# Quick expression on geodesic equation

Tags:
1. Apr 7, 2015

### unscientific

Taken from Hobson's book:

How did they get this form?

$$\dot u^{\mu} = - \Gamma_{v\sigma}^\mu u^v u^\sigma$$
$$\dot u^{\mu} g_{\mu \beta} \delta_\mu ^\beta = - g_{\mu \beta} \delta_\mu ^\beta \Gamma_{v\sigma}^\mu u^v u^\sigma$$
$$\dot u_{\mu} = - \frac{1}{2} g_{\mu \beta} \delta_\mu ^\beta g^{\mu \gamma} \left(\partial_v g_{\sigma \gamma} + \partial_\sigma g_{v\gamma} - \partial_\gamma g_{v\sigma} \right)$$
$$= - \frac{1}{2} \delta_\mu ^\beta \delta_\beta ^\gamma \left(\partial_v g_{\sigma \gamma} + \partial_\sigma g_{v\gamma} - \partial_\gamma g_{v\sigma} \right)$$
$$= -\frac{1}{2} \left( \partial_v g_{\sigma \mu} + \partial_\sigma g_{v\mu} - \partial_\mu g_{v\sigma} \right)$$

Why are the first two terms zero?

2. Apr 7, 2015

### stevendaryl

Staff Emeritus
Since $u_\mu = g_{\mu \nu} u^\nu$, then
$\dot{u}_\mu = \dot{g}_{\mu \nu} u^\nu + g_{\mu \nu} \dot{u}^\nu$

Using the chain rule, or whatever the rule is, $\dot{g}_{\mu \nu} = u^\sigma \partial_\sigma g_{\mu \nu}$. So we have:

$\dot{u}_\mu = u^\sigma u^\nu \partial_\sigma g_{\mu \nu} + g_{\mu \nu} \dot{u}^\nu$

Using $\dot{u}^\nu = - \Gamma^\nu_{\sigma \lambda} u^\sigma u^\lambda$, we have:

$\dot{u}_\mu = u^\sigma u^\nu \partial_\sigma g_{\mu \nu} - g_{\mu \nu} \Gamma^\nu_{\sigma \lambda} u^\sigma u^\lambda$

Finally, using $g_{\mu \nu} \Gamma^\nu_{\sigma \lambda} = \Gamma_{\mu \sigma \lambda} = \frac{1}{2}(\partial_\sigma g_{\mu \lambda} + \partial_\lambda g_{\mu \sigma} - \partial_\mu g_{\sigma \lambda})$, we can write this as:

$\dot{u}_\mu = u^\sigma u^\nu \partial_\sigma g_{\mu \nu} - \frac{1}{2} u^\sigma u^\lambda (\partial_\sigma g_{\mu \lambda} + \partial_\lambda g_{\mu \sigma} - \partial_\mu g_{\sigma \lambda})$

After some cancellations, you get their result.

3. Apr 7, 2015

### unscientific

Brilliant, thank you!