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Let $(M,g)$ be a Riemannian manifold and $\gamma\subset M$ a curve in $M$. Suppose I would like to show that $\gamma$ satisfies the geodesic equations $$\ddot{x}^k=-\Gamma_{ij}^k\dot{x}^i\dot{x}^j.$$

If $\gamma$ is not parametrized by arc length, $\gamma$ may not satisfy the geodesic equations, as this example demonstrates: Why is $\gamma(t)=(0,t)$ a geodesic in the hyperbolic plane?

My question is why a geodesic, which is not parametrized by arc length, may not satisfy the geodesic equations (as in the above example).

Do Carmo derives the geodesic equations on page 62 and states that a curve in $M$ is a geodesic iff it satisfies the geodesic equation. Where does he assume that the curve is parametrized by arc length?

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Let $\nabla$ be the Levi-Civita connection of $(M,g)$ and $c$ be a geodesic of $(M,g)$, then by definition, one has: $$\overline{\nabla}_{\dot{c}}\dot{c}=0,$$ where $\overline{\nabla}$ stands for $c^*\nabla$, then since $g$ is $\nabla$-parallel, the induced metric $\overline{g}:=c^*g$ is $\overline{\nabla}$-parallel, so that: $$\frac{\mathrm{d}}{\mathrm{d}t}\overline{g}(\dot{c},\dot{c})=2\overline{g}(\nabla_{\dot{c}}\dot{c},\dot{c})=0,$$ therefore $c$ is parametrized at constant speed that is proportionally to the arc length. I want to insist on this last point, a geodesic is not necessarily parametrized at speed $1$. This little discussion established that geodesic are not purely geometric objects.

The geodesic equation stated by Do Carmo is simply the coordinate-dependent version of $\overline{\nabla}_{\dot{c}}\dot{c}=0$. Indeed, by the very definition of the Christoffel $(2,1)$-tensor, one has $\nabla=\nabla^0+\Gamma$, where $\nabla^0$ is the flat connection of the given chart of $M$. Therefore, as we saw, being a solution of the geodesic equation implies being parametrized proportionally to the arc length, so no need to assume that the curve satisfies this condition.

Feel free to let me know if I totally missed the point of your question.

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