Geodesic equation and mysterious conservation equation

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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?

Staff Emeritus
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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##.

George Keeling
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It's only valid for affine parameters ##\lambda##, but you can show that you always can parametrize geodesics with affine parameters.

George Keeling
Gold Member
@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!

$$L=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}.$$
$$\mathrm{D}_{\lambda} x^{\mu}=\ddot{x}^{\mu} + {\Gamma^{\mu}}_{\nu \rho} \dot{x}^{\nu} \dot{x}^{\rho}$$
$$\Gamma_{\mu \nu \rho}=\frac{1}{2} (\partial_{\nu} g_{\mu \rho} + \partial_{\rho} g_{\mu \nu} -\partial_{\mu} g_{\nu \rho}), \quad {\Gamma^{\sigma}}_{\nu \rho} =g^{\mu \sigma} \Gamma_{\mu \nu \rho}.$$