I Coulomb gauge implies instantaneous radial electric field?

[Bob44]

Scalar and vector potentials in Coulomb gauge

Assume Coulomb gauge so that
$$\nabla \cdot \mathbf{A}=0.\tag{1}$$
The scalar potential $\phi$ is described by Poisson's equation
$$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$
which has the instantaneous general solution given by
$$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$
In Coulomb gauge the vector potential $\mathbf{A}$ is given by
$$\nabla^2\mathbf{A}-\frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2}=-\mu_0\mathbf{J}+\frac{1}{c^2}\nabla\frac{\partial\phi}{\partial t}.\tag{4}$$
We decompose the vector potential $\mathbf{A}$ into its transverse and longitudinal components
$$\mathbf{A}=\mathbf{A}_\perp+\mathbf{A}_\parallel.\tag{5}$$
We define the transverse current density $\mathbf{J}_\perp$ by subtracting off the longitudinal component $\nabla \partial \phi/\partial t$ from the total current density $\mathbf{J}$ to obtain
$$\mathbf{J}_\perp=\mathbf{J}-\varepsilon_0 \nabla \frac{\partial \phi}{\partial t}.\tag{6}$$
Therefore the transverse component of the vector potential $\mathbf{A}_\perp$ obeys the wave equation
$$\nabla^2\mathbf{A}_\perp-\frac{1}{c^2}\frac{\partial^2\mathbf{A}_\perp}{\partial t^2}=-\mu_0\mathbf{J}_\perp\tag{7}$$
which has the retarded general solution
$$\mathbf{A}_\perp=\frac{\mu_0}{4\pi}\int \frac{\mathbf{J}_\perp(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|}d^3r'\tag{8}$$
where
$$t_r=t-\frac{\mathbf{r}-\mathbf{r}'}{c}.\tag{9}$$
The longitudinal component of the vector potential $\mathbf{A}_\parallel$ obeys the Poisson equation
$$\nabla^2 \mathbf{A}_\parallel=-\frac{1}{c^2} \nabla \frac{\partial \phi}{\partial t}\tag{10}$$
which has the instantaneous general solution
$$\mathbf{A}_\parallel=\frac{1}{4\pi c^2}\int \frac{\nabla'(\partial\phi/\partial t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{11}$$
Point charge at the origin connected to a wire

Let us consider a time-dependent point charge $q(t)$ located at the origin, which is connected to a wire that carries current to or from the charge.

At any point from the origin the radial electric field $\mathbf{E}_\parallel$ is given by
$$\mathbf{E}_\parallel=-\nabla \phi - \frac{\partial \mathbf{A}_\parallel}{\partial t}\tag{12}.$$
We have
$$
\begin{eqnarray}
\phi(\mathbf{r},t)&=&\frac{1}{4\pi\varepsilon_0}\frac{q(t)}{r},\tag{13}\\
-\nabla \phi&=&\frac{1}{4\pi\varepsilon_0}\frac{q(t)}{r^2}\hat{\mathbf{r}},\tag{14}\\
\frac{\partial\phi}{\partial t}(\mathbf{r}',t)&=&\frac{1}{4\pi\varepsilon_0}\frac{\dot{q}(t)}{r'},\tag{15}\\
\nabla'\frac{\partial \phi}{\partial t}(\mathbf{r}',t)&=&-\frac{1}{4\pi\varepsilon_0}\frac{\dot{q}(t)}{r'^2}\hat{\mathbf{r}}'.\tag{16}
\end{eqnarray}
$$
Substituting eqn.$(16)$ into eqn.$(11)$ and using the standard Poisson integral of a radial gradient we obtain
$$
\begin{eqnarray}
\mathbf{A}_\parallel(\mathbf{r},t)&=&\frac{1}{4\pi\varepsilon_0}\frac{\dot{q}(t)}{c^2 r}\hat{\mathbf{r}},\tag{17}\\
-\frac{\partial \mathbf{A}_\parallel}{\partial t}&=&-\frac{1}{4\pi\varepsilon_0}\frac{\ddot{q}(t)}{c^2 r}\hat{\mathbf{r}}.\tag{18}
\end{eqnarray}
$$
Therefore by substituting eqn.$(14)$ and eqn.$(18)$ into eqn.$(12)$ we find that the radial electric field $\mathbf{E}_\parallel$ is given by
$$\mathbf{E}_\parallel=\frac{1}{4\pi\varepsilon_0}\left(\frac{q(t)}{r^2}-\frac{\ddot{q}(t)}{c^2r}\right)\hat{\mathbf{r}}.\tag{19}$$
Does eqn.$(19)$ violate causality since it is an instantaneous expression?

 

[QuarkyMeson]

Longitudinal electric field isn't an observable, E is. (Save some electrostatic near field stuff when E is approx $E_{\parallel}$ and complete far field where E is approx $E_{\perp}$). The question seems a bit ill framed in it's foundations, can a math equation violate causality? Sure. Can physics? No, not when you account for Maxwell + charge conservation + retarded boundary conditions.

Here, when you look at the total field $$E = E_{\perp} + E_{\parallel}$$ any instantaneous bit from the longitudinal electric field is cancelled exactly by the transverse field, so there is nothing causality violating in the observable E.

 

[Dale]

Does eqn.(19) violate causality since it is an instantaneous expression?

As was pointed out by @QuarkyMeson, no, because it is only part of the field.

However, Maxwell’s equations admit causality-reversed solutions. So you can get non-physical solutions. These can be found using advanced potentials.

However, these solutions have a light-speed delay, just to the past instead of the future. So if you find an “instantaneous” transfer of information in the field then you made a mistake.

 

[Bob44]

Here, when you look at the total field $$E = E_{\perp} + E_{\parallel}$$ any instantaneous bit from the longitudinal electric field is cancelled exactly by the transverse field, so there is nothing causality violating in the observable E.

But how could the retarded transverse field cancel the instantaneous longitudinal field as it takes a finite time for the transverse field to get to the observation point whereas the longitudinal field is there already?