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?

## Best Answer

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

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 havenotshown is that theconnection coefficientsvanish 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.