# Does Larmor radiation show that the fundamental laws are not necessarily time-reversible?

It is often said that the fundamental laws of physics are time-reversal symmetric, and therefore the arrow of time is an emergent phenomenon. An example that is often given is the elastic collision of two particles - in principle, the argument goes, you would not be able to distinguish a movie of the collision run forwards from one run backwards.

But consider a Coulomb collision of two like-signed charged particles. As the particles interact they accelerate, and hence generate Larmor radiation. A movie of the collision in the forward direction would show radiation carried away from the collision (or if the radiation were not visible in the movie, then it would show the particles with less energy after the collision), but if the movie were run backwards it would show the radiation coming in from far away leading up to the collision, and the particles would get a boost from it as they collide. Not only would the movie be different in the forward and backward directions, but it would be easy for an observer to tell which was the forward direction, since the laws of physics would predict outward radiation in the time-forward direction.

Doesn't this example show that the fundamental laws are not necessarily time-reversible? Or is the production of Larmor radiation not to be viewed as a "fundamental" process, and if not, why not? Where does time-irreversibility enter into the production of Larmor radiation? (And incidentally, since most interactions between material objects are electromagnetic, doesn't it show that there is no such thing as an elastic collision, even in the ideal limit?)

EDIT: The expression for the power radiated by an accelerated charged particle is $$\begin{equation*} P_{rad} = \dfrac{1}{4\pi\epsilon_0}\dfrac{2 q^2 a^2}{3 c^3}\dfrac{\left(1 - \beta^2\sin^2\psi \right)}{\left(1-\beta^2\right)^3} \end{equation*}$$ where $$\begin{eqnarray*} q&=&\text{ particle charge}\\ a&=&\text{ particle acceleration}\\ \beta&=&v/c\\ \psi&=&\text{ angle between velocity and acceleration} \end{eqnarray*}$$ For the special case $$\mathbf{v}$$ parallel to $$\mathbf{a}$$ ($$\psi = 0$$, head-on collision), this comes from integrating power per steradian in the radiation field: $$\begin{equation*} \dfrac{dP_{rad}}{d\Omega} = \dfrac{1}{4\pi\epsilon_0}\dfrac{q^2a^2}{4\pi c^3}\dfrac{\sin^2\theta}{\left(1-\beta\cos\theta\right)^5} \end{equation*}$$ where $$\theta$$ is the polar angle between the velocity/acceleration and the normal to the surface of integration. (ref: G. L. Pollack and D. R. Stump, Electromagnetism, San Francisco: Addison-Wesley (2002), sec. 15.6.3)

• Maxwell's equations are time reversal invariant but the Larmor formula isn't because it assumes an asymmetric boundary condition, i.e. that the particles are producing radiation that goes out instead of absorbing radiation that comes in. Commented May 23 at 18:29
• Can you please include the equation for Larmor radiation in the question. Preferable at the most detailed level, e.g. $d^3P/(d\Omega d\nu)$? It makes it much easier to answer.
– JEB
Commented May 23 at 18:36
• @JEB The text I'm looking at only gives the differential form for the case where velocity and acceleration are parallel, then states the general integrated form without deriving it. I've added them.
– pwf
Commented May 23 at 19:04
• Theory is not fundamental to physics: it must follow facts obtained from experiments and observations. That past and future are different things is the most fundamental fact of physics. It doesn't emerge from the theory: the theory struggles to accommodate it, more than other fundamental facts. Commented May 26 at 11:42

The Larmor formula for energy flow away from the particle is derived based on some assumptions, one of which is that field of the accelerated particle is given by the retarded solution to Maxwell's equations. This is a boundary condition on their solutions. This boundary condition selects solutions where field changes move out from the accelerated charged particle to infinity. This is the opposite of the advanced fields, where field changes are moving in from the infinity to the particle. If fields were advanced, the Larmor formula would be incorrect, and the correct formula would have minus in front of the expression there, thus the power would not come out of the region around the particle, but would come into it.

Maxwell's equations and equations of motion of the charged particles (the usual ones, without friction or self-force terms) are time-symmetric. Thus the fundamental laws in EM theory, written in equation form, are time-symmetric.

The boundary condition that selects the retarded solution isn't time-symmetric, but usually, this condition isn't considered part of the fundamental laws. It isn't treated as a generally valid equation, but as a condition on the solutions of the other equations, that is assumed valid in some range of situations, but not universally. For example, an accelerated particle in a macroscopically described perfectly reflecting cavity is surrounded by electric field whose tangential component on the walls vanishes, thus isn't the retarded solution for the particle in empty space.

On the microscopic level, where there is no perfect reflection, all classical experiments are much easier to explain assuming all individual particle fields are retarded, so the boundary condition on the walls above is a result of superposition of the retarded field of the particle, and many different retarded fields from the walls. But advanced fields existing in some way are hard to disprove, because they can superpose in such a way as to imitate retarded fields.

A time-symmetric solution of Maxwell's equations for one accelerated particle in empty space can be found as well, e.g. a sum of one half retarded and one half advanced solution. But this is a time-symmetry of solutions, not of laws.

The situation with the asymmetric boundary condition is similar to one in classical models of irreversible thermodynamics. The microscopic equations of motion are time-symmetric, but with appropriate initial condition, they do have a solution that describes establishment of an equilibrium state starting from a non-equilibrium state, which isn't a time-symmetric solution. The initial condition is what introduces the asymmetry.

Thus in general, even if all physics laws, written as equations, are time-symmetric, some of their solutions are not necessarily time-symmetric, and those that apply to our world often aren't.

• Yes, that's what I was thinking, but could you point out where in the derivation the advanced-time solutions are thrown out? If a correct prediction from Maxwell's equations requires an arbitrary step of throwing out the advanced time solution, then I would argue the physical law is not "Maxwell's equations," but "Maxwell's equations, excluding time-advanced solutions." IMO that means the physical law is not time-symmetric.
– pwf
Commented May 23 at 22:04
• The Larmor formula derivation works with the assumption that density of energy flux is given by the Poynting vector $\mathbf E\times\mathbf B/\mu_0$, and the assumption that this vector, on surface of a space region containing the accelerated particle, points away towards the open space (not inside the region). This is true for retarded EM fields. Advanced fields are like waves coming in from the infinity, their Poynting vector points to the inside of the region, and thus net surface integral would come out negative. Commented May 23 at 22:20
• Larmor's formula is unfortunately not necessarily always correct prediction. It is hard to measure energy lost through arbitrary surface. It's a theoretical descriptive quantity which is nice to think about, but it is not measurable. But fields being retarded is an experimentally well founded hypothesis, thus one can regard it as part of physical laws, at least valid in many cases, if not always. When people say physics laws are time-symmetric, they don't mean all the laws, they only mean the equations which manifest that symmetry. Commented May 23 at 22:22
• It really boils down to what we consider part of physical laws. Boundary conditions are traditionally not, and this allows the claim laws are time-symmetric, even if solutions are not. But I agree this viewpoint is by no means necessary, and in some cases, it seems like we observe persistent preference of Nature for certain type of boundary conditions, and this preference may be called a physical law. Commented May 23 at 22:33

OK, the formula is: $$\begin{equation*} P_{rad} = \dfrac{1}{4\pi\epsilon_0}\dfrac{2 q^2 a^2}{3 c^3}\dfrac{\left(1 - \beta^2\sin^2\psi \right)}{\left(1-\beta^2\right)^3} \end{equation*}$$

under time reversal:

$$P_{rad} \rightarrow -P_{rad}$$ $$q \rightarrow +q$$ $$a \rightarrow +a$$ $$\beta \rightarrow +\beta$$ $$c \rightarrow -c$$ $$\theta \rightarrow +\theta$$ $$\Omega \rightarrow +\Omega$$

(Note: $$\beta$$ is dimensionless, and it is $$c$$ that changes sign)

making the equation:

$$\begin{equation*} (-1)P_{rad} = \dfrac{1}{4\pi\epsilon_0}\dfrac{2 (+1)^2q^2 (+1)^2a^2}{3 (-1)^3c^3}\dfrac{\left(1 - (+1)^2\beta^2\sin^2(+1)\psi \right)}{\left(1-(+1)^2\beta^2\right)^3} = -P_{rad} \end{equation*}$$

so it is manifestly time odd on both sides.

A time reversal error looks something like:

$$F = \alpha \frac{dv}{dx}$$

(where $$\alpha$$ is a constant with the right units). Under time reversal:

$$F \rightarrow +F$$ $$v \rightarrow -v$$ $$x \rightarrow +x$$

and the equation transforms as:

$$\alpha \frac{dv}{dx}\rightarrow -\alpha \frac{dv}{dx}$$

which is time odd (while $$F$$ is even), which means:

$$\alpha = 0$$

or $$T$$ symmetry is violated.

• Yes, that's my point. The time-reversed solution would have radiation going outward just like the time-forward one, but the movie played backwards would have radiation coming in and could be distinguished. (You could, for example, set up detectors at an increasing distance from the collision, and the distant detectors would light up after the close ones. If you reversed time, their order should also reverse.) There seems to be a paradox.
– pwf
Commented May 23 at 20:26
• @pwf $\vec S = \vec E \times \vec H$ is T-odd, so I guess $P_{rad}$ becomes $P_{abs}$?
– JEB
Commented May 24 at 14:08
• $P_{rad}$ is a scalar. The Poynthing energy flow $\vec P = E \times B$ is a vector. Reversing the "movie" will have the photons converging on the particles from infinity. Thus $\overrightarrow{P} \rightarrow -\overrightarrow{P}$ but the magnitudes (which you calculate) are the same, and the overall physics is symmetric in time, but looks strange since entropy decreases as photons converge from infinity with just the right trajectories to coherently nudge the charged particles. (You end up flipping the sign of B in the wave description of the convergent fields; $B \propto v$ so is time-odd.) Commented May 24 at 16:22
• @SarahMesser hmm, good point. Power is $E/[T]=[M][L/T]^2/[T]=[M][L]^2/[T]^3$ which is indeed odd. $\epsilon_0$ goes a $[T]^2$, leaving a $1/c^3$ in $P_{rad}$ formula...which is ofc odd, but does $c \rightarrow -c$ under $T$? I'll have to edit my answer later.
– JEB
Commented May 24 at 17:11
• Derivation of the Larmor radiation formula here explicitly takes the magnitude of the power. $c$ is a scalar number: Speed of light is isotropic and conserved under $T$ Commented May 24 at 20:35

You could argue that the Larmor formula is in the realm of statistical physics. It looks as if you are only looking at one particle. But in fact, you are looking at the ensemble of the particle and all the photons it has created. If you time-reversed this entire ensemble, the particle would accelerate.

By creating photons, the particle is increasing entropy. The second law of thermodynamics, which says that entropy should always increase, is one of the few physical laws that breaks time-symmetry.