Why do accelerated charges emits a photon?

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Is it only when the acceleration is negative? If yes, when it is positive it absorbs a photon?
 

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  • #2
Demystifier
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Perhaps you should first understand why classical (not quantum) charge emits electromagnetic radiation.
 
  • #3
ZapperZ
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This thread has been moved from QM sub-forum to the classical physics sub-forum. I agree that the OP needs to first understand this from the classical E&M perspective.

Zz.
 
  • #4
Andrew Mason
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This is an important question and, remarkably, one that has no definitive answer as far as I can tell.

A charged particle that is accelerated by gravity does not emit an em wave/photon. The only other way to accelerate a charged particle is to apply a mechanical or electro-magnetic force to it over some distance. And a mechanical force is nothing but an em force. Thus, the observational evidence amounts to:
a charged particle does not emit an electro-magnetic wave or photon except when it receives electro-magnetic energy.​

So this leads to a reformulation of the question: does the charged particle emit a photon because it accelerates? Or does it emit a photon in response to receiving a photon, which incidentally causes it to accelerate?

The question seems to be still unresolved: See this site, for example.

AM
 
  • #5
olgranpappy
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This is an important question and, remarkably, one that has no definitive answer as far as I can tell.

This very cute problem is addressed to some extent by Peierls in his little book "Surprises in Theoretical Physics".
 
  • #6
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A charged particle that is accelerated by gravity does not emit an em wave/photon.
What is your basis for that assertion?
 
  • #7
olgranpappy
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What is your basis for that assertion?

Apparently, the principle of equivalence...

Consider a man standing on a scale on earth. The scale reads, say, 170 pounds.

Consider a man in completely empty space in a rocketship accelerating at 9.8m/s^2. The man is standing on a scale. The scale reads 170 pounds.

Now, consider the exact same situation but now with a point charge stapled to the man's nose.

If the principle of equivalence holds then... well, shouldn't the stationary point charge in a gravitational field be radiating? Regardless of whether or not it "should", it does not.
 
  • #8
Andrew Mason
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What is your basis for that assertion?
Further to what olegranpappy has said, according to the principle of equivalence, a charged particle accelerating in freefall in a gravitational field is locally equivalent to a charged particle at rest in an inertial frame of reference. Since a charged particle at rest in an inertial frame does not radiate, the same must hold true for the charged particle in freefall. This is consistent with all observation.

A charged particle that is stationary in a gravitional field is locally equivalent to a charged particle accelerating in a gravity-free space. So it should radiate. This is a bit of a problem, because as olegranpappy points out, it is not observed. Here is a paper that explains why this occurs - it suggests that the radiation is a relative phenomenon that depends on the relative acceleration between the observer and the charge:

From "Radiation from a Charge in a Gravitational Field" said:
It is found that the “naive” conclusion from the principle of equivalence - that a freely falling charge does not radiate, and a charge supported at rest in a gravitational field does radiate - is a correct conclusion, and one should look for rdiation whenever a relative acceleration exists between an electric charge and its electric field. The electric field which falls freely in the gravitational field is accelerated relative to the static charge. The field is curved, and the work done in overcoming the stress force created in the curved field, is the source of the energy carried by the radiation. This work is done by the gravitational field on the electric field, and the energy carried by the radiation is created in the expence of the gravitational energy of the system.

AM
 
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  • #9
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Further to what olegranpappy has said, according to the principle of equivalence, a charged particle accelerating in freefall in a gravitational field is locally equivalent to a charged particle at rest in an inertial frame of reference. Since a charged particle at rest in an inertial frame does not radiate, the same must hold true for the charged particle in freefall. This is consistent with all observation.
AM
But I thought that principle of equivalence contradicts special relativity, and by consequence, classical electromagnetic theory as well.
 
  • #10
olgranpappy
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But I thought that principle of equivalence contradicts special relativity, and by consequence, classical electromagnetic theory as well.

Hopefully not too much, since special relativity and classical electromagnetic theory are correct.
 
  • #11
Andrew Mason
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But I thought that principle of equivalence contradicts special relativity, and by consequence, classical electromagnetic theory as well.
The General Theory and the Special Theory of Relativity are perfectly consistent. The principle of equivalence is the cornerstone of GR. Maxwell's equations are consistent with both - certainly with SR. In fact, the belief that Maxwell's equations are valid in all inertial frames of reference was one of the things that drove Einstein to develop SR.

AM
 
  • #12
Andrew Mason
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A charged particle that is stationary in a gravitional field is locally equivalent to a charged particle accelerating in a gravity-free space. So it should radiate.
I should have added after: "So it should radiate"... "IF accelerating charges radiate because they are accelerating.

AM
 
  • #13
olgranpappy
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Oh, I finally found that book I was talking about earlier. Here's some highlights:

Although the formula for the rate at which a small object loses energy via radiation is
[tex]
\frac{dW}{dt}=\frac{2e^2}{3c^3}a^2
[/tex]
where a is acceleration. And this gives the right answer for total loss. It can not be the right expression for instantaneous rate of loss of energy.

blah blah blah. For periodic or other types of motion we can use instead the expression
[tex]
\frac{dW}{dt}=\frac{-2e^2}{3c^2}v\dot a
[/tex]
where v is velocity and [tex]\dot a[/tex] is "jerk".

Unfortunately, further thought shows that this rewriting does not resolve the paradox... actually, it gets a bit complicated. A careful discussion was given by Rohrlich and Fulton in Ann. Phys. 9, 499 (1960). But the paradox was not truly resolved until arguments by D. Boulware came along (unpublished as of 1979, but given by Peierls in his book).
 
  • #14
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what about a positive charge?

Hi,

I was reading this thread and I have an additional question to ask. If an electron and a proton were accelerating at the same rate, would they emit the same photon?
 
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  • #15
Andrew Mason
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Hi,

I was reading this thread and I have an additional question to ask. If an electron and a proton were accelerating at the same rate, would they emit the same photon?
The force on, or rate of momentum change of, the proton would be 1800 times greater than that for the electron. This means that the electromagnetic interaction needed to cause this (ie. the momentum of the photon) would have to be 1800 times greater.

AM
 
  • #16
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So, it would take really high frequencies of light to accelerate a proton? And would the proton re-emit this frequency of light as well?

A related question:
If we were to accelerate a proton using a very weak magnetic field, would it emit radiation?
 
  • #17
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This is a bit of a problem, because as olegranpappy points out, it is not observed.

You're saying that if an apparatus stands stationary to the Earth, it can not detect radiation from co-supported charges (this much seems sensible, else where would the energy come from: no other part of the system has given up change). But just to clarify:

If a charge is released to fall freely past the apparatus (and vice-versa), is radiation then detected? If the apparatus is placed on (say) an accelerating train, will it continue not to detect radiation from co-accelerated charges?

Actually, my understanding was that the field lines of a constantly accelerated charge merely "droop". This isn't radiation if you can stay stationary with respect to charge and the direction of acceleration. I think the equivalence principle here is unscathed (until you add boundary conditions, but that feels like cheating).
 
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  • #18
olgranpappy
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Hi,

I was reading this thread and I have an additional question to ask. If an electron and a proton were accelerating at the same rate, would they emit the same photon?

classically you would use the same formula to determine the power radiated, there's no reference to mass in that formula.
 
  • #19
Andrew Mason
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If a charge is released to fall freely past the apparatus (and vice-versa), is radiation then detected? If the apparatus is placed on (say) an accelerating train, will it continue not to detect radiation from co-accelerated charges?
There is a difference between a charge in free fall in a gravitational field and a charge accelerating due to a non-gravitational force (ie. an electromagnetic force). In the first case it does not radiate. In the second, it does.

Actually, my understanding was that the field lines of a constantly accelerated charge merely "droop". This isn't radiation if you can stay stationary with respect to charge and the direction of acceleration. I think the equivalence principle here is unscathed (until you add boundary conditions, but that feels like cheating).
I am not sure I understand what you mean here.

AM
 
  • #20
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There is a difference between a charge in free fall in a gravitational field and a charge accelerating due to a non-gravitational force (ie. an electromagnetic force). In the first case it does not radiate. In the second, it does.
Do you have evidence of that? How do you define "radiate"?
I am not sure I understand what you mean here.
I say that to a classical detector-apparatus, "radiation" means any time-varying field at the apparatus.

Now, if an apparatus inside an elevator is stationary with respect to some charge, regardless of whether the elevator is "standing in a planet's gravitational field" or "accelerating constantly in space" the apparatus will detect no radiation (because in both cases the field at the apparatus is time-independent). Hence I disagree with your statement that there is some problem with the equivalence principle.

(As for the rough shape of the field inside the elevator, it happens to droop -- in both cases as per the equivalence principle -- as illustrated here. Actually if you're interested in the exact shape, you should be able to distinguish the two cases due to the differing global geometry.)
 
  • #21
Andrew Mason
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Do you have evidence of that?
Sure. An electron whipping around a curved path radiates (synchrotron radiation). Microgravity experiments have shown that a falling electron does not radiate: .
How do you define "radiate"?
Emit a photon.

I say that to a classical detector-apparatus, "radiation" means any time-varying field at the apparatus.

Now, if an apparatus inside an elevator is stationary with respect to some charge, regardless of whether the elevator is "standing in a planet's gravitational field" or "accelerating constantly in space" the apparatus will detect no radiation (because in both cases the field at the apparatus is time-independent). Hence I disagree with your statement that there is some problem with the equivalence principle.
I never said that there was a problem with the principle of equivalence. There is no problem. The problem may be with the theory that a charge radiates because it accelerates. Outside a gravitational field, a charge cannot accelerate unless it interacts with a an em field ie. a photon. Perhaps a charge radiates because it interacts with a photon which, because it has momentum, causes the charge (which has mass) to accelerate.

AM
 
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  • #22
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I'm trying to avoided muddying the issue with photons, since this thread was tossed to the classical forum.

So you're saying you think charges only radiate if they are driven by an electromagnetic force (not allowed to fall)? So then, if the moon was given an electric charge, you think it also would not radiate (presuming one could detect light-month wavelengths)?
 
  • #23
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As I understand it, you have to be *very* careful about things. We're talking about 3 different things here -- the charged particle, the gravitational field, and the observing equipment. They can be all moving relative to each other. Assuming that we've got a (quasi-)uniform gravitational field, there are a few interesting cases:

1. The charge is not moving relative to the measuring equipment, but both are free-falling in the gravitational field. In this case, you get the same answer as in flat space, not moving.

2. The charge is not moving relative to the measuring equipment. They are also stationary relative to the gravitational field. In this case, you get the same answer as if you were all accelerated, as required by the equivalence principle. Thus, you get the "drooping" referred to in the papers so far cited.

3. The measuring instrument is stationary wrt gravitational field. The charge free-falls. We get the same pattern as if a uniformly accelerating charge moves past.

I think that's what the equations say... Corrections anyone?
 
  • #24
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The measuring instrument is stationary wrt gravitational field
I presume you mean 'supported in a grav. field' as in
charges only radiate if they are driven by an electromagnetic force (not allowed to fall)?
 
  • #25
Andrew Mason
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So you're saying you think charges only radiate if they are driven by an electromagnetic force (not allowed to fall)? So then, if the moon was given an electric charge, you think it also would not radiate (presuming one could detect light-month wavelengths)?
Yes. If the moon had net charge it would not radiate merely because it was orbiting the earth. It would be a charge in free-fall in a gravitational field and according to the princple of equivalence, it is locally indistinguishable from a charge in an inertial frame of reference.

AM
 

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