Do Maxwell's Equations Follow from the Continuity Equation?

This is the somewhat amended script of a short talk I gave for a class.

What's the logical relationship between Maxwell's equations and the continuity equation? The continuity equation, recall, is the statement of local charge conservation. It reads:

\[\nabla \cdot \bm J = -∂_t \rho,\]

or to say it in words, the net flow of current density out of a point is minus the rate of change of the charge density at that point.

It's well known that Maxwell's equations imply continuity. We can see this by taking the divergence of both sides of the Ampère-Maxwell Law and recalling that the divergence of a curl is always zero. We get

\[0 = \nabla \cdot (\nabla \times \bm B) = \mu_0 \,\nabla\cdot (\bm J + \epsilon_0 ∂_t\bm E) \\[5pt] \implies \nabla \cdot \bm J = -\epsilon_0 \nabla \cdot (∂_t\bm E) = -∂_t \big(\epsilon_0 \nabla \cdot \bm E\big) = -∂_t\rho.\]

The usual interpretation of this result is that continuity is a consistency condition for the Maxwell equations. If charge isn't conserved, Maxwell's equations don't yield consistent solutions for \(\bm E\) and \(\bm B.\)

How about the other way? Can we get Maxwell's equations from the continuity equation? You might think we cannot since intuitively, there's a lot more content in Maxwell's four equations than there is in just the one continuity equation. That thought is correct, but we can derive something surprisingly close to the Maxwell equations from continuity alone.

José Heras (2007) proved that if \(\rho\) and \(\bm J\) satisfy the continuity equation, then the \(\bm E\)- and \(\bm B\)-like retarded fields defined from \(\rho\) and \(\bm J\) by the Jefimenko equations satisfy four Maxwell-like equations.^[1] The proof is a bit of a slog and uses obscure vector calculus identities. We won't present the whole thing, but we will look at the most important steps and offer some reflections.

Highlights of the proof

We start by defining notations. Let \(\bm r = \bm x - \bm x'\) be the displacement vector from the source point to the field point as ususal. Because there are going to be scads of retarded times in all of our equations, let's denote them with a square bracket to keep things neat. A bracketted quantity is that quantity evaluated at the retarded time, so

\[ [\text{blah}] = \text{blah}\lvert_{t' = t- r/c}.\]

Let's also call our \(\bm E\)- and \(\bm B\)-like fields \(\bm F\) and \(\bm G\) respectively:

\[\bm F = \frac{\alpha}{4π} \iiint \bigg(\frac{\hat {\bm{r}}}{r^2} [\rho] + \frac{\hat{\bm{r}}}{rc}[∂_t\rho] - \frac{1}{rc^2}[∂_t\bm J]\bigg)\,d\cal V'\] \[\bm G = \frac{\beta}{4π} \iiint \bigg([\bm J]\times\frac{\hat{\bm{r}}}{r^2} + [∂_t\bm J]\times \frac{\hat{\bm{r}}}{rc}\bigg)\,d\cal V'\]

where \(\alpha\) and \(\beta\) are adjustable scalars that define our unit system.

Now our starting assumption is that the continuity equation holds everywhere and always, so in particular, it holds at the source point at the retarded time:

\[\nabla'\cdot \bm J (\bm x', t - r/c) = -\big(∂_{t'} \rho(\bm x', t') \big)\big\lvert_{t' = t-r/c},\]

or in our compact notation:

\[[\nabla' \cdot J] = -[∂_t \rho].\]

We take appropriate derivatives of this equation and invoke appropriate identities. The main subtlety to watch out for is that \([\rho]\) and \([\bm J]\) are functions of \(\bm x'\) explicitly and implicitly via \(r\), so we have to remember the chain rule each time we differentiate a retarded quantity. Eventually we wind up with

\[\nabla \times \bigg([\bm J]\times\frac{\hat{\bm{r}}}{r^2} + [∂_t\bm J]\times \frac{\hat{\bm{r}}}{rc}\bigg) - ∂_t \bigg(\frac{\hat {\bm{r}}}{r^2} [\rho] + \frac{\hat{\bm{r}}}{rc}[∂_t\rho] - \frac{1}{rc^2}[∂_t\bm J]\bigg) \\[5pt] = 4π[\bm J]\delta (\bm x - \bm x') - \nabla (\nabla' \cdot [\bm J]/r) \quad \text{and}\] \[\nabla \cdot \bigg(\frac{\hat {\bm{r}}}{r^2} [\rho] + \frac{\hat{\bm{r}}}{rc}[∂_t\rho] - \frac{1}{rc^2}[∂_t\bm J]\bigg) = 4π[\rho]\delta (\bm x - \bm x') + \nabla'\cdot \bigg(\frac{[∂_t\bm J]}{rc^2}\bigg).\]

We recognize the stuff on the left hand sides of these equations from the definitions of \(\bm F\) and \(\bm G,\) which strongly suggests we should integrate over all space WRT the source coordinates. Think about what happens to the rightmost term in each equation when we do that. Each of these terms is a divergence of something WRT the primed coordinates, so by the divergence theorem, we get surface integrals of the something over a surface at infinity. But assuming we're dealing with a localized source, \(\rho\) and \(\bm J\) are both zero at infinity, so the surface integrals vanish. We're left with

\[\nabla \times \bigg(\frac{4π}{\beta} \bm G\bigg) - ∂_t\bigg(\frac{4π}{\alpha}\bm F\bigg) = 4π\iiint [\bm J]\delta (\bm x - \bm x')\,d\cal V'\quad \text{and}\] \[\nabla \cdot \bigg(\frac{4π}{\alpha}\bm F\bigg) = 4π\iiint [\rho]\delta (\bm x - \bm x')\,d\cal V'.\]

The delta functions on the right hand sides pick out \(\bm x' = \bm x,\) at which point \(r=0,\) so retarded and unretarded quantities are the same. Therefore

\[\nabla \times \bm G - \frac{\beta}{\alpha}∂_t \bm F = \beta \bm J \quad \text{and} \quad \nabla \cdot \bm F = \alpha \rho.\]

These equations are just Ampère-Maxwell and Gauss spelled funny, and the other two Maxwell-like equations can be derived with a remix of the same moves.

Reflections

So there is Heras's theorem. Is it a proof of Maxwell's equations? Not quite. Maxwell's equations are a statement about \(\bm E\) and \(\bm B\), the electric and magnetic fields, whereas Heras's theorem is about these other fields \(\bm F\) and \(\bm G.\) We cannot leap to identify \(\bm F\) with \(\bm E\) and \(\bm G\) with \(\bm B\) because we haven't proven that these Heras fields play the roles of the EM fields in the Lorentz force law:

\[\bm F_\text{EM} = q(\bm E + \bm v \times \bm B).\]

This force law is the operational definition of \(\bm E\) and \(\bm B\). That is to say, something is the electric field just in case it makes particles accelerate the way that the Lorentz force law says \(\bm E\) does, and something is the magnetic field just in case it makes particles accelerate in the way the Lorentz force law says \(\bm B\) does. We can't know a priori that the Heras fields cause particles to accelerate in that way.^[2] They might have accelerated particles according to some different force law. Or they might not have accelerated particles at all. \(\bm F\) and \(\bm G\) might have just been mathematical epiphenomena that didn't grab ahold of observables in any way.

An example helps to make this point less confusing. In free space, Poynting's theorem tells us the field energy density \(\cal U\) and the Poynting vector \(\bm S\) satisfy a continuity equation:

\[\nabla \cdot \bm S = -∂_t \cal U.\]

It therefore follows that we can define Heras fields from \(\cal U\) and \(\bm S\), and those fields will satisfy four Maxwell-like equations. Are they the electric and magnetic fields? Can you toss them into the Lorentz force law and get out correct predictions? Of course not. These Heras fields are good for nothing. You can keep track of them as an accounting exercise, but they won't tell you how particles accelerate.

It's a remarkable fact that the Heras fields defined from \(\rho\) and \(\bm J\) are good for something. They appear in a simple and empirically correct force law. That's more than we can say for most Heras fields.

But even if Heras's theorem does not constitute a proof of Maxwell's equations, it does help to explain why Maxwell's equations are the way they are and not some other way. Maxwell's equations are true because Jefimenko's equations are true and charge is conserved. I find this explanation pretty satisfying. For one, we know the explanation is valid—that was the content of Heras's theorem. And further, the two explanatia seem to me more fundamental, less in doubt, closer to bedrock than the explanandum. Our faith in charge conservation is founded on Noether's theorem plus hard-to-vary symmetry assumptions. And although I confess I don't feel every term of Jefimenko's equations in my bones, something a lot like it has to be true if we are to avoid FTL signaling and respect some geometric intuitions.

If we accept Heras's theorem as an explanation for the Maxwell equations, it could serve as a test case for philosophy of science debates about non-causal explanations. Most of the explanations science provides are causal in the sense that they tell us how the explanandum was caused by other events out in the world. For example, the boats rose because the moon exerted a gravitational pull on the sea on which the boats were floating. Causal explanations like this one are the normal case. Some philosophers have gone so far as to claim that every genuine explanation must be causal, but cases like Heras's theorem cast doubt on that claim. It's hard to see how laws of nature, such as Maxwell's equations, could be caused by events under any normal understanding of causation. And moreover, the explanation of Maxwell's equations we gave above didn't seem to involve any causal claims. It seemed to involve something more like logical or mathematical entailment. So if Heras's theorem does explain Maxwell's equations, it explains them in a spooky non-causal way.^[3]

[1]	Why are these equations merely "Maxwell-like"? How are they not the Maxwell equations? I'll explain that later.

[2]

Notably, some other mathematical derivations of the Maxwell equations close this gap. For example, Feynman proved that Newton's 2nd Law and the fundamental commutation relations of QM jointly entail the existence of fields satisfying Maxwell's equations and the Lorentz force law. By definition, the fields whose existence Feynman proved really are the electric and magnetic fields.

[3]	There's a lot more to be said on this topic. If you're curious, I recommend Lange's (2017) Because Without Cause, which gives an account of non-causal explanation and illustrates it with dozens of beatuiful examples from all branches of science.