Radiation reaction


The radiation reaction (force) is the force an accelerating point charge applies onto itself due to emission of electromagnetic radiation. It is also called Abraham-Lorentz force or radiation damping force. Assuming non-relativistic speeds, where vcv\ll c, the reaction is

Frad=μ0q26πca˙\mathbf{F}_\text{rad}=\frac{\mu_{0}q^{2}}{6\pi c}\dot{\mathbf{a}}

where qq is the charge, cc is the speed of light and a˙\dot{\mathbf{a}} is its jerk, the time derivative of acceleration.

Derivation

Conservation of energy demands that any energy being radiated by an accelerating charge has to be removed from somewhere else. That somewhere can only be the kinetic energy of the charge, meaning that a radiating particle accelerates less that if it didn't radiate, or equivalently less than a neutral particle. Basically, radiating energy enacts a sort of recoil force back on the charge itself, which counteracts and reduces the acceleration: we call this the radiation reaction.

For a nonrelativistic particle, vcv\ll c, the total radiant power radiated is given by the Larmor formula:

P=μ0q2a26πcP=\frac{\mu_{0}q^{2}a^{2}}{6\pi c}

As mentioned before, this energy has got to go somewhere, so all this energy is removed from the fields. But which fields? We know from the Liénard-Wiechert potentials that an accelerating point charge has not one, but two fields attached to it, the velocity and acceleration fields. The Larmor formula is derived from the radiation of the acceleration field, as that's the only one that actually causes radiation to be emitted. It does therefore miss a piece: the velocity field is still energy and still needs to be counted to avoid accidentally tearing a hole in conservation of energy. The velocity and acceleration fields constantly interchange energy as the particle accelerates and decelerates. Some of that energy is lost by radiation (Larmor formula), but energy can be lost by other ways. Were we to find the force by only the Larmor power, we'd be missing a piece. The energy lost in any given time interval must therefore be the one lost to Larmor power, plus the one transferred to the velocity field by all other phenomena1.

If we consider a time interval in which the system returns to its initial state (periodic motion), the energy in the velocity field is the same at both the start and end of the motion. The only overall loss is due to radiation. In this context, we can use the Larmor formula to state that the radiation reaction force Frad\mathbf{F}_\text{rad} is given entirely by the energy lost to radiation:

U=t1t2Fradvdt=μ0q26πct1t2a2dtU=\int_{t_{1}}^{t_{2}}\mathbf{F}_\text{rad}\cdot \mathbf{v}dt=- \frac{\mu_{0}q^{2}}{6\pi c}\int_{t_{1}}^{t_{2}}a^{2}dt

This is, again, assuming that the system is in identical states in t1t_{1} and t2t_{2}. In other words, we must integrate over an entire period. The integral is solved with integration by parts:

t1t2a2dt=t1t2dvdtdvdtdt=(vdvdt)t1t2t1t2d2vdt2vdt\int_{t_{1}}^{t_{2}}a^{2}dt=\int_{t_{1}}^{t_{2}} \frac{d\mathbf{v}}{dt}\cdot \frac{d\mathbf{v}}{dt}dt=\left.{\left( \mathbf{v}\cdot \frac{d\mathbf{v}}{dt} \right)}\right|_{t_{1}}^{t_{2}}-\int_{t_{1}}^{t_{2}} \frac{d^{2}\mathbf{v}}{dt^{2}}\cdot \mathbf{v}dt

The boundary term drops because the state is identical in t1t_{1} and t2t_{2}. Then

U=t1t2Fradvdt=μ0q26πct1t2a˙vdtt1t2(Fradμ0q26πca˙)vdt=0U=\int_{t_{1}}^{t_{2}}\mathbf{F}_\text{rad}\cdot \mathbf{v}dt=\frac{\mu_{0}q^{2}}{6\pi c}\int_{t_{1}}^{t_{2}}\dot{\mathbf{a}}\cdot \mathbf{v}dt\quad\Rightarrow \quad \int_{t_{1}}^{t_{2}}\left( \mathbf{F}_\text{rad}- \frac{\mu_{0}q^{2}}{6\pi c}\dot{\mathbf{a}} \right)\cdot \mathbf{v}dt=0

This is certainly true if

Frad=μ0q26πca˙\boxed{\mathbf{F}_\text{rad}=\frac{\mu_{0}q^{2}}{6\pi c}\dot{\mathbf{a}}}

This is known as the Abraham-Lorentz formula for radiation reaction. Sort of, at least. As it stands, this formula is on thin ice. It is valid only in a very particular form of motion (perfectly periodic) and we don't really have any information on the components of v\mathbf{v} that are perpendicular to v\mathbf{v} as those will get deleted by the Scalar product anyway and therefore never matter in the integral. The actual proof of this form is available, but not with this premise; see > Mechanism below.

What's more urgent is the consequences of this law. Take Newton's second law:

Frad=μ0q26πca˙=maF_\text{rad}=\frac{\mu_{0}q^{2}}{6\pi c}\dot{a}=ma

under the assumption that the charge is not subject to any other force. This is a first-order linear Ordinary differential equation with a simple exponential solution2:

a(t)=a0et/τa(t)=a_{0}e^{t/\tau}

where

τμ0q26πmc\tau\equiv \frac{\mu_{0}q^{2}}{6\pi mc}

If you don't intuitively catch the problem, allow me to make you notice that this is the acceleration of the charge, which is increasing exponentially, by itself, with no external force. This is called a runaway solution and has been studied extensively due to its absurdity. As it happens, you can just pretend these don't exist by arbitrarily setting a0=0a_{0}=0, but if you do, you'll just get yourself in an even worse problem: if a0a_{0}, whenever an external force is applied, the particle ends up responding before the force is applied by a time τ\tau, an evidently nonsensical phenomenon called acausal preacceleration. This is the point where I'd like to tell you that there's a niche, specific fact hidden somewhere that we can use to solve the problem at once (sort of like how we solve advanced potentials by just claiming they aren't physically valid). I have no such fact. Not even relativity solves it, and we don't really know if quantum physics does. Either way, there's clearly something broken here and we don't know what it is, though it may have something to do with point charges being an approximation (extended distributions don't have this issue) and/or with a flaw in the classical non-quantum theory specifically.

Mechanism

If the radiation reaction is generated by the fields acting back on the charge, then there must be some way to calculate this force from the fields themselves. The issues is that fields blow up at point charges, so actually calculating the force seems unreasonable. To solve the issue, we'll instead use a charge distribution, so that there are no singularities in the fields, then take the limit so that the volume of the distribution goes to zero. The important part here is to look at Newton's third law and the reactions within internal pieces of the distribution. In general, the electromagnetic force between two elements AA and BB of the distribution do not balance each other out, in that the force from AA to BB is not equal and opposite to the force from BB to AA. If you integrate over all the distribution, you'll then be left with a residual internal force that is never fully cancelled out: a self-force. It is this self-force (and the corresponding failure of Newton's third law) that causes the radiation reaction.

For the sake of simplicity, in order to calculate this imbalance we'll use a shape of this sort:

Diagram Radiation reaction mechanism.svg

The charge distribution is two "blobs" of charge, each containing half of the total charge, separated by a distance dd. These of course move over time to incur the radiation reaction. Of course, this is not really a realistic model, but it does exhibit the behavior that we are looking for and does so in a simple manner. When d0d\to 0 we return to a point charge and we get the Abraham-Lorentz formula again.

Say the charges are instantaneously at rest (v=0\mathbf{v}=0) at the retarded time. The electric field on (1)(1) due to (2)(2) is

E1=q/24πε0r(ru)3[(c2+ra)u(ru)a]\mathbf{E}_{1}=\frac{q/2}{4\pi \varepsilon_{0}} \frac{\mathfrak{r}}{(\boldsymbol{\mathfrak{r}}\cdot \mathbf{u})^{3}}[(c^{2}+\boldsymbol{\mathfrak{r}}\cdot \mathbf{a})\mathbf{u}-(\boldsymbol{\mathfrak{r}}\cdot \mathbf{u})\mathbf{a}]

where u=cr^\mathbf{u}=c \hat{\boldsymbol{\mathfrak{r}}} and r=lx^+dy^\boldsymbol{\mathfrak{r}}=l \hat{\mathbf{x}}+d \hat{\mathbf{y}} so that ru=cr\boldsymbol{\mathfrak{r}}\cdot \mathbf{u}=c\mathfrak{r}, ra=la\boldsymbol{\mathfrak{r}}\cdot \mathbf{a}=la and r=l2+d2\mathfrak{r}=\sqrt{ l^{2}+d^{2} }. We're only interested in the xx component, since the yy component will cancel out when we add the forces on the two ends (same goes for the magnetic forces). We have ux=cl/ru_{x}=cl/\mathfrak{r}, so

E1,x=q8πε0c2lc2ad2(l2+d2)3/2E_{1,x}=\frac{q}{8\pi \varepsilon_{0}c^{2}} \frac{lc^{2}-ad^{2}}{(l^{2}+d^{2})^{3/2}}

The system is symmetrical, so E1,x=E2,xE_{1,x}=E_{2,x}, so the net force must be

Fself=q2(E1+E2)=q28πε0c2lc2ad2(l2+d2)3/2x^\mathbf{F}_\text{self}=\frac{q}{2}(\mathbf{E}_{1}+\mathbf{E}_{2})=\frac{q^{2}}{8\pi \varepsilon_{0}c^{2}} \frac{lc^{2}-ad^{2}}{(l^{2}+d^{2})^{3/2}}\hat{\mathbf{x}}

We now want to expand in powers of dd, as when the charge's size goes to zero, all positive powers of dd will go to zero. Using a Taylor series about trt_{r} the position becomes

x(t)=x(tr)+x˙(tr)(ttr)+12x¨(tr)(ttr)2+13!x...(tr)(ttr)3+x(t)=x(t_{r})+\dot{x}(t_{r})(t-t_{r})+ \frac{1}{2}\ddot{x}(t_{r})(t-t_{r})^{2}+ \frac{1}{3!}\dddot{x}\,(t_{r})(t-t_{r})^{3}+\ldots

By defining TttrT\equiv t-t_{r} for brevity, we have

l=x(t)x(tr)=12aT2+16a˙T3+l=x(t)-x(t_{r})=\frac{1}{2}aT^{2}+ \frac{1}{6}\dot{a}T^{3}+\ldots

TT is determined by the retarded time condition

(cT)2=l2+d2(cT)^{2}=l^{2}+d^{2}

so that

d=(cT)2l2=cT1(aT2c+a˙T26c+)2=cTa28cT3+()T4+d=\sqrt{ (cT)^{2}-l^{2} }=cT\sqrt{ 1- \left( \frac{aT}{2c}+ \frac{\dot{a}T^{2}}{6c}+\ldots \right)^{2} }=cT- \frac{a^{2}}{8c}T^{3}+(\ldots)T^{4}+\ldots

where we omitted the coefficient of T4T^{4} since we won't need it. This equation is d(T)d(T), but we want T(d)T(d). We need to invert the series to do so, a process formally known as reversion of series. That's a bit complicated for the very simplistic model we're using, so we'll use more heuristic argument just to prove the point. We'll use something that comes in perturbation theory a few times, which is to replace dd with its first order approximation3. To first order, dcTd\simeq cT and so Td/cT\simeq d/c. Then, replacing this TT in the cubic term

dcTa2d38c4Tdc+a2d38c5d\simeq cT- \frac{a^{2}d^{3}}{8c^{4}}\quad\Rightarrow \quad T\simeq \frac{d}{c}+ \frac{a^{2}d^{3}}{8c^{5}}

and so on. Thus

T=1cda28c5d3+()d4+T= \frac{1}{c}d- \frac{a^{2}}{8c^{5}}d^{3}+(\ldots)d^{4}+\ldots

Back to the series for ll, we can now write it in terms of dd:

l=a2c2d2+a˙6c3d3+()d4+l=\frac{a}{2c^{2}}d^{2}+ \frac{\dot{a}}{6c^{3}}d^{3}+(\ldots)d^{4}+\ldots

Put this in the self-force equation to eventually get

Fself=q24πε0[a4c2d+a˙12c3+()d+]x^\mathbf{F}_\text{self}=\frac{q^{2}}{4\pi \varepsilon_{0}}\left[ - \frac{a}{4c^{2}d}+ \frac{\dot{a}}{12c^{3}}+(\ldots)d+\ldots \right]\hat{\mathbf{x}}

These accelerations are evaluated at the retarded time, but we express them in terms of the current time by yet another Taylor series:

a(tr)=a(t)+a˙(t)(trt)+=a(t)a˙(t)T+=a(t)a˙(t)dc+a(t_{r})=a(t)+\dot{a}(t)(t_{r}-t)+\ldots=a(t)-\dot{a}(t)T+\ldots=a(t)-\dot{a}(t) \frac{d}{c}+\ldots

and so

Fself=q24πε0[a(t)4c2d+a˙(t)3c3+()d+]x^\mathbf{F}_\text{self}=\frac{q^{2}}{4\pi \varepsilon_{0}}\left[ - \frac{a(t)}{4c^{2}d}+ \frac{\dot{a}(t)}{3c^{3}}+(\ldots)d+\ldots \right]\hat{\mathbf{x}}

The first term, the one in a(t)a(t), can be equated with Newton's second law to get the mass of the charges:

m=2m0+14πε0q24dc2m=2m_{0}+ \frac{1}{4\pi \varepsilon_{0}} \frac{q^{2}}{4dc^{2}}

where m0m_{0} is the mass of each individual charge on either side. In practice, the term adds an effective mass term to the system. In the context of classical physics, this seems really weird, but when confronted with relativity, it is rather sensible, as since E=mc2E=mc^{2}, and

E=q2V=14πε0(q/2)2dm=Ec2=14πε0(q/2)2dc2E=\frac{q}{2}V= \frac{1}{4\pi \varepsilon_{0} } \frac{(q/2)^{2}}{d}\quad\Rightarrow \quad m=\frac{E}{c^{2}}=\frac{1}{4\pi \varepsilon_{0}} \frac{(q/2)^{2}}{dc^{2}}

The second term is something we've already discovered: the radiation reaction.

Fradint=μ0q2a˙12πcF_\text{rad}^{\text{int}}=\frac{\mu_{0}q^{2}\dot{a}}{12\pi c}

where we added int\text{int} as a superscript to indicate that this is the interaction between (1)(1) and (2)(2). These two terms are the only ones that don't scale with positive dd, which means that in the d0d\to 0 limit, they are the only ones that remains, hence our previous results. Well, actually d0d\to 0 certainly has an effect on the mass adjustment, since 1/d1/d\to \infty. Of course, this is absurd, as it begs to imply the this internal force somehow adds infinite mass to the system, a fact that is manifestly false. This problem is only resolved in quantum electrodynamics by renormalizing mass so that this does not happen. For now, we'll just ignore the problem. Also, the radiation reaction is a little different from the one we know: it's half of it (1212 at the denominator instead of 66). This is because, as mentioned before, this is the interaction between (1)(1) and (2)(2). We are missing the interaction of (1)(1) with itself, which comes out identical (the charge is the same), and summing them (i.e. "integrating" over the whole distribution) gives the familiar result

Frad=μ0q2a˙6πcF_\text{rad}=\frac{\mu_{0}q^{2}\dot{a}}{6\pi c}

In conclusion, the radiation reaction is a force that manifests itself due to unbalanced internal forces that leave a net force from a charge on itself.

Footnotes

  1. Technically, there's three energy "storage locations". The electric field is E=Evel+Eaccel\mathbf{E}=\mathbf{E}_\text{vel}+\mathbf{E}_\text{accel}, but the energy depends on the square E2=Evel2+2EvelEaccel+Eaccel2E^{2}=E^{2}_\text{vel}+2\mathbf{E}_\text{vel}\cdot \mathbf{E}_\text{accel}+E^{2}_\text{accel}. The cross term is, for simplicity, considered to be part of the velocity field, as it goes like r3\sim \mathfrak{r}^{-3} and therefore does not affect radiation.

  2. If you're trying to solve for position, then it's third order since a=x¨a=\ddot{x} and a˙=x...\dot{a}=\dddot{x}, but we don't care about position here.

  3. For a real-world example, see Fermi gas > Virial expansion of the equation of state.