GR analysis of Cavendish experiment

[Dale]

TL;DR

What does it measure?

In GR the gravitational force is a fictitious force. Fictitious forces cannot actually be measured, only inferred from the motion with respect to the chosen reference frame.

So, what does the Cavendish experiment measure? To be clear, I know that it is the torsion in the string, but what GR quantity is being inferred. It doesn't seem like geodesic deviation is the right thing, since the masses are not in free fall. For a similar reason it seems that the expansion tensor isn't the right thing because that is also about the behavior of geodesics.

 

[Nugatory]

TL;DR: What does it measure?

t doesn't seem like geodesic deviation is the right thing,

How about:

We can consider the separation between two intersecting worldlines as as a function of proper time along them, even if they aren't geodesics. In the Cavendish setup we are comparing the actual worldline of one the balls with the counterfactual but physically reasonable worldline that the ball would follow if it were not subject to the force from the torsion of the wire.

 

[Dale]

In the Cavendish setup we are comparing the actual worldline of one the balls with the counterfactual but physically reasonable worldline that the ball would follow if it were not subject to the force from the torsion of the wire.

So I guess that is a proper acceleration then. Since proper acceleration measures the amount a geodesic worldline turns away from a geodesic.

 

[Ibix]

So I guess that is a proper acceleration then. Since proper acceleration measures the amount a geodesic turns away from a geodesic.

I would say so, yes. Or proper acceleration times the mass of the test masses, anyway. As far as I'm aware, the proper acceleration needed to hover is the general relativistic interpretation of the Newtonian gravitational acceleration.

 

[Mike_bb]

So I guess that is a proper acceleration then. Since proper acceleration measures the amount a geodesic turns away from a geodesic.

I saw such interpretation: Proper acceleration does not deviate from a geodesic—it is, in itself, a physical indicator that a body has deviated from its natural geodesic motion. Is this true?

 

[Dale]

I would not say:

Proper acceleration does not deviate from a geodesic

But this seems fine:

it is, in itself, a physical indicator that a body has deviated from its natural geodesic motion

But I would like to stick specifically with the Cavendish experiment for this thread.

 

[Dale]

Or proper acceleration times the mass of the test masses, anyway.

Yes. Which does have units of force, but I think it is still probably better thought of as proper acceleration than something more force-like. Such as the Christoffel symbols or the stress energy tensor.

 

[Ibix]

Yes. Which does have units of force, but I think it is still probably better thought of as proper acceleration than something more force-like. Such as the Christoffel symbols or the stress energy tensor.

Pedantically, you're measuring a strain of some sort, and you connect that through the Young's modulus (or whatever its twist strain equivalent is), the bar length and the test body masses to their proper accelerations. And therefore through the covariant derivative you're measuring some (invariant) combination of the Christoffel symbols, in which the only unknown in this setup is $G$.

Edit: presumably that final Christoffel combination is expressible in terms of some contraction of the Riemann tensor, which will be more obviously an invariant..

 

[Dale]

presumably that final Christoffel combination is expressible in terms of some contraction of the Riemann tensor, which will be more obviously an invariant

That is what I would like to determine. But I think that combination must be proportional to the proper acceleration in the end.

 

[Ibix]

That is what I would like to determine. But I think that combination must be proportional to the proper acceleration in the end.

Ideally, I think your test masses move radially towards the large masses, so the proper acceleration is in the $r$ direction. So I think if you work in Schwarzschild coordinates you measure the $r$ component of $\nabla_au^a$, where $u^t=1/\sqrt{g_{tt}}$ and $u^a=0$ otherwise.

 

[renormalize]

In GR the gravitational force is a fictitious force. Fictitious forces cannot actually be measured, only inferred from the motion with respect to the chosen reference frame.
So, what does the Cavendish experiment measure?

Dieter Brill offers a nice discussion of this question here: The Cavendish Experiment in General Relativity.

 

[Matterwave]

So I guess that is a proper acceleration then. Since proper acceleration measures the amount a geodesic turns away from a geodesic.

Do you mean world line?

 

[A.T.]

So, what does the Cavendish experiment measure?

Is there a difference to asking what a spring scale measures?

 

[Matterwave]

Is there a difference to asking what a spring scale measures?

The spring scale measures gravitational attraction to the Earth whereas the Cavendish experiment is measuring the gravitational attraction between test masses. The geometry is more difficult to deal with, but at a purely conceptual level I would say both should be measuring "the same kind of thing" (worldline deviation from geodesics, i.e. proper acceleration, as has already been identified in this thread). Unless I'm missing something.

 

[Dale]

Do you mean world line?

oops, yes

 

[Dale]

Is there a difference to asking what a spring scale measures?

A spring scale measures a force, but the gravitational force doesn’t exist in GR.

 

[PeterDonis]

what does the Cavendish experiment measure?

The simplest answer in the GR context is that it is measuring the coupling constant between stress-energy and spacetime curvature. Or, to look at it another way, it is measuring whether, and to what extent, it is correct to adopt "geometric units" in which the Newtonian gravitational constant $G$ is set equal to 1, which is equivalent to assuming that the factor relating the Einstein tensor to the stress-energy tensor in the Einstein Field Equation is the same everywhere in spacetime. In GR this is possible; but in other theories of gravity it can't be done because the coupling constant involved is not the same everywhere in spacetime. See Misner, Thorne & Wheeler, Section 40.8.

 

[Matterwave]

which is equivalent to assuming that the factor relating the Einstein tensor to the stress-energy tensor in the Einstein Field Equation is the same everywhere in spacetime.

Wouldn't a statement about "everywhere in spacetime" require an experiment which could actually at least test some different areas of space time? Like an astronomical experiment?

It's hard for me to imagine how a single isolated Cavendish experiment would test this. Except for the points of spacetime local to where that experiment is taking place.

 

[PeterDonis]

Wouldn't a statement about "everywhere in spacetime" require an experiment which could actually at least test some different areas of space time? Like an astronomical experiment?

Yes, you would need to run Cavendish experiments in significantly different regions of spacetime to really test this.

It's hard for me to imagine how a single isolated Cavendish experiment would test this.

You could think of a run of the experiment here on Earth, say, as calibrating what conversion factor between curvature and stress-energy you expect to see elsewhere. But as above, to actually test that you would need to run the experiment elsewhere.

Of course the experiment is really only testing the conversion factor in the weak field limit, which is why the Newtonian interpretation that it is measuring the "gravitational force" works as well as it does.

 

[A.T.]

A spring scale measures a force, but the gravitational force doesn’t exist in GR.

Yes, but the force that the spring scale applies to the test mass exists in GR too. Similarly, the tangential horizontal forces applied to the test masses by the bar in the Cavendish experiment exist in GR too.

The difference is, that in the Cavendish experiment done on Earth the vertical and radial force components by the apparatus on the test masses are of no interest. But you could do a similar experiment in space, where you measure the total force needed to keep two masses apart. This would be pretty much what you do with a spring scale on Earth.

So I would say that in the GR interpretation, both experiments still measure a real interaction force applied to a test mass. Which provided you know the mass can be used to get its proper acceleration. For the Cavendish experiment done on Earth it is just the tangential horizontal components which are of interest.

 

[Dale]

Yes, but the force that the spring scale applies to the test mass exists in GR too. Similarly, the tangential horizontal forces applied to the test masses by the bar in the Cavendish experiment exist in GR too.

The difference is, that in the Cavendish experiment done on Earth the vertical and radial force components by the apparatus on the test masses are of no interest. But you could do a similar experiment in space, where you measure the total force needed to keep two masses apart. This would be pretty much what you do with a spring scale on Earth.

So I would say that in the GR interpretation, both experiments still measure a real interaction force applied to a test mass. Which provided you know the mass can be used to get its proper acceleration. For the Cavendish experiment done on Earth it is just the tangential horizontal components which are of interest.

Yes, but what is the gravitational quantity here? In Newtonian mechanics it is the gravitational force that is being balanced by the mechanical force. But not in GR.

I like the idea that it is measuring G, since that applies to both theories. The proper acceleration has also been suggested. So has some unknown contraction of the curvature tensor.