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In General Relativity, the following condition hold: $\nabla_\rho g_{\mu\nu}=0$, where $g_{\mu\nu}$ is the metric of spacetime which has to do with measuring distances and angles and $\nabla$ is the covariant derivative which is torsion-free. What does this condition intuitively mean for an observer that moves on a manifold and measures distances on it?

I will give it a try but forgive me if I am a bit imprecise on some points as I try to see this intuitively and I feel that I don't get this quite right.
I start with $\partial_\rho g_{\mu\nu}=0$ (everywhere) which says that the metric is constant all over the manifold in question. Now, since the covariant derivative is an intrinsic derivative to the manifold, which roughly means that it is the change that a local observer on the manifold would experience, I would conclude that $\nabla_\rho g_{\mu\nu}=0$ means that an observer living on the manifold does not experience any change in the metric. I guess this seems pretty intuitive since an observer using a measuring tape to measure distances, won't experience any local change in the distances of objects around him/her.

Is this way of thinking about it right or is it imprecise? Any additional input would be appreciated.

Also, if an interpretation like the one above holds (or something similar), why do we need to have a torsion-less covariant derivative in order for the interpretation to have the meaning that it has?

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The condition $\nabla_{a}g_{bc} = 0$ is just pure mathematics. Every metric admits a torsion-free (for one defintition, one that satisfies $\nabla_{[a}\nabla_{b]}f = 0$ for every function on the manifold) connection that satisfies this condition.

That general relativity is formulated using this connection is a statement that gravity obeys the equivalence principle -- a freely falling observer is parallel translated along the geodesics of $\nabla$ relative to $g$. And the fact that this is a parallel translation is encoded in that condition in the metric.

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  • $\begingroup$ Thanks for the answer. The second paragraph is illuminating although I find it a bit unsatisfactory with respect to the specifics of my question. The metric has to do with angles and distances. So, there surely must be a way to interpret the condition from the point of view of observers that want to measure distances on a manifold. $\endgroup$ – TheQuantumMan Jul 31 '18 at 17:29
  • $\begingroup$ @TheQuantumMan: Yes, vectors are parallel transported. I said that. If you have a vector, it's length will be constant and it's angle(s) with respect to the geodesic will also be preserved as translated along the geodesic. $\endgroup$ – Jerry Schirmer Jul 31 '18 at 19:04
  • $\begingroup$ So, can we say that an observer holding a rod and moving on a manifold, does so by parallel transporting himself and $\nabla_\rho g_{\mu\nu}=0$ means that s/he always sees constant angles and distances for the rod? If so, why, intuitively speaking, would a torsion-full covariant derivative break this physical picture? $\endgroup$ – TheQuantumMan Jul 31 '18 at 19:07
  • $\begingroup$ @TheQuantumMan because stepping forward in time and then doing the parallel translation in space along the geodesic are no longer commutative operations. Geodesic translation becomes "twisted". You also have effects where test spinors don't travel along geodesics anymore. $\endgroup$ – Jerry Schirmer Jul 31 '18 at 19:27
  • $\begingroup$ Great. What about my question in my previous comment on translating a rod? $\endgroup$ – TheQuantumMan Jul 31 '18 at 19:36

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