4
$\begingroup$

Consider an electric dipole, with total mass $M$, consisting of charges $q$ and $-q$, separated by a distance $d$. The total mass $M$ includes the mass defect due to the negative electrostatic energy associated with the opposite charges.

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).

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

Applying Newton's second law of motion to the dipole we have: gravitational force (gravitational mass times field strength) plus electric force equals the inertial mass times acceleration

$$Mg+F_e=M a$$

where, following the equivalence principle, we assume gravitational and inertial masses are equal.

Therefore the acceleration $a$ of the dipole is given by

$$a=g\large(1-\frac{e^2}{Mc^2d}\large)^{-1}.$$

Thus the dipole accelerates faster than gravity. An observer falling with the dipole sees it move away from him whereas in deep space the observer does not see the dipole move away.

Surely this contradicts the equivalence principle?

P.S. I think that if advanced EM fields are somehow included in the calculation of the electric force $F_e$ then $F_e=0$ and the equivalence principle is obeyed.

$\endgroup$
  • $\begingroup$ Related paper: arxiv.org/abs/1311.5798 $\endgroup$ – kkm Jan 20 at 21:59
  • 1
    $\begingroup$ Possible duplicate of Does a charged particle accelerating in a gravitational field radiate? $\endgroup$ – Aaron Stevens Jan 20 at 23:56
  • 2
    $\begingroup$ @AaronStevens: How is this a duplicate? Radiation is something measured far away, while force is a result of local action on a body. Nevermind the fact that charge is not a dipole. This is complementary question for a related problem but not a duplicate. $\endgroup$ – A.V.S. Jan 21 at 14:02
  • 1
    $\begingroup$ @A.V.S. Yeah sorry I should have read more into things. Don't know what I was thinking :) Good thing it takes multiple votes to mark a question as duplicate. $\endgroup$ – Aaron Stevens Jan 21 at 14:18
  • $\begingroup$ @JohnEastmond, you have changed the question. Now it is a mistake to include into equation of motion additional force acting on the dipole due to inernal electric forces. This force is already accounted for by using the decreased mass $M$. $\endgroup$ – Ján Lalinský Jan 22 at 20:47
4
$\begingroup$

In your calculation you assume that gravitational mass $M_G$ of the system is $2m$ where $m$ is rest mass of a single particle, thus you assume it is independent of the mutual distance between the charged particles $d$. In other words, you do not take into account the force of gravity acting on the system due to the concentrated bound negative energy of EM field near the charged particles. However, since the system has lower inertial mass, it should also have lower gravitational mass.

It is well known that systems with negative potential EM energy have inertial mass defect. In this case, the dipole is such a system, so it will have lower inertial mass than $2m$, thanks to its negative electrostatic potential energy $-\frac{e^2}{d}$.

This "mass defect" effect comes from the forces of "acceleration electric fields" acting (in this case) to speed up the charged particles. This you have taken into account by including force $F_{em/self} = \frac{e^2}{c^2d}a$, which is the electromagnetic self-force acting on the dipole.

But defect in inertial mass should mean also defect in gravitational mass. Heuristically/naively, the gravitational mass to use in the formula $F_G = M_G g$ should correspond to total energy of the system via Einstein's formula

$$ E = M_Gc^2 $$

where $E$ is total energy of the system, including its internal potential energy. Using the Coulomb formula for potential energy, $$ E = 2mc^2 - \frac{e^2}{d} $$ and so the gravitational mass of the dipole should be taken as $$ M_{G} = 2m -\frac{e^2}{c^2d}. $$ Then, the Newtonian equation of motion turns out as follows. We have $$ M_G g + F_{em/self} = 2ma; $$ using the above expression for $M_G$ and $\frac{e^2}{c^2d}a$ for $F_{em/self}$, we obtain

$$ \left(2m - \frac{e^2}{c^2d}\right)g = 2ma - \frac{e^2}{c^2d}a $$ which always implies $$ a = g, $$ confirming that if $M_G\neq 0$, the dipole will move in accordance with the equivalence principle.

$\endgroup$
  • $\begingroup$ By your argument surely the inertial mass of the dipole on the RHS of Newton's equation of motion should be $M_I=M_G=2m-e^2/c^2d$ instead of just $M_I=2m$? $\endgroup$ – John Eastmond Jan 21 at 9:24
  • $\begingroup$ The term $2ma$ on the RHS of the equation of motion above is inertial mass of the dipole when the EM self force is considered explicitly as another contributing force to total force. However, because this contribution is proportional to acceleration, it is possible to discard it and hide the contribution as a change of inertial mass. When that is done (the penultimate equation), there is no more self force in the equation, just gravitational force on decreased gravitational mass acting on decreased inertial mass. $\endgroup$ – Ján Lalinský Jan 21 at 9:32
3
$\begingroup$

The paper that claims this result was written in reply to An electric dipole in self-accelerated transverse motion, which claims that a dipole in zero gravitational field really can accelerate itself, indefinitely. So if you believe both of the results of these papers, the equivalence principle is satisfied; dipoles can have an extra, weird contribution to their acceleration in both a gravitational field and in a freely falling frame.

However, the analysis of point charges in electromagnetism, and especially their self-interaction, is full of subtleties. More recently, the paper Nonexistence of the self-accelerating dipole and related questions has claimed that both of these results are incorrect, but since the author of this paper is active on Physics.SE I'm going to refrain from attempting to summarize the paper, because I'll probably get it wrong!

From a more general perspective, we already know postulating perfectly pointlike charges leads to a ton of subtleties, even before bringing relativity into the mix. In the modern formulation of quantum field theory in curved spacetime, everything is manifestly covariant from the start, so the equivalence principle is satisfied by construction. Of course it's interesting to see how it can come about in a less fundamental theory like classical electromagnetism, but issues with that aren't going to bring all of relativity crashing down.

$\endgroup$
  • 1
    $\begingroup$ In addition to the points you raise, there are likely to be a couple of vexing issues: (1) that a boosted dipole isn't a dipole (related to Mansuripur's paradox); (2) there is no universally accepted way to define what is a radiation field in this kind of context. $\endgroup$ – Ben Crowell Jan 20 at 22:18
  • $\begingroup$ The weird self-acceleration is merely a special case for extremely small separations of the two charges. John Eastmond's observation and question is valid even for dipoles that do not have such strange solutions. $\endgroup$ – Ján Lalinský Jan 21 at 1:21
0
$\begingroup$

You can’t apply the gravitational force to a free falling body.

What is the difference of a free falling body near the surface of the earth (your dipole) and a body free floating in space or orbiting the earth? There is no difference, both following their geodesic path and both are not feeling any acceleration. Being you in these positions, you will feel weightless.

$\endgroup$
0
$\begingroup$

Instead of doing explicit calculations, let me point out why (naïvely) applying equivalence principle to a point charge (or a dipole with a small but fixed separation between the charges) in accelerated reference frame is a wrong idea. Equivalence principle is a statement about local physics. Expressions attempting to derive the self-consistent forces / laws of motion for a charge are non-local, because instead of treating only local entities (that would be the charge/dipole and electromagnetic field in its vicinity) those expressions remove the degrees of freedom of the EM field and present only expressions depending on the motion of the charge. This could be done, of course, but as a result we are loosing locality, since EM field is a long-range, these expressions would carry information about the long-distance structure of space-time.

A simple example from electrostatic of point charges and extended conducting bodies. A force acting on a point charge: $$ \mathbf{F}= q \,\mathbf{E}_\text{reg}(\mathbf{r}_q), $$ (where $\mathbf{E}_\text{reg}$ is a regularized, with divergent part removed, electric field at the position of a charge), is a local expression. One needs the value of an electric field only in the small vicinity of a charge (to perform regularization) to compute the force. The expression is applicable for a wide variety of problems, including moving charges, fields of radiation the only requirement is it must be applied in the reference frame where the charge is a (momentarily) at rest.

A force acting on a point charge at a distance $d$ from a conducting plane: $$ \mathbf{F} = - \frac{q^2 }{4d^2} \mathbf{n}, $$ is a non-local expression, since it encodes a structure of the boundary conditions on electric field over the whole domain of a problem. The appearance of the quantity $d$ is an indicator of non-local character of the equation. The expression however is easier to use but is applicable only to a specific problem.

Likewise, if we consider self-action of a dipole in a gravitational field, or in a constant acceleration, expressions for forces like the one in the body of a question are non-local by their origin, so the equivalence principal does not directly apply to them. The analogy with my electrostatic example above could be made more explicit if we remember, that in a reference frame following a constantly accelerating dipole there is a Rindler horizon. This surface provides a specific boundary conditions for electrostatic field of a dipole that could be interpreted as an induced surface charge, that would provide additional force on the dipole.

$\endgroup$
  • $\begingroup$ I am not sure why you're thinking that equivalence principle does not apply to internal electric forces. An hydrogen atom has a mass defect because of the force OP mentioned, and since both the force action and the corresponding negative electrostatic energy are localized near the pair, the equivalence principle should hold. Also, why do you think the approximation of the self-force by the leading term $\propto a$ is not applicable? We can assume, for the sake of easier analysis, that motion is non-relativistic. $\endgroup$ – Ján Lalinský Jan 21 at 18:52
  • $\begingroup$ It may help to notice that Griffiths is talking about a very special case of small separation distance. However, the question here does not require such special case; the acceleration electric field is always present and changes motion of the pair. $\endgroup$ – Ján Lalinský Jan 21 at 18:55
  • $\begingroup$ > "For the dipole it should be a higher power of a..." The total dipole field does fall off faster than monopole field, but how does that have any relevance to the self-force due to mutual interaction of point particles? The self-force is a sum of interaction forces proportional to $a/d$ (radiation electric fields of the two particles). These forces are in the same direction as acceleration (as opposed to a case of two particles of same charge, where the force would oppose the acceleration.) The rate of decay of total field does not enter the calculation of self-force. $\endgroup$ – Ján Lalinský Jan 21 at 20:20
  • $\begingroup$ If the self force leading term was a higher power of $a$, the correction to mass would be zero for small enough $a$ and we wouldn't be able to recover the property that bound system of negative potential energy has lower inertial mass to external forces than sum of masses of its constituents. $\endgroup$ – Ján Lalinský Jan 21 at 20:27
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – A.V.S. Jan 21 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.