# Foucault's pendulum shows absolute motion?

1. Sep 18, 2013

### profgemelli

In the 19th century they believed in the absolute motion and the ether, and they were searching for a scientific proof of the motion of the Earth. When Foucaults experiment was done, people still believing that the earth was still wasfinally convinced that it moves instead.

But it is well known that in the 20th, thanks to Einstein, this belie in absolute motion has been abandoned: motions are relative, there is no eather and no absolute motion... Yet everybody talks without problems of the motion of the Earth, estabilished by Foucaults pendulum experiment, in terms of the 19th century, even in textbooks.

As for me, I have a problem in conciliating modern belief in relativity and relative motions with Foucaults pendulum experiment. Anyone can help? Thank you in advance.

2. Sep 18, 2013

### D H

Staff Emeritus
In a sense, acceleration is still absolute in relativity. Newton's inertial frames exist, unchanged, in special relativity. Inertial frames also exist (but drastically changed from Newton's concept) in general relativity.

3. Sep 18, 2013

### harrylin

With GR, Einstein attempted to make all motion relational (Machian). However that objective wasn't realized. Consequently he came back on his earlier claims, admitting that acceleration (incl. rotation) is in a certain sense "real" or "absolute".

4. Sep 21, 2013

### profgemelli

Einstein came back to absolute motion? Really? This is new to me! Where and when did he write it?

5. Sep 21, 2013

### Staff: Mentor

Not absolute motion, absolute acceleration. Non-gravitational (there are some subtleties about exactly what that term means here) acceleration can be detected and measured without reference to any external body elsewhere in the universe.

It is somewhat startling that the first derivative of position, namely velocity, is relative but the second derivative, acceleration, is absolute. But that's how it is.

6. Sep 21, 2013

### Staff: Mentor

The acceleration that is absolute is not the second derivative of position; that's still relative, because it changes when you change coordinates. Absolute acceleration, or proper acceleration is it's usually called, is what you measure with an accelerometer; mathematically, it's the path curvature of the object's worldline, which is invariant--it's the same regardless of which coordinates you use.

7. Sep 21, 2013

### Staff: Mentor

Quite true - that's a better way of stating it.

8. Sep 21, 2013

### WannabeNewton

Rotation is also absolute in the sense described above i.e. it can be detected locally by using (ideal) gyroscopes and/or by the presence (or lack thereof) of Coriolis forces. This kind of rotation can be written down covariantly, in a manner similar in spirit to the covariant expression for proper acceleration. So just because "motion" is relative doesn't mean all kinematical quantities are as well.

9. Sep 21, 2013

### D H

Staff Emeritus
That "locally" is key. Not quite absolute, which (to me) connotes some notion of universality. Newton's inertial frames are absolute, global in extent. General relativity still has a concept of inertial frames, but they are local.

10. Sep 21, 2013

### Staff: Mentor

Rotation in flat spacetime implies (proper) acceleration, so is a particular case of PeterDonis's comment, is it not? Indeed, this thread started with Foucault's pendulum which specifically detects the rotation of the earth about its axis.

11. Sep 21, 2013

### WannabeNewton

Yes certainly I agree. I suppose I was referring more to the point harrylin brought up about Einstein trying to initially push in the Machian view of rotation.

In that context yes indeed. I just meant that you can equally well use operational techniques to detect rotation in and of itself.

12. Oct 1, 2013

### profgemelli

Proper acceleration? Wait a moment. In general relativity a particle which is in free fall, follows a geodesics, i.e. its "proper" acceleration (derivative of the unit velocity vector, tangent to the path curve, with respect of the proper time parameter) is zero. So you are not talking about that. I can't see what you mean with proper acceleration. Is it something linked with spacetime curvature?

13. Oct 1, 2013

### WannabeNewton

Well for starters, proper acceleration is defined for all test particles, not just for freely falling test particles; yes in the special case of freely falling particles the proper acceleration vanishes. Secondly, the point being made by multiple people above is that proper acceleration is absolute. Motion not being absolute does not imply that all kinematical quantities are also not absolute, with one of the simplest counter-examples being proper acceleration.

14. Oct 1, 2013

### profgemelli

Thanks for your help, everybody, but I am still trying to understand what Foucaults pendulum measures, according to Einstein gravitation. At the moment, this is my humble reasonment and (probably wrong) conclusion:

Earth in a sense is a free falling test particle, so it should travel on a geodesic path in curved spacetime and have zero acceleration.

But under another point of view it is certainly not a particle but a continuum system, so... there is relative motions of the Earth's particles: which should lead to the geodesic deviation equation!

Is it geodesic deviation what we are talking about? That is linked to the Riemann curvature of the spacetime, that is absolute. As a matter of fact, I thougth that geodesic deviation equation deals with deformable bodyes, otherwise the relative distances of he particles should be constant, but maybe I am wrong about it: all in all, there are no really uncompressible solids in general relativity. So, is it Coriolis force generated, in general relativistic terms, by some component of the Riemann tensor? Foucaults experiment is a local measure of the curvature of the spacetime?

15. Oct 1, 2013

### WannabeNewton

Hi again gemelli! So there's sort of two levels to what you're talking about. First if we consider an object with negligible extended size compared to the characteristic size of the source (so that the effects of tidal gravity are irrelevant), the Coriolis forces vanish if the frame of the object is Fermi-Walker transported; if the object is freely falling then Fermi-Walker transport of its frame corresponds to parallel transport. If the frame of the object is not Fermi-Walker transported then there will be Coriolis forces. This is not related to the Riemann curvature.

Now imagine we have an extended non-spherical body whose center of mass is in free fall in an inhomogeneous gravitational field and whose characteristic size is small compared to the variations of space-time curvature. The inhomogeneity of the object's shape as well as that of the gravitational field will induce a torque on this object and cause it to rotate. More precisely, if the center of mass has 4-velocity $u^{\mu}$ and we attach a spin 4-vector $S^{\mu}$ to the worldline of the center of mass, then along the worldline the spin $S^{\mu}$ satisfies $u^{\mu}\nabla_{\mu}S^{\alpha} = \epsilon^{\alpha\beta\mu\nu}u_{\nu}u^{\sigma}u^{\lambda}t_{\beta\rho}R^{\rho}{}{}_{\sigma\mu\lambda}$ where $t_{\mu\nu}$ is the reduced quadrupole moment tensor of the object and $\epsilon^{\alpha\beta\mu\nu}$ is the volume element. Notice that if the gravitational field is uniform (so that the Riemann curvature tensor vanishes identically) then the spin gets parallel transported along the geodesic worldline of the center of mass.

Last edited: Oct 1, 2013
16. Oct 3, 2013

### profgemelli

Ok, dipole, quadrupole... this means something to me. Thank you. I also remember the papapetrou's scheme for a spinning particle, which also involves the Riemann tensor, and maybe that could be another way of seeing the Earth: a spinning particle, i.e. a particle which is not a point test particle, but has some structure. It still seems to me that geodesic deviation equation could be involved too, if we don't assume a polar expansion. In any case when there is a structure the Riemann tensor is involved in the laws of motion for the particle, so this happens for the Earth too.

But still you have not answered me to this question: is this (Riemann curvature of the spacetime) the general relativistic correspondent of Coriolis force that deviates the pendulum in Foucault's experiment?

17. Oct 3, 2013

### Bill_K

Funny, I thought the whole point of the Foucault experiment was that the pendulum did NOT deviate, rather it's the Earth (better, the reference frame of the Earth) that turns. And it takes place just as well in flat space, where the Riemann tensor is zero, so that cannot be involved.

The Foucault deviation is caused by the use of a rotating frame/coordinate system, and the Coriolis force is due to the nonvanishing Christoffel symbols in this frame.

18. Oct 3, 2013

### profgemelli

Yes, in flat spacetime, i.e. in special relativity, there are inertial frames, and the Earth is not inertial, so you can say that the Earth rotates in absolute terms, and so fictional forces like Coriolis are legitimate. Fictional forces are apparent, and appear only in non-inertial frames.

As far as I know, things should be diferent in general relativity: there are no inertial frames, all reference frames are equivalent, and gravitational forces are equivalent to fictional forces. So there should be no asimmetry between one reference frame and another. Yet rotation, as showed by Foucaults, is, in a sense, absolute... That is to puzzle me. All in all Earth is gravitating object so no absolute acceleration should there be, in general relativistic terms. So which is Einstein's explaination of Foucault's experiment? My possible explaination at the moment is that it is originated by tidal forces, in some way, i.e. in other words Foucaults experiment is a sort of local measure of the curvature of the spacetime.

19. Oct 3, 2013

### Staff: Mentor

GR does, however, have a coordinate-independent and frame-independent notion of a geodesic. Foucault's pendulum is trying to follow a geodesic; the trajectory of a point that is stationary in the rotating coordinate system is not a geodesic. This is true and sufficient to explain the precession regardless of the curvature of space near the earth.

20. Oct 3, 2013

### profgemelli

Don't think I dont appreciate your effort to explain me this thing, but I still doubt it is so simple, and I just rely to einstein: since in GR all reference frames are equivalent, the force which acts on the pendulum, as seen by an observed on the surface of the earth, in GR in my opinion should be a kind of gravitational force. That's why I try to interpret it in terms of Riemann tensor.

Moreover, if a point which is stationary with respect to the surface of the Earth, like the point where the pendulum is suspended, does not follow a geodesic, as you say, in my opinion the pendulum also should not, due to being suspended to it.