In curved space, you can define the Riemann curvature tensor at a point. In Riemann normal coordinates with that point as origin, the Riemann curvature tensor can be expressed as depending only on the derivatives of christofell symbols at the points (since the christofell symbols themselves vanish at origin, assuming no torsion). I think the derivatives of christoffel symbols at origin indicate the variation in gravitational forces at origin. If so, could you measure the Riemann curvature tensor by placing a small rotating body at the origin. The tidal forces developed in the rotating body should instantaneously slow down the rotation and simultaneously generate some heat. Could you use information about the change in angular velocity and/or rate of heat generation to estimate the Riemann curvature tensor? I used space and space-time interchangeably. Please let me know if one or other is more appropriate here. Please note that I'm just studying this stuff. So, I won't be able to understand extremely complicated answers.
1 Answer
The Riemann curvature tensor can be broken down into a tidal part, which is due to distant matter, and a non-tidal part, which is due to matter that is present at that point. If you're willing to trust the Einstein field equations, then it may not be that exciting to measure the non-tidal part, since it's probably much easier if you just examine what matter is present.
The tidal part can be measured by a tensor called the Weyl tensor, and I think the answer is that by orienting your rotating body along the three coordinate axes, you could measure three numbers, which would tell you something about the Weyl tensor. However, the Weyl tensor has 10 degrees of freedom, not just 3, so this method of measurement isn't going to tell you everything about the Weyl tensor.
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$\begingroup$ Oh okay. The Weyl tensor is a few lectures ahead. I'll keep this in mind during the lecture. Thanks. $\endgroup$– Matrix23Dec 24, 2017 at 23:38
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$\begingroup$ I also updated my question to specify that the curvature tensor I was talking about was the Riemann curvature tensor, just in case of ambiguity. $\endgroup$– Matrix23Dec 25, 2017 at 0:02