3

Both sides of Maxwell's equations are equal to each other, so each of these equations connects quantities simultaneous in time, and as a consequence none of these equations can represent a causal relation:

\begin{align} \mathbf\nabla\cdot\mathbf{E}(\mathbf{r}, t) &= 4\pi\rho(\mathbf{r}, t) \label{Diff I}\\ \mathbf\nabla\times\mathbf{B}(\mathbf{r}, t) &= \dfrac{4\pi}{c} \mathbf{J}(\mathbf{r}, t)+\dfrac{1}{c}\dfrac{\partial\mathbf{E}(\mathbf{r}, t)}{\partial t} \label{Diff IV}\\ \mathbf\nabla\times\mathbf{E}(\mathbf{r}, t) &= -\dfrac{1}{c}\dfrac{\partial\mathbf{B}(\mathbf{r}, t)}{\partial t} \label{Diff III}\\ \mathbf\nabla\cdot\mathbf{B}(\mathbf{r}, t) &= 0 \label{Diff II} \end{align}

But the solutions of this equation (Jefimenko's equations) reflects the "causality", because of right hand sides involve "retarded" time:

\begin{equation} \mathbf{E}(\mathbf{r}, t) = \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\rho(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2}\frac{1}{c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t} - \frac{1}{|\mathbf{r}-\mathbf{r}'|}\frac{1}{c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \mathrm{d}^3 \mathbf{r}', \end{equation}

\begin{equation} \mathbf{B}(\mathbf{r}, t) = -\frac{1}{с} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} \times \mathbf{J}(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2} \times \frac{1}{c}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \mathrm{d}^3 \mathbf{r}', \end{equation}

where $\mathbf{r}'$ is a point in the charge distribution, $\mathbf{r}$ is a point in space, and $t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}$ is the retarded time.

The question can be purely technical, how does the “retardation” arise in the solutions, if it was not in the original equations?

Sergio
  • 2,705
  • 1
  • 23
  • 34

1 Answers1

1

You are correct that Maxwell's equations do not impose causality. In the Lorenz gauge, Maxwell's equations in terms of $A^\alpha$ is a wave equation $$\Box^2 A^\alpha = - \mu_0 J^\alpha$$

The solution to this can be found using the Green's function of the operator $\Box^2$. There are two possible Green's functions: The retarded one $$G_1 = -\frac{\delta\left(t-t' - \left|\mathbf{r}-\mathbf{r}'\right|/c\right)}{4\pi \left|\mathbf{r}-\mathbf{r}'\right|}$$ and the advanced one $$G_2 = -\frac{\delta\left(t-t' + \left|\mathbf{r}-\mathbf{r}'\right|/c\right)}{4\pi \left|\mathbf{r}-\mathbf{r}'\right|}$$

These two solutions are purely mathematical and can be derived using Fourier transforms. No physics has been involved yet. Therefore, in principle, $A^\alpha$ can be written as a linear combination of both solutions: $$A^\alpha = \frac{\mu_0}{4\pi} \int \left(aG_1 + bG_2\right) J^\alpha (\mathbf{r}',t')\text{d}^3{\mathbf{r}'} \text{d}t'$$

It is when we apply this to physical situations that $G_2$ is discarded because it violates causality.

Interestingly, in the 1940's, Feynman and Wheeler pointed out that there are certain physical situations that could involve both $G_1$ and $G_2$, known as Wheeler-Feynman absorber theory. It was originally proposed to explain the self-force of an accelerating charge. They propose that emitters have $a=b=1/2$ while absorbers have $a = 1/2$ and $b= -1/2$, such that the combined reaction on a test charge is purely retarded. A simple discussion can be found here.

Vincent Thacker
  • 15,416