It is well known that an accelerometer (or any other object for that matter) in a gravitational orbit will register nearly zero acceleration. According to this answer, this is because the object is, in a sense, not accelerating with respect to local spacetime, and is instead following its geodesic.

However, assume that instead of an accelerometer orbiting a massive body, we have a uniformly (say negatively) charged accelerometer in space orbiting a positively charged, but relatively small, object. Assuming that radiative and gravitational effects are negligible in this case, the two situations are mathematically equivalent from a Newtonian perspective, and so the accelerometer should register zero. However, since there is negligible local spacetime curvature, the four-acceleration $\mathbf A$ (as defined in the linked answer) should depend nearly entirely on the $\frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2}$ term, and so the four-acceleration should significantly deviate from zero.

Where have I gone wrong? Is it that:

  1. The charged accelerometer will in fact register a significant acceleration?
  2. The four-acceleration is significant but the accelerometer registers zero?
  3. The four-acceleration is nearly zero and I used the wrong equation (or otherwise messed up)?
  4. I'm an idiot and missed something obvious?
  5. Something else?
  • 2
    $\begingroup$ No, you're not an idiot and this is a well asked question. $\endgroup$ – Selene Routley Jul 16 '16 at 15:36

There is indeed a nett force on the body owing to the electrostatic attraction / repulsion. Therefore, there is nonzero four acceleration, and the body will have a different orbit from the ones defined by the spacetime geodesics for the metric describing the massive body's neighborhood. From the standpoint of an observer stationary with respect to the massive body's surface, the body's orbital period will be longer (slower orbit) than that of a freefalling body in an equivalent altitude orbit, if the electrostatic force is repulsive, otherwise it will be a shorter period if the force is attractive. The accelerometer, as long as its parts are uncharged and therefore not interfered with by the electrostatic force, will therefore register a nonzero acceleration.

If, however, the parts are charged, then the reading will depend on the charge distribution if we are talking about a mechanical mass-on-spring accelerometer. If the charge is perfectly uniform and of the same density as the that of the spacecraft carrying it, then it will read zero. However, it is not then working as a true accelerometer; one would need to account for the electrostatic force on it to glean a four-acceleration reading from it. In principle, you would draw a free body diagram of the accelerometer's plunger and work out whether the electrostatic force on it is enough - through Newton's second law - to account for the plunger's four-acceleration given the four acceleration of the spaceship calculated from its total charge. If so, the spring would need to impart zero force and so the accelerometer would read zero; in general it would read something different from its true four-acceleration.

If, however, the accelerometer were a laser based system inferring acceleration through deviation of light paths, then charge would not interfere with its working. It would register an accurate reading of the body's true four-acceleration.

A comment from the OP, which sums up much of what I am trying to say in a very elegant phrase:

Just a small insight I gleaned from your answer - if the accelerometer uses some kind of physical (e.g. mass-and-spring) system to measure acceleration, then once it is charged it is no longer an accelerometer. Other influences beside its acceleration now affect the accelerometer reading.

I can't add anything to this: this is a beautifully written paragraph.

User ChrisWhite adds

Regarding your second and third sentences, I think the OP was asking about a massless but charged primary body affecting a test accelerometer. That is, there is case A -- massive primary and accelerometer held in orbit by gravity -- and case B -- charged, massless primary and accelerometer held in orbit by Coulomb. Just because the path in A and B looks the same, it's only a spacetime geodesic in case A.

I missed that part about the orbitted object's being small. You're exactly right- the path in both case would be the exactly the same but the two spacetime metrics are different, so the path is only a geodesic in A. It sounds as though the OP has gleaned useful insight notwithstanding my clumsiness, so I think I'll leave this answer and its trail of preserved comments and corrections as is as it seems useful to read the lot. The general principles involved in answering this question stay the same, it's just that the scenario as stated by the OP is a cleaner one than I addressed originally.

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  • $\begingroup$ "If, however, the accelerometer were a laser based system inferring acceleration through deviation of light paths," - wouldn't it be easier to measure the Doppler shift? $\endgroup$ – John Dvorak Jul 16 '16 at 19:04
  • $\begingroup$ @JanDvorak Yes, of course! We'll going into business building laser accelerometers as rivals companies, and your company is going to slay mine wholesale! $\endgroup$ – Selene Routley Jul 16 '16 at 23:44
  • $\begingroup$ By your definition, no accelerometer is "working as a true accelerometer", as every accelerometer has gravitational mass. A "true accelerometer" would have inertial mass but no gravitational mass, and thus read a non-zero acceleration in gravitational orbit. I think the OP's question came down to "if I replace the gravitational force by an analogous electromagnetic force, will it have the same property?" $\endgroup$ – T.C. Proctor Jul 17 '16 at 1:22
  • $\begingroup$ Just a small insight I gleaned from your answer - if the accelerometer uses some kind of physical (e.g. mass-and-spring) system to measure acceleration, then once it is charged it is no longer an accelerometer. Other influences beside its acceleration now affect the accelerometer reading. $\endgroup$ – Samuel Li Jul 17 '16 at 1:25
  • $\begingroup$ @SamuelLi Exactly. I've taken the liberty of putting your comment as a footnote to my answer, because it is a beautifully written summary of what I am trying to say. Incidentally, I recalled last night after answering this question, another answer of mine here. Take a look at the last paragraph. A body force that accelerates all points of a body equally (as an electrostatic field on a uniformly charged body causes the body to undergo an isometry and there is no strain anywhere in the body. When my daughter was going through her .... $\endgroup$ – Selene Routley Jul 17 '16 at 2:33

The charged accelerometer will register a non vanishing acceleration. The reason why the setup you are proposing (interaction via electric charge) gives a physically distinct result from the setup in the answer you linked (interaction via gravity), even though they can be described by the same mathematical force, is because the Equivalence Principle applies for the latter but not for the former.

The Equivalence Principle requires that inertial (the mass $m_i$ in Newton's second law) and gravitational masses (the mass $m_g$ in the Universal Gravitation law) equal each other. Then all free falling bodies will have the same acceleration, namely $$m_ia=m_g g\Rightarrow a=g.$$

On the other hand the same does not happen for a charged particle "falling freely" under action of an electric field $E$ (let us neglect gravity). The acceleration of the body depends on its electric charge $q$ as well as on its inertial mass, $$m_ia=qE\Rightarrow a=\frac{qE}{m_i}.$$ Now imagine your accelerometer consists of an electrically charged bob in the center of an electrically neutral box. When the system is put in orbit (or in "free fall") around a fixed charge, then the bob will move with respect to the box and this will give the acceleration reading.

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  • $\begingroup$ The OP said he had a "uniformly charged accelerometer". Assuming that both mass and charge are uniformly distributed, both the mass and bob will have the same acceleration, and there will be no acceleration reading. $\endgroup$ – T.C. Proctor Jul 17 '16 at 2:31
  • $\begingroup$ @T.C.Proctor As you already pointed out, only if the bob and box has the same mass. Moreover, the box would have to be small enough for the tidal forces be negligible. $\endgroup$ – Diracology Jul 17 '16 at 2:48
  • $\begingroup$ If both charge and mass are uniformly distributed, charge will be proportional to mass, ie $ q_i = c m_i $. Then, from your derivation, we have $ a = c E $ for both the bob and the box. $\endgroup$ – T.C. Proctor Jul 17 '16 at 13:20
  • $\begingroup$ As for the tidal forces, the same goes for an accelerometer in normal gravitational orbit, right? $\endgroup$ – T.C. Proctor Jul 17 '16 at 13:24

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