Proof of geodesic has constant speed

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.

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 $$\gamma$$."

• "I know the definition of geodesic", you should provide the definition that you are working with. Commented Dec 14, 2019 at 14:21
• I added it to the question above. Commented Dec 14, 2019 at 14:31
• Okay, I’ve edited my answer to remove the more general context.
– cmk
Commented Dec 14, 2019 at 14:47
• Also see the first page of maths.dur.ac.uk/users/pavel.tumarkin/past/fall16/DG/… Commented May 10 at 9:16
• Also see proposition 1 of faculty.sites.iastate.edu/jia/files/inline-files/geodesics.pdf Commented May 10 at 9:35

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, $$0=2\alpha'\cdot\alpha''=\frac{d}{dt}\left(\|\alpha'\|^2\right).$$
• 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).