Was Maxwell's displacement current the only way to fix up Ampère's Law?

Question

It is well-known that Maxwell added the displacement current term to Ampère's Law to make electrodynamics whole. As it is taught in the modern context (I am currently reading Griffiths's text, Introduction to Electrodynamics), we can motivate the addition of the displacement current term by noting that its addition to Maxwell's equations means that Maxwell's equations imply the continuity equation. However, as Griffiths remarks, this nicety (the fact that the continuity equation falls out of Maxwell's equations) is not incontrovertible evidence that the addition of the specific form of the displacement current term is necessarily correct. Indeed, he says that there "might, after all, be other ways to doctor up Ampère's Law". My question is, therefore, twofold:

(1) Is it true, as Griffiths says, that there are conceivably other ways to "fix" Ampere's Law? That is, can we let $$\nabla \times \mathbf{B}=\mu_{0}\mathbf{J}+\mathbf{v}$$ for some arbitrary vector function $\mathbf{v}$ and still develop a consistent theory? I'm not sure how to define "a consistent theory" here but, perhaps, we can roughly say that a consistent theory would mean no contradictions with the other three Maxwell equations (mathematically-speaking). At least to me, I would suspect that the answer is "yes" since the problem (at least as it is understood in the more modern language of vector calculus, as compared to what Maxwell was doing) with Ampere's Law without Maxwell's correction is that the divergence of the right-hand side does not in general vanish, as it must. Thus we would be requiring that (using continuity and Gauss's Law) $$\nabla \cdot \mathbf{v}=-\nabla \cdot(\mu_{0}\mathbf{J})=\mu_{0}\frac{\partial\rho}{\partial t}=\mu_{0}\nabla \cdot(\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t})$$ but, of course, the divergence of a vector function does not fully specify that vector function. However, assuming we choose $\mathbf{v}$ to satisfy the above, and putting aside experimental verification for the moment, would choosing something else for $\mathbf{v}$ break the structure of Maxwell's theory somewhere else?

(2) Moving now to consider experimental verification, Griffiths says that Hertz's discovery of EM waves confirmed Maxwell's choice for the displacement current term. I understand that Maxwell's equations imply wave solutions that were observed experimentally, but perhaps someone can (at a high-level, even) explain why any other choice of the displacement current term would have yielded inconsistencies with experiment (assuming that my attempt at answering (1) above was correct for, if there are mathematical inconsistencies, then we are done).

Answer 1

The correct, comprehensive, and incontrovertible way to explain the term is using special relativity. You are right that without experiment and special relativity v can be anything.

Once you consider special relativity v must be $\partial E / \partial t$ and there is no other theory to fully explain it with mathematical consistency.

Special relativity plays a very important role in Maxwell equation because if you have a moving charge which creates some magnetic field you can always go to a reference frame in which B is zero.

From the conservation laws and special relativity we have:

$\partial_\mu F^{\mu \nu} = \mu_o J^\nu $

where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ and $A_\mu$ is the vector potential. The $F^{\mu i}$ term is the equation you are after.

Answer 2

Maxwell was firmly steeped in the Galilean transform of non-relativistic theory and devoted an entire discussion in his treatise to the matter of how the fields transform under the transform.

Let $𝐮$ be the boost velocity of the transform. Then, the charge density $ρ$ would remain invariant - so $ρ' = ρ$ - the current density would transform as $𝐉' = 𝐉 - ρ𝐮$, the gradient operator remains invariant $∇' = ∇$ and the time derivative operator transforms as $(∂/∂t)' = ∂/∂t + 𝐮·∇$.

The equation Maxwell actually posed for the Ampère Law was $∇×𝐇 = 𝐂$. By your suggestion, we should stipulate that $𝐂 = 𝐉 + 𝐕$, with a slight change from your notation. Ok. We'll do that. But now: what's its transform?

Upon substitution, we have $$∇'×𝐇' = 𝐂' = 𝐉' + 𝐕' \hspace 1em⇒\hspace 1em ∇×𝐇' = 𝐂' = 𝐉 - ρ𝐮 + 𝐕'. \label{1}\tag{1}$$ But, now we also have the Maxwell equation $∇·𝐃 = ρ$. Under transform, this becomes: $$∇'·𝐃' = ρ' \hspace 1em⇒\hspace 1em ∇·𝐃' = ρ = ∇·𝐃 \hspace 1em⇒\hspace 1em ∇·(𝐃'-𝐃) = 0,$$ which strongly suggests that $𝐃' = 𝐃$ (as Maxwell determined). We'll leave that aside for now.

However, what we do have, from ($\ref{1}$) is $$∇×𝐇' = 𝐉 - ∇·𝐃𝐮 + 𝐕', \hspace 1em ∇×𝐇 = 𝐉 + 𝐕. \label{2}\tag{2}$$ We can rewrite the first of ($\ref{2}$) as: $$∇×(𝐇' + 𝐮×𝐃) = 𝐉 - 𝐮·∇𝐃 + 𝐕',$$ and combine it with the second of ($\ref{2}$), to get: $$∇×(𝐇' - 𝐇 + 𝐮×𝐃) = -𝐮·∇𝐃 + 𝐕' - 𝐕.$$ See that residual $𝐮·∇$? It's the shadow of $∂/∂t$, strongly suggesting that $𝐕 = ∂𝐃/∂t$.

So, we'll just define $$𝐖 = 𝐕 - \frac{∂𝐃}{∂t}, \hspace 1em \hat{𝐉} = 𝐉 + 𝐖,$$ so that we can write Maxwell's $𝐂$ as $$𝐂 = \hat{𝐉} + \frac{∂𝐃}{∂t}.$$

Define: $$𝐡 = 𝐇' - (𝐇 - 𝐮×𝐃), \hspace 1em 𝐝 = 𝐃' - 𝐃, \hspace 1em \hat{𝐣} = \hat{𝐉'} - \left(\hat{𝐉} - ρ𝐮\right).$$ Then, we get the following: $$∇'·𝐝 = 0, \hspace 1em ∇'×𝐡 - \left(\frac{∂}{∂t}\right)'𝐝 = \hat{𝐣}.$$ So, the transform factors into two stages: one that is purely kinematic $$\left(∇, \frac{∂}{∂t}, \hat{𝐉}, ρ, 𝐃, 𝐇\right) → \left(∇, \frac{∂}{∂t} + 𝐮·∇, \hat{𝐉} - ρ𝐮, ρ, 𝐃, 𝐇 - 𝐮×𝐃\right),$$ and another that is internal $$\left(∇, \frac{∂}{∂t}, \hat{𝐉}, ρ, 𝐃, 𝐇\right) → \left(∇, \frac{∂}{∂t}, \hat{𝐉} + \hat{𝐣}, ρ, 𝐃 + 𝐝, 𝐇 + 𝐡\right),$$ subject to the condition $$∇·𝐝 = 0, \hspace 1em ∇×𝐡 - \frac{∂𝐝}{∂t} = \hat{𝐣},$$ that allows us to identify the second stage as just an instance of superposition by a second charge-free field. That adds nothing new. The most we could say is that if the two stages were not independent, then a Galilean transform would be associated with the appearance of a second charge-free field.

But, in any case, with the redefinition of $𝐉$, we're back to Maxwell's formula for $𝐂$. It's all coming out of the transform behavior of the fields. We'll just call $\hat{𝐉}$ the actual $𝐉$ and remove the diacritic. That's similar to what Heaviside did in his formulation of the electromagnetic field equations.