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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 $M$"). $\endgroup$ Commented Nov 20, 2013 at 20:10

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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.

On an arbitrary surface $M$, a curve $\beta$ satisfying the geodesic equation is necessarily constant speed. It suffices to prove that the square of the speed has vanishing derivative: $$ \begin{aligned} \frac{d}{ds}\left(|\beta'|^2\right)&=2\langle\beta'',\beta'\rangle \\&= \langle L_{ij} {\alpha^i}' {\alpha^j}' n \,, {\alpha^k}' x_k \rangle \\&=L_{ij} {\alpha^i}' {\alpha^j}' {\alpha^k}' \langle n, x_k \rangle \\&= 0. \end{aligned} $$

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  • $\begingroup$ Hi, I know this post is super old. But I really like to see the proof of the link. But the page is not active (it doesnt load). Do you have the proof still ? Thank you in advance $\endgroup$
    – AdrinMI49
    Commented Jan 20 at 23:00
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    $\begingroup$ @AdrinMI49, it works for me. If you try again and it still does not work, I can send you the pdf. $\endgroup$ Commented Jan 21 at 9:41
  • $\begingroup$ Unfortunately, it's not working, I don't understand why. It says bad gateway. If it is not a problem, you would help me with the pdf. $\endgroup$
    – AdrinMI49
    Commented Jan 21 at 20:02
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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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