So I'm working with differential geometry. So my book claim that "any geodesic has constant speed".

And the proof is left as an exercise and I found the exercise in the book.

Exercise: "Prove that any geodesic has constant speed and so a very simple unit-speed reparametrization."

I know the definition of geodesic, but I don't know how to work it out.

Thanks in advance

Edit: The definition,

"Let $\gamma$ be a curve on a surface $\textbf{S}$. Then the curve is called a $\textit{geodesic}$ if $\ddot{\gamma}$ is zero or orthogonal to the tangent plane of the surface at the point $\gamma$."


1 Answer 1


Since geodesics have acceleration normal to the surface, the acceleration is also normal to the velocity (which is tangent to the surface). So, if our curve is $\alpha,$ then $\alpha'\cdot\alpha''=0.$ Hence, $$0=2\alpha'\cdot\alpha''=\frac{d}{dt}\left(\|\alpha'\|^2\right).$$

  • $\begingroup$ So this is actually the proof or? The last part of the exercise "and so a very simple unit-speed reparametrization." does not make sense for me, how does it work? $\endgroup$ Commented Dec 15, 2019 at 17:51
  • 2
    $\begingroup$ That's the entire proof that the speed is constant. Regarding the parameterization question, if $\|\alpha'\|=c,$ then you can simply reparameterize via $\beta(s)=\alpha(s/c)$ (which now has unit speed). $\endgroup$
    – cmk
    Commented Dec 15, 2019 at 20:10

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