Falling electric dipole contradicts the equivalence principle?

[jcap]

Consider an electric dipole consisting of charges $q$ and $-q$, both of mass $m$, separated by a distance $d$.

If the dipole is given an acceleration $a$ perpendicular to its moment the total electric force on it, due to each charge acting on the other, is given approximately by
$$F_e=\frac{e^2a}{c^2d}$$
where we introduce $e^2 \equiv q^2/4\pi\epsilon_0$ for clarity. The exact expression is given in Electrostatic Levitation of a Dipole Eq(5) (http://inspirehep.net/record/206900/files/slac-pub-3529.pdf).

Now suppose the dipole, initially oriented horizontally, is dropped in a vertical gravitational field of strength $g$.

Applying Newton's second law to the dipole as a whole we have: gravitational force (gravitational mass times field strength) plus electric force must equal the inertial mass times acceleration
$$2mg+F_e=2m a$$
Therefore the acceleration $a$ of the dipole is given by
$$a=g\large(1-\frac{e^2}{2mc^2d}\large)^{-1}$$
Thus the dipole is accelerating faster than gravity. An observer falling with the dipole will see it move away from him whereas in deep space the observer would not see the dipole move away.

Surely this contradicts the equivalence principle?

 

[Ibix]

Surely this contradicts the equivalence principle?

No. I gather that one way to look at it is to note that the EM field of the dipole covers a large region, large enough that curvature is obvious. The equivalence principle is only true locally (i.e., over small enough region that curvature is negligible).

A recent discussion on the topic:
https://www.physicsforums.com/threads/do-electrons-radiate-when-in-free-fall.957277/#post-6070086

 

[PAllen]

Actually, if you read the reference, there is no need for global effects, The reference itself uses the POE to predict a gravitation effect, which they then explain.

1) The dipole spontaneously accelerates far away from gravity, in an inertial frame.
2) In a rocket whose acceleration matches the the dipole’s, the dipole will appear to float.

is equivalent to the following on a planet of the right mass and size, over a small region, for a short time (the locality of the POE):

1) The dipole spontaneously accelerates in a free fall frame.
2) The dipole floats in a stationary frame.

The key is that the dipole under discussion is not just neutral matter, it is analogous to a small rocket, with an intrinsic acceleration.

The reference also goes on to state why this theroretical effect can never be achieved in practice.

 

[sweet springs]

$$-\frac{e^2}{dc^2}$$ in your Fe forluma is electromagnetic interaction energy/c^2. So mass of the system M is
$$M=2m - \frac{e^2}{dc^2} < 2m$$ of separate charges. If not dipole but pair of charges of same signature, $$M=2m + \frac{e^2}{dc^2} > 2m$$.

May I regard your case same as if a single particle of mass M falls in the gravitational field?