trajectory in polar coordinates

Question

I need to prove that for a given a smooth trajectory $\vec{z}(t)$ in $\mathbb{R}^2$ defined on some interval $I$ and such that $\|\vec{z}(t)\|=1$, there exists a unique twice differentiable $\varphi(t)$ defined on $I$ such that $\varphi(0)=0$ and \begin{equation} \vec{z}=\begin{pmatrix} \cos\varphi \\ \sin\varphi \end{pmatrix}. \end{equation}

I suppose I'm free to assume that $\vec{z}(0)=( 1 ~~0)$, otherwise the problem doesn't make sense. I tried constructing the function $\varphi$ (adding or subtracting $2\pi$ whenever the trajectory crosses {$\{ (x,y)\in\mathbb{R}^2~:~x<0\}$} so that $\varphi$ is continuous and then proving it's twice differentiable) but the professor wasn't happy with that solution. Can someone point me in the right direction?

Answer 1

I assume that $z$ is twice differentiable else $\varphi$ can not exist. The quicker answer uses covering spaces theory. Indeed, $z$ is a $\mathcal{C}^2$ function from $\mathbb{R}$ to the circle $\mathbb{S}^1$ which admits for universal covering space, $$ p : \left\{\begin{array}{rcl} \mathbb{R} & \rightarrow & \mathbb{S}^1 \\ t & \mapsto & e^{it} \end{array}\right. $$ Covering spaces satisfy the lifting property that says that onc you fix $\varphi(0) = 0$, $z$ lifts into a unique continuous $\varphi : \mathbb{R} \rightarrow \mathbb{R}$ such that $z = p \circ \varphi : t \mapsto e^{i\varphi(t)}$, which is what you want.

To get the twice differentiability of $\varphi$, use the fact that $p$ is locally a smooth diffeomorphism. Indeed, for all $t \in \mathbb{R}$, there is a small neighborhood $U$ of $\varphi(t)$ such that $p_{|U} : U \rightarrow p(U)$ is a smooth diffeomorphism. Therefore, in $U$, $\varphi = p_{|U}^{-1} \circ z$ is twice differentiable and it is true around any point so $\varphi$ is twice differentiable on $\mathbb{R}$.

However, if you didn't see covering spaces, your handmade proof seems to be a good idea.