Geodesic equation and mysterious conservation equation

  • #1
George Keeling
Gold Member
156
39
TL;DR Summary
Mysterious alternative 4-velocity conservation equation "from geodesic equation". Normal equation being ##U_\nu U^\nu=-1##
I'm still on section 5.4 of Carroll's book on Schwarzschild geodesics

Carroll says "In addition, we always have another constant of the motion for geodesics: the geodesic equation (together with metric compatibility) implies that the quantity $$
\epsilon=-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}
$$is constant along the path."

I don't see how that comes from the geodesic equation. But it is very similar to ##U_\nu U^\nu=-1## which comes from the metric equation:$$
-d\tau^2=g_{\mu\nu}dx^\mu dx^\nu\Rightarrow-1=g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=g_{\mu\nu}U^\mu U^\nu=U_\nu U^\nu
$$So ##\epsilon## is just a constant of proportionality between the affine parameter ##\lambda## and the proper time ##\tau##.

What have I missed?
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
20,004
10,660
He is treating both null geodesics and timelike geodesics. For timelike geodesics you can take ##\lambda = \tau## and get ##\epsilon = 1##, but not for null geodesics. For null geodesics, ##\epsilon = 0##.
 
  • Like
Likes George Keeling
  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,471
13,398
It's only valid for affine parameters ##\lambda##, but you can show that you always can parametrize geodesics with affine parameters.
 
  • Like
Likes George Keeling
  • #4
George Keeling
Gold Member
156
39
@vanhees71 and @Orodruin are right and I forgot to explicitly say that for null paths ##d\tau=0##. So both variants of the equation are correct with ##\epsilon = 1, \epsilon = 0## for timelike and null paths and they still follow from the metric equation. In full:$$
-d\tau^2=g_{\mu\nu}dx^\mu dx^\nu\Rightarrow-\frac{d\tau^2}{d\lambda^2}=g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}
$$Timelike: ##\lambda=\tau## $$
-1=g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}
$$Null: ##d\tau=0##$$
0=g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}
$$
I still don't need the geodesic equation to get to these!
 
  • #5
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,471
13,398
This also follows from the fact that one possible Lagrangian (afaik the most convenient one) for the geodesic equation is
$$L=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}.$$
Since this is not explicitly dependent on the world-line parameter ##\lambda## (derivatives wrt. ##\lambda## are denoted with a dot), the corresponding "Hamliltonian" is again ##L## itself, i.e., ##L=\text{const}## in this form of the action principle for geodesics you get automatically a parametrization with an affine parameter. For timelike (spacelike) geodesics you can simply set ##L=\pm 1## and for lightlike ones ##L=0##.

The Euler-Lagrange equations are the usual geodesic equations for an affine parametrization,
$$\mathrm{D}_{\lambda} x^{\mu}=\ddot{x}^{\mu} + {\Gamma^{\mu}}_{\nu \rho} \dot{x}^{\nu} \dot{x}^{\rho}$$
with
$$\Gamma_{\mu \nu \rho}=\frac{1}{2} (\partial_{\nu} g_{\mu \rho} + \partial_{\rho} g_{\mu \nu} -\partial_{\mu} g_{\nu \rho}), \quad {\Gamma^{\sigma}}_{\nu \rho} =g^{\mu \sigma} \Gamma_{\mu \nu \rho}.$$
 

Suggested for: Geodesic equation and mysterious conservation equation

Replies
13
Views
1K
  • Last Post
Replies
18
Views
716
  • Last Post
Replies
11
Views
684
Replies
25
Views
1K
  • Last Post
Replies
9
Views
870
Replies
0
Views
289
  • Last Post
Replies
2
Views
562
  • Last Post
2
Replies
65
Views
2K
Replies
12
Views
756
Replies
8
Views
2K
Top