The following is an improved version of my previous post Falling electric dipole contradicts equivalence principle?

Consider the following system comprising a particle on the left with charge $+q$ that is a large distance $d$ away from two oppositely charged particles on the right, with charges $+q$,$-q$, held apart by a spring of length $L$ and spring constant $k$. Falling charges Let us assume that the left-hand particle is sufficiently far from the right-hand particles so that the "static" horizontal component of the electric field that it produces near those particles is negligible.

To start with consider the system at rest in empty space. The only electric forces acting on the right-hand particles are "static" attractive forces that are balanced by the compressed spring.

Now let us assume that the whole system is falling in a gravitational field with acceleration $g$. According to the equivalence principle this situation should be locally indistinguishable from the system in empty space.

But now as the left-hand charge $+q$ has an acceleration $g$ it should produce a "radiative" vertical component to the electric field in the vicinity of the right-hand particles.

Each right-hand particle is subjected to an extra vertical electromagnetic force whose magnitude is given by

$$F_{EM}=\frac{q^2}{4\pi\epsilon_0 d c^2}g.$$

As the forces on the differently-charged right-hand particles point in opposite directions, the spring is stretched by an amount given by

$$\Delta L=\frac{2F_{EM}}{k}.$$

Thus a local observer can tell that he is falling in a gravitational field which contradicts the equivalence principle.

What's gone wrong?

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    $\begingroup$ Is it self-consistent to talk about a "local" observer and at the same time require the dipole and the point charge to be "sufficiently far apart"? (Honest question — it's just the first potential loophole I spotted.) $\endgroup$ – Michael Seifert Jun 3 at 18:22
  • $\begingroup$ You could have two positive point charges connected by a rigid horizontal rod that bisects the vertical dipole. The rod could be as short as you like provided that the vertically accelerating charges on its ends induce some vertical radiative component to the forces on the opposite charges in the dipole. $\endgroup$ – John Eastmond Jun 4 at 14:03
  • $\begingroup$ physics.stackexchange.com/q/557110/217867 please answer this $\endgroup$ – Ali Raza Jun 4 at 14:54
  • $\begingroup$ You are assuming that electric vertical force in the Earth frame is present also in the falling frame, where you can use the Hooke relation for spring. That may not be correct, because force is not invariant with respect to coordinate transformations. Try to go the other way - assume in the falling frame the charges and electric forces are completely static and calculate how the forces look like in the Earth frame. Maybe the radiation force $\propto g/d$ is explainable as a result of transformation of electrostatic forces in the system back to frame where gravity has strength $g$ ... $\endgroup$ – Ján Lalinský Jun 6 at 22:58
  1. Yes, this falling EM system does contradict (Einstein) equivalence principle. There is nothing wrong with it, because this system is sensitive to non-local effects.

  2. No, the additional force stretching dipole's spring is not given by (just) the expression $F_\text{EM}$, at the very least not in any realistic gravitational field. There would be many other, additional terms contributing to the forces on the dipole charges, potentially much more important.

Now let us elaborate on those two points. Einstein equivalence principle says (quoted from Wikipedia):

The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

The essential word here is local. The field of a charge has infinitely long range. The effects like “radiation reaction” are based on the long range behavior of fields and are thus entirely non-local and depend on the structure of space time on scales of at least $\ell_\text{R}= c^2/g$ (which could be interpreted as a distance to Rindler horizon). But if it is the gravitational field that causes the acceleration, then the spacetime on such scales would be noticeably curved, and the EM field of a charge system would be modified by this curvature. As a result, if the observer can measure variations of the force at the level of radiative corrections, then this observer can indeed distinguish between empty flat Minkowski spacetime and a curved spacetime.

Let us do a simple reality check: consider as a constant gravitational field for our experiment the surface of the Sun (gravitational acceleration $g=274\, \text{m}/\text{s}^2$), which is the strongest gravitational field available within the Solar system. The length scale $\ell_\text{R}$ is then about $2200\,\text{au}$, much larger than the size of Solar system. So any expressions for “radiative forces” that assume static homogeneous gravitational field (such as originating from the paper by Rohrlich, 1963) would not be valid for any experiment conceivable for a Solar system, even if we assume that it could be shielded from external EM fields.

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  • $\begingroup$ Let's assume we have uniform field of strength $g$ in a region of size $c^2/g$. Why would the radiation forces disappear then? $\endgroup$ – Ján Lalinský Jun 6 at 22:36
  • $\begingroup$ @JánLalinský: I did not say radiation forces would disappear. If we have such a region, then the difference in measured forces between charges while in that region of gravitational field (if we specify boundary conditions the right way) and the same in flat space would be much lower. $\endgroup$ – A.V.S. Jun 7 at 5:37
  • $\begingroup$ I don't understand. Are you saying that radiation force in the frame where things accelerate due to gravity will be much lower than $Cq^2 \frac{a}{r_{ret}}$ where $r_{ret}$ is retarded distance? What would be the reason for this? How does this decrease behave when the region of uniform field is expanded to infinity? $\endgroup$ – Ján Lalinský Jun 7 at 14:00
  • $\begingroup$ I am saying the following: If the distance $d$ at which the force is measured (by dipole's spring) is smaller than $c^2/g$ then dipole is within the near field region. Separation of fields/forces into radiative and bound parts is then impossible to do objectively (independent of choice of frame). So if one writes the system motion and forces in Rindler coordinates, there could be radiative terms on the force, but they would be summed with frame-dependent forces from bound fields that would have comparable magnitudes. $\endgroup$ – A.V.S. Jun 7 at 14:24
  • $\begingroup$ when the region of uniform field is expanded to infinity what do we mean by uniform gravitational field? If it is Rindler spacetime then it is just a Minkowski spacetime with the interpretation that our system of charges emerges from past horizon reaches some finite point and then falls back into the future horizon. Forces measured by dipole would be the same as in flat space, just interpretation is different. $\endgroup$ – A.V.S. Jun 7 at 14:36

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