What would a uniformly charged accelerometer read when responding to an electrostatic force?

• Karl Coryat
Karl Coryat
Suppose we have an accelerometer carrying a charge. The charge density everywhere in the instrument is uniform, or at least what I mean to say is, the charge on any component is proportional to that component's mass. Now, in an inertial reference frame, we place the accelerometer in an electric field. The accelerometer will accelerate. But will it register this acceleration?

The context of this question is a discussion about gravity. An accelerometer in freefall does not register an acceleration. In general relativity, this is explained by gravity being geometry rather than a force acting at a distance, so the accelerometer does not experience any forces unless it's being pushed upward by, say, a table. But a couple people are telling me that an accelerometer's zero reading during freefall can be explained via Newtonian gravity as well: The accelerometer reads zero because all of the components of the accelerometer, including the little tuning forks, are being acted on equally by the Newtonian force of gravity.

Thus the question about the accelerometer with a charge: If the electrostatic force is acting equally on all of the instrument's components, including the little tuning forks inside, will the accelerometer register zero, or something else, when the instrument is accelerated in the electric field?

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Summary:: Asking if a uniformly charged accelerometer would measure an acceleration when placed in an electric field.

Suppose we have an accelerometer carrying a charge. The charge density everywhere in the instrument is uniform, or at least what I mean to say is, the charge on any component is proportional to that component's mass. Now, in an inertial reference frame, we place the accelerometer in an electric field. The accelerometer will accelerate. But will it register this acceleration?

Interesting question! First, if you (who is not freely falling) create an electrostatic field through which the accelerometer falls, the accelerometer will 'see' both an electric and magnetic field, because the fields also transform. Beyond that, I can't say much, but I suppose once you determine the transformed fields (or better- the stress-energy tensor), you simply plug it in as a stress-energy component of the usual general relativity field equation and go from there.

If you are thinking of a situation where the charged accelerometer is sitting on a table (or hanging from an insulating thread), and then you turn on a electrostatic field, The accelerometer should register the change in applied forces. Interesting idea for a undergrad lab... although in practice, I'm not sure you can accumulate charge so easily without creating problems with the electrical components. Maybe if the accelerometer was enclosed within a charge-carrying spherical shell?

Karl Coryat
Thanks! I'm trying to keep gravity out of the thought experiment (the experiment being an analogy for gravity); you can just imagine the accelerometer in a frictionless environment, perhaps in deep space, and being accelerated by the electric field only.

The heart of this question, given the hypothetical boundary condition of the "little tuning forks" being charged uniformly like the chassis of the device, is: Those little tuning forks should experience the same electrostatic force as the rest of the device, per unit mass. If they do, one assumes the reading will be zero -- even though it'd be hard to argue that the accelerometer is not being acted on by a real force and is not experiencing a proper acceleration.

But — if you physically stopped the device from accelerating (gravity equivalent: accelerometer sitting on a table), the tuning forks will deflect and will show a force — even though in this case, the accelerometer is not accelerating! I need ibuprofen.

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Staff Emeritus
I don't think you have specified the problem well enough (or, alternatively, this involves fictional materials that can't be realized in the real world) to answer. It matters how the accelerometer works. Suppose it uses piezoelectrics. For that to meet your requirement, electrons and protons would need the same charge to mass ratio. But then you wouldn't have atoms.

hutchphd
Karl Coryat
Fair enough. Let me try to strip it down as much as I can.

We'll model our accelerometer as a 1kg mass with three 1g masses attached by springs. In deep space we’ll charge all of the masses, and the springs, uniformly (it's all the same material with uniform charge density).

Now we bring the accelerometer close to a large, flat surface that is carrying an electric charge.

My assumption has to be that none of the smaller masses will move relative to the large mass, even as the system accelerates toward or away from the surface (say, in plane with the wall to neglect any "tidal" forces). In other words, the accelerometer won't measure an acceleration under these conditions, despite experiencing a proper acceleration.

And, I assume that if the accelerometer were attracted to the wall and then stopped when it struck an insulated protuberance, then the smaller masses would deflect and stay deflected — thereby registering what would seem to be a proper acceleration, despite the accelerometer not accelerating.

Again, the motivation behind this is to see if it's like a Newtonian interpretation of why a real-world accelerometer is unable to measure acceleration in freefall, but does measure acceleration when resting on a surface. (I am firmly on Team Einstein, BTW...I just find this thought experiment interesting.)

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Staff Emeritus
As I understand what you have described is something that measures the differential force on two ends of a spring and thus infers the acceleration. You then allow an additional electrostatic force into the system, and now your device behaves differently.

Do I have this right?

Karl Coryat
Yeah. I realize it's a contrived boundary condition. I'm just asking if my assumptions are correct regarding what this system would physically do in these two situations (free to accelerate and stopped from accelerating), and if perhaps I'm overlooking something, as I often do.

Homework Helper
Gold Member
Your situation is somewhat equivalent to a falling standard accelerometer. An accelerometer does not measure acceleration but something called 'specific force' which is acceleration minus gravity.

The reason is that gravity force F acts on both the accel's body and proof mass; F/m is the same for both. Think of a spring (constant k) and mass m attached on one end and the other end attached to a box which is your accel's body. An arrow on the box aligns with the spring's major axis, indicating acceleration direction. If you drop the box, arrow pointing down or up (or any other directin for that matter) it's obvious that there is no spring stretching while it's falling. But if you hold the box, arrow pointing up or down, the spring along with the mass will stretch by an amount mg/k since now you plus gravity set ##\Sigma F##/m = 0 on the box while the spring plus gravity set the same ##\Sigma F##/m = 0 on the proof mass. So the spring stretches which is your measure of specific force.

If the charge/mass ratio is the same for the body (minus the proof mass) as for the proof mass, you're again equating F/m for the box and for the proof mass; the accel would read zero. Any other proportion would give you a finite reading.

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Karl Coryat
Mentor
In the absence of any non-electrical forces, the different parts of the accelerometer (weight, springs, case, ...) will each initially accelerate at their own rate determined by their individual ratios of charge to mass. You could in principle separately charge each component in such a way that they all had the same charge-to-mass ratio and furthermore size and shape each component so that the electrostatic forces between them all balance so they accelerate together with no deflection of the springs. (Of course we cannot do this in practice, and even if we could any small change would disrupt the delicate balance).

This situation is best understood as a malfunctioning accelerometer. The electric charge on the test mass is causing a false reading in the same way that we consider a compass to be malfunctioning when it no longer points north because of a nearby mass of iron.

Karl Coryat
Staff Emeritus
Yeah. I realize it's a contrived boundary condition.

More than that. If you are adding another force to the system, why would you expect it to still be a good accelerometer?

Karl Coryat
Why would you expect it to still be a good accelerometer?
I never said I did.

Staff Emeritus
I never said I did.

Then what is the point of this thread? I think we could all agree a bad measuring device is confusing without going any further.

Karl Coryat
The point of the thread was to ask if an accelerometer that had been treated in a certain way, under certain conditions, would mimic an ordinary accelerometer's null result in freefall.

If you really want to know, I've been battling relativity deniers who don't like the proposition that an accelerometer in freefall demonstrates that gravity is not a force. A couple of them suggested this thought experiment to suggest why Newtonian gravitational force cannot be detected in freefall. All I really wanted to do was verify the conditions under which a such-prepared model accelerometer would measure nothing, so that I can respond accurately.