- #1
bueller11
- 17
- 0
I'm trying to show that [tex]\frac{d}{dt}\; g_{\mu \nu} u^{\mu} v^{\nu} = 0[/tex] in the context of parallel transport (or maybe not zero), and I'm rather insecure about the procedure. This is akin to problem 3.14 in Hobson's et al. book (General Relativity an introduction for physicists).
As a guess, I tried the take the time derivative:
[tex]\frac{d}{dt}\; g_{\mu \nu} u^{\mu} v^{\nu}+g_{\mu \nu} \dot{u^{\mu}} v^{\nu}+g_{\mu \nu} u^{\mu} \dot{v^{\nu}}=0[/tex]
I was assuming a stationary metric, so the first part would be zero, leaving
[tex]g_{\mu \nu} \dot{u^{\mu}} v^{\nu}+g_{\mu \nu} u^{\mu} \dot{v^{\nu}}=0[/tex]
From there I can substitute in for [tex]\dot{u^{\mu}}[/tex] and [tex]\dot{v^{\nu}}[/tex].
Is this the right path to take? It seems there's then some index trickery involved to solve this.
Thanks!
As a guess, I tried the take the time derivative:
[tex]\frac{d}{dt}\; g_{\mu \nu} u^{\mu} v^{\nu}+g_{\mu \nu} \dot{u^{\mu}} v^{\nu}+g_{\mu \nu} u^{\mu} \dot{v^{\nu}}=0[/tex]
I was assuming a stationary metric, so the first part would be zero, leaving
[tex]g_{\mu \nu} \dot{u^{\mu}} v^{\nu}+g_{\mu \nu} u^{\mu} \dot{v^{\nu}}=0[/tex]
From there I can substitute in for [tex]\dot{u^{\mu}}[/tex] and [tex]\dot{v^{\nu}}[/tex].
Is this the right path to take? It seems there's then some index trickery involved to solve this.
Thanks!