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How do you prove that $g(T, T)$ is constant along a geodesic, where $g$ is a metric and $T$ is the tangent vector to the geodesic?

PPR
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  • Hint: consider the derivative of the norm (squared). – Berci Dec 04 '13 at 21:59
  • If $t$ is the parameter of the curve, $C$ is the curve and $\Psi$ is the chart of some section of the curve and $\pi^\alpha$ is the projection map to the $\alpha$th component, you mean $\frac{d}{dt}(g_{\alpha\beta}\frac{d}{dt}(\pi^\alpha\circ\Psi\circ C)\frac{d}{dt}(\pi^\beta\circ\Psi\circ C))$? I'm not sure what's the derivative of $g_{\alpha \beta}$, nor where I would go with this.. – PPR Dec 04 '13 at 22:39
  • Do you know about the covariant derivative along a curve, usually written $D_t$? – Tim kinsella Dec 05 '13 at 00:42
  • No. I know a covariant derivative in general. So a covariant derivative along a curve is a covariant derivative taking the tangent vector field as its argument? I.e. $D_t = \nabla_T$ where $T$ is the tangent vector field? – PPR Dec 05 '13 at 00:56
  • @Psycho_pr: When $\gamma$ is a geodesics and $T$ is its tangent vectors, what is $\nabla_T T$? –  Dec 05 '13 at 02:26
  • @Psycho_pr That's right. You have to do a little more work to define $D_t$ and then the fact you want is immediate from the formula $\frac{d}{dt}|_p <\gamma', \gamma'> = <(D_t \gamma')_p, \gamma'>+<\gamma', D_t \gamma'>$ because the right side is zero since $\gamma$ is a geodesic (so $D_t \gamma' =0$). If you don't want to use $D_t$ I think you can just extend the velocity vector field in a neighborhood of a point (since $\gamma$ is a local imbedding) on the curve and then use a similar "derivation" formula for $\nabla$. Riemmanian Geom is not my strong suit so there might be a better way. – Tim kinsella Dec 05 '13 at 07:55
  • @John: by the affinely-parametrized geodesic equation, that would be zero. – PPR Dec 05 '13 at 17:29
  • @Psycho_pr: Note that it suffices to compute $\frac{d}{dt} g(T, T)=0$. –  Dec 05 '13 at 20:24
  • I guess it is true only if the geodesic is affinely parametrized? – PPR Dec 08 '13 at 16:23
  • @PPR: Yes, it's only true for affinely parametrized geodesics. – LFH May 23 '17 at 20:57

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