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The following is in the context of Riemannian geometry, but the pseudo-Riemannian case is just as important to me (not directly addressed here). I want to understand what freedom we have to parametrize geodesics. This is well known stuff, but I haven't seen all of the following questions treated together in one post. To some extent, my whole question is summarized in the last paragraph.

A geodesic is a (let's say smooth) curve $\gamma: \mathbb{R} \to M$ such that $\nabla_{\dot{\gamma}} \dot{\gamma} = 0$. It follows immediately that $g(\dot{\gamma},\dot{\gamma})$ is constant along a geodesic.

Does this already imply that the parameter is an affine function of the curve length? And then we can always perform the affine reparametrization that will make $\dot{\gamma} = 1$? Since is curve is an equivalence class of its reparametrizations, all geodesics then have unit speed? i.e. we would not consider $\gamma(2\lambda)$ to be a different geodesic than $\gamma(\lambda)$ (or even a geodesic at all?)

But this whole process has severely restricted what the curve parameter can be, right?. How is that when we solve for geodesics, like for example in General Relativity, that almost anything can be chosen to parametrize geodesics, like one of their coordinates? These will be in general non-affine parametrizations? Yet in the (physics) literature, this is rarely ever touched on.

A good example of what I'm trying to understand: when we have geodesic flow which acts on $V \in TM$ by $G^t(V) = \dot{\gamma}(t)$ via the exponential map, is the parameter here necessarily affine? And if we restrict to the unit tangent bundle which the flow preserves, is the parameter $t$ necessarily arc length? Can I take $t$ to be one of the coordinates of the geodesic?

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  • $\begingroup$ "Does this already imply that the parameter is an affine function of the curve length?" Have you tried to verify this yourself? It's a straightforward exercise. $\endgroup$
    – Deane
    Commented Jul 14, 2023 at 18:45
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    $\begingroup$ I prefer to call a curve $\gamma$ that satsfies $\nabla_{\dot\gamma}\dot{\gamma}=0$ a constant speed geodesic. I say geodesic, when the curve is a geodesic with an arbitrary parameterization. $\endgroup$
    – Deane
    Commented Jul 14, 2023 at 18:46
  • $\begingroup$ You can answer your last question using the definition of the exponential map. $\endgroup$
    – Deane
    Commented Jul 14, 2023 at 18:47
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    $\begingroup$ In the pseudo-Riemannian case, it’s no longer true, because you can’t define arclength in general (in a useful way atleast). Null geodesics have zero arclength, yet can have an affine parametrization (an affine parameterization is one for which the equation $\nabla_{\dot{\gamma}}\dot{\gamma}$ is satisfied rather than $\nabla_{\dot{\gamma}}\dot{\gamma}=f\dot{\gamma}$ for some smooth function $f:I\to\Bbb{R}$). $\endgroup$
    – peek-a-boo
    Commented Jul 14, 2023 at 18:47
  • $\begingroup$ Thank you both for your patience. Deane: So for the last question it will be no, correct? As the parameter (by a similar argument to why the parameter must be affine for auto-parallel transport) is necessarily the arc length itself. $\endgroup$ Commented Jul 14, 2023 at 19:06

2 Answers 2

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$\newcommand\R{\mathbb{R}}$ Given a complete pseudo-Riemannian manifold $M$, $p \in M$, and $v \in T_pM$, let $$ \gamma_v: \R \rightarrow M$$ be the parameterized curve satisfying \begin{align*} \gamma_v(0) &= p\\ \dot\gamma_v(0) &= v\\ \nabla_{\dot\gamma}\dot\gamma &= 0. \end{align*} The exponential map at $p\in M$ is defined to be the map \begin{align*} \exp_p: T_pM &\rightarrow M\\ v &\mapsto \gamma_v(1). \end{align*}

Now assume $M$ to be a Riemannian manifold. Given $v \in T_pM$, the curve $$ \gamma_v: \R \rightarrow M $$ has constant speed $|\dot\gamma_v| = |v|$. If $v \ne 0$, let $$ u = \frac{v}{|v|}. $$ Then $\gamma_u$ has unit speed. It is not hard to show that $$ \gamma_u(|v|) = \gamma_v(1). $$ More generally, $$ \gamma_u(t|v|) = \gamma_v(t). $$ So the exponential map satisfies $$ \exp_p tv = \gamma_{tv}(1) = \gamma_v(t) = \gamma_u(|v|t). $$

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Let $v ∈ T_pM$ be a tangent vector to the manifold at $p$. Then there is a unique geodesic $\gamma_v$ satisfying $\gamma_{v}(0) = p$ with initial tangent vector $\dot{\gamma}_{v}(0) = v$. The corresponding exponential map is defined by $\exp(v) = \gamma_{v}(1)$. Note by the property of parallel transport, for every parameter value along the curve, $|\dot{\gamma}_v| = |v|$. We know that parameter must therefore be affine wrt to length.

On the other hand, the geodesic flow is defined as follows: $G^t(v) = \exp(tv)$. Therefore it takes the vector $tv$ for $t \in \mathbb{R}$ to $\gamma_{tv}(1)$. Likewise we know for every parameter value, $|\dot{\gamma}_{tv}| = |tv|=t|v|$. Note the arclength $$s = \int_{0}^{1} t|v| d\lambda$$ So $\lambda = \frac{s}{t|v|}$. In the case of the unit tangent bundle $\lambda = \frac{s}{t}$.

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  • $\begingroup$ Rescale parameter linearly. $\endgroup$
    – Deane
    Commented Jul 14, 2023 at 20:26

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