Why Is Thomas Precession Critical in General Relativity Calculations?

[WannabeNewton]

Hello! I was wondering if anyone knew of any good references (texts or whatever) to learn about the Thomas Precession. None of the current textbooks I own talk about it (maybe there is a chapter on it in Wald? I can't seem to find anything). In particular, I am dealing with a calculation involving the derivation of the lens thirring effect and while I have done it in the standard coordinates of the metric perturbation centered on the origin for a slowly rotating spherical shell, I would like to try it by considering a frame that is co - moving with the free - falling test particle that is precessing. There seems to be a complication involving the transformation to such a co - moving frame since said frame doesn't have to be non - rotating with respect to the original frame but we want to make sure the co - moving frame is non - rotating so that the observer in this frame can properly measure the precession of the falling object. It would seem the Thomas Precession would give a way of doing this with regards to making the proper corrections in the coordinate transformations so if anyone has a reference in a text or something that elaborated on this specific situation I would be much obliged. Thanks!

EDIT: I should probably say that I know what to do in order to correct it i.e. in terms of subtracting off the changes in angle of the orientation of the co - moving frame as seen from the original frame between one instant of time and the next instant of time but what I really want to see is why the correction needs to be made - why this complication arises in the first place. Sorry for my poor wording beforehand. Thanks again.

 

[bcrowell]

http://www.lightandmatter.com/html_books/genrel/ch02/ch02.html#Section2.5

Not all GR books will discuss it, since it's a special-relativistic effect.

 

[Bill_K]

...the changes in angle of the orientation of the co - moving frame as seen from the original frame between one instant of time and the next instant of time.

Do you recognize this perhaps as a derivative?

I'm always amazed that so many of the discussions you find on Thomas Precession talk nervously about "an infinite sequence of infinitesimal Lorentz transformations", instead of just writing down the differential equation and solving it. Never let it be said that relativity is too mathematical a subject!

 

[pervect]

There seems to be a complication involving the transformation to such a co - moving frame since said frame doesn't have to be non - rotating with respect to the original frame but we want to make sure the co - moving frame is non - rotating so that the observer in this frame can properly measure the precession of the falling object.

The technique I have seen used, (for instance in MTW, pg 171) and would suggest for that problem is to Fermi-Walker transport the basis vectors.

Online, there's http://relativity.livingreviews.org/open?pubNo=lrr-2004-6&page=articlesu18.html

It basically involves satisfying a differential equation as Bill mentions.

Let u^{\alpha} be the worldline along which the basis vectors are to be Fermi-walker translated (the worldline of your infalling particle), and let a^{\alpha} = du/d\tau be the proper acceleration along the worldline.

The vector v^{\beta} is Fermi Walker transported if:

eq1
<br /> \frac{d v^{\beta}}{d\tau} = \left( a^{\alpha} v_{\alpha} \right) u^{\beta} - \left( u^{\alpha} v_{\alpha}\right) a^{\beta}<br />

If you fermi-walker transport all your basis vectors, your local "frame of reference" won't rotate.

How was this derived? First off the general idea of a 4-d rotation is needed, which must be linear , anti-symmetric, and preserve the length of 4-vectors.

This is dv^{\beta}/d\tau = -\Omega^{\beta\alpha}v_{\alpha}

Eq1 can be written in this form, it specifies the components of some particular rotation \Omega.

When you accelerate, you have an unavoidable rotation in the a^u plane, ^ being the wedge product. But you want to insure that there's no rotation in any other plane. The differential equations of the Fermi Walker transport ensures that the derivative with respect to \tau of a fermi-walker transported vector is zero if it's orthogonal to both a and u, so a vector that's orthogonal to a and u doesn't change with time

Furthermore, the fermi walker transport of u^{\alpha} must be u^{\alpha}. So if you let v=u, you need du/d\tau = a. You can check that this happens in eq1.

So Fermi-Walker transport is the unique 4-d rotation which does not rotate vectors perpendicular to the a^u plane, and which transforms the 4-velocity u into itself as an identity along the specified worldline.

I recently did a longish solution involving fermi-walker transport in the Supplee paradox thread. It was a bigger hassle than I thought it would be. The issue (a common one with fermi-walker transport or fermi-normal coordinates) is a lack of closed form solutions to the differential equations, meaning that one winds up with series methods.