I Does Poisson's equation hold due to vector potential cancellation?

[Bob44]

Imagine that two charged particles, with charge $+q$, start at the origin and then move apart symmetrically on the $+y$ and $-y$ axes due to their electrostatic repulsion.

The $y$-component of the retarded Liénard-Wiechert vector potential at a point along the $x$-axis due to the two charges is
$$
\begin{eqnarray*}
A_y&=&\frac{q\,[\dot{y}]_{\mathrm{ret}}}{4\pi\varepsilon_0 c^2[(x^2+y^2)^{1/2}+y\dot{y}/c]_{\mathrm{ret}}}\tag{1}\\
&+&\frac{q\,[-\dot{y}]_{\mathrm{ret}}}{4\pi\varepsilon_0 c^2[(x^2+y^2)^{1/2}+(-y)(-\dot{y})/c]_{\mathrm{ret}}}\\
&=&0.
\end{eqnarray*}
$$
Along the $x$-axis we have
$$A_x=A_y=A_z=0\tag{2}$$
and
$$\nabla \cdot \mathbf{A}=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}=0.\tag{3}$$
Gauss's law is given by
$$\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}\tag{4}.$$
Using the expression for the electric field in terms of potentials
$$\mathbf{E}=-\nabla \cdot \phi - \frac{\partial \mathbf{A}}{\partial t}\tag{5}$$
we substitute eqn.$(5)$ into eqn.$(4)$ to obtain
$$\nabla^2 \phi + \frac{\partial}{\partial t} \nabla \cdot \mathbf{A} = -\frac{\rho}{\varepsilon_0}.\tag{6}$$
Now by substituting eqn.$(3)$ into eqn.$(6)$ it seems that we can assert that along the $x$-axis Poisson's equation holds
$$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{7}$$
with the solution for the pair of charged particles
$$\phi=\frac{q}{2\pi\varepsilon_0(x^2+y^2)^{1/2}}.\tag{8}$$
By substituting eqn.$(2)$ into eqn.$(5)$ we find that the $x$-component of the electric field along the $x$-axis is given by
$$
\begin{eqnarray*}
E_x &=& -\frac{\partial \phi}{\partial x}\tag{9}\\
&\approx& \frac{q}{2\pi\varepsilon_0 x^2} - \frac{3\,q\,y^2}{4\pi\varepsilon_0 x^4}.
\end{eqnarray*}
$$
I began with the Lorenz gauge, but the symmetry of the experimental setup appears to enforce the Coulomb gauge along the $x$-axis.

 

[Bob44]

I set up an electrodynamics experiment with an axial symmetry which causes the vector potential to vanish.

The Lorenz gauge
$$\nabla \cdot \mathbf{A} = - \frac{1}{c^2} \frac{\partial \phi}{\partial t}$$
is usually employed in electrodynamic situations so that the scalar and vector potentials obey wave equations.

But in the above example the divergence of the vector potential along the x-axis vanishes.

This implies that the time variation of the scalar field $\phi$ also vanishes.

Thus we end up with Poisson’s equation with an instantaneous solution for the scalar potential $\phi$ along the x-axis which must be wrong.

Is there a conceptual error here?

 

[TSny]

In your example in post #1, $\large \frac{\partial A_y}{\partial y} \neq 0$ for points along the x-axis. So, $\nabla \cdot \mathbf{A} \neq 0$ along the x-axis.

You can verify this explicitly for the case where the two charges move with constant speed. Expressions for $\mathbf{A}(x, y, z, t)$ and $\phi(x, y, z, t)$ in the Lorenz gauge for a uniformly moving particle can be found in standard E&M textbooks. Applying these to your two-particle example, you can show that on the x-axis, $\large \frac{\partial A_y}{\partial y} \neq 0$, $\large \frac{\partial \phi}{\partial t} \neq 0$, and $\nabla \cdot \mathbf{A} = \large -\frac{1}{c^2} \frac{\partial \phi}{\partial t}$.

 

[tech99]

I presume the charges are accelerating and so would seem to radiate in the directions 45 degrees to the axes and zero along the axes.