49
$\begingroup$

The Abraham-Lorentz force gives the recoil force, $\mathbf{F_{rad}}$, back on a charged particle $q$ when it emits electromagnetic radiation. It is given by:

$$\mathbf{F_{rad}} = \frac{q^2}{6\pi \epsilon_0 c^3}\mathbf{\dot{a}},$$

where $\mathbf{\dot{a}}$ is the rate of change of acceleration.

If a particle has a constant acceleration, $\mathbf{\dot{a}}=0$, then there is no reaction force acting on it. Therefore the particle does not lose energy.

Does this mean that a constantly accelerating charged particle does not emit electromagnetic radiation?

$\endgroup$
41
$\begingroup$

This is an old, hard, controversial question. It is in some sense not well defined, because there are subtle ways in which it can be difficult to pin down the distinction between a radiation field and a non-radiative field. Perhaps equivalently, there are ambiguities in the definition of "local." If an accelerating charge did radiate, it would cause a problem for the equivalence principle.

There are arguments by smart people who claim that an accelerating charge doesn't radiate (Harpaz 1999; Feynman's point of view is presented at http://www.mathpages.com/home/kmath528/kmath528.htm ). There are arguments by smart people who claim that an accelerating charge does radiate (Parrott 1993). There are other people who are so smart that they don't try to give a yes/no answer (Morette-DeWitt 1964, Gralla 2009, Grøn 2008). People have written entire books on the subject (Lyle 2008).

A fairly elementary argument for the Feynman point of view is as follows. Consider a rigid blob of charge oscillating (maybe not sinusoidally) on the end of a shaft. If the oscillations are not too violent, then in the characteristic time it takes light to traverse the blob, all motion is slow compared to c, and we can approximate the retarded potentials by using Taylor series (Landau 1962, or Poisson 1999). This procedure will lead us to compute a force and therefore the lower derivatives (x'') from the higher derivatives (x'''); but this is the opposite of how the laws of nature normally work in physics. Even terms in the Taylor series are the same for retarded and advanced fields, so they don't contribute to radiation and can be ignored. In odd terms, x' obviously can't contribute, because that would violate Lorentz invariance; therefore the first odd term that can contribute is x'''. Based on units, the force must be a unitless constant times $kq^2x'''/c^3$; the unitless constant turns out to be 2/3; this is the Lorentz-Dirac equation, $F=(2/3)kq^2x'''/c^3$. The radiated power is then of the form $x'x'''$. This is nice because it vanishes for constant acceleration, which is consistent with the equivalence principle. It's not so nice because you get nasty behavior such as exponential runaway solutions for free particles, and causality violation in which particles start accelerating before a force is applied.

Integration by parts lets you reexpress the radiated energy as the integral of $x''x''$, plus a term that vanishes over one full cycle of periodic motion. This gives the Larmor formula $P=(2/3)kq^2a^2/c^3$, which superficially seems to violate the equivalence principle.

Note that starting from the expression $x'x'''$ for the radiated power, you can integrate by parts and get $x''x''$ plus surface terms. On the other hand, if you believe that $x''x''$ is more fundamental, you can integrate by parts and get $x'x'''$ plus surface terms. So this fails to resolve the issue. The surface terms only vanish for periodic motion.

In a comment, Michael Brown asks the natural question of whether the issue can be resolved by experiment. I don't know that experiments can resolve the issue, since the issue is really definitional: what constitutes radiation, and how do we describe the observer-dependence of what constitutes radiation? In particular, if observers A and B are accelerated relative to one another, it's not obvious that what A calls a radiation field will also be a radiation field according to B. We know that bremsstrahlung exists and that it's the process responsible for the x-rays that produce an image of my broken arm. There doesn't seem to be much controversy over whether the power generated by the x-ray tube can be calculated according to $x''x''$. What about the frame of the decelerating electron, in which $x''=0$? The question then arises as to whether this frame can be extended far enough to encompass the photographic film or CCD chip that forms the image.

It gets even tougher when we deal with gravitational accelerations. To a relativist, a charge sitting on a tabletop has a proper acceleration of 9.8 m/s2. Does this charge radiate? How about a charge orbiting the earth (Chiao 2006) or free-falling near the earth's surface? Lyle 2008 has this clear-as-mud summary (gotta love amazon's Look Inside! feature):

To a first approximation, remaining close enough to the charge for curvature effects to be negligible, in the sense that the metric components remain roughly constant, GR+SEP tells us that there should not be electrogravitic bremsstrahlung for a charge following a geodesic, although there will when the charge follows curves [satisfying the equations of motion], due to its deviation from the geodesic.

Unfortunately, calculations show that the electromagnetic radiation from a free-falling charge, if it exists as suggested by the Larmor $x''x''$ formula, would be many, many orders of magnitude too small to measure.

Chiao, http://arxiv.org/abs/quant-ph/0601193v7

Gralla, http://arxiv.org/abs/0905.2391

Grøn, http://arxiv.org/abs/0806.0464

Harpaz, http://arxiv.org/abs/physics/9910019

Landau and Lifshitz, The classical theory of fields

Lyle, "Uniformly Accelerating Charged Particles: A Threat to the Equivalence Principle," http://www.amazon.com/Uniformly-Accelerating-Charged-Particles-Equivalence/dp/3540684697/ref=sr_1_1?ie=UTF8&qid=1373683154&sr=8-1&keywords=Uniformly+Accelerating+Charged+Particles%3A+A+Threat+to+the+Equivalence+Principle

C. Morette-DeWitt and B.S. DeWitt, "Falling Charges," Physics, 1,3-20 (1964); copy available at http://www.scribd.com/doc/100745033/Dewitt-1964 (may be illegal, or may fall under fair use, depending on your interpretation of your country's laws)

Parrott, http://arxiv.org/abs/gr-qc/9303025

Poisson, http://arxiv.org/abs/gr-qc/9912045

$\endgroup$
  • 1
    $\begingroup$ What about the experimental side of the question? Presumably this is impractical to test in a linear accelerator, otherwise the theorists wouldn't be arguing. $\endgroup$ – Michael Brown Jul 13 '13 at 3:27
  • 1
    $\begingroup$ @MichaelBrown: I don't know that experiments can resolve the issue, since the issue is really definitional: what constitutes radiation, and how do we describe the observer-dependence of what constitutes radiation? $\endgroup$ – Ben Crowell Jul 13 '13 at 3:38
  • 3
    $\begingroup$ @MichaelBrown: Impractical indeed. Jackson includes a calculation for linear accelerators in 14.2, where he shows the accelerating field needs to be on the order of 2x10^{14} MeV/meter to get significant electron radiation loss, a field well beyond the state of the art. $\endgroup$ – Art Brown Jul 13 '13 at 4:36
  • 1
    $\begingroup$ @MichaelBrown if we accept that angular acceleration is acceleration, the synchrotron radiation is experimental proof of "accelerating charges radiate" en.wikipedia.org/wiki/Synchrotron_radiation . The numbers for linear acceleration are why we are discussing an ILC after LHC en.wikipedia.org/wiki/International_Linear_Collider . $\endgroup$ – anna v Jul 13 '13 at 4:47
  • 2
    $\begingroup$ @annav I'm aware of synchrotron radiation, but the question is about uniform linear acceleration, so these arguments wouldn't apply. A rotating frame is not required to be equivalent to a nonrotating one due to the equivalence principle, and a rotating frame does not have an event horizon like a uniformly accelerating frame does. $\endgroup$ – Michael Brown Jul 13 '13 at 5:02
14
$\begingroup$

The Abraham-Lorentz equation does not apply to a constantly accelerating charge.

From the Lienard Wiechert fields, a constantly accelerating charge produces a field which falls off as the inverse distance, the very definition of radiation.

Where's the disconnect? In the Wikipedia derivation of the A-L force (and likewise in Jackson section 17.2), there's a step that assumes periodic motion, in order to make a boundary term vanish and so give the result you quoted. The formula in question (derived from the Larmor power formula) for the reaction force is:

$$ \int_{t_1}^{t_2} \mathbf{F}_\text{rad} \cdot \mathbf{v} \;dt = - \int_{t_1}^{t_2} \frac{2}{3} \frac{e^2}{c^3} \mathbf{\dot v} \cdot \mathbf{\dot{v}}\; dt = \frac{2}{3} \frac{e^2}{c^3} \int_{t_1}^{t_2} \mathbf{ \ddot{v}}\cdot \mathbf v\;dt - \frac{2}{3} \frac{e^2}{c^3} \left (\mathbf{\dot{v}} \cdot \mathbf{v}) \right|_{t_1}^{t_2} $$

Clearly, a constantly accelerating charge does not satisfy the periodic motion assumption, so the boundary term does not go away, and the quoted A-L formula does not apply in this case.

$\endgroup$
  • 1
    $\begingroup$ +1, but you have a typo, it is $\boldsymbol{ \ddot{v}\cdot {v}}dt$ and not $\boldsymbol{ \ddot{v}\cdot \dot{v}}dt$, in the right side of the equation. $\endgroup$ – Trimok Jul 13 '13 at 6:54
5
$\begingroup$

I'm not entirely sure of this answer, comments appreciated

The Abraham-Lorentz equation is not a "cause-effect" equation. It doesn't say that "an acceleration of $\dot a$ will give rise to a force $F_{rad}$". Rather it says that "if a charged particle has a jerk of $\dot a$, it will coexist with a recoil force of $F_{rad}$".

On the other hand, the Larmor formula1 calculates the power of radiation given the acceleration.

Note that these two formulae are not contradictory. Even though one knows the power, one doesn't know the distribution of emitted photons. Which means that the exact recoil force experienced is unknown unless you solve from first principles. This is where the Abraham-Lorentz force comes in.

1. Which may or may not apply onl to an oscillating system. Feynman claims it is a special case of a more general power series, but I'll need to decipher his words.

$\endgroup$
1
$\begingroup$

Charged particle is accompanied with EM radiation (has field that falls with distance as $1/r$) when it moves with acceleration. This can be shown to be a consequence of Maxwell's equations, well-verified and reliable part of physics. It is immaterial whether the charged particle is a point or an extended body.

None of that depends on the value that Larmor's or the Lorentz-Abraham formula give for energy or force.

Larmor's formula is based on the energy interpretation of integral Poynting theorem. In macroscopic EM theory the theorem is valid for extended charged bodies, but in microscopic theory of point particles, it isn't, because the expression

$$ \mathbf j\cdot\mathbf E $$

is not defined at the point where the particle is. The Poynting expressions can be integrated over any region that is free of charged points, but then it cannot be used to infer anything on energy.

Consequently, Larmor's formula is correctly derived and well working for macroscopic charged bodies. It is neither correctly derived nor well working for point particles.

The Lorentz-Abraham (LA) force can be derived in several ways:

  • calculate the force on a charged sphere due to its own parts - gives the LA force (this is due to Lorentz and Abraham);

  • for oscillating charged body long-term energy balance suggest the LA force should be introduced (due to Lorentz, I think).

Both methods give the same expression for the additional force (strictly, the first one gives more complicated formula that gives value very close to the LA force). However, both assume the body is extended with charge density finite at all points; the first one to integrate interactions of parts, the second one to apply Larmor's formula.

Summary: all charged bodies radiate when accelerated, this is based on Maxwell's equations that are well established and work well both for point and extended charged particles. The Larmor and the Lorentz-Abraham are well established only for extended bodies. It is groundless to apply them to point particles.

$\endgroup$
0
$\begingroup$

Comment on radiation emission from uniformly accelerated charged particle Referring to the commonly asked question as to whether there is radiation emission from uniformly accelerated charged particle. If emission is there, how to rationalize it by referring to the momentum and energy balance equations that express respectively the steady growth of momentum and kinetic energy of the charge accelerated by the application of constant electric field. It is important to note that the dynamic properties that are referred in the momentum and energy balance equations are canonically averaged values of the respective properties. Under increased acceleration of the charged particle in the external field, coherent interaction of field and particle progressively deviates from the nonlocal regularity of having virtual exchange of radiation. The evolution becomes exceedingly nonstationary with incessant radiation emission and absorption, and at the onset of decoherence the indicated energy fluctuation reaches criticality. The dynamics at the onset is best represented by canonically averaged properties. Under the nonstationary evolution the energy balance expresses the general growth rate of kinetic energy, where the jerk related power in the wake of radiation emission and absorption contributes. The power manifesting out of jerk is constituted of the two complementary rates respectively due to radiation loss and radiation recoil based kinetic retardation over the characteristic time period of 2q2/3m¬¬0c3. The jerk related power is thereby expressed in canonically averaged form over the characteristic time. For the uniformly accelerating charge this average power representing the overall effect from the energy-momentum fluctuation is characteristically null with the result that the kinetic energy of charge grows steadily under the constant external field. Though the recoil momentum does not figure in the momentum balance of the uniformly accelerated case, the field of the emitting radiation imparts osculation effect on the dynamic course of the accelerated charge. The osculation is registered as canonically averaged torsion of the quantum trajectories that represent the dynamic course in the path intergral approach due to Feynman. Considering that the radiative motion is executed with the measure of minimum self energy loss of the charge, it is possible to prove that the upper most limit of torsion perturbation beyond which the ever present nonlocal defense critically fails. The critical torsion is expressible as , being phase velocity of the matter waves and is light-like unit vector. The expression shows that the matter waves phase velocity ( ) is pushed down to its limiting value of the signal speed in order that the nonlocal stress reaches its upper limit of defense before it yields. The radiation emission from uniformly accelerated is thus implied by the recoiling torsion effect described above. As noted in inertial frame, an observer in noninertial frame will also consider that the radiation emission marks the failure of coherent interaction in the field particle system and he will conclude that uniformly accelerating charge radiates. An observer on earth surface will distinguish free falling charge from another charge which is at rest with him. To him the free falling charge unlike the other one is in nonstationary state leading to the steady growth of its average kinetic energy and constant loss of radiation. With the noted distinction of energy states of the two charged particles it is inappropriate to justify for their identical energy using the equivalence principle. The quantum states of the field-particle evolution need not be redefined by using the principle.

$\endgroup$
-6
$\begingroup$

The question seems to be: Does the Larmor formula only hold for oscillating charges?

The Larmor formula says the power $P$ radiated by a system with electron charge $e$ and acceleration $a$ is given approximately by:

$$P \approx \frac{e^2}{\epsilon_0 c^3} a^2$$

Let us investigate what happens if we seek to maximize the power $P$ radiated. We need to maximize the acceleration $a$. This can be done by assuming a relationship:

$$a = \frac{c}{\Delta t}$$

Thus we have:

$$\frac{\Delta E}{\Delta t} \approx \frac{e^2}{\epsilon_0 c^3} \left(\frac{c}{\Delta t}\right)^2$$

$$\Delta E \approx \frac{e^2}{\epsilon_0 c} \frac{1}{\Delta t}$$

Now the definition of the fine structure constant $\alpha\approx1/137$ is given by:

$$\alpha = \frac{e^2}{4 \pi \epsilon_0 \hbar c}$$

Thus we have approximately:

$$\frac{e^2}{\epsilon_0 c} \approx h$$

Thus, finally we have:

$$\Delta E \ \Delta t \approx h.$$

Thus if we look for a system that emits maximum power then the Larmor formula leads us to a quantum uncertainty relationship indicating an oscillating system. Of course we do not expect such a system to actually radiate by the principles of quantum mechanics.

Even so this fact seems to indicate that the formula applies to oscillating systems (up to the limit of quantum oscillating systems) rather than those with a linear motion.

$\endgroup$
  • 3
    $\begingroup$ You lost me at $a=c/\Delta t$. Why? $\endgroup$ – Ben Crowell Jul 16 '13 at 4:37
  • $\begingroup$ Well I was just pushing the Larmor formula to the limit by putting in a value for maximum acceleration. $\endgroup$ – John Eastmond Jul 16 '13 at 5:55
  • 3
    $\begingroup$ I don't think the physical content of this answer holds water. The last two paragraph don't make sense to me. Everything before that looks like formal manipulations without any physical justification. $\endgroup$ – Ben Crowell Jul 16 '13 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.