What does the metric condition $\nabla_\rho g_{\mu\nu}=0$ in General Relativity intuitively mean for an observer measuring distances?

Question

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?

Answer 1

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.

Answer 2

It says that the metric behaves as a constant with respect to covariant differentiation, meaning that vectors have constant magnitude under parallel displacement. In other words, if you displace a metre rule, it is still a metre rule, and if you displace a clock which measures one second per second (without disturbing the mechanism), it will still measure one second per second.

Answer 3

First of all, note that the covariant derivative of a vector is a tensor. This means that if we have two vectors $A_\mu$ and $A^\nu$, then their covariant derivatives $\nabla_\rho A_\mu$ and $\nabla_\rho A^\nu$ are tensors. These two tensors satisfy the transformation law: $$(\nabla_\rho A_\mu)=g_{\mu\nu}(\nabla_\rho A^\nu)$$ The same transformation can be applied to the two vectors $A_\mu$ and $A^\nu$ as follows: $$A_\mu=g_{\mu\nu}A^\nu$$ and take a covariant derivative of the transformation to obtain $$\nabla_\rho A_\mu=(\nabla_\rho g_{\mu\nu})A^\nu+g_{\mu\nu}(\nabla_\rho A^\nu)$$ Now compare this with the first equation to observe that $$\nabla_\rho g_{\mu\nu}=0$$

Intuitive Meaning:

The metric tensor is thus conserved during covariant differentiation. However, this doesn't mean that the partial derivatives are all zero. One can express the partial derivatives of $g_{\mu\nu}$ in terms of the Christoffel connections and can make it zero only at a local point in the space-time. Using this process, one can show that the Christoffel connections are functions of the first derivatives of metric tensor.

One can obtain the expression for the Riemann curvature tensor which is a function of the first derivatives of the Christoffel connections (i.e., a function of the second derivatives of the metric tensor) to find that the Riemann tensor cannot be reduced to zero a local point. This means that all the second derivatives of the metric tensor cannot be reduced to zero at any local point of the space-time. The number of surviving second derivatives is $\frac{1}{12}D^2(D^2-1)$, which for $D=4$ dimensions is 20. This is the number of independent components of the Riemann tensor.

The condition $\nabla_\rho g_{\mu\nu}=0$ intuitively means that all the first derivatives of the metric tensor can be zero at a local point in the space-time, but all the second derivatives cannot be identically zero at such a point. And the number of surviving second derivatives of the metric tensor determines the curvature of space-time. This also defines the very notion of locally-inertial frames.

Answer 4

I believe the best way to understand this is by the equivalence principle using a ruler and a clock. The equivalence principle says that within certain limits of distance and time, there is no way to determine whether or not you are freely falling toward a massive body, or floating in deep space far removed from all massive celestial bodies.

If the metric weren't a covariant constant, then your ruler and/or your clock would change their measurements. So, for example, you might move past an identically manufactured ruler lying perpendicular to the direction of travel, and your ruler would have a different length when you lay it parallel to the other one as you pass.

If you traveled around some closed path, your clock and ruler might not measure the same as when you left.

One of the arguments I encountered was that the scalar product should be invariant. But that just begs the question, why?

Answer 5

I think I have a fairly intuitive answer that is grounded in geometry and physics.

If the covariant derivative acting on the metric tensor vanishes, it means that during the parallel transport of any two vectors $u$, $v$ (that is, vectors living on the tangent space at a point of the manifold) along any curve, the inner product between them is covariantly constant (which in general does not mean numerically contant), i.e.
$$<v,u>=constant$$ This also means that under parallel transport of vector $v$, its own length is also covariantly preserved (set $u=v$ above).

Now, since the generalization of straight lines are geodesic curves (which are locally perceived as straight lines to a travelling observer), this condition mean that parallel transporting the tangent vector to a curve $\gamma$ means that its length is preserved. Now, a tangent vector to a curve in real space, means that the vector is the velocity vector. So the above is the "curved generalization" of the statement that when the sum of external forces acting on a body is zero, then its velocity is constant (locally it is indeed constant -not only in lenght- when viewed by a local observer since he/she views his path along a geodesic as following a straight line. Although again, it is not numerically constant).

Note: There is a bit more to the story since General Relativity needs a covariantly constant metric and a torsion-free connection. Torsion tends to "twist" vectors, but the main idea in my answer is, I believe, correct.