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Is there any operational way to parallel transport vectors in a curved spacetime?

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For example, we can operationally parallel transport a tangent vector of a geodesic along the geodesic by letting a particle free fall. I'm wondering if there's any clever way to parallel transport vectors in more general scenarios, say, a non-tangent vector along a geodesic, or even non-geodesic? By "operational" I mean establishing some experimental set up and making nature do the parallel transport, instead of knowing the metric information beforehand and calculating the parallels.

asked Apr 3, 2015 in Experimental Physics by Jia Yiyang (2,465 points) [ revision history ]
recategorized Apr 9, 2015 by dimension10

1 Answer

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Gyroscopes can be used as with gravity probe B.

answered Apr 9, 2015 by E (30 points) [ no revision ]

Thanks for the reply, but could you elaborate your answer? What observables(angular momenta?) of a gyroscope are parallel transported and by what mechanism?

Let's say that we have a flat/normal space and a curved space inside that space. If an arrow moves along a path in the curved space, it will change directions from the point of view of the overall space. Even if it ends in the same place it starts it can change directions with respect to both spaces.

Gravity probe B used gyroscopes to confirm that spacetime was curved by orbiting around the Earth many times and detecting small changes in direction of angular momentum that were not otherwise accounted for.

Yes I know the mathematical relation between parallel transport and curvature. What's unclear to me is that why the angular momenta will be parallel transported, given the GR dynamics.

Why should angular momentum be different than any other vector? If there is no force on it, it should stay the same, which in a curved space corresponds to parallel transport.

@tyler: Because the gravitational field on a moving object can in principle act to produce a torque on a spin independent of the process of parallel transport. This was the obvious answer, but it is not clearly correct. There is an operational definition of parallel transport in Einstein's old papers of 1911-14, using parallel geodesics, and matching the corresponding equal length points, it is just not immediate that gyroscopes follow the parallel transport law when free-falling, as there are gravitomagnetic forces acting, which might shift the spin differently from parallel transport.

There is an operational definition of parallel transport in Einstein's old papers of 1911-14

@RonMaimon, what's the title of the publication? Where can I read it?

@JiaYiyang: I don't remember the paper, but I remember the argument--- it's the old 19th century definition of parallel transport using geodesics. The main idea is that if you want to transport a vector v along a vector w, you send out a geodesic from x in the direction of w, and you send out all possible geodesics from x+\epsilon v. You then pick the unique geodesic where the distance between corresponding points (defined by the geodesic linking equal proper time increments) is neither growing nor shrinking to linear and quadratic order, and the result is that you draw a "square" of parallel transport vectors. You can interpret the square as parallel transport of v along w or as the parallel transport of w along v, the fact that the box closes to the appropriate leading infinitesimal order is the condition of no-torsion. This operational definition is also the easiest way to see that placing an infinitesimal Kerr solution somewhere induces torsion, leading to Einstein Cartan theory for spinning sources. The best equivalent definition today might be using a vierbein, and demanding that the geodesics shot out along the transported vierbein make the best-possible local approximation to rectangular coordinates, something like Riemann normal coordinates. The operational definition of parallel transport by geodesics explains the best interpretation of the symmetry of the Christoffel symbols.

For this question today, however, with 100 years of hindsight, gyroscope transport is probably the best correct answer on equivalence principle grounds, because if you go to a free-falling frame, the gravitational field vanishes, and the gyroscope should maintain direction, and then this should turn into parallel transport if you go to a non-free-falling frame. But perhaps there's some crazy higher order effect that leads the gyroscope to slowly spin independent of parallel transport, I don't think so, but I personally need to sit down and think carefully, although it might be obvious that gyroscopes work for this, and it might be obvious to OP. I'll sort it out and write an answer, I meant to earlier.

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