Quick expression on geodesic equation

[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?

 

[stevendaryl]

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?

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.

 

[unscientific]

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.

Brilliant, thank you!