Electric dipole equivalence principle paradox?

[jcap]

Imagine an electric dipole with charges $+q$, $-q$, mass $m$ and size $d$.

Assume this dipole is oriented horizontally and is sitting at rest on top of some weighing scales on earth.

As the charges are at rest the weight of the dipole is just $-mg$.

But by the principle of equivalence this situation is equivalent to the case where there is no gravitational field but that the dipole is being accelerated upwards by an amount $g$.

But in the case where the dipole is being accelerated there seems to be an additional electromagnetic self-force due to each charge acting on the other one.

If one takes an inertial frame in which the dipole is instantaneously at rest then one finds that after a time $d/c$, in which the dipole has moved upwards slightly, each charge experiences an electric field with two vertical components. The first component is due to the Coulomb field of the other charge. The second component is due to the radiative field due to the acceleration of the other charge. These components are in opposite directions but the radiative component is approximately twice as strong as the Coulomb component so that the dipole experiences an overall electric force in the upwards direction given by:

$$F_{elec} \approx \frac{q^2}{4 \pi \epsilon_0 c^2 d} g$$

Thus in the accelerating situation if one weighs the dipole on the scales one will get the weight $-mg+F_{elec}$.

There seems to be a problem here. The calculation of the electrical self-force is fairly involved using Lienard-Wiechert fields so I haven't included it in this post. One might argue that I have got the calculation wrong and that the vertical components of the Coulomb field and the radiative field cancel exactly. But even without doing the calculation one would expect that an exact cancellation is very unlikely.

 

[PeterDonis]

The simple answer to this is that the equivalence principle only applies locally (within the confines of a single local inertial frame), and the electromagnetic field associated with the dipole is not local--you have to take into account the behavior of the field going out to infinity. But there has been plenty of discussion among physicists about whether that simple answer is really the "right" answer, etc., etc.