# Locally Flat Coordinates and Christoffel Symbols

We know in general relativity (or rather in differential geometry) that you can have some locally flat coordinates (I think they are called Riemann normal coordinates) at a point $P$ in our manifold (space time). At this point $P$, the metric is Euclidean up to second order deviation, i.e. $$g_{\tau \mu} \approx \eta_{\tau \mu} + B_{\tau \mu \ ,\lambda \sigma} \ x^\lambda x^\sigma + ...$$ where $B_{\tau \mu \ ,\lambda \sigma}$ are just the Taylor coefficient terms (second order in $g$).

Now I was led to believe that the Christoffel symbols should vanish at this point $P$ in locally flat coordinates, but under the definition of them, I get

\begin{split} \Gamma_{\rho \nu}^\lambda & \equiv \frac{1}{2} g^{\lambda \tau} (\partial_\rho g_{\nu \tau} + \partial_{\nu} g_{\rho \tau} - \partial_{\tau} g_{\rho \nu}) \\ & = \eta^{\lambda \tau}(B_{\tau \nu \ , \kappa \rho} + B_{\tau \rho \ , \kappa \nu} - B_{\rho \nu \ , \kappa \tau})\ x^\kappa + ... \end{split}

This is a non vanishing Christoffel symbol. If I am misunderstanding, then when exactly do the symbols vanish?

• ...and setting $x=0$ you see that they vanish. Jul 30, 2016 at 4:33
• The point is that Cristoffel symbols are not tensors. If tensor vanishes in one coordinate system, it vanishes in any other. But the Cristoffel symbols physically describe the acceleration, and the system of coordinates you choose is equivalent to going to a locally "free falling" reference frame. The linear term you see describes the tidal forces. In fact, $B$ is related to Riemann tensor, which is something standardly used to describe the tidal forces. Jul 30, 2016 at 4:37
• @Peter Kravchuk why set $x=0$? Yes, I do understand that the Christoffel symbols are not tensors, and for this reason we get an extra term under a charge of coordinates which leads us to find the definition of the covariant derivative. All I'm asking is why they are not vanishing in this free falling frame as ou say Jul 30, 2016 at 5:18
• you cannot make them vanish identically. Think about the equivalence principle, for example the Earth rotating around the sun. There is gravitational pull towards the sun, but we do not experience it since we are not in an inertial system, but rather an accelerating one; more precisely, we are freely falling in Sun's gravity. Because of that, say, in the center of the Earth the gravitational pull towards the Sun is zero. But it is not zero everywhere on the Earth -- these non-zero forces are the tidal forces (however the tidal forces from the Moon are stronger). Jul 30, 2016 at 5:35

Your expression for the Cristoffel's symbols seems to be correct. In any case, it is definitely true that they should only vanish for $x=0$. The reason is as follows:

By choosing a coordinate system, you label the different points in a piece of your manifold by a set of numbers $x^\mu$. By construction, the point $P$ has the coordinates $x=0$, and non-zero $x$ correspond to points around $P$. The statement that in Riemann normal coordinates around $P$, the Cristoffels's symbols vanish at $P$ means that the symbols vanish for $x=0$.

If the Cristoffel's symbols were to vanish for $x$ in some neighbourhood of $0$, this would mean the curvature tensor vanishes in that neighbourhood. This is only true if the manifold is actually flat at $P$.

In some coordinate systems they do vanish, I think, which to me makes sense as you have the freedom to choose whichever system they vanish in.

Vanishing of Christoffel Symbols

Question : Are the values of the christoffel symbols the same for all coordinate systems on a surface/manifold? I would love to see an example for the cone in two different parametrizations.

Answer : The answer is no. The reason is that the Christoffel symbols are not scalar fields, nor tensor fields, they might vanish completely in a coordinate system and yet be non-vanishing in another. As a simple example consider the plane in cartesian coordinates: All Christoffel symbols vanish. Now consider polar coordinates, there will be some which are non-vanishing