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

[Garth]

I set up that thought experiment in #32 to tease out your understanding.

I also do not think the freely falling charge should emit, it is the one sitting on the laboratory bench I worry about, the one of which you said:

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

I am querying in what sense does it radiate, can this radiation be detected by a bench-frame observer, both one supported by the bench and the one 'just' freely-falling, and if so what is source of the power of such radiation?

Basically I am querying the concept of the Equivalence Principle as normally understood.

Garth

 

[Demystifier]

1. I also do not think the freely falling charge should emit, it is the one sitting on the laboratory bench I worry about, the one of which you said:
I am querying in what sense does it radiate, can this radiation be detected by a bench-frame observer, both one supported by the bench and the one 'just' freely-falling, and if so what is source of the power of such radiation?

2. Basically I am querying the concept of the Equivalence Principle as normally understood.

1. I think I have already answered this question. Are you satisfied with my answer?

2. The equivalence principle is valid only locally. Radiation is not a purely local concept.

 

[Garth]

1. I think I have already answered this question. Are you satisfied with my answer?

No I am confused as to under what circumstances you say the charged particle emits a radiation that can be received by an observer.

As far as the supported stationary charge is concerned you said "The energy for radiation comes from the supporting force." That to me speaks of a 'free lunch'.

2. The equivalence principle is valid only locally. Radiation is not a purely local concept.

Yet in the thought experiment I set up in #32 both freely falling charges and observers can be arbitrarily close and therefore 'local'.

Garth

 

[Demystifier]

No I am confused as to under what circumstances you say the charged particle emits a radiation that can be received by an observer.

As far as the supported stationary charge is concerned you said "The energy for radiation comes from the supporting force." That to me speaks of a 'free lunch'.

But in #25, #27, #29, and #31 I provided a more precise answer. Why aren't you satisfied with this?

 

[Demystifier]

Yet in the thought experiment I set up in #32 both freely falling charges and observers can be arbitrarily close and therefore 'local'.

You missed the point. How much are you familiar with general relativity? According to your comments on this thread, not much.

 

[Garth]

You missed the point. How much are you familiar with general relativity? According to your comments on this thread, not much.

Be careful, that kind of comment can rebound on you.

But in #25, #27, #29, and #31 I provided a more precise answer. Why aren't you satisfied with this?

Let us take #27

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.

Of course; GR conserves energy-momentum and not in general energy. The violation of the local conservation of energy was a perplexity for Einstein and others until Klein encouraged Emmy Noether to demonstrate that GR fell the category of his improper energy theorems. The gauge invariant symmetry group of GR resulted in the conservation of energy-momentum, but not in general energy. This result is to be expected from the 'no preferred frames' principle of relativity in the presence of a gravitational field.

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

Then we agree, the theory leads us to expect the static supported charged particle should radiate.

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.

How can this be the wrong question? It is asking whether experimentally one should be able to detect this radiation according to the principles of GR and electro-magnetic theory. If practical one might then set up the experiment to see if this prediction is verified.

I am still confused, having read your posts and paper, as to exactly under what circumstances you predict radiation to be detected.

Go back to my gedanken experiment.
In a sufficiently close region in a gravitational field away from the centre of mass of the gravitating body, i.e. on the surface of the Earth, two charged particles A and B are observed by two observers A' and B' on a laboratory bench.

A and A' fall off at the same moment and are momentarily stationary wrt the bench and in free-fall. B and B' remain stationary supported by the bench.I know I am a little slow but I would be grateful if you could clearly answer yes/no the following questions: According to your understanding of those principles:

1. Does A' detect radiation coming from A?
2. Does A' detect radiation coming from B?
3. Does B' detect radiation coming from A?
4. Does B' detect radiation coming from B?

By 'detect' I mean in the same sense that synchrotron radiation is detected.

Garth

 

[pervect]

I think it might be helpful to address how the question of how, experimentally, one would detect particles given the definition proposed by Demystifier in http://arxiv.org/abs/gr-qc/0111029.pdf.

[add]To keep things really simple, one might specialize the answer to only the electromagnetic field rather than a more general field.

This might also help clarify the rather abstract mathematical definition (which I can't quite follow in detail, frankly, (though http://en.wikipedia.org/wiki/Wightman_axioms does give me some "feel" for what the paper is about), and would make the mathematical defintion more "physical" by giving it a more direct tie to experiment.

I also think this thread highlights some of the confusion generated by the original question - what constitutes "radiation"? Different authors seem to take different, plausible definitions of the term, and come up with different answers to the question.

 

[Garth]

Whereas there is disagreement in the theoretical literature on the subject, involving differing understanding of the concepts of 'particle' and 'radiation', one step towards 'de-mystification' is to concentrate on the observables predicted by theory.

Hence my question about the detection of radiation as normally understood; I do not understand why the question is so difficult to answer plainly (as a thought experiment).

Of course another possible answer to each or any of my four questions is simply "Don't know", that itself would tell us something about the state of the theory on the subject.

Garth

 

[pmb_phy]

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

There is no paradox. Whenever there is a relative acceleration between charge and observer then the observer will detect the presence of radiation. There are several journal articles on this. See a list of what I could find at

http://www.geocities.com/physics_world/falling_charge.htm

If you'd like to read any of them then let me know and I'll get it to you somehow (i.e. I'll send it to you in e-mail or post it on my website and you can download it).

Note: Demystifier posted the following comment

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

This accertion runs contradictory to the results arrive at in the journals I mentioned. So long as there is a relative acceleration between charge and detector then radiation will be detected.

"Detected" is the key word here.

Pete

 

[magnetar]

Thank you very much "pmb phy"!

You can post it on your website ,so
anybody who interested in this field can find it!

 

[Garth]

Demystifier your answers to my four questions that I find in your posts above and your papers are as follows, am I correct? :

In a sufficiently close region in a gravitational field away from the centre of mass of the gravitating body, i.e. on the surface of the Earth, two charged particles A and B are observed by two observers A' and B' on a laboratory bench.

A and A' fall off at the same moment and are momentarily stationary wrt the bench and in free-fall. B and B' remain stationary supported by the bench. Is radiation detected by A' and B', where 'detect' is used in the same sense that synchrotron radiation is detected?

From your papers you say both A' and B' observe the same as each other in each case,

so:
1. Does A' detect radiation coming from A? No; A is following a geodesic.
2. Does A' detect radiation coming from B? Yes; B "does not move along a geodesic, so it radiates"
3. Does B' detect radiation coming from A? No; A is following a geodesic.
4. Does B' detect radiation coming from B? Yes; B "does not move along a geodesic, so it radiates"

Now my next question, which you did not answer to my satisfaction is that if, in case 4, B', which co-moves with B, detects radiation, then where does the power for that radiation come from?

In post #21 you say

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

and in post #25

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.

so, in the B' bench frame, does B travel and 'do work'? You change the meaning of 'radiate'. It fact you appear to be saying that B' does not detect radiation from B in the sense that synchrotron radiation is detected, are you not?

I which case, you state in your eprint gr-qc/9909035

In particular, this implies that in classical physics the notion of radiation is observer independent, contrary to the conclusion of some existing papers.

I find this misleading to say the least.

My point in this discussion is to recognise that this experiment does identify a preferred frame of reference, that co-moving with the centre of mass in which a charged particle accelerating wrt the freely falling frame does not radiate detectable energy (in the sense that synchrotron radiation is detected.) This understanding I find contrary to the spirit of the Equivalence Principle and is one of the points of departure for my own work.

Garth

 

[Demystifier]

Garth, now I see that your understanding of general relativity is fine. Now I think I also see where is the source of conflict between you and me. I hope that the following remarks will help you to understand my point:

I cannot say with confidence if this or that observer will detect radiation, because I do not understand sufficiently well how the radiation detectors work. However, I am not even particularly interested in this question. I am not intersted in detectors, but in fundamental properties of nature that do not depend on detectors. Thus, I am interested in question whether there *is* (rather than detected) radiation with respect to certain coordinate frame (rather than certain observer). From this theorethical point of view (which you may find not sufficiently physical), it only makes sense to talk about electromagnetic field transforming as a tensor under coordinate transformations. My answers refer to this point of view, from which it may not be obvious to deduce how a realistic detector will respond.

 

[Demystifier]

My point in this discussion is to recognise that this experiment does identify a preferred frame of reference, that co-moving with the centre of mass in which a charged particle accelerating wrt the freely falling frame does not radiate detectable energy (in the sense that synchrotron radiation is detected.) This understanding I find contrary to the spirit of the Equivalence Principle and is one of the points of departure for my own work.

I agree with the part I have not cited, but not with this cited part above. I do not agree that this frame of reference is really preferred, so I do not see a conflict with the equivalence principle.

 

[Garth]

Thanks, I understand, for me the crucial question is, "Can energy be extracted by this process?"

The question of possible preferred frames of reference is intimately connected to the question of the local conservation, or otherwise, of energy. This is, of course, because energy is a frame dependent concept.

Garth

 

[pmb_phy]

Thank you very much "pmb phy"!

You can post it on your website ,so
anybody who interested in this field can find it!

You've noticed that there is already a link to it since that's how I showed it to you. However I would hazard to guess that you're looking for a link to click on my web site. That has already been done. I.e. go to

http://www.geocities.com/physics_world/

then scroll down to the Selected Literature section and see item # 7

Best Regards

Pete

 

[Garth]

I agree with the part I have not cited, but not with this cited part above. I do not agree that this frame of reference is really preferred, so I do not see a conflict with the equivalence principle.

The gravitational field of the Earth is weak and the argument becomes rather academic.

However, if we move to a quasar or AGN engine then we are dealing with strong fields surrounding a massive black hole.

These bodies radiate strongly, both from their jets and accretion disks, in different regions of the spectrum up to X-ray emission. Some of the powerful emission is thought to be synchrotron radiation. (Jester S et al. 2006 ApJ 648 900).

Here energy is being extracted from 'the process' of charged particles being accelerated from their freely falling frames by powerful electro-magnetic fields.

In this case the forces acting on the particle, which is moving in the BH's rest frame, are doing work. However, consider the rather artificial situation where the electrostatic and magnetic forces exactly balance the gravitational force and keep a charged particle suspended stationary over the BH.

In this hypothetical scenario a powerful force is acting on the particle accelerating it from its geodesic path, yet no work is being done. Will this particle radiate with synchrotron radiation as all the others are doing?

My answer in this case is no, for otherwise where would the power of this radiation be coming from? All forces are balanced and the particle is at rest in the BH's rest frame. However does not GR suggest that it should be powerfully radiating as it is being accelerated from its geodesic path?

IMHO in this case the rest frame of the BH is a 'preferred frame' because only when at rest in that frame are charged particle's not radiating when under the influence of powerful magnetic and electrostatic forces.

This I understand to be in conflict with the principle of relativity and therefore by inference also with the equivalence principle.

If such a particle actually does powerfully radiate then that would be in accordance with the principle of relativity but contrary to the conservation of energy. This would not be inconsistent with the principles of relativity but it would represent a 'free lunch'!

Garth

 

[Demystifier]

This would not be inconsistent with the principles of relativity but it would represent a 'free lunch'!

There is even a simpler way to get a free lunch from general relativity. Assume, for example, that the matter density of the universe exceeds the critical one, so that one day the universe will start to shrink. This means that it will become hotter and hotter as time passes. A kind of free lunch, isn't it? But of course, this increase of the temperature will go at the expense of decreasing gravitational "energy". I suggest you to think if your example can be reinterpreted in a similar way.