The scalar potential $\phi$ and vector potential $A$ at a distance $r$ from a charge $q$ are given approximately by

$$\phi = \frac{q}{r}$$

$$\mathbf{A} = \frac{q\mathbf v}{r}$$

where the constants have been suppressed.

The corresponding electric and magnetic fields are given by

$$\mathbf{E} = -\nabla \phi - \frac{\partial \mathbf A}{\partial t}$$

$$\mathbf{B} = \nabla \times \mathbf A$$

Now the gradient and curl terms fall off as $1/r^2$ so that at large distances from the charge $q$ we only have an electric field $E$ given approximately by

$$\mathbf{E} = - \frac{\partial \mathbf A}{\partial t}\ \ \ \ \ \ \ \ \ (1)$$

where

$$\mathbf{A} = \frac{q\mathbf v}{r}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$

Thus at large distances an accelerating charge produces an electric field that is proportional to its acceleration. This is the standard (retarded) radiation field of an accelerated charge.

But perhaps this is not the only solution to equation (1). Imagine that the charge is fixed with respect to an inertial frame and that the observer is moving with velocity $-v$.

In the rest frame of the observer the charge is moving with velocity $v$. This implies that in the observer's rest frame there is an $A$-field.

Now if the observer changes his velocity then in his comoving frame he experiences a changing $A$-field.

My proposal is that Equation (1) still applies in this situation so that this "apparently" changing $A$-field induces an electric field in the observer's accelerating frame. This field causes any charges that are co-moving with the accelerating observer to feel a kind of electromagnetic inertial force.

I realise that this is non-standard as Maxwell's equations are usually only applied in an inertial frame. My hypothesis is that the retarded solutions apply to an inertial frame whereas the previously unused advanced solutions apply to accelerated frames.

-