# Christoffel symbols in Riemann normal coordinates

I'm studying General Relativity using Misner, Thorne and Wheeler textbook called Gravitation. And I'm stuck on an exercise after chapter 11.6 which is about Riemann normal coordinates. Exercise is the following, one must show that in Riemann normal coordinates at point $$\mathcal{P}_0$$, for the Christoffel symbols we have the following: $$\Gamma^{\alpha}_{\beta\gamma,\mu} = -\frac{1}{3}(R^\alpha_{\beta\gamma\mu} + R^\alpha_{\gamma\beta\mu})$$

But the only thing I can get (following the textbook's hint), is: $$\Gamma^{\alpha}_{\beta\gamma,\mu} = -\frac{1}{2}R^\alpha_{\gamma\beta\mu}$$ For the next step I tried to use Riemann tensor symmetries, such as: $$R^\alpha_{\beta\gamma\delta} = R^\alpha_{\beta[\gamma\delta]}$$ and $$R^\alpha_{[\beta\gamma\delta]} = 0$$ But they didn't bring me to the answer.

• The expression you derived $\Gamma^\alpha_{\beta\gamma,\mu}=-\frac{1}{2}R^\alpha_{\ \ \gamma\beta\mu}$ is definitely incorrect, since the left hand side is symmetric under exchanging $\beta$ and $\gamma$, while the right hand side is not. So I would focus on identifying the error in deriving that expression, rather than trying to manipulate it further. Commented Sep 19, 2022 at 15:25
• My first idea would be to expand the Riemann curvature tensors on the RHS in terms of Christoffel symbols ($\Gamma \cdot \Gamma$) and its derivatives ($\partial \Gamma$). Then, at $P_0$ in normal coordinates, the Christoffel symbols vanish, but its derivatives remain. On the RHS, you should be left with four derivative terms $\partial \Gamma$, which you need to simplify. It might also be helpful to see your derivation.
– psm
Commented Sep 19, 2022 at 20:54
• Andrew, thanks for noticing my mistake. Commented Sep 20, 2022 at 5:18
• psm, well the idea of the derivation is using the geodesic deviation formula $(\nabla_{\vec{u}}\nabla_{\vec{u}}\vec{N}_{(\mu)})^\alpha + R^\alpha_{\beta\gamma\delta}u^\beta N^\gamma_{(\mu)} u^\delta = 0$, where $\vec{u} = \frac{\partial}{\partial\lambda}$ - tangent to geodesic, and $\vec{N}_{(\mu)} = \frac{\partial}{\partial v^{\mu}}$ - deviation vector Commented Sep 20, 2022 at 5:31
• @Filippo, okay thanks) Commented Sep 20, 2022 at 13:52

## 1 Answer

I'll use the same notations as yours.

In Riemann normal coordinates, the point is $$x^{\alpha}=\lambda\, n^{\alpha}$$, the tangent vector is $$u^{\alpha} = n^{\alpha}$$, the deviation vectors are $$N_{(\mu)}^{\alpha} = \lambda \,\delta_{(\mu)}^{\alpha}$$.

The geodesic deviation equation is $$$$\tag{1} \frac{\nabla^2 N_{(\mu)}^{\alpha}}{d\lambda^2} = {R^{\alpha}}_{\beta\gamma\delta} u^{\beta} u^{\gamma} N_{(\mu)}^{\delta}.$$$$ Expanding Eq. (1) using the definition of covariant derivative $$$$\tag{2} \frac{\nabla N_{(\mu)}^{\alpha}}{d\lambda} = \frac{dN_{(\mu)}^{\alpha}}{d\lambda} + \Gamma^{\alpha}_{\beta\gamma} u^{\beta} N_{(\mu)}^{\gamma},$$$$ and the geodesic equation $$d u^{\alpha}/d\lambda + \Gamma^{\alpha}_{\beta\gamma} u^{\beta} u^{\gamma} = 0$$, we obtain $$$$\tag{3} \frac{d^2 N_{(\mu)}^{\alpha}}{d\lambda^2} + 2 \Gamma^{\alpha}_{\gamma\delta} u^\gamma \frac{dN_{(\mu)}^{\delta}}{d\lambda} + \Bigl( \Gamma^{\alpha}_{\gamma\delta,\beta} - \Gamma^{\alpha}_{\mu\delta} \Gamma^{\mu}_{\beta\gamma} + \Gamma^{\alpha}_{\beta\mu} \Gamma^{\mu}_{\gamma\delta} - {R^\alpha}_{\beta\gamma\delta} \Bigr) u^{\beta} u^{\gamma} N_{(\mu)}^{\delta} = 0.$$$$ Substituting $$n^{\alpha}$$ and $$N_{(\mu)}^{\alpha}$$ into the above equation, we find $$$$\tag{4} 2 \Gamma^{\alpha}_{\gamma\delta} n^{\gamma} + \lambda \Bigl( \Gamma^{\alpha}_{\gamma\delta,\beta} - \Gamma^{\alpha}_{\mu \delta} \Gamma^{\mu}_{\beta\gamma} + \Gamma^{\alpha}_{\beta\mu} \Gamma^{\mu}_{\gamma\delta} - {R^{\alpha}}_{\beta\gamma\delta} \Bigr) n^{\beta} n^{\gamma} = 0.$$$$ We expand $$\Gamma^{\alpha}_{\gamma\delta}$$ in powers of $$\lambda$$: $$\begin{equation*} \Gamma^{\alpha}_{\gamma\delta} = \Gamma^{\alpha}_{\gamma\delta} \big|_{\mathcal{P}_0} + \lambda \Gamma^{\alpha}_{\gamma\delta,\mu} u^{\mu} + O(\lambda^2) = \lambda \Gamma^{\alpha}_{\gamma\delta,\beta} n^{\beta} + O(\lambda^2). \end{equation*}$$ Substituting this expression back into (4), dividing through by $$\lambda$$ and then evaluating on $$\mathcal{P}_0$$, we arrive at $$\begin{equation*} \bigl( 3 \Gamma^{\alpha}_{\gamma\delta,\beta} - {R^{\alpha}}_{\beta\gamma\delta} \bigr)\big|_{\mathcal{P}_0} n^{\beta} n^{\gamma} = 0. \end{equation*}$$ After symmetrizing in the indices $$\beta$$ and $$\gamma$$, we get $$$$\tag{5} \bigl( \Gamma^{\alpha}_{\gamma\delta,\beta} + \Gamma^{\alpha}_{\beta\delta,\gamma} \bigr) \big|_{\mathcal{P}_0} = \frac{1}{3} \bigl( {R^{\alpha}}_{\beta\gamma\delta} + {R^{\alpha}}_{\gamma\beta\delta} \bigr) \big|_{\mathcal{P}_0}.$$$$ By cycling the indices $$(\gamma\delta\beta)$$ to get two other equations $$A$$ and $$B$$, adding $$A$$ to and substracting $$B$$ from (5), we finally get $$$$\tag{6} \Gamma^{\alpha}_{\beta\delta,\gamma} \big|_{\mathcal{P}_0} = \frac{1}{3} \bigl( {R^{\alpha}}_{\beta\gamma\delta} + {R^{\alpha}}_{\delta\gamma\beta} \bigr) \big|_{\mathcal{P}_0}.$$$$ If you like, you may also write it as $$$$\tag{7} \Gamma^{\alpha}_{\beta\gamma,\delta} \big|_{\mathcal{P}_0} = - \frac{1}{3} \bigl( {R^{\alpha}}_{\beta\gamma\delta} + {R^{\alpha}}_{\gamma\beta\delta} \bigr) \big|_{\mathcal{P}_0}.$$$$

• Thanks a lot for your solution! Commented Sep 20, 2022 at 13:52