How to verify that $\beta$ can't be a reparametrization of $\alpha$?

Question

I am reading some notes from a differential geometry course. It says the following:

If $\beta$ is a reparametrization of $\alpha$, then $\alpha,\beta$ have the same trace. The converse is not true. Example: Consider the following parametrizations

$\alpha:[0,2\pi]\to \Bbb{R}^2$ with $\alpha(t)=(R\cos t , R \sin t)$

$\beta:[0,4\pi]\to \Bbb{R}^2$ with $\beta(t)=(R\cos t , R \sin t)$

These parametrizations have the same trace but $\beta$ is not a reparametrization of $\alpha$.

I am a bit confused as to how we can verify this. I know that if $\beta$ is a reparametrization of $\alpha$, then there exists $h$ such that $\beta=\alpha \circ h$. How can we check that $h$ can't exist?

Answer 1

A reparametrisation must be bijective. But there is one parameter that gives $(0,R)$ on $\alpha$ and two parameters that give the same point on $\beta$. Hence the connecting function $h$ cannot be bijective and there is no reparametrisation.

Answer 2

Here is an alternative answer to the other one. The arclength of a curve is independant of any paramterisation, but it is a direct computation to show that $\mathrm{length}(\alpha) = 2\pi$ while $\mathrm{length}(\beta) = 4\pi$.