Existence of reparametrization by arc length

Question

I have got some issues in understanding the proof of the existence of reparametrization by arc length for any regular parameterized curve $\alpha :I\rightarrow R^n$.

I know the key is to construct a diffeomorphism $\phi :J \rightarrow I$ such that the speed of $\alpha \circ\phi$ is identially 1.

Why the following function's inverse is the diffeomorphism $\phi$ ? $$\psi(t)=\int_{t_0}^t \left\lVert \alpha '(u)\right\rVert du $$

where $t_0$ is a point in I

By computation, $\psi '(t)=\left\lVert \alpha '(t)\right\rVert$ and it is always greater than 0 so there is no change of sign implies $\left\lVert \alpha '(t)\right\rVert$ is smooth. Is it suffice to say that $\left\lVert \alpha (t)\right\rVert$ is also smooth? Is this fact also implies that the inverse of $\psi$ is smooth?

Answer 1

The derivative of $\psi^{-1}$ is given by $$ (\psi^{-1})'(s) = \frac{1}{\psi'(\psi^{-1}(s))} $$ whenever $\psi'(\psi^{-1}(s))$ is non-zero. Note that $\psi'(\psi^{-1}(s)) = \|\alpha'(\psi^{-1}(s))\|$ is non-zero because of the regularity of $\alpha$. So smoothness of $\|\alpha'(t)\|$ along with regularity of $\alpha$ will imply that $\psi^{-1}$ is smooth.

Answer 2

You can calculate it: Let $\tau:(0,L(\gamma))\mapsto (a,b) $ be the inverse of $\psi$ s.t. $\psi(\tau(s))=s$ for all $s$. By differenttiating we get $$\frac{d}{ds}(\psi(\tau(s)) = \psi'(\tau(s))~\tau(s)=1. $$

Now we define $$\gamma:=\alpha(\tau)$$ then by the chain rule we get $$\gamma(s)'=\alpha'(\tau(s))~\tau'(s).$$ And there for $$|\gamma'(s)|=|\alpha'(\tau(s))|~\tau'(s)=\psi'(\tau(s))\tau'(s)=1$$

So the function $\tau$ is bijective and $\gamma$ is parametrises $\alpha(I)$ by arclength.