Why do accelerated charges emits a photon?

In summary: The belief that Maxwell's equations are valid in all inertial frames of reference was one of the key insights of Einstein's general theory of relativity.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. 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:There is a paper that suggests
  • #1
Fisix
2
0
Is it only when the acceleration is negative? If yes, when it is positive it absorbs a photon?
 
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  • #2
Perhaps you should first understand why classical (not quantum) charge emits electromagnetic radiation.
 
  • #3
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
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
Andrew Mason said:
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
Andrew Mason said:
A charged particle that is accelerated by gravity does not emit an em wave/photon.
What is your basis for that assertion?
 
  • #7
lugita15 said:
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
lugita15 said:
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 gravitational 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
Andrew Mason said:
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
lugita15 said:
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
lugita15 said:
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 said:
A charged particle that is stationary in a gravitational 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
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
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
idea2000 said:
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
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
Andrew Mason said:
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
idea2000 said:
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
cesiumfrog said:
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
Andrew Mason said:
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"?
Andrew Mason said:
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 http://links.jstor.org/sici?sici=0950-1207%2819271101%29116:775%3c720:OEPIGF%3e2.0.CO%3b2-K&origin=ads due to the differing global geometry.)
 
  • #21
cesiumfrog said:
Do you have evidence of that?
Sure. An electron whipping around a curved path radiates (synchrotron radiation). http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V3S-4N68NMP-7&_user=10&_coverDate=12%2F31%2F2007&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=ea92b6de2f38ef51359814f6390924f5have 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
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
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
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
cesiumfrog said:
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
 
  • #26
OK, so you are insisting the electrically charged moon would not radiate. Nonetheless, do you agree that there would be a periodic variation in the electric field measured from earth? And from any distant star?
 
  • #27
cesiumfrog said:
OK, so you are insisting the electrically charged moon would not radiate.
The principle of equivalence requires that an electrically charged moon not radiate.

Nonetheless, do you agree that there would be a periodic variation in the electric field measured from earth? And from any distant star?
Since the sun and the Earth and moon are moving relative to a distant star, the field of a charged moon measured from a distant star would be constantly changing.

From the earth, the electric field would change direction periodically.

AM
 
  • #28
Andrew Mason said:
The principle of equivalence requires that an electrically charged moon not radiate. [But] the field of a charged moon measured from a distant star would be constantly changing.

The principle of equivalence requires no such double-speak.
 
  • #29
Andrew Mason said:
An electron whipping around a curved path radiates (synchrotron radiation). http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V3S-4N68NMP-7&_user=10&_coverDate=12%2F31%2F2007&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=ea92b6de2f38ef51359814f6390924f5have shown that a falling electron does not [..] Emit a photon.

The article you cited does not appear to gives any evidence as to whether or not falling charges radiate. It discusses one particular apparatus for an Eotvos experiment (for charged particles), analyses several error sources, and proposes that it might obtain higher accuracy from in orbit.

As for the issue of emitting photons, one lesson from attempting to join QM with relativity is that the existence of particles is subjective (eg. Unruh radiation); I think genneth spoke for the classical equations correctly.

Back to the charged moon scenario, you agreed this produces an equivalent field as to a synchrotron. Classical electrodynamics will show both transmit energy to infinity, conservation implies the moon will be slowed, and ... I assume this deviation from geodesic motion can be attributed somehow to the electromagnetic self-force of a charge upon itself in curved geometry?
 
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  • #30
I think one issue here is the mis-identification of classical fields with radiation of photons. In the classical situations I presented, none of the fields oscillate in an obvious wave fashion. The fields are simply a bit complicated, and non-zero, extending out to infinity. We don't say that the classical field due to a stationary electron correspond to emission of photons (maybe virtual photons, but I'd hesitate to interpret any QFT particles as ontology, especially in time-evolution processes). Thus, uniformly accelerated charges don't really *emit* a photon. The field they set up is a coherent state involving essentially infinite number of photons.
 
  • #31
Can you recommend any literature that supports this approach? That the only electromagnetically accelerated charge emits electromagnetic radiation?
 
  • #32
I just got a marvelous book on this very fascinating (and very complicated) issue of classical charged particles within the Maxwell theory. I highly recommend it since it's a very clear exposition of classical electromagnetics and on top solves (at least it claims so) the age-old problem of the selfconsistent dynamics of classical point charges and electromagnetic fields:

Rohrlich, F.: Classical charged particles, 3 edition, World Scientific Pub Co Inc, 2007
 
  • #33
Didn't know about his book. He also has a http://www.sciencedirect.com/science/article/pii/S037596010100264X" [Broken] in which he describes his method. I found some threads in PF and some papers in arXiv. It seems that:
  • as long as the particle is not too small, the semi-classical approach can be used. This is called the Abraham-Lorentz-Dirac equation see e.g. http://arxiv.org/abs/gr-qc/9912045" [Broken].
  • otherwise you should use the QED approach which is called Abraham-Lorentz-Dirac-Lengevin (ALDL) equation.
I was also checking the movement of charge particle in the gravitational field. There are many papers trying to derive equations to determine whether and when the particle does and doesn't radiate. For example http://iopscience.iop.org/0264-9381/21/16/R01". But in none of the sources I could see any hint, that confirms the post #21 and #4 in this thread.

Any ideas?
 
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1. Why do accelerated charges emit a photon?

Accelerated charges emit a photon because of their interaction with the electromagnetic field. As the charge accelerates, it creates a disturbance in the field which propagates outward as a photon.

2. How does an accelerated charge emit a photon?

An accelerated charge emits a photon by releasing energy in the form of electromagnetic radiation. This energy is carried away by the photon, which is a quantum of light.

3. Is there a specific speed at which an accelerated charge emits a photon?

Yes, there is a specific speed at which an accelerated charge emits a photon. According to classical electromagnetism, the acceleration must be non-uniform and changing in order for a photon to be emitted. This means that the charge must be constantly changing its speed and direction.

4. Can an accelerated charge emit multiple photons?

Yes, an accelerated charge can emit multiple photons. The number of photons emitted is directly proportional to the acceleration of the charge. This means that a higher acceleration will result in the emission of more photons.

5. What is the significance of an accelerated charge emitting a photon?

The emission of a photon by an accelerated charge is significant because it is the fundamental mechanism behind the production of light and other forms of electromagnetic radiation. It also plays a crucial role in many important phenomena, such as electricity, magnetism, and the behavior of atoms and molecules.

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