Solving Geodesic Eq.: Mysterious Conservation Eq. (Sec. 5.4 Carroll)

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SUMMARY

The discussion focuses on the conservation equation for geodesics as presented in section 5.4 of Sean Carroll's book on general relativity. The key takeaway is that the quantity $$\epsilon=-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$$ remains constant along geodesics, with specific values for timelike ($\epsilon = 1$) and null geodesics ($\epsilon = 0$). The relationship between the affine parameter $$\lambda$$ and the proper time $$\tau$$ is clarified, emphasizing that the geodesic equation is not strictly necessary to derive these results. The discussion also highlights the use of the Lagrangian $$L=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}$$ for geodesics, reinforcing the connection between the Hamiltonian and the action principle.

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  • Familiarity with the metric tensor and its properties
  • Knowledge of Lagrangian mechanics and the Euler-Lagrange equations
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Students and researchers in theoretical physics, particularly those focusing on general relativity, geodesics, and the mathematical foundations of spacetime. This discussion is beneficial for anyone seeking to deepen their understanding of the relationship between geodesics and conservation laws in curved spacetime.

George Keeling
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TL;DR
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?
 
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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##.
 
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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.
 
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@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!
 
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
$$\Gamma_{\mu \nu \rho}=\frac{1}{2} (\partial_{\nu} g_{\mu \rho} + \partial_{\rho} g_{\mu \nu} -\partial_{\mu} g_{\nu \rho}), \quad {\Gamma^{\sigma}}_{\nu \rho} =g^{\mu \sigma} \Gamma_{\mu \nu \rho}.$$
 

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