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Consider a Riemannian manifold $(M,g)$ with the Levi-Civita connection $\nabla$. If $D_t$ is the covariant derivative along curves descending from $\nabla$, a geodesic is a curve $\gamma: I\subseteq\mathbb R\longrightarrow M$ such that $D_t\gamma'=0$ where $\gamma'$ is the velocity vector field along $\gamma$. In coordinates (respect the coordinare frame $\frac{\partial}{\partial x^1},\ldots\frac{\partial}{\partial x^n}$) a geodesic is a curve that satisfies the following equation(s):

$$\ddot \gamma^k(t)+ \dot\gamma^j(t)\dot\gamma^i(t)\Gamma^k_{ij}(\gamma(t))=0$$

It is evident that there aren't constraints for the parameter $t$, but on the book "Carroll - Spacetime and relativity" I read the following mysteriuous phrases:


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Notation: Here by the parametrization with the "proper time" $\tau$, he means the parametrization with the arclength parameter. The equation $3.44$ is the same that I've just written above and moreover the others cited equations deal with the variational approach to geodesics.

So I don't understand why, according to Carroll, if a curve satisfies the above equation(s), then its parametrization should be affine. The author doesn't explain this point.

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    $\begingroup$ IMO the author was only trying to say that this "affine change of parameter" preserves the geodesic. That is, if $\gamma(t)$ is a geodesic, so is $\tilde{\gamma}(t):=\gamma(at+b)$, if the right-hand side makes sense, of course. Btw, it actually works the other way round as well. The main point probably is the fact that the parametrization matters (and so, geodesic is a curve, not "just its image on $M$"). $\endgroup$ Nov 20, 2013 at 20:10

2 Answers 2

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To elaborate on the comment by @PavelC, the geodesic equation $\ddot \gamma^k(t)+ \dot\gamma^j(t)\dot\gamma^i(t)\Gamma^k_{ij}(\gamma(t))=0$ actually implies that the curve is constant speed. This is what the author seems to be saying: the parameter is necessarily affine with respect to the unit speed parametrisation. The proof of this can be found in my course notes here, Lemma 8.3.5 on page 67.

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If $\gamma(t)$ is a geodesic then $$\frac{d}{dt}\left<\gamma'(t),\gamma'(t)\right>=2\left<D_t\gamma'(t),\gamma'(t)\right>=0,$$ thus the magnitude of the velocity field (i.e., the speed) is necessarily constant. The "proper time" parameter for a geodesic is then the parameter $\tau$ such that the constant magnitude of its velocity field is equal to 1.

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