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.

-

You are working in Coulomb gauge, which is non-standard in radiation problems, so it's a little difficult for me to give you an in-depth analysis of your claims, but I can point out some preliminary mistakes straight away. Firstly, your assumption that the grad and curl terms fall off faster than $1/r^2$ is not a valid assumption in the time-dependent case. It's easy to see why this is the case - if we assume that the curl term falls off faster than $1/r^2$ then in the radiation zone, i.e. at large distances, you find that the magnetic field vanishes. However, we know the magnetic field due to an accelerating point charge does not vanish at large distances - either from empirical results regarding radiation or from something like the Liénard-Wiechert potentials. Now, you may be tempted to argue that the Liénard-Wiechert potentials are derived in Lorenz gauge, and we don't expect the results to agree with Coulomb gauge. This is certainly true for $\phi$ and $\vec A$, but it is definitely not true for $\vec E$ and $\vec B$, which are physically measurable quantities and must be independent of choice of a particular gauge. Thus, we already see that there is a problem with your argument.
Now, if one wanted to, with some care one could derive expressions for $\phi$ and $\vec A$ in the radiation zone in Coulomb gauge. However, it is unclear what one would hope to gain from this. The electric and magnetic fields gotten from this $\phi$ and $\vec A$ would be the same as in Lorenz gauge, and hence we would not expect any new physics to appear from there. It is possible to gain new ways of understanding the physics in Coulomb gauge. But in this case, whatever theory you have about inertial vs non-inertial frames, go ahead and try to see if it holds with the standard expressions for $\vec E$ and $\vec B$ in the radiation zone derived in Lorenz gauge, and if it does, perhaps you have something. If it doesn't, switching to Coulomb gauge isn't going to change a thing, and moreover, it will only make your life more difficult.
Without going to the explicit L-W potentials we can just note that $\mathbf B = 0$ won't verify Faraday's law. – Robin Ekman Aug 24 at 21:00