# Thomas precession

1. Jan 23, 2010

### Sweet16

Hi guys,

Is there a Thomas precession for a motion along a geodesic line in a curved spacetime? If yes, consider a little giroscope moving along a circular orbit in the Schwarzschild metric. What is the precession after a full cycle?

2. Jan 23, 2010

### bcrowell

Staff Emeritus
Hi, Sweet16 -- welcome to Physics Forums!

Yes, there is such a precession, as confirmed by Gravity Probe B. The actual observed precession, which goes by the name of the geodetic effect, is the sum of a contribution from spacetime curvature and one from Thomas precession. Here is a calculation of the effect: http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 [Broken] .

Last edited by a moderator: May 4, 2017
3. Jan 23, 2010

### jambaugh

I quote from Misner, Thorne and Wheeler p1118-1119

The Thomas precession comes into play for a gyroscope on the surface of the Earth (a=Newtonian acceleration of gravity), but not for a gyroscope in a freely moving satellite.

So to answer your question, no there is no Thomas precession. There is however precession due to frame dragging in a Kerr metric of a rotating mass.

[Edit... Oops! apparently there is! Hmmm...]

4. Jan 23, 2010

### bcrowell

Staff Emeritus
I find this topic of summing the Thomas and curvature effects confusing myself, and would love it if someone could give a very clear conceptual explanation.

Rindler's Essential Relativity (1969 edition, p. 141), says, "The total effect, geometrical and Thomas, gives the well-known Fokker-de Sitter precession of $3\pi m/r$, in the same sense as the orbit." This superficially appears to contradict the MTW quote, but I'm sure this is just because they're referring to slightly different things.

In the calculation I linked to, which is my own, I certainly need to sum the two effects in order to reproduce the actual data from Gravity Probe B.

I think MTW compute this somewhere using a local coordinate basis that twists around as you rotate it along, and I think this is equivalent to subsuming the Thomas effect into this change of basis. Rindler, in his longer book, which I don't have handy, has his own way he likes to handle this kind of thing, by putting the metric into a certain canonical form, and then interpreting one piece of it as the spatial geometry. My own calculation has the virtue of using only ordinary tensor gymnastics, but summing the two effects is then clearly a completely ad hoc thing, which I'm not satisfied with.

Last edited: Jan 23, 2010
5. Jan 23, 2010

### Sweet16

Thanks for the answers, I will look at the paper.

The reason for asking is exactly the book quoted that says there is no precession in a free moving satellite. Well, obviously for the observers co-falling with the gyroscope that observe it, there is no precession. But this is obious, isn't it? You also dont observe any precession if you co-move with an electron on the atomic orbit.

The question is about a static observer that looks at an orbiting gyroscope every full period.

I personally bet that the precession should be given by the same, special relativistic formula:

$$\boldsymbol{\Omega} = -\frac{1}{v^2}\left(\frac{1}{\sqrt{1-v^2/c^2}} - 1\right) \boldsymbol{v}\times\dot{\boldsymbol{v}}$$

but the velocity should be replaced by:

$$v^2 = -\frac{g_{ij}\mbox{d}x^i \mbox{d}x^j}{g_{00}\mbox{d}x^0 \mbox{d}x^0}$$

(with i,j = 1,2,3) which is local "real" velocity in a curved spacetime and:

$$\dot{\boldsymbol{v}} = \frac{\mbox{d}v}{\sqrt{g_{00}}\mbox{d}x^0}$$

Exactly as for the Lorentz contraction, which is given by the special-relativistic formula with velocity defined according to the above expression.

Does it agree with your calculations? What do you think?

Last edited: Jan 23, 2010
6. Jan 23, 2010

### Sweet16

bcrowell:

BTW, I looked at the text you posted and there is a quote I'd like to comment on:

"The existence of event horizons in general relativity violates unitarity, because it allows information to be destroyed. If a particle is thrown behind an event horizon, it can never be retrieved."

I believe this is a typical misunderstanding. Suppose there is an entangled state of AB. If A is outside horizon and B is falling into it, at A's reference frame it will NEVER pass the horizon. It will only get closer and closer, but will never cross it. So in the A's frame the entanglement is always present and the evolution is unitary.

The same from the point of view of the falling B: it passes through the horizon, but even when it is already below it it sees A being outside. So also from the B's point of view there is no paradox, because it is still entangled with A.

So I don't know what is meant by the conflict of GR with unitarity of quantum mechanics.

7. Jan 23, 2010

### jambaugh

I originally was about to write "Yes" and then started checking MTW and was surprised by their answer. I'm going to have to read up on this.

My instinct says the (strong?) equivalence principle (used in reverse of its typical app.) would allow you to treat gravitation as a physical force and hence invoke Thomas precession.

It may be a matter of perspective, calling it a gravitomagnetic effect or kinematic Thomas precession. Some interesting preprints pop up in xxx.lanl.org when I searched.

8. Jan 23, 2010

### bcrowell

Staff Emeritus
Hi, Sweet16 -- Thanks for your comments. Well, a huge amount has been written about the black hole information paradox, black hole complementarity, etc. I don't think it has ever been resolved in a way that satisfies everybody. There are some things about your argument that don't quite work for me. One is that it deals with entanglement between two particles, but it's not clear to me that entanglement between two particles naturally encapsulates everything involved. I'm also not so sure about the interchangeable way you're referring to particles and observers; in MWI we do have a picture of the observer becoming entangled with what's observed, but it's not so clear (to me, at least) that this can be a literal model of what's going on, since it's not clear whether coherence can really be maintained between a subatomic particle and a a human brain. This is something I'd be interested in discussing further, although if you want to do that we should probably start a separate thread. --Ben

9. Jan 23, 2010

### bcrowell

Staff Emeritus
Sweet16, do you think you could edit #5 and surround all the latex math with tex tags, so it will get rendered as actual tex math? Please have pity on my bleeding eyeballs, I just got my first pair of bifocals :-) For a sample of how to do the tex tags, click the Quote button on this post, and look at how I did this:

$$E=mc^2$$

For inline equations, use itex rather than tex, like this: $F=ma$.

10. Jan 23, 2010

### Sweet16

bcrowell: thanks, done :-) About the observer falling onto the horizon: forget about any quantum stuff. If you drop an apple, it will never cross the horizon, so it will always be accessible, right? The same from the point of view of the apple: you can send signals and they will reach the apple even if it is already under the horizon. So you don't loose any information and the apple doesn't loose any information. So where is the paradox?

11. Jan 23, 2010

### bcrowell

Staff Emeritus
Do you want to start a separate thread about this? Really, I think that would make more sense. For one thing, others who are interested in that topic would be more likely to notice the thread.

12. Jan 23, 2010

### bcrowell

Staff Emeritus
Hmm...I don't know. It seems like MTW are trying to separate the Thomas and curvature parts. But a co-falling observer won't observe either one of those, relative to himself -- as you point out, that would amount to trying to detect the deflection of one gyroscope relative to another gyroscope that had followed the same world-line.

Re the equations, my off-the-cuff reaction is just that it doesn't look manifestly covariant, so it scares me. Personally, I'm chicken to write down anything in GR that isn't in manifestly covariant form -- it's just way too easy to get lost. That doesn't mean that it must be nonsense if it's not manifestly covariant; it just means that I don't consider myself smart enough to be able to tell the difference between non-manifestly-covariant sense and non-manifestly-covariant nonsense.

13. Jan 23, 2010

### Sweet16

I hoped that my approach might be a manifestly-covariant nonsense and could be easily disproved if it was wrong :-)

Here is my line of thought.

Consider an inertial obserwer A seeing two uniformly moving coordinate systems: B and C. Suppose that B and A have parallel spatial axes. The same for C and A. If B and C coincide at some instant then their axes will also coincide at that instant. This means that all points of the axes will coincide simultaneously from the point of view of A. From the point of view of B the coincidence of consecutive points will not be simultaneous and that's because C is rotated in respect to B. This is the Thomas rotation.

Let us now consider an inertial frame A and noninertial, Rindler observer B accelerating towards x. B is defined by an infinite set of accelerating clocks in every point of space, each moving along a hyperbola. Each clock has also a cartesian coordinate system attached, all of them parallel to A's axes. If A sees a freely moving object C, which has a constant velocity along y relative to A and also parallel axes, from the point of view of C B's axes will not be parallel to those of C (the same argument as before). The angle of rotation will gradually change in C's time. Also B will notice that C's axes are gradually rotating. That's the Thomas precession. Since the noninertial Rindler frame is equivalent to Schwarzschild metric close to the horizon, I suppose that an observer freely falling onto the horizon will undergo a normal Thomas precession. Since the formulas in the Rindler case are just ordinary special-relativistic equations, I'd expect the same in the Schwarzschild solution.

Does it make sense?

Last edited: Jan 23, 2010
14. Jan 23, 2010

### Sweet16

Sure, just let us finish with the Thomas precession problem :-)

15. Jan 27, 2010

### George Jones

Staff Emeritus
Instead of parallel transporting $\mathbf{L}$ along the timelike worldline of the orbiting gyroscope, you have parallel transported $\mathbf{L}$ along a spacelike curve for which $t$ is constant. The tangent 4-vector to the satellite's worldline is its 4-velocity, so the correct parallel transport law (see MTW) involves a directional covariant derivative,

$$0 = \nabla_{\mathbf{u}} \mathbf{L},$$

or, in component notation,

$$0 = u^\alpha \nabla_\alpha L^\beta.$$

If $r$ and $\theta$ are constant (plane circle), then the 4-velocity has the form

$$\mathbf{u} = \left(u^t, u^r, u^\theta, u^\phi \right) = \left(u^t, 0, 0, u^\phi \right).$$

Setting $u^t = 0$ gives your two equations, but, since $\mathbf{u}$ is timelike, the time component of 4-velocity is never zero.

I suspect that parallel transporting $\mathbf{L}$ along the correct $\mathbf{u}$ will give the correct precession, so Thomas precession will not need to be added (for geodesics).

16. Jan 27, 2010

### bcrowell

Staff Emeritus
Hi, George -- Thanks for the suggestion! I always find your posts on GR to be very knowledgeable and helpful. I actually did worry about whether the omission of the time coordinate was the problem, so I tried again with 2+1 dimensions, and still got the same answer. However, it's quite possible that I did the 2+1 calculation wrong. I will definitely give it another shot and see if that does the trick. --Ben

[EDIT] By the way, if the 2+1 calculation does turn out to eliminate the need to add the Thomas effect by hand, then it would give a very nice physical interpretation and explanation of the seeming contradiction between Rindler and MTW. The effect from the calculation with 2 spatial dimensions would be interpeted as the purely spatial-geometry contribution, while the difference between the 2 and 2+1 results would be interpreted as the Thomas effect that you *would* have gotten if you'd whirled the gyroscope around in a circle on the end of a rope, instead of letting it orbit on a geodesic.

Last edited: Jan 27, 2010
17. Jan 29, 2010

### bcrowell

Staff Emeritus
Aha! Thanks, George Jones, for kicking me in the right direction. I'd indeed made a mistake before, which convinced me that going from 2 spatial dimensions to 2+1 didn't change the result. In fact it does change the result, and by exactly the right amount, i.e., the same amount you get from calculating Thomas precession. I've got both versions of the calculation online now, at the URL I posted above. (I think it's actually nicer to present both, since it makes it more physically clear what's going on, and the purely spatial calculation is a lot simpler.)

One thing I need to think about some more is that there are actually three eigenfrequencies, and only one of them is physically correct. You get $\Omega$, $-\Omega$, and 0. The positive/negative issue is present in the calculation with two spatial dimensions, and the zero-frequency one pops up when you go to 2+1.

[EDIT] Ah, I get it. The zero-frequency one has to be there because if you parallel-transport your own velocity four-vector, it can't precess.

Last edited: Jan 29, 2010
18. Jan 30, 2010

### bcrowell

Staff Emeritus
There's another point that I hadn't understood correctly before. If you take Rindler's view that the geodetic effect for a satellite like Gravity Probe B is explained by a combination of spatial curvature and a Thomas precession, then you have to say that the Thomas precession has the opposite sign in the case of rotation produced by a gravitational force than a nongravitational force. This seems extremely awkward to me, so I prefer something more like MTW's point of view, which is that there's no point in trying to discuss the Thomas precession as a separate effect in the case of a satellite.