Does a uniformly accelerating charge radiate?

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Main Question or Discussion Point

It is a well-known consequence of classical electromagnetism that any accelerating charged particle must radiate (not necessarily visible) light. My question is, does the same hold true for a charged particle experiencing a uniform acceleration?
If a uniformly accelerating charge DID radiate, then by Einstein's equivalence principle, so would a charged particle at rest in a uniform gravitational field. But this is patent nonsense; if you have a particle that's just sitting on the earth, it will obviously not radiate light.

Any help would be greatly appreciated.
Thank You in Advance.
P.S. Let us restrict this to classical electromagnetism, not quantum theory.
 

Answers and Replies

Redbelly98
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This is a good question. I am certain that accelerated charges radiate, even uniformly accelerated ones. And as you said, this seems to imply a charge at rest in a gravitational field should also radiate. But I'm at a loss to explain where the energy comes from.

Hmmm ...
 
Shooting Star
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If a uniformly accelerating charge DID radiate, then by Einstein's equivalence principle, so would a charged particle at rest in a uniform gravitational field. But this is patent nonsense; if you have a particle that's just sitting on the earth, it will obviously not radiate light.
Likewise, a charge freely falling in a uniform gravitational field should radiate, but it can be thought of as the standing still in an IFR...

This question has not been resolved yet, according to most people. Feynman said, or rather showed, that a uniformly accelerated charge should not radiate, though his deduction is open to question according to many people. His point was that the formula for the power radiated was deduced for oscillating or bounded motions. However, the Larmor formula can also be deduced by other means, so this argument was a bit confusing for me.

I remember coming across a book "Surprises in Theoretical Physics" by Peierls, where it was definitely shown that a uniformly accelerating charge does not radiate. At that time, the equations were beyond my grasp. Get the book if you can.

One paper which is oft-quoted is "Principle of Equivalence, F. Rohrlich, Ann. Phys. 22, 169-191, (1963), page 173."

There are thousands of discussion in the net. Have a look at http://www.mathpages.com/home/kmath528/kmath528.htm" [Broken].
 
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I don't think accelerating charges always radiate. For example, if you run a DC current through a loop of wire then you have accelerating charges without radiation. I don't know under what conditions they do radiate, but the mere fact of acceleration is not enough.
 
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I don't think accelerating charges always radiate. For example, if you run a DC current through a loop of wire then you have accelerating charges without radiation. I don't know under what conditions they do radiate, but the mere fact of acceleration is not enough.
well.. under DC, charges travel with a constant velocity i.e. the 'drift velocity'. afaik, the assumption of a constant drift velocity is quite a good approximation and for all practical purposes, we can consider it to hold true.

in the opposing case i.e. of AC, the charges a accelerated due to fluctuating electric fields across the ends of the conductor and hence they do emit EM radiation which is a cause for the 'skin effect'.
 
Shooting Star
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I don't think accelerating charges always radiate. For example, if you run a DC current through a loop of wire then you have accelerating charges without radiation. I don't know under what conditions they do radiate, but the mere fact of acceleration is not enough.
When a point moves in a circular loop or a curve, its acceleration is not uniform. Even if the speed is constant, the centripetal acceleration is continually changing direction. So, the charges in a DC loop do not undergo uniform acceleration, but variable

According to classical EM, an accelerated point charge must radiate. That was the whole reason why the Rutherford model of an atom could not be justified, because the electron should radiate and spiral into the nucleus. The Bohr model had to be invented to account for this. The rest is history.

On the other hand, how a free point charge radiates and loses KE is seen in a bubble chamber or a cloud chamber. We have all seen photos of the typical spiral path

The electrons in a DC loop do accelerate when going through a curved potion. Even though the drift speed is constant, an individual electron is undergoing random motion and accelerating. Why they do not radiate is a QM phenomenon, and not within the bounds of classical EM.
 
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The electrons in a DC loop do accelerate when going through a curved potion.
sorry.. i missed the part where he said a 'loop'.

Why they do not radiate is a QM phenomenon, and not within the bounds of classical EM.
could you please tell me something as in what I should be searching for to know more about this QM phenomena?
 
Hurkyl
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(Classical) radiation can superimpose and just make a lump in the electromagnetic field. For example, if we happened to have a continuous matter distribution in a closed loop and a constant current through the loop, then (from some viewpoint), every point on the loop is radiating EM waves -- but they superimpose and just give you a boring magnetostatic field.

As I understand it, the way in which an EM wave slows down in (continuous) matter can be viewed in the same way: all the reradiated energy cancels out the component of the EM field that lies beyond the (slower than light-speed) wavefront. (again, this is the classical viewpoint!)

A charged particle with a constant velocity is also a situation with changing charge and current distributions -- ostensibly there would be EM radiation produced, but they combine together and you just get a boring EM field that travels along with the electron. (Still talking classically!)

So I don't find it surprising that an electron uniformly accelerated in its direction of travel doesn't generate an EM field that looks like radiation. (Similarly, I wouldn't find it surprising if it actually did generate such a field)
 
Shooting Star
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I'm sorry -- I should have been more precise. When I said it is a QM phenomenon, I was thinking about electrons, which are localized charges, the net drifting of which makes up direct current. If we think of current from the classical point of view, it would give rise just magnetostatics. However, there must be a correspondence between the two views in the case of many electrons.

could you please tell me something as in what I should be searching for to know more about this QM phenomena?
You can look up any basic physics (or statistical physics) book. The physics in a wire with DC is not much different from that of the wire with no current. A free electron in metal is continually radiating and absorbing energy, but this is nothing but the thermal radiation with which the electron gas is in equilibrium (considering the lattice atoms to be stationary). This crude picture of the the interaction between the electrons and the radiation is ultimately explained by QM.
 
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So, the charges in a DC loop do not undergo uniform acceleration, but variable

According to classical EM, an accelerated point charge must radiate.
Then the charges in a DC loop should have variable radiation, but they should still radiate.

(Classical) radiation can superimpose and just make a lump in the electromagnetic field. For example, if we happened to have a continuous matter distribution in a closed loop and a constant current through the loop, then (from some viewpoint), every point on the loop is radiating EM waves -- but they superimpose and just give you a boring magnetostatic field.

As I understand it, the way in which an EM wave slows down in (continuous) matter can be viewed in the same way: all the reradiated energy cancels out the component of the EM field that lies beyond the (slower than light-speed) wavefront. (again, this is the classical viewpoint!)

A charged particle with a constant velocity is also a situation with changing charge and current distributions -- ostensibly there would be EM radiation produced, but they combine together and you just get a boring EM field that travels along with the electron. (Still talking classically!)

So I don't find it surprising that an electron uniformly accelerated in its direction of travel doesn't generate an EM field that looks like radiation. (Similarly, I wouldn't find it surprising if it actually did generate such a field)
I like this explanation very much, thanks. I guess the key thing would be to determine exactly what the field of a uniformly accelerating charge should look like and then see if a charge at rest on Earth has that field.

I haven't finished reading http://www.mathpages.com/home/kmath528/kmath528.htm" [Broken], so I don't know the conclusion, but it seemed relevant.
 
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jtbell
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On the other hand, how a free point charge radiates and loses KE is seen in a bubble chamber or a cloud chamber. We have all seen photos of the typical spiral path
That loss of energy is mostly caused by the particle scattering from the atoms of the bubble chamber fluid (multiple scattering), not by radiation caused by the curvature of the path. When I was a grad student, I worked on bubble-chamber experiments. As I recall, the energy loss was modeled using multiple scattering equations.
 
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Shooting Star
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Then the charges in a DC loop should have variable radiation, but they should still radiate.
No, the radiation is proportional to the square of the magnitude of the acceleration, but that is beside the point here. The point is the reconciliation of the equivalence principle and classical EM theory. Till date, no body has given a satisfactory resolution.

That loss of energy is mostly caused by the particle scattering from the atoms of the bubble chamber fluid (multiple scattering), not by radiation caused by the curvature of the path. When I was a grad student, I worked on bubble-chamber experiments. As I recall, the energy loss was modeled using multiple scattering equations.
That was a sort of a “slip of the mind”. I unconsciously correlated two different things just because both pertain to losing energy by particles. (If the particle does not ionise, how shall we see the track? I am embarrassed.)

But low energy electrons lose energy mostly through bremsstrahlung, without ionising any atom, when they pass close to a nucleus. How do we see their tracks, if at all?
 
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Yes, Charges under uniform radiation do radiate, but because of the radiation reaction, uniform acceleration cannot be sustained during the radiation. It follows readily from the Abraham-Lorenz equation.
 
Shooting Star
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Yes, Charges under uniform radiation do radiate,
I know it's a long typo, but by "radiation", do you mean acceleration?

but because of the radiation reaction, uniform acceleration cannot be sustained during the radiation. It follows readily from the Abraham-Lorenz equation.
What about where uniform acceleration is maintained in spite of the radiation reaction? We will pour in whatever amount of energy needed.

Have you thought about what happens when a point charge is in free fall in a uniform gravitational field, and why this question is meaningful to ask? That is what this discussion is about. The original question was, if a point charge is sitting in a g-field, would it radiate? By the principle of equivalence, it should.
 
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if a point charge is sitting in a g-field, would it radiate? By the principle of equivalence, it should.
Sure, except for a false dichotomy (before even involving gravity): how do you define "whether it radiates" to be observer independent?
 
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Redbelly98
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Can somebody post a number? 1.6x10^-19 Coulombs, 9.8 m/s^2. What is the radiated power in Watts?

cesiumfrog wrote:
Sure, except for a false dichotomy (before even involving gravity): how do you define "whether it radiates" to be observer independent?
Good point. In other words, for a charge at rest in a gravitational field, is radiation detected by (1) an observer also sitting in a g-field, and/or (2) an observer in free-fall?
 
Redbelly98
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Can somebody post a number? 1.6x10^-19 Coulombs, 9.8 m/s^2. What is the radiated power in Watts?
Okay, dimensional analysis time:
Classical electromagnetism, so no h or G allowed.
To get a value in Watts:
1/(4*pi*e_0) * q^2 * a^2 / c^3

I get 8 * 10^-50 Watts. Numerical factors aside, we are talking about an incredibly tiny amount of power!
 
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I have been thinking about this a little more. I don't think that the equivalence principle is in any way incompatible with accelerating charges radiating. After all, the equivalence principle doesn't say that a uniform gravitational field is the same as accelerating reference frame, it just says that you cannot make any experiment that will detect the difference.

So a charge at rest in a gravity field next to an antenna at rest should detect nothing. Similarly the same charge and same antenna uniformly accelerating together should also detect nothing. Other configurations (e.g. dropped charge w/ resting antenna) should also produce similar pairs of equivalent experimental results.

Whether it is due to radiation of the charge or to the antenna's motion through the charge's field is frame-dependent in the same way that the electric and magnetic fields are frame-dependent in special relativity. As long as the experimental result is the same in each case the equivalence principle holds.

This is all a guess on my part and I have certainly not worked out the math to justify it.
 
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I know it's a long typo, but by "radiation", do you mean acceleration?
Oh yes - my mistake


Have you thought about what happens when a point charge is in free fall in a uniform gravitational field, and why this question is meaningful to ask? That is what this discussion is about. The original question was, if a point charge is sitting in a g-field, would it radiate? By the principle of equivalence, it should.
Well, I tried to solve Newtons 2nd law along with the Abraham-Lorenz equation. ie:

[tex]
k\frac{d}{dt}a(t)+ma(t)=mg
[/tex]

The answer baffled me, to be honerst:

[tex]
a(t)=g\left(1-e^{-\frac{mt}{k}}\right)
[/tex]

Well so much for messing with the Abraham Lorenz formula...
 
Shooting Star
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Sure, except for a false dichotomy (before even involving gravity): how do you define "whether it radiates" to be observer independent?
Make a sphere out of photographic film, as small as desired. Put it around the charge. If a photon is emitted, it will leave a mark on the film which can be examined by any observer. This will be an absolute event and will remove any “dichotomy”.


[tex]
k\frac{d}{dt}a(t)+ma(t)=mg
[/tex]

The answer baffled me, to be honerst:

[tex]
a(t)=g\left(1-e^{-\frac{mt}{k}}\right)
[/tex]
(EDITED)

I'm thinking about whether this is another runaway solution to the Abraham-Lorentz formula. (Look up Jackson for runaway solutions.)
 
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Make a sphere out of photographic film, as small as desired. Put it around the charge. If a photon is emitted, it will leave a mark on the film which can be examined by any observer. This will be an absolute event and will remove any “dichotomy”.
No, this is still not observer independent. In this experiment, the photographic film functions as the observer, and the way in which it is moving (i.e. whether it is accelerating or not) affects whether or not a mark will be made on the film.
 
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As mentioned already, this question is discussed by Peierls in Suprises in Theoretical Physics.
Chapter 8, Relativity
8.1 Radiation in Hyperbolic Motion
pages 160-166
some clever use of Google Book search should provide the whole text, see below for search terms:
pg 160: "it is therefore essential to use relativistic kinematics"
pg 161: "is interested in the emission of radiation"
pg 162: "involve the surprising finding that uniform acceleration"
pg 163: "Fulton and Rohrlich calculate the retarded potential"
pg 164: "This leaves the paradox with the equivalence principle"
pg 165: "radiation which Fulton and Rohrlich showed" - not available ?
pg 166: "It might be added that the paradox is somewhat academic"

Peter
 
atyy
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The equivalence principle gives the wrong answer here. An accelerating charge does radiate.

General Relativity by J L Martin says that this is due to tidal forces, which is the signature of real gravity, as opposed to fake gravity produced by an accelerated frame.

There's an interesting comment in General Relativity by Wald that a second order approximation of GR is needed to reproduce Newtonian physics. I suppose since the equivalence principle is a first order approximation of GR, it will fail to predict some things that Newtonian physics gets right.
 
atyy
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One of the most outspoken critics was J. L. Synge, who famously wrote (in his 1960 text on general relativity)

I have never been able to understand this Principle… Does it mean that the effects of a gravitational field are indistinguishable from the effects of an observer’s acceleration? If so, it is false. In Einstein’s theory, either there is a gravitational field or there is none, according as the Riemann tensor does not or does vanish. This is an absolute property; it has nothing to do with any observer’s world-line. Spacetime is either flat or curved… The Principle of Equivalence performed the essential office of midwife at the birth of general relativity, but… I suggest that the midwife be now buried with appropriate honors and the facts of absolute space-time faced.

From <http://www.mathpages.com/home/kmath622/kmath622.htm> [Broken]
 
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Shooting Star
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First: my apologies to Troels. I had abruptly disappeared in the middle of a conversation, so to say. (I was out of circulation for a few months due to health problems and then some personal issues. It feels very good to be back.)

Hi Troels,

I didn’t notice the error in the sign in your equation the first time, and the result left me quite confused initially. The actual equation would be (refer post #19):

[itex]

ma(t)= k\frac{d}{dt}a(t) + mg

[/itex]

which gives the result a=g if the particle is dropped from rest, which will eventually make the speed greater than c. This is only to be expected from a hodgepodge of classical electrodynamics and Newtonian gravity. The Abraham-Lorentz formula is anyway valid for IFRs, which this system is not because of the superimposed constant gravitational field.

The Equivalence Principle was arrived at by considering falling frames in what is called a Symmetric Homogeneous Gravitational Field (SHGF), which is the typical scenario that is used when considering projectile motion – that is, a constant gravitational field pointing in one direction superimposed on an IFR. The specialty of such a gravitational field is that it creates no tidal stress in a body in free fall. Such a field is not compatible with special relativity, and that is why we consider events happening for a vanishingly short time over a vanishingly small volume of space in a falling frame in this field, while deriving the Equivalence Principle. A long-term analysis of a falling charge in this scenario is meaningless.

No, this is still not observer independent. In this experiment, the photographic film functions as the observer, and the way in which it is moving (i.e. whether it is accelerating or not) affects whether or not a mark will be made on the film.
When in the original question you asked about an accelerating charge, I presumed that you meant the acceleration was wrt an IFR. The sphere or enclosure made of photographic film or some such device I had mentioned is obviously meant to be static wrt the IFR, and the charge is inside that enclosure and accelerating. If EM radiation is emitted by the charge, it has to fall on the film somewhere, thus giving an absolute indication of whether radiation has been emitted by the charge or not. There may be ( I am not very sure in this context) certain frames in certain kinds of motion wrt the charge where the news of this radiation does not reach at all, but that would not prevent the accelerating charge from radiating, if it does at all, as the above thought experiment elucidates.

There are claims that the co-moving frame will see only a static electric field and not radiation, but I haven’t gone through the derivation. (American Journal of Physics -- February 2006 -- Volume 74, Issue 2, pp. 154-158.)
 

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