Geodesic Eq: Deriving 2nd Term on RHS

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 1K views
peterpang1994
Messages
36
Reaction score
0
As the geodesic equation in a form of
ed2e808ce2b6aa1859eb947f21f23ec0.png

is quite familiar for me. But I still cannot derive it in terms of time coordinate parameter;
a82eae864b04bc27b468fc0becfabe9d.png

I can't get the second term on the right hand side
what I can get is
½{d[lngαβ(dxα/dt)(dxβ/dt)]/dt}dxμ/dt

How can I obtain that term?
 
Physics news on Phys.org
I suggest you do not work in terms of the metric. The equation follows directly from the variable substitution to parameterise the curve with ##t## instead of ##s## and looking at how the geodesic equation transforms under this change. Keep in mind that
$$
\frac{d^2 t}{ds^2} = - \Gamma^0_{\alpha\beta} \dot x^\alpha \dot x^\beta .
$$
 
  • Like
Likes   Reactions: PAllen