When is the radiation from an accelerated charge a matter of controversy?

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In summary: The short answer is "when it accelerates". However, there are some very interesting controversies involved.
  • #1
Rob Woodside
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The short answer is "when it accelerates". However, there are some very interesting controversies involved.

One involves the equivalence principle and has been much argued over in the literature. I'm really asking about the current state of the arguments.

Loosely the equivalence principle says if you are locked in an Einstein cage and cannot look outside, you cannot perform experiments inside the cage to tell whether you are at rest on a very large homogeneous, spherical planet with a downward gravitational acceleration g or whether you are lost in space with a rocket accelerating you upwards at g. If that is so, then why aren't you blinded by the radiation from a charged body sitting on a table in a uniform gravitational field. The short answer is that the radiation at 10 m/s^2 is too small to measure. But others have argued that the radiation is not even there! Any one know the current status of this?

I apologize if this has been discussed before, but the search engine here could not find any postings involving 'radiation, accelerated charge'.
 
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  • #2
Rob,

radiates what exactly, may I ask ?



roger
 
  • #3
roger said:
Rob,

radiates what exactly, may I ask ?



roger
I had in mind an electric charge and electromagnetic radiation.
 
  • #4
It's not a "relativity" question so much as a "quantum theory" question. An accelerating charged particle does NOT radiate energy. It radiates energy only when the change in energy is an integral multiple of the quantum.
 
  • #5
HallsofIvy said:
It's not a "relativity" question so much as a "quantum theory" question. An accelerating charged particle does NOT radiate energy. It radiates energy only when the change in energy is an integral multiple of the quantum.

The best theory so far on radiating charges is quantum electrodynamics. My question is not about this or any quantum theory of radiation. It is simply about classical electromagnetism and the equivalence principle. Once this is understood then you can look for quantum gravity by blending the equivalence principle and QED. I leave that to the HallsofIvy
 
  • #6
Rob Woodside said:
The short answer is "when it accelerates". However, there are some very interesting controversies involved.

One involves the equivalence principle and has been much argued over in the literature. I'm really asking about the current state of the arguments.

I'm not up on the latest in the literature, but I had the impression that the answer was that notion of "radiating" was observer dependent.
 
  • #7
Rob Woodside said:
The best theory so far on radiating charges is quantum electrodynamics. My question is not about this or any quantum theory of radiation. It is simply about classical electromagnetism and the equivalence principle. Once this is understood then you can look for quantum gravity by blending the equivalence principle and QED. I leave that to the HallsofIvy

Ron I believe I understand that the question is not a quantum mechanics issue but could be about a macroscopic charged ball, sitting on the table beside you (perched on an insulated stand)


you and the ball are either in the cabin of an accelerating rocket ship or you are in your livingroom on Earth in one gee.

in neither case do you see any EM radiation because the ball is not
accelerating relative to you

you and the ball are either both in the one gee gravity field
or you are both in the accelerating rocket ship

somebody NOT on the rocket ship might be able to detect the EM waves made by the accelerating charged ball. assuming the cabin walls did not shield it. because they would see the charge accelerating


====
there is a peculiar radiation associated with acceleration called "unruh radiation" after a guy at Univ. BC Vancouver named Bill Unruh
but you don't need a CHARGE to be accelerated for that---it is something different
====

anyway if you had a trillion extra electrons on a ball at the end of a wand and someone waved the wand back and forth it would make EM waves, like radio waves, because of the acceleration

and as long as you were not also at the end of a wand being waved back and forth then you would be able to detect these waves.

in principle, except for they're probably being very weak and low frequency so hard to detect.
 
  • #8
Rob Woodside said:
I apologize if this has been discussed before, but the search engine here could not find any postings involving 'radiation, accelerated charge'.

there's been sporadic mention of the Caltech animation of
radiation from an (accelerating) moving charge

the Caltech MovingCharge applet
http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html


for instance in this thread:
https://www.physicsforums.com/showthread.php?t=54404

I don't recall anyone raising just this very question tho
 
  • #9
Detecting the radiation should have nothing to do with whether you are in the same accelerating frame or not - the radiation is going to travel relative to space at c - and since the velocity of the comoving observer is nill compared to c, this observer will detect the radiation if it exists.

In the case of a static G field, the question arises as to where the energy is derived to produce the radiation. Interesting question!
 
  • #10
Rob Woodside said:
If that is so, then why aren't you blinded by the radiation from a charged body sitting on a table in a uniform gravitational field. The short answer is that the radiation at 10 m/s^2 is too small to measure. But others have argued that the radiation is not even there! Any one know the current status of this?

I apologize if this has been discussed before, but the search engine here could not find any postings involving 'radiation, accelerated charge'.

The latest research that I know of on this subject is

Radiation from a Uniformly Accelerated Charge, David G. Boulware, Annals of Physics: 124, 169-188 (1980)
Abstract: The electromagnetic field associated with a uniformly accelerated charge is studied in some detail. The equivalence principle paradox that the co-accelerating observer measures no radiation while a freely falling observer measures the standard radiation of an accelerated charge is resolved by noting that all the radiation goes into the region of space time inaccessible to the co-accelerating observer.
See also
http://www.geocities.com/physics_world/falling_charge.htm

Classical Radiation from a Uniformly Accelerated Charge, Thomas Fulton, Fritz Rohrlich, Annals of Physics: 9, 499-517 (1960)
Radiation Damping in a Gravitational Field, Bryce S. DeWitt, Robert W. Brehme, Annals of Physics: 9, 220-259 (1960)
Principle of Equivalence, F. Rohrlich, Annals of Physics: 22, 169-191, (1963)
Radiation from an Accelerated Charge and the Principle of Equivalence, A. Kovetz and G.E. Tauber, Am. J. Phys., Vol. 37(4), April 1969

Pete
 
  • #11
Rob Woodside said:
The short answer is "when it accelerates". However, there are some very interesting controversies involved.

One involves the equivalence principle and has been much argued over in the literature. I'm really asking about the current state of the arguments.

Loosely the equivalence principle says if you are locked in an Einstein cage and cannot look outside, you cannot perform experiments inside the cage to tell whether you are at rest on a very large homogeneous, spherical planet with a downward gravitational acceleration g or whether you are lost in space with a rocket accelerating you upwards at g. If that is so, then why aren't you blinded by the radiation from a charged body sitting on a table in a uniform gravitational field. The short answer is that the radiation at 10 m/s^2 is too small to measure. But others have argued that the radiation is not even there! Any one know the current status of this?
If a charge sitting in a gravitational field radiated energy, one could create a perpetual energy source without doing any work. The fact is that no one has ever detected radiation from a charge due to gravity.

The notion that an accelerated charge should produce EM radiation is not required by Maxwell's equations. It is derived from the notion that an accelerated charge must interact with its own field.

There have been controversial attempts to explain why gravitational acceleration of a charge should not produce radiation by showing that radiation is a function of the third time derivative of position (ie. non-uniform acceleration). The fact is, however, as anyone who tries to make electrons go in a circle using magnets knows, uniformly changing the direction of a moving electron with magnetic force produces radiation.

It seems that uniform acceleration caused by electromagnetic force does produce radiation, whereas uniform acceleration caused by gravity does not. I think that the explanation for this is still a live issue. There is an interesting discussion in the "Speed of Gravity Controversy" thread relating to this.

AM
 
  • #12
Andrew Mason said:
The notion that an accelerated charge should produce EM radiation is not required by Maxwell's equations. It is derived from the notion that an accelerated charge must interact with its own field.

How would a charge know whether the field it was feeling was generated by itself or not? I don't see any problem with this assumption.

I think Pete's link answered the question pretty well, but I have a bit more to say about it. Although showing things rigorously is difficult, the apparent paradox in this problem is due to a couple of misunderstandings. First of all, a charge is not a localized object. It is inevitably connected to its field, which makes it act like an extended body. The equivalence principle is therefore not applicable.

Also, since it is effectively an extended body, its material portion doesn't necessarily fall on a geodesic in the absence of external electromagnetic fields.
 
  • #13
roger said:
Rob,

radiates what exactly, may I ask ?



roger
Sorry for dismissing your question, there is substance to it. What is electromagnetic radiation in Maxwell's theory (F=dA, d*F=*J)? Surely a photon is like an electron or an atom in that it exists independently of any observer's state of motion. Its existence begins at emission and ends in absorbtion. If a Maxwell radiation field is really a cloud of photons then we should expect the radiation to have an independent existence starting at emission and ending in absorbtion. The older literature defined e/m radiation only in the far field zone. Radiation left behind the bound fields of the emitter and was the only thing that survived at large distances. Work by Teitlebaum in the 70's split the Lennard Wiechert field of an arbitrarily accelerated charge into a bound field (associated with the charge) and a free field (radiation field). This permitted splitting to total field into bound and radiating pieces all the way down to the emitter itself. This still does not answer the question, I'll say more in my next replies.
 
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  • #14
Stingray said:
How would a charge know whether the field it was feeling was generated by itself or not? I don't see any problem with this assumption.

I think Pete's link answered the question pretty well, but I have a bit more to say about it. Although showing things rigorously is difficult, the apparent paradox in this problem is due to a couple of misunderstandings. First of all, a charge is not a localized object. It is inevitably connected to its field, which makes it act like an extended body. The equivalence principle is therefore not applicable.
That is quite correct. However the equivalence principle will hold in a uniform gravitational field. At least when the field can be considered to be restricted within the region which is uniform.
Also, since it is effectively an extended body, its material portion doesn't necessarily fall on a geodesic in the absence of external electromagnetic fields.
Unless the field is uniform.

Pete
 
  • #15
pervect said:
I'm not up on the latest in the literature, but I had the impression that the answer was that notion of "radiating" was observer dependent.

This is very disturbing, but possibly true. Real radiation, like light, ought to be observer independent. You may think you can red shift light out of existence, but its angular momentum is a Lorentz invariant. If Maxwell radiation is independent of the observer then there will be an invariant definition that does not involve the observer. The source free Maxwell equations (F=dA, d*F=0) produce two types of fields. When *F is closed and not exact, there is a topological charge found by integrating *F over a closed surface that encloses a missing point. This is how the charged black holes get their charge. Here r=0 is the missing point and the flat space-time analogue has a missing timelike line. There is another possibility. Instead of removing a timelike line in the flat space analogue, one could remove a null line and get a topological charge traveling at light speed. Like the magnetic monopole these have never been observed. But unlike the magnetic monopole these monsters are permitted by Maxwell's equations as they stand. To eliminate them one could require for radiation fields that *F was globally exact. In a similar way one bans magnetic monopoles by requiring F to be globally exact. Fooling with Maxwell's equations is not to be taken lightly. However I would suggest taking F=dA, d*F=0 , *F closed and not exact, for bound fields and F=dA, *F=dB for radiation fields with B an analogue of the vector potential. All the radiation fields I have checked have an F that is globally exact and globally coexact. Such a refinement nicely explains the duality rotation of one radiation field into another with A and B rotating into each other. This invariant characterization of radiation fields, leaves no room for the observer. I would be interested to see any observer dependent definition of a radiation field. I haven't yet checked this refinement against Teitlebaum's work. If this refinement is correct then there should be consistency between the two.
 
  • #16
marcus said:
Ron (sic) I believe I understand that the question is not a quantum mechanics issue but could be about a macroscopic charged ball, sitting on the table beside you (perched on an insulated stand)


you and the ball are either in the cabin of an accelerating rocket ship or you are in your livingroom on Earth in one gee.

in neither case do you see any EM radiation because the ball is not
accelerating relative to you

you and the ball are either both in the one gee gravity field
or you are both in the accelerating rocket ship

somebody NOT on the rocket ship might be able to detect the EM waves made by the accelerating charged ball. assuming the cabin walls did not shield it. because they would see the charge accelerating

Watch out! Acceleration is absolute, the inertial forces tell you so. Your argument leads to a charge in free fall with no acceleration radiating as it appears to accelerate past you at rest on earth. With no acceleration between a freely falling observer and freely falling charge, this observer would see no radiation. Clearly your radiation is observer dependant. I find this very hard to swallow.

marcus said:
====
there is a peculiar radiation associated with acceleration called "unruh radiation" after a guy at Univ. BC Vancouver named Bill Unruh
but you don't need a CHARGE to be accelerated for that---it is something different
====

Here you are excavating particles out of the quantum background as you accelerate through it. The work done to make the particles is supplied by whatever is accelerating the detector. I may be wrong, but I don't think this observer dependent radiation is what we are talking about.
 
  • #17
marcus said:
there's been sporadic mention of the Caltech animation of
radiation from an (accelerating) moving charge

the Caltech MovingCharge applet
http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html


for instance in this thread:
https://www.physicsforums.com/showthread.php?t=54404

I don't recall anyone raising just this very question tho

Thank you Marcus. The moving charge applet is great. You can see the field lines bunching to make the radiation field. In MTW they have a great picture (due to JJ Thompson?) of the field lines from a charge given a short acceleration. One might think that the charge radiates in order to fight the acceleration directly. If so then the radiation would be in the forward direction, but it is perpendicular to the acceleration! In the applet at high speed you can see the bunched field lines leaving the charge directly opposite to the acceleration, again giving radiation perpendicular to the acceleration. Great video!
 
  • #18
yogi said:
Detecting the radiation should have nothing to do with whether you are in the same accelerating frame or not - the radiation is going to travel relative to space at c - and since the velocity of the comoving observer is nill compared to c, this observer will detect the radiation if it exists.

In the case of a static G field, the question arises as to where the energy is derived to produce the radiation. Interesting question!

Thanks Yogi, I share your belief that radiation ought to be observer independent

Your second point is fascinating. In an accelerated cage, it is clearly the accelerating agent that supplies the energy for the radiation. How on Earth could a static gravitational field produce the energy for the radiation?
 
  • #19
pmb_phy said:
The latest research that I know of on this subject is

Radiation from a Uniformly Accelerated Charge, David G. Boulware, Annals of Physics: 124, 169-188 (1980)

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

Classical Radiation from a Uniformly Accelerated Charge, Thomas Fulton, Fritz Rohrlich, Annals of Physics: 9, 499-517 (1960)
Radiation Damping in a Gravitational Field, Bryce S. DeWitt, Robert W. Brehme, Annals of Physics: 9, 220-259 (1960)
Principle of Equivalence, F. Rohrlich, Annals of Physics: 22, 169-191, (1963)
Radiation from an Accelerated Charge and the Principle of Equivalence, A. Kovetz and G.E. Tauber, Am. J. Phys., Vol. 37(4), April 1969

Pete

Pete thank you very much. I'll get these and have a look at them. I'm surprised that the most recent reference is almost 25 years old. May be people think this has been flogged to death.
 
  • #20
Rob Woodside said:
Pete thank you very much. I'll get these and have a look at them. I'm surprised that the most recent reference is almost 25 years old. May be people think this has been flogged to death.
There are a few reasons for that. Here are two that I can think of

(1) I couldn't find more recent ones. :rolleyes:
(2) People don't think its worth redoing a calculation they can't find something wrong with.

Pete

ps - The AJP article is short enough to scan and e-mail if you'd like?
 
  • #21
Andrew Mason said:
If a charge sitting in a gravitational field radiated energy, one could create a perpetual energy source without doing any work. The fact is that no one has ever detected radiation from a charge due to gravity.

This is disturbing. In the accelerated cage the accelerating agent provides the energy, but in the uniform gravitaional field? Pete's reference to Boulware claims that the radiation is in a sector of space-time not available to the co-accelerating observer to make a perpetual motion machine. I must look at this. At first sight it seems as though the gravitational field is sqeezing out the radiation like tooth paste, but what is the source for its energy? The mass of the planet whose gravitational field the charge sits in?

Andrew Mason said:
The notion that an accelerated charge should produce EM radiation is not required by Maxwell's equations. It is derived from the notion that an accelerated charge must interact with its own field.

Aren't Maxwell's equations about the interaction of charges and fields. I'm missing something here. One can remove the charge entirely by cutting the point where the charge lives out of the space and be left with a topological charge only. This topological charge has only a coulomb field and I would expect it to radiate just like a real spherical charge on acceleration.

Andrew Mason said:
There have been controversial attempts to explain why gravitational acceleration of a charge should not produce radiation by showing that radiation is a function of the third time derivative of position (ie. non-uniform acceleration). The fact is, however, as anyone who tries to make electrons go in a circle using magnets knows, uniformly changing the direction of a moving electron with magnetic force produces radiation.

Yes, I made this mistake. It comes from the Lorentz-Dirac force law. Teitlebaum derived this very cleanly with Lennard Wiechert fields and I need to have a look at this as I am currently in some confusion about the divergence of the Maxwell stress-energy tensor and this law.

There is also another attempt to rule out any radiation with the 15 parameter conformal group which leaves Maxwell's equations invariant. In addition to the 10 parameter Poincare group, it has one parameter for scaling space-time and 4 parameters for accelerated frames. If uniform acceleration was one of these, there would be no radiation in a uniform gravitational field. I haven't worked the details, but I doubt it.

Andrew Mason said:
It seems that uniform acceleration caused by electromagnetic force does produce radiation, whereas uniform acceleration caused by gravity does not. I think that the explanation for this is still a live issue. There is an interesting discussion in the "Speed of Gravity Controversy" thread relating to this.

AM

This is hard to swallow. From the posts here it does seem like a live issue, even though there have been no publications on it for the last 25 years.

I essentially stayed out of the "speed of gravity controversy" because there seemed to be a lot of confusing noise that was beyond my pedagogical duties or abilities and the others seemed to be handling it well. Someone appeared to be doubting the Lennard Wiechert fields because the accelerating agent was left out. Yes, the only way we know to accelerate an electron is with an electromagnetic field (Wigner said he was asked by a student at a seminar how such a small thing as an electron could be accelerated and he replied, "With compressed air!"). However, as others have pointed out, waving around a charged wand should produce radiation and the accelerating agent is quite arbitrary, just as in Lennard Wiechert fields.
 
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  • #22
Stingray said:
How would a charge know whether the field it was feeling was generated by itself or not? I don't see any problem with this assumption.

Right!

Stingray said:
I think Pete's link answered the question pretty well, but I have a bit more to say about it. Although showing things rigorously is difficult, the apparent paradox in this problem is due to a couple of misunderstandings. First of all, a charge is not a localized object. It is inevitably connected to its field, which makes it act like an extended body.

Right again! In fact you can cut out the point charge completely and be left with a Coulomb field only. When the singularity is accelerated it is the field around it that gets screwed up and produces radiation as can be seen in the applet.

Stingray said:
The equivalence principle is therefore not applicable.

Here I was careful to say "loosely" in my original post. Mathematically one is hard pressed to find a uniform gravitational field without any tidal forces, but physically it is a truly great approximation. Throwing it out does not answer the question about whether a charged body sitting at rest on the Earth will radiate or not.

Stingray said:
Also, since it is effectively an extended body, its material portion doesn't necessarily fall on a geodesic in the absence of external electromagnetic fields.

Yes, and this could be a reason for a freely falling charge to radiate. However we are perilously close to violating Galileo's principle of relativity. What would the radiation look like in different inertial frames. Could we use it to define preferred frames or (shudder) an aether? Where would the energy come from? It seems safer to ban such radiation.
 
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  • #23
pmb_phy said:
There are a few reasons for that. Here are two that I can think of

(1) I couldn't find more recent ones. :rolleyes:
(2) People don't think its worth redoing a calculation they can't find something wrong with.

Pete

ps - The AJP article is short enough to scan and e-mail if you'd like?

Thanks Pete that is very kind of you. I'll have to get to the library to check the citations of these. Is the physics citation index on line yet and how far back does it go? I'll Google this.

(1) is valid and I thank you for your efforts. (2) This is surprising as there are lots of published contradictions here. (3) "The bright young people are all doing string theory and only cranks and crack pots could be interested in classical electromagnetism" is possibly a widespread attitude.
 
  • #24
Rob Woodside said:
This is hard to swallow. From the posts here it does seem like a live issue, even though there have been no publications on it for the last 25 years.
That is not exactly true. There has been quite a bit written in the last 25 years. Here is a list of some of the published material on the subject in the last 25 years that I have found:

D. Boulware, "Radiation from a uniformly accelerated charge", Annals of Physics 124 , 169-187 (1980)

Kirk T. Mcdonald, "Hawking-Unruh Radiation and Radiation of a Uniformly Accelerated Charge", http://puhep1.princeton.edu/mcdonald/accel/ (1998)

S. Parrott, "Radiation from a particle uniformly accelerated for all time", General Relativity and Gravitation 27 1463-1472 (1995)

S. Parrott, "Radiation from Uniformly Accelerated Charge and the Equivalence Principle", http://arxiv.org/abs/gr-qc/9303025, (2001)

S. Parrott, "Relativistic Electrodynamics and Differential Geometry", New York: Springer Verlag, 1987.

A. Shariati, and M. Khorrami, "Equivalence Principle and Radiation by a Uniformly Accelerated Charge", Found. Phys. Lett. 12 427-439 (1999)

A. K. Singal, "The Equivalence Principle and an Electric Charge in a Gravitational Field", General Relativity and Gravitation 27 953-967 (1995)

A. K. Singal, "The Equivalence Principle and an Electric Charge in a Gravitational Field II. A Uniformly Accelerated Charge Does Not Radiate", General Relativity and Gravitation 27 1371-1390 (1997)

It seems to be quite a controversial issue. One of the great unsolved problems in physics - or at least a problem for which no one has provided a solution that is without controversy. This is, perhaps, not all that surprising given the difficulty in measuring the gravitational effect on a charge.
Yes, the only way we know to accelerate an electron is with an electromagnetic field (Wigner said he was asked by a student at a seminar how such a small thing as an electron could be accelerated and he replied, "With compressed air!"). However, as others have pointed out, waving around a charged wand should produce radiation and the accelerating agent is quite arbitrary, just as in Lennard Wiechert fields.
There are three ways that a charge can be accelerated:

1. being situated in (or moving through) an e/m field

2. absorption or emission of a photon

3. being situated in or moving through a gravitational field.

Mechanical force is ultimately, of course, an e/m force.

As far as I can tell, experimental evidence confirms that a charge will radiate only in the following circumstances:

1. motion in an e/m field.

2. in a response to absorbing a photon

3. nuclear interactions (weak force)

I don't think there are any other. There is no experimental evidence that gravitational force or acceleration produces e/m radiation.

AM
 
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  • #25
Rob Woodside said:
[Originally Posted by Stingray]
How would a charge know whether the field it was feeling was generated by itself or not? I don't see any problem with this assumption

Right!
What assumption? The assumption that an accelerating electron must interact with its own field? Or the assumption that a charge does not 'know' whether the field was its own? I don't understand what it means for a charge to 'know' something about the field with which it is interacting.

In fact you can cut out the point charge completely and be left with a Coulomb field only. When the singularity is accelerated it is the field around it that gets screwed up and produces radiation as can be seen in the applet.
How does the charge accelerate? The only proven way that a charge can accelerate is through interacting with another electromagnetic field. I don't see that field anywhere here so this applet cannot be showing the correct field around an accelerating charge. Also, this applet is based on the concept of the retarded potential propagating as a force field at speed c. While that is the conventional way to analyse the field from moving charges, there is no experimental evidence to confirm that, at least as far as I have been able to determine.

AM
 
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  • #26
My knee jerk response to the whole thing is where is the energy coming from? All mass possessing bodies sacrifice energy for mass, and vice-versa. Charge distribution is nicely explained by Maxwell equations. Gravity and EM do not interact. That has been well known for a century or more.
 
  • #27
Had a thought in reading one of the previous posts - can't find it now - maybe I am merely restating someone else's idea - with that apology - the idea is that perhaps there is a fundamental difference between an electron in a G field and an acceleration field - when we derive the classical radiation formula we tacitly assume it is the electron as a particle that is rapidly accelerated - the field adjusts to the acceleration rather than being simultaneously wafted along by the accelerating force - thus causing a crowding of the lines and a discontinuity that is explained as radiation. But in a G field, not only would the electron itself be in free fall, but the field itself would also be acted upon equally (both in free fall) - so there is no field distortion since there is no relative acceleration between the particle and the field; ergo, no radiation.
 
  • #28
O yes - it was a line of thought started by stingray - and commented upon by Woodside - don't know if my post above adds anything or not
 
  • #29
HallsofIvy said:
It's not a "relativity" question so much as a "quantum theory" question. An accelerating charged particle does NOT radiate energy. It radiates energy only when the change in energy is an integral multiple of the quantum.
This is an electrodynamics question which can be addressed purely classically.

Pete
 
  • #30
Rob Woodside said:
Watch out! Acceleration is absolute, ...
4-acceleration is absolute, coordinate acceleration is relative.
Your argument leads to a charge in free fall with no acceleration radiating as it appears to accelerate past you at rest on earth.
Sounds good to me.
With no acceleration between a freely falling observer and freely falling charge, this observer would see no radiation. Clearly your radiation is observer dependant. I find this very hard to swallow.
All the literature that I've read so far on this concludes that the detection of radiation is observer dependant. All that is required is that there exist a relative acceleration of observer and charge. E.g. ("at rest" means that the observer is, say, sitting on the surface of the Earth)

(1) Observer at rest, charge at rest - Observer does not detect radiation.
(2) Observer at rest, charge in free fall - Observer detects radiation.
(3) Observer in free fall, charge at rest - Observer detects radiation.
(4) Observer in free fall, charge in free fall - Observer does not detect radiation.

Pete
 
  • #31
Andrew Mason said:
That is not exactly true. There has been quite a bit written in the last 25 years. Here is a list of some of the published material on the subject in the last 25 years that I have found:

D. Boulware, "Radiation from a uniformly accelerated charge", Annals of Physics 124 , 169-187 (1980)

Kirk T. Mcdonald, "Hawking-Unruh Radiation and Radiation of a Uniformly Accelerated Charge", http://puhep1.princeton.edu/mcdonald/accel/ (1998)

S. Parrott, "Radiation from a particle uniformly accelerated for all time", General Relativity and Gravitation 27 1463-1472 (1995)

S. Parrott, "Radiation from Uniformly Accelerated Charge and the Equivalence Principle", http://arxiv.org/abs/gr-qc/9303025, (2001)

S. Parrott, "Relativistic Electrodynamics and Differential Geometry", New York: Springer Verlag, 1987.

A. Shariati, and M. Khorrami, "Equivalence Principle and Radiation by a Uniformly Accelerated Charge", Found. Phys. Lett. 12 427-439 (1999)

A. K. Singal, "The Equivalence Principle and an Electric Charge in a Gravitational Field", General Relativity and Gravitation 27 953-967 (1995)

A. K. Singal, "The Equivalence Principle and an Electric Charge in a Gravitational Field II. A Uniformly Accelerated Charge Does Not Radiate", General Relativity and Gravitation 27 1371-1390 (1997)

It seems to be quite a controversial issue. One of the great unsolved problems in physics - or at least a problem for which no one has provided a solution that is without controversy. This is, perhaps, not all that surprising given the difficulty in measuring the gravitational effect on a charge.

Thank you so much Andrew, my work is cut out for me now. I'll get these and have a good read. I think you nailed the problem. We are trying to argue on principle a very delicate effect without any direct experimental evidence. This borders on philosophy and controversies should be expected. It is truly amazing that Einstein danced this kind of dance and came up with verifiable ideas and predictions.
 
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  • #32
Andrew Mason said:
What assumption? The assumption that an accelerating electron must interact with its own field? Or the assumption that a charge does not 'know' whether the field was its own? I don't understand what it means for a charge to 'know' something about the field with which it is interacting.

We should ask Stingray what the assumption was. I took it mean that currents should react to e/m fields in the same way, independently of whether it was a self field or an external field. My old boss Maurice Pryce used to say, "You've got to think like an electron." Pure metaphor, but occasionally effective.

Andrew Mason said:
How does the charge accelerate? The only proven way that a charge can accelerate is through interacting with another electromagnetic field. I don't see that field anywhere here so this applet cannot be showing the correct field around an accelerating charge. Also, this applet is based on the concept of the retarded potential propagating as a force field at speed c. While that is the conventional way to analyse the field from moving charges, there is no experimental evidence to confirm that, at least as far as I have been able to determine.

AM
I appreciate your empirical skepticism. Since Maxwell's equations are linear, why can't we add in an external field that produces the acceleration and then subtract it out to be left with the Lennard Wiechert fields? I really don't think this is an issue.
 
  • #33
Andrew Mason said:
What assumption? The assumption that an accelerating electron must interact with its own field? Or the assumption that a charge does not 'know' whether the field was its own? I don't understand what it means for a charge to 'know' something about the field with which it is interacting.

I was talking about the assumption that charged objects interact with their own fields. By using the word 'know,' I wasn't implying any kind of consciousness. What I meant was that a classical charged object (which means it's not a point) in Maxwell's theory interacts via well-defined rules which do not decompose the fields at a point into 'self' and 'external.' I also don't think it would make any sense at all to modify this part of the theory.

I don't think this question requires an experiment either (although it would of course be nice to have one). Most of us probably accept GR and Maxwell electrodynamics. If so, then those theories give an unambiguous way of answering any question of this nature that you'd like to ask. The only problems are of (1) calculational complexity and (2) definitions. I think that most of the historical issues have come up due to either inconsistent or incomplete definitions and/or poor approximations. In principle, all you need to do is specify your charge model, and then evolve it via Maxwell's equations + stress-energy conservation with the appropriate boundary conditions. You'll then have the electromagnetic field everywhere.

A related problem to all of this is the equation of motion of a charge in a gravitational field. This was worked out by Dewitt and Brehme (quoted in Pete's link), although it's missing a term due to some trivial algebraic error. It's since been rederived by others as well. Anyway, their equation makes it quite clear that GR does predict a nontrivial modification to the equation of motion when the particle has charge.
 
  • #34
Chronos said:
My knee jerk response to the whole thing is where is the energy coming from? All mass possessing bodies sacrifice energy for mass, and vice-versa.

The equivalence principle suggests that a charge at rest in a uniform gravitational field radiates and it is a very good question as to where this energy comes from. There's been a lot of rubbish written about energy in general relativity and I am reluctant to add to it. But here goes. A universe contains only a charge sitting at rest near a planet's surface and radiates. At spatial infinity one sees the planet's mass M and at null infinity the radiation. One would have to conclude that the M was decreasing as the radiation persisted. Of course this should be calculated, but it fits with the non locality of energy in general relativity. More local attempts with pseudo-tensors and preferred frames seem to be coalescing in recent literature. The weak field approximation using one of these pseudo tensors might give some indication as to where the energy is coming from.


Chronos said:
Charge distribution is nicely explained by Maxwell equations. Gravity and EM do not interact. That has been well known for a century or more.

Chronos, when solving Maxwell's equations you specify the currents and boundary conditions. The charge distribution is contained in the current distribution which is specified. How on Earth can a set of equations explain an arbitrary input?

Read Einstein's 1916 paper "The foundations of General Relativity" in "The Principle of Relativity", a Dover paperback. There you will find the original Einstein Maxwell theory that links gravity and electromagnetism. This writes the total stress energy tensor as the sum of a current piece and Maxwell's stress-energy tensor for the e/m field. Then Einstein's and Maxwell's equations are solved simultaneously for the metric and the e/m field. Perhaps you could supply a reference for the "well known" fact that gravity and EM do not interact?
 
  • #35
Here's a very simple-minded look: QFT says, without equivovcation, that any charged particle will emit and absorb photons whenever, and however pertubed. It makes good physical sense that this should be true: for an accelerating or disturbed charge, the radiation field is the adjustment that makes today's field tomorrow's field. It also makes sense that classical and quantum theory should agree on this point: the Leonard-Weichart potentials are rigorous solutions of Maxwell's eq.s for arbitrarily moving sources, and stem from the same interaction term, the four product of current and vector potential. (Gauge Invariance undelies all of the above.)

Under the circumstances under which I can transform my "g mu nu" to diagonal form, then standard QFT, standard physics applies locally. So, "I'm at rest and the train station is accelerating. Because there are all manner of interactions between me and the station -- I can see it, hear it, etc -- means I'm absorbing photons emitted by the station. Bingo. That means I'm going to radiate, the station is going to radiate because we are undergoing mutual acceleration. Seems to me that all the time, everything is radiating.

So I conclude that only under exceptional circumstances can radiation be effectively repressed. Something that forces every emitted photon back to the source for absorbtion, for example. The black hole that eats everybody's lunch.

Regards,
Reilly Atkinson
 

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