electromagnetism - Vector Potential Oscillating E Field of the "Null" Field of a Hertzian Dipole?

The vector potential of a Hertzian dipole falls off spherically as $1/r$. The polar axis of the dipole is a "Null" field -- meaning no electric and magnetic field. The absence of magnetic field is clear enough since there is no curl in the vector potential there. However, the potentials description of the E field involves the equation:

$$\mathbf E = -∇Φ - ∂\mathbf A/∂t$$

which seems to indicate there will be an oscillating electric field in the far field of the Hertzian dipole -- even in the "Null" of the dipole, where the where there should be no electric field.

Clearly, as there is no poynting vector in the "Null", an electron placed there can not oscillate in response to an oscillating electric field as there is no energy locally available, otherwise we'd encounter a violation of conservation of energy or a violation of locality.

How is this conundrum resolved?

Answer

The vector potential of an oscillating dipole (using the usual electric dipole approximation) can be written as $$ \vec{A} =\frac{\mu_0 I_0 l}{4\pi r} \cos \omega(t-r/c)\ \hat{z},$$ where the dipole is of length $l$, with a current $I_0 \cos \omega t$ and $\hat{z}$ is a unit vector along the z-axis of the dipole.

Using the Lorenz gauge one can then calculate a corresponding scalar potential $$\phi = \frac{\mu_0 I_0 l c^2\cos \theta}{4\pi} \left( \frac{\cos \omega(t-r/c)}{cr} + \frac{\sin \omega(t-r/c)}{\omega r^2} \right),$$ where $\theta$ is the usual polar angle to the z-axis.

So now the electric field is given by $$\vec{E} = -\frac{\partial \vec{A}}{\partial t} - \nabla \phi$$

If you discard all the terms that have a denominator of $r^2$ or higher from the right hand term (i.e. just consider the far-field), you find the gradient of the scalar potential only has a radial component $$\vec{E} \simeq \frac{\mu_0 I_0 l \omega}{4\pi r} \sin \omega(t-r/c) \hat{z} - \frac{\mu_0 I_0 l \omega \cos \theta}{4\pi r} \sin \omega(t-r/c) \hat{r} $$

But when $\theta =0$ (or $\pi$) along the z-axis, then $\hat{r} \equiv \hat{z}$ and the two terms cancel leaving no far-field.

I am still pondering whether there is a more qualitative way of explaining this.