A free particle moves along geodesics, one form being

\begin{split} \ddot x^\mu &= -\Gamma^{\mu}_{\sigma \rho} \dot x^\sigma \dot x^\rho \\ &= -\frac{1}{2}g^{\mu \nu}(\partial_\sigma g_{\rho \nu} + \partial_\rho g_{\sigma \nu} - \partial_\nu g_{\rho \sigma}) \dot x^\sigma \dot x^\rho \end{split}

Applying the lowering operator, we have

\begin{split} \ddot x_\mu &= g_{\mu \alpha} \ddot x^\alpha \\ &= -\frac{1}{2}\delta^\mu_\nu(\partial_\sigma g_{\rho \nu} + \partial_\rho g_{\sigma \nu} - \partial_\nu g_{\rho \sigma}) \dot x^\sigma \dot x^\rho \end{split} Hence $$\tag{1} \ddot x_{\mu} = -\frac{1}{2}(\partial_\sigma g_{\rho \mu} + \partial_\rho g_{\sigma \mu} - \partial_\mu g_{\rho \sigma}) \dot x^\sigma \dot x^\rho $$

Equivalently, using the intrinsic derivative,

$$ \frac{Dt_\mu}{D\tau} = 0 $$ for a free particle, so $$ \frac{dt_\mu}{d\tau} = \Gamma^\sigma_{\mu \rho} \dot x_\sigma \dot x^\rho $$


$$ \ddot x_\mu = \frac{1}{2}g^{\sigma \nu}(\partial_\mu g_{\rho \nu} + \partial_\rho g_{\mu \nu} - \partial_\nu g_{\rho \mu}) \dot x_\sigma \dot x^\rho $$ Hence $$\tag{2} \ddot x_\mu= \frac{1}{2}\partial_\mu g_{\sigma \rho}\dot x^\sigma \dot x^\rho $$ (The summation on $\sigma$ and $\rho$ in the last two terms cancels)

Clearly (1) $\neq$ (2). Does anyone know where my reasoning has gone astray?

Much Appreciated


1 Answer 1


The covariant derivative of $g$ is zero, but it's ordinary derivative isn't. You need to know what is $\ddot{x_\mu}$,

\begin{equation} \ddot{x}_{\mu} = \frac{d^2}{dt^2}(g_{\mu\nu} x^{\nu} ) = g_{\mu\nu} \ddot{x}^{\nu} + \ddot{g}_{\mu\nu} x^{\nu} + 2\dot{g}_{\mu\nu} \dot{x}^{\nu} \ne g_{\mu\nu} \ddot{x}^{\nu} \end{equation} That's the mistake you made in equation (1).

To get agreement, you need to know what is the meaning of vector $t_{\mu}$ in your equation. I think that's actually the components of the dual of the velocity vector, \begin{equation} t_{\mu} = g_{\mu\nu} \dot{x}^{\nu} \end{equation} notice that $g_{\mu\nu}$ is not inside the time derivative, that's how we lower indices by musical isomorphism.

Plug in this definition of $t_{\mu}$ and you will get the standard geodesic equation. In fact, equation (2) is the Euler-Lagrangian equation for the action, \begin{equation} S = \int d\tau \frac{1}{2} g_{\mu\nu} \dot{x}^{\mu} \dot{x}^{\nu} \end{equation} whose solution gives the distance minimizing curve(with affine parameterization): the geodesic.

  • $\begingroup$ Thanks for your help. It is important to distinguish between the ordinary and covariant derivative of the metric. $\endgroup$
    – Pierre
    Commented Jan 21, 2014 at 22:37

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