# Does a constantly accelerating charged particle emit EM radiation or not?

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?

• That's not what the Larmor formula states. What you should be asking is why the arising of an electromagnetic self-force that depends of the jerk, whilst classical physics is based on the, very experimentally tested, idea that the movement of a body can be described by, only, knowing the position and velocity of the body at a given instant...
– PML
Jul 12, 2013 at 20:11
• "The radiation of a uniformly accelerated charge is beyond the horizon: a simple derivation": arxiv.org/abs/physics/0506049 Jul 12, 2013 at 20:37
• "Does a uniformly accelerated charge radiate?": mathpages.com/home/kmath528/kmath528.htm Jul 12, 2013 at 20:37
• Jul 12, 2013 at 22:08
• Reminds me of the words of my fourth year EM theory professor: "Classical electrodynamics is a solved problem. Well, except for this." Jul 13, 2013 at 3:44

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.

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 https://journals.aps.org/ppf/abstract/10.1103/PhysicsPhysiqueFizika.1.3

• 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. Jul 13, 2013 at 3:27
• @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?
– user4552
Jul 13, 2013 at 3:38
• @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. Jul 13, 2013 at 4:36
• @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 . Jul 13, 2013 at 4:47
• @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. Jul 13, 2013 at 5:02

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.

• +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. Jul 13, 2013 at 6:54

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.

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.

Despite all the ink spilled over this problem, I think the answer is pretty simple:

• A uniformly accelerated charge in flat spacetime radiates continuously (the Larmor formula is correct).
• There is no backreaction on a uniformly accelerated charge (the Abraham-Lorentz formula is correct).
• There is no inconsistency because the cost of the radiation is paid at the start and end of the acceleration.

It seems impossible that brief reaction forces at the start and end can pay for an arbitrary amount of reactionless radiation in between, but they do, as shown below.

### Consistency of the Abraham-Lorentz and Larmor formulas

The manifestly covariant version of the Abraham-Lorentz force, called the Lorentz-Dirac force, is (in $$c=4π\epsilon_0=1$$ units and the $$+{-}{-}-$$ metric convention):

$$\dot{\mathbf p} = \frac23 q^2 \left( \ddot{\mathbf v} + \lVert\dot{\mathbf v}\rVert^2 \mathbf v \right)$$

where $$\mathbf p = m \mathbf v$$ and the dots are proper-time derivatives.

If the object moves inertially before $$τ_i$$ and after $$τ_f$$, and accelerates uniformly in between, then at $$τ_i$$ and $$τ_f$$ there is a reaction impulse given by

$$Δ\mathbf p = \frac23 q^2 \left( Δ\dot{\mathbf v} + \int \lVert\dot{\mathbf v}\rVert^2 \mathbf v \; dτ \right) \approx \frac23 q^2 Δ\dot{\mathbf v}$$

where the approximation is good if the period of nonzero jerk is short (and exact if it's instantaneous).

So the impulses at the beginning and end are proportional to the acceleration. The key point is that they're proportional to the four-acceleration, which is not constant during uniform acceleration: its value is $$\dot{\mathbf v} = g(\mathrm{\hat z}\cosh gτ + \mathrm{\hat t}\sinh gτ)$$ for acceleration in the $$\mathrm{tz}$$ plane at the scalar rate $$g$$. The sum of the initial and final impulse, representing the total "cost" of the acceleration, therefore depends on the elapsed time. In fact, the length of the sum is exponential in the elapsed proper time, but that shouldn't be surprising since the coordinate distance and time traveled are also exponential in the proper time.

Because $$\mathbf x = g^{-1}(\mathrm{\hat z}\cosh gτ + \mathrm{\hat t}\sinh gτ) = \dot{\mathbf v}/g^2$$ (for an appropriately chosen origin), you can actually express the cost in terms of the spacetime distance traveled:

$$Δ\mathbf p_i + Δ\mathbf p_f = -\frac23 q^2 g^2 (\mathbf x_f - \mathbf x_i)$$

The $$\mathrm{\hat t}$$ component of that in any inertial frame is $$ΔE = -\frac23 q^2 g^2 Δt$$ which is (the negative of) the Larmor power.

### Consistency with the equivalence principle

A stationary charge in a static gravitational field (e.g., resting on the surface of the Earth) can't radiate: if it did, it would be a source of unlimited free energy. The equivalence principle implies that this setup isn't locally distinguishable from a uniformly accelerating charge in Minkowski space. I said above that the charge in Minkowski space does radiate. Is there an inconsistency here? I don't think so.

The derivation of the Larmor power above wasn't local. It depended on a global inertial frame that doesn't exist in this case. Neither the sum $$Δ\mathbf p_i + Δ\mathbf p_f$$ nor the distance $$\mathbf x_f - \mathbf x_i$$ is meaningful in the static gravitational field.

The equivalence principle does imply that if detectors freefall past the stationary charge, and are close enough that spacetime is roughly flat, they should detect radiation from the charge at the rate given by the Larmor formula. That isn't a contradiction because it isn't a static configuration. If absorbing the radiation causes the orbits of the detectors to decay quickly enough, they will harvest no more energy than was put into their orbits. It's one thing to say that and another to prove it – but until someone shows that you can build a perpetual-motion machine this way, I am going to assume that you can't.

### What if you can't afford to stop?

Parrott objects that it doesn't make sense that half of the cost is paid in the indefinitely postponable future. He asks what happens if you shed enough mass (rocket fuel) that the final Lorentz-Dirac impulse is unphysical, because it would result in a spacelike four-momentum. Do you have to accelerate forever?

If the jerk is brief then the four-impulse is perpendicular to the four-velocity, so the four-momentum remains timelike if $$\frac23 q^2 g < m$$. Spreading the jerk out over time seems to make the problem worse.

I think you're saved here by the fact that you can't have charge without mass. The self-energy of a uniformly charged ball of radius $$r$$ is $$\frac35 q^2/r$$. If you plug that into the inequality, the charge cancels and you get $$g r < \frac9{10}$$. Even if uncharged, the ball has to satisfy $$gr<1$$ (where $$g$$ is the acceleration at its center) or else it will fall through the Rindler horizon. The precise value $$\frac9{10}$$ is meaningless – the factor $$\frac35$$ depends on the charge distribution and the $$\frac23$$ assumes no significant variation of the acceleration across the object – but the fact that the stopping criterion resembles $$gr<1$$ at all suggests to me that this is the right resolution.

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.

• You lost me at $a=c/\Delta t$. Why?
– user4552
Jul 16, 2013 at 4:37
• Well I was just pushing the Larmor formula to the limit by putting in a value for maximum acceleration. Jul 16, 2013 at 5:55
• 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.
– user4552
Jul 16, 2013 at 15:48