Arc length parametrisation is reparametrisation

Question

I was trying to show that if $\gamma : [a,b] \to \mathbb R^3$ is a curve and $$ p(t) = \int_{t_0}^t |\gamma'(\tau)|d\tau$$ where $t_0 \in [a,b]$ then $p^{-1}: [c,d] \to [a,b]$ is a reparametrisation of $\gamma$. This is exercise 2 here.

But I am not even sure what I need to show:

1. $p^{-1}$ is a bijection

2. and $(p^{-1})'(s) \neq 0$ for all $s$

3. and $\gamma ( p^{-1}(s)) = \gamma (s)$ for all $s$?

I tried to show 3. but the problem is that it does not even have same domain on both side. How to show that $p^{-1}$ is a reparametrisation of $\gamma$? Thank you for help.

Answer 1

You must show that $p^{-1}$ is continuous and injective. To achieve this it is enough to show that $p$ is strictly increasing and continuous.

some details

You have a $C^1$ function $\gamma\colon [a,b]\to \mathbb R^3$ with $|\gamma'(t)|\neq 0$ for all $t$ and define $$ p(t) = \int_a^t |\gamma'(\tau)|\, d\tau. $$ Let $L=p(b)$. L is the length of the curve $\gamma$. The function $$ p\colon [a,b] \to [0,L] $$ is differentiable and increasing because by the fundamental theorem of calculus: $$ p'(t) = |\gamma'(t)| > 0. $$ Hence $p$ is a bijection. It is injective because it is strictly increasing. It is surjective in view of the intermediate values theorem since $p(0)=0$, $p(b)=L$. So $p$ is invertible. Since $p'(t)\neq 0$ the inverse is also differentiable and $$ (p^{-1})'(s) = \frac{1}{p'(p^{-1}(s))}> 0. $$

Answer 2

Using definition and notations from Wikipedia:

Let $\gamma_1 = \gamma$, $i: \mathbb R \to \mathbb R^3 , x \mapsto (x,0,0)$, $\gamma_2 = i \circ p^{-1}$ and $\phi = \mathrm{id}$.

Then $\phi' \neq 0$ and $\gamma_2(\phi(x))= \gamma_1 (x)$ are immediate. To finish proof all that remains to be shown is that $\gamma$ and $p^{-1}$ have the same properties. Here it means one must verify that $p^{-1}$ is well-defined and continuous (= is a continuous function because $\gamma$ is assumed to be a curve). (Or maybe also differentiable, as assumed in Emanuele's answer)

To show that it is well-defined, one must show $p$ is invertible. A map $p$ is invertible if it is bijective.

To show that it is continuous use that differentiable => continuous (Emanuele's answer)