such as "time dependent", "energy", etc, are not well defined in curved spacetime.

 

[Garth]

My question really asks whether a laboratory observer should be able to detect this radiation.

Garth

 

[Demystifier]

My question really asks whether a laboratory observer should be able to detect this radiation.

As I said, first define radiation! The observer will certainly observe strong electromagnetic fields. If you call it radiation (as I do), then the answer is yes.

 

[Garth]

As I said, first define radiation! The observer will certainly observe strong electromagnetic fields. If you call it radiation (as I do), then the answer is yes.

I would only call it radiation if the electromagnetic field oscillated.

Garth

 

[Demystifier]

I would only call it radiation if the electromagnetic field oscillated.

By oscillation, you probably mean periodic change with time. In this case, a static observer (with respect to coordinates in which metric is time-independent) will not observe radiation. But of course, time is an observer dependent concept, and so is oscillation.

 

[Garth]

Two electrons are stationary on a laboratory bench, then one falls off.

For an instance their velocities are zero relative to the bench, but one, A, is freely falling and the other, B, is not.

According to classical physics A is accelerating wrt the bench frame and according to Maxwell's equations should radiate whereas B is not.

According to the Equivalence Principle A is not accelerating whereas B is accelerating upwards at g, and therefore should radiate?

They are both, for an instance, at rest in the same frame of reference and there is no time dilation between them.

Which actually radiates in the bench frame?

Garth

 

[pervect]

This is why I believe that the usual definition(s) of particles, including the Wigner definition, is not a physically correct definition of particles. For the alternative(s) I propose see
http://arxiv.org/abs/gr-qc/0111029
...

I've only gotten as far as the first paper, but I think I have a slightly better idea of what you are saying now.

Unfortunately, because of my unfamiliarity with Wightman functions it's only a slightly better idea.

 

[tehno]

If a "charged particle" is free falling in the earth`s gravitational field, it will emit electromagnetic waves?

(1)If a person stands on the ground, he will say:Yes ! because,the charged particle is accelerated by the gravitational field!

(2)If another person is free falling along with that particle,he will say:No! because according to the "equivalent principle" he and the particle have the same amount of acceleration, from his point of view,the particle has no acceleration,it can not emit electromagnetic waves!

(3)Paradox! I want to ask: who is right?

.

Both.

 

[Demystifier]

According to classical physics A is accelerating wrt the bench frame and according to Maxwell's equations should radiate whereas B is not.

Here is the place you are making the mistake. The Maxwell equations in flat spacetime are not the same as Maxwell equations in curved spacetime. Consequently, the solutions are also not the same.
See http://arxiv.org/abs/gr-qc/9909035
especially the discussion around Eq. (11).