# Covariant Derivative of Metric Tensor

I'm an amateur studying General Relativity. I'm reading some notes of lectures by Susskind. In them, it is written that

"we know that [the covariant derivative of the metric tensor] is zero. Why? Because the ordinary derivative of the metric tensor in Gaussian coordinates is zero. So, in any coordinate system, we have [that the ordinary partial derivatives of the metric tensor in arbitrary coordinates minus the two Chrisoffel correction terms] = 0."

I don't see how this follows. Could someone explain?

• Sure, they're here: lapasserelle.com/general_relativity/lesson_4.pdf – davidp Jun 17 at 3:33
• Does this answer your question? Why is the covariant derivative of the metric tensor zero? – NarcosisGF Jun 17 at 4:37
• I don't think this question is a duplicate. Susskind puts forth a specific argument which on its face seems to demonstrate that the covariant derivative of the metric is zero without needing to impose it as a demand. Asking for clarification seems different from asking why we would make such an imposition in the first place. – J. Murray Jun 17 at 6:01

Susskind's argument is that by the definition of the normal coordinates centered at a point $$p$$,

1. The connection coefficients vanish, $$\Gamma^i_{jk}(p) = 0$$
2. The first derivatives of the metric components vanish, $$\partial_\mu g_{\alpha\beta} = 0$$

This straightforwardly implies that $$\nabla_\mu g_{\alpha\beta} = \partial_\mu g_{\alpha\beta} - \Gamma^{\rho}_{\mu \alpha} g_{\rho\beta} - \Gamma^\rho_{\mu \beta} g_{\alpha \rho} = 0 - 0 - 0 = 0$$

The reason this argument is not so straightforward, however, is the following. To construct Gaussian normal coordinates $$y^\mu$$ centered at a point $$p$$, we begins with a tangent vector $$\mathbf a$$ attached to $$p$$. Next, construct the unique geodesic whose tangent vector at $$p$$ is equal to $$\mathbf a$$. Finally, we follow the geodesic by for a path length $$s$$.

The Gaussian normal coordinate of the resulting spacetime point is $$y^\mu= s a^\mu$$. One can show that in a sufficiently small neighborhood of $$p$$ these coordinates are well-defined. They fail to be well-defined if the geodesics ever cross, which is why the neighborhood may end up being quite small.

From there, one can plug these coordinates into the geodesic equation $$\frac{d^2y^\mu}{ds^2} + \frac{1}{2}g^{\alpha\rho}\big(\partial_\beta g_{\gamma\rho}+\partial_\gamma g_{\beta\rho} - \partial_\rho g_{\beta\gamma}\big)\frac{dy^\beta}{ds}\frac{dy^\gamma}{ds} = 0$$

to yield $$\frac{1}{2}g^{\alpha\rho}\big(\partial_\beta g_{\gamma\rho}+\partial_\gamma g_{\beta\rho} - \partial_\rho g_{\beta\gamma}\big)a^\beta a^\gamma=0$$ for all $$\mathbf a$$. This implies that $$\partial_\beta g_{\gamma\rho}+\partial_\gamma g_{\beta\rho} - \partial_\rho g_{\beta\gamma}=0$$ This further implies that all first derivatives of the metric vanish. To see this, note that \begin{align}\partial_\rho g_{\beta\gamma} &= \partial_\beta g_{\gamma \rho} + \partial_\gamma g_{\beta \rho}\\ &=\big(\partial_\gamma g_{\beta\rho} + \partial_{\rho}g_{\beta\gamma}\big)+\partial_\gamma g_{\beta\rho}\\ &=\partial_\rho g_{\beta\gamma} + 2\partial_\gamma g_{\beta\rho}\end{align} $$\implies \partial_\gamma g_{\beta\rho} = 0$$

for all $$\gamma,\beta,\rho$$.

So we've shown that in Gaussian normal coordinates, the metric derivatives vanish. However - what we have not shown is that the connection coefficients vanish in these coordinates, and for a generic connection they do not.

The standard choice in GR is to use the Levi-Civita connection, which we can read off from the geodesic equation; the connection coefficients are simply

$$\Gamma^{\alpha}_{\ \ \beta\gamma} = \frac{1}{2}g^{\alpha\rho}\big(\partial_\beta g_{\gamma\rho}+\partial_\gamma g_{\beta\rho} - \partial_\rho g_{\beta\gamma}\big)$$

Clearly if we choose this connection then the connection coefficients vanish in our Gaussian normal coordinates, and so Susskind's argument holds.

To summarize, Susskind's argument is either physically motivated or mathematically misleading, depending on how you want to look at it. The linchpin of his approach is that he assumes on physical grounds the existence of Gaussian normal coordinates in which the connection coefficients vanish.

This is physically reasonable, as it amounts to demanding that the equivalence principle hold in sufficiently small patches of spacetime (i.e. that spacetime be "locally Minkowski"). However, this crucial assumption does not generally hold for arbitrary choices of connection, and is ultimately equivalent (along with his earlier assumption that the connection is torsion-free) to assuming that we're using the Levi-Civita connection in the first place.

• Thank you! This is very helpful. – davidp Jun 17 at 7:09
• This argument doesn't have to be misleading mathematically if one is clear about this being a choice. I.e. one can define a connection by demanding that in Riemannian normal coordinates, infinitesimal parallel transport keeps the components constant. Then one obtains the LC connection. This is also the path followed by Weinberg in his GR book. – Bence Racskó Jun 17 at 10:56
• @BenceRacskó Right, I’m wholly in agreement. I said that because as far as I can tell, Susskind says “let’s use these coordinates” without mentioning that in doing so, one is already assuming something about the connection. – J. Murray Jun 17 at 13:19

This is simply by definition of connection. The metric behaves as a constant under covariant derivative along any vector. This condition is part of the definition of the connection and is called compatibility with the metric or metricity.