0
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Your equation is not correct. It is not even second order in s. $\endgroup$
    – magma
    Jul 27, 2023 at 1:32
  • 1
    $\begingroup$ @magma Maybe, but I took the equation from a propsed exercise in V. F. Mukahnov (2005): Physical Foundations of Cosmology, p. 57 $\endgroup$
    – Davius
    Jul 27, 2023 at 9:32
  • 3
    $\begingroup$ Your $x$ should be $u\,.$ What do you get when you differentiate $u_\alpha=g_{\alpha\mu}u^\mu$ w.r.t. $s$ ? $\endgroup$
    – Kurt G.
    Jul 27, 2023 at 9:38
  • 3
    $\begingroup$ 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... $\endgroup$ Jul 27, 2023 at 13:06

1 Answer 1

3
$\begingroup$

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.

[1] V. Mukhanov, Physical Foundations of Cosmology.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.