# Another identity with Christoffel symbols

The equations of geodesics typically are written as:

$$\frac{\text{d}u^\alpha}{\text{d}s} + \Gamma^\alpha_{\beta\gamma}u^\beta u^\gamma = 0$$

but, due to the symmetries of the Christoffel symbols of second kind $$\Gamma^\alpha_{\beta\gamma}$$, it seems it can be written as:

$$\frac{\text{d}u_\alpha}{\text{d}s} -\frac{1}{2}\frac{\partial g_{\beta\gamma}}{\partial x^\alpha} u^\beta u^\gamma = 0$$

How can we check that this second form correct?

My attempt: apparently we need to compute the covariant 4-velocity $$u_\alpha = g_{\alpha\mu} u^\mu$$ and the Christoffel symbols of first kind $$[\beta\gamma,\alpha] = g_{\alpha\mu}\Gamma^\mu_{\beta\gamma}$$, then apply the definition $$[\beta\gamma,\alpha]:= (g_{\alpha\gamma,\beta} + g_{\beta\alpha,\gamma} - g_{\beta\gamma,\alpha})/2$$, but this is insufficient and it does not work as expected.

• Your equation is not correct. It is not even second order in s. Commented Jul 27, 2023 at 1:32
• @magma Maybe, but I took the equation from a propsed exercise in V. F. Mukahnov (2005): Physical Foundations of Cosmology, p. 57 Commented Jul 27, 2023 at 9:32
• Your $x$ should be $u\,.$ What do you get when you differentiate $u_\alpha=g_{\alpha\mu}u^\mu$ w.r.t. $s$ ? Commented Jul 27, 2023 at 9:38
• In addition to the correction $x\to u$, notice that $g_{\alpha\mu}\frac{du^\mu}{ds}\neq \frac{d}{ds} g_{\alpha\mu} u^\mu$ as instead you are assuming, I think. Here is the sourse of the apparently missed derivative of the metric... Commented Jul 27, 2023 at 13:06

There is a problem with your notation. The book [1] by V. Mukhanov where you have that from writes on p. 57 \begin{align} &\frac{d\color{red}{u}^\alpha}{ds}+\Gamma^\alpha_{\beta\gamma}u^\beta u^\gamma=0\,,\tag{2.53}\\[2mm] &\frac{d\color{red}{u}_\alpha}{ds}-\frac{1}{2}\frac{\partial g_{\beta\gamma}}{\partial x^\alpha}u^\beta u^\gamma=0\,.\tag{2.54} \end{align} The first equation is well-known and Mukhanov wants us to show its equivalence to the second equation. By the chain rule $$\frac{d}{ds}g_{\alpha\mu}=g_{\alpha\mu,\nu}\frac{dx^\nu}{ds}=g_{\alpha\mu,\nu}\,u^\nu\,.$$ This and the product rule yield \begin{align} \frac{du_\alpha}{ds}&=\frac{d}{ds}(g_{\alpha\mu}u^\mu)=\frac{dg_{\alpha\mu}}{ds}u^\mu+g_{\alpha\mu}\frac{du^\mu}{ds}=g_{\alpha\mu,\nu}\,u^\nu u^\mu\underbrace{\,-\,g_{\alpha\mu}\Gamma^\mu_{\nu\rho}u^\nu u^\rho}_{(2.53)}\\ &=g_{\alpha\mu,\nu}\,u^\nu u^\mu-\Gamma_{\alpha\nu\rho}u^\nu u^\rho=g_{\alpha\mu,\nu}\,u^\nu u^\mu-\frac{1}{2}\Big(g_{\alpha\mu,\nu}+g_{\alpha\nu,\mu}-g_{\nu\mu,\alpha}\Big)u^\nu u^\mu\\ &=\frac{1}{2}\Big(g_{\alpha\mu,\nu}-g_{\alpha\nu,\mu}+g_{\nu\mu,\alpha}\Big)u^\nu u^\mu=\frac{1}{2}\Big(g_{\alpha\mu,\nu}\,u^\nu u^\mu-g_{\alpha\nu,\mu}\,u^\nu u^\mu+g_{\nu\mu,\alpha}\,u^\nu u^\mu\Big)\\ &=\frac{1}{2}\Big(\underbrace{g_{\alpha\nu,\mu}\,u^\nu u^\mu-g_{\alpha\nu,\mu}\,u^\nu u^\mu}_{0}+g_{\nu\mu,\alpha}\,u^\nu u^\mu\Big) \end{align} which shows (2.54). The reverse uses that same chain of equations.