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It is well-known that Maxwell added the displacement current term to Ampรจre'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 Ampรจre'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).

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  • $\begingroup$ See this question. for a classic case of the displacement current for charging a capacitor. $\endgroup$
    – wyphan
    Dec 4, 2020 at 21:06
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    $\begingroup$ Hi @wyphan, and thank you for the response. Indeed, I did see that question, but it gets at something quite different from what I am asking. I'm not asking why the displacement current term is necessary, nor how it "manifests" itself physically in the special case of a capacitor charging. This question is about whether or not it was possible, in theory, to have patched up Ampere's Law with a different vector function. They are related, but nevertheless very different questions. $\endgroup$
    – EE18
    Dec 5, 2020 at 17:01
  • $\begingroup$ Related: physics.stackexchange.com/questions/224256/… $\endgroup$ Dec 7, 2020 at 14:32
  • $\begingroup$ Are you asking a question about history, so that answers are only allowed to use experimental data that was known to Maxwell? Or are you asking a pedagogical question about what constraints we could deduce on $\mathbf{v}$ today? And are you asking a quantitative question, like how much of a deviation from the usual displacement term would exceed the experimental uncertainties? Or are you asking a qualitative question, like in what sense the usual displacement term is the simplest choice compatible with experiments? $\endgroup$ Dec 7, 2020 at 23:17
  • $\begingroup$ @ChiralAnomaly I will admit that I suppose I am asking both questions. Part (1) of my question above is a mathematical one asking about the mathematical constraints on $\mathbf{v}$ and, in part (2), I am asking about the experimental confirmation of the choice for the displacement term. $\endgroup$
    – EE18
    Dec 8, 2020 at 12:58

2 Answers 2

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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.

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    $\begingroup$ Thanks very much for your answer. I've not yet studied special relativity, so I have no idea whether or not what you've written constitutes an answer (my fault, not yours!). Is there any way to answer without invoking special relativity? If not, no worries. $\endgroup$
    – EE18
    Dec 9, 2020 at 17:45
  • $\begingroup$ $\partial_\mu F^{\mu \nu}$ is kinda a conservation law. You can drive the Maxwell equations from these conservation laws. Note that conservation laws come from the symmetry of the Lagrangian. In this case it is a U(1) symmetry. Although i did not use special relativity to answer your question, I had to use more advance topics. Bottom line, it has to do with the conservation laws related to the Maxwell equations $\endgroup$ Dec 11, 2020 at 17:25
  • $\begingroup$ @KianMaleki $\partial_\mu F^{\mu\nu}$ is not a conservation law, it is just an expression. The complete equation is $\partial_\mu F^{\mu\nu} = j^\nu/\epsilon_0$, which is the equation of motion of the field tensor, which turns into the wave equation of the four potential in the Lorenz gauge. $\endgroup$
    – my2cts
    Dec 28, 2020 at 16:20
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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 $๐ฎ$ be the boost velocity of the transform. Then, the charge density $ฯ$ would remain invariant - so $ฯ' = ฯ$ - the current density would transform as $๐‰' = ๐‰ - ฯ๐ฎ$, the gradient operator remains invariant $โˆ‡' = โˆ‡$ and the time derivative operator transforms as $(โˆ‚/โˆ‚t)' = โˆ‚/โˆ‚t + ๐ฎยทโˆ‡$.

The equation Maxwell actually posed for the Ampรจre Law was $โˆ‡ร—๐‡ = ๐‚$. By your suggestion, we should stipulate that $๐‚ = ๐‰ + ๐•$, with a slight change from your notation. Ok. We'll do that. But now: what's its transform?

Upon substitution, we have $$โˆ‡'ร—๐‡' = ๐‚' = ๐‰' + ๐•' \hspace 1emโ‡’\hspace 1em โˆ‡ร—๐‡' = ๐‚' = ๐‰ - ฯ๐ฎ + ๐•'. \label{1}\tag{1}$$ But, now we also have the Maxwell equation $โˆ‡ยท๐ƒ = ฯ$. Under transform, this becomes: $$โˆ‡'ยท๐ƒ' = ฯ' \hspace 1emโ‡’\hspace 1em โˆ‡ยท๐ƒ' = ฯ = โˆ‡ยท๐ƒ \hspace 1emโ‡’\hspace 1em โˆ‡ยท(๐ƒ'-๐ƒ) = 0,$$ which strongly suggests that $๐ƒ' = ๐ƒ$ (as Maxwell determined). We'll leave that aside for now.

However, what we do have, from ($\ref{1}$) is $$โˆ‡ร—๐‡' = ๐‰ - โˆ‡ยท๐ƒ๐ฎ + ๐•', \hspace 1em โˆ‡ร—๐‡ = ๐‰ + ๐•. \label{2}\tag{2}$$ We can rewrite the first of ($\ref{2}$) as: $$โˆ‡ร—(๐‡' + ๐ฎร—๐ƒ) = ๐‰ - ๐ฎยทโˆ‡๐ƒ + ๐•',$$ and combine it with the second of ($\ref{2}$), to get: $$โˆ‡ร—(๐‡' - ๐‡ + ๐ฎร—๐ƒ) = -๐ฎยทโˆ‡๐ƒ + ๐•' - ๐•.$$ See that residual $๐ฎยทโˆ‡$? It's the shadow of $โˆ‚/โˆ‚t$, strongly suggesting that $๐• = โˆ‚๐ƒ/โˆ‚t$.

So, we'll just define $$๐– = ๐• - \frac{โˆ‚๐ƒ}{โˆ‚t}, \hspace 1em \hat{๐‰} = ๐‰ + ๐–,$$ so that we can write Maxwell's $๐‚$ as $$๐‚ = \hat{๐‰} + \frac{โˆ‚๐ƒ}{โˆ‚t}.$$

Define: $$๐ก = ๐‡' - (๐‡ - ๐ฎร—๐ƒ), \hspace 1em ๐ = ๐ƒ' - ๐ƒ, \hspace 1em \hat{๐ฃ} = \hat{๐‰'} - \left(\hat{๐‰} - ฯ๐ฎ\right).$$ Then, we get the following: $$โˆ‡'ยท๐ = 0, \hspace 1em โˆ‡'ร—๐ก - \left(\frac{โˆ‚}{โˆ‚t}\right)'๐ = \hat{๐ฃ}.$$ So, the transform factors into two stages: one that is purely kinematic $$\left(โˆ‡, \frac{โˆ‚}{โˆ‚t}, \hat{๐‰}, ฯ, ๐ƒ, ๐‡\right) โ†’ \left(โˆ‡, \frac{โˆ‚}{โˆ‚t} + ๐ฎยทโˆ‡, \hat{๐‰} - ฯ๐ฎ, ฯ, ๐ƒ, ๐‡ - ๐ฎร—๐ƒ\right),$$ and another that is internal $$\left(โˆ‡, \frac{โˆ‚}{โˆ‚t}, \hat{๐‰}, ฯ, ๐ƒ, ๐‡\right) โ†’ \left(โˆ‡, \frac{โˆ‚}{โˆ‚t}, \hat{๐‰} + \hat{๐ฃ}, ฯ, ๐ƒ + ๐, ๐‡ + ๐ก\right),$$ subject to the condition $$โˆ‡ยท๐ = 0, \hspace 1em โˆ‡ร—๐ก - \frac{โˆ‚๐}{โˆ‚t} = \hat{๐ฃ},$$ 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 $๐‰$, we're back to Maxwell's formula for $๐‚$. It's all coming out of the transform behavior of the fields. We'll just call $\hat{๐‰}$ the actual $๐‰$ and remove the diacritic. That's similar to what Heaviside did in his formulation of the electromagnetic field equations.

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