Let $\alpha(t)$ be a regular curve. Prove that if $\beta(s)$ and $\gamma(\bar{s})$ are two reparameterizations by arclength, then $s = \pm\bar{s} +a$, where $a \in \mathbb{R}$ is a constant.
I know that the following must be true:
$||(\alpha(\beta(s)))'||=||\alpha'(\beta(s))\cdot\beta'(s)|| = 1$
$||(\alpha(\gamma(\bar{s})))'||=||\alpha'(\gamma(\bar{s}))\cdot\gamma'(\bar{s})|| = 1$
But I don't know where to go from there. Where do I start? I'd appreciate any help.
EDIT: Well, from what I wrote above, it's now obvious to me that:
$|\beta'(s)| = |\gamma'(\bar{s})| \Rightarrow \beta'(s) = \pm \gamma'(\bar{s}) \Rightarrow \beta(s) = \pm\gamma(\bar{s}) + a$
Is this correct? It doesn't appear to be in the exact way the exercise wanted me to get to, but they are equivalent, right?
EDIT 2: I now see that I've done exactly what the exercise asked. It's just a matter of notation.
Implicit in the statement $s=\bar s + a$ is that they lead to the same point in $t$ parameter space, $t=\beta(s) = \gamma(\bar s)$. At such points we indeed have $|\beta'(s)| = |\gamma'(\bar s)| = 1$ for all such $s， \bar s $, by your computation.
As regular parameterisations cannot change sign, the sign of one $s,\bar s$ pair determines the sign of the derivative, and $\{\beta'(s),\gamma'(s')\} = \{ 1\},\{-1\},$ or $\{1,-1\}$. This accounts for the $±$ but they are all proven the same way, so say $\beta'(s)=\gamma'(\bar s) = 1$. Then for any $s$, integration reveals $$ \beta(s) = s + c_1$$ and for any $\bar s$, $$ \gamma(\bar s) = \bar s + c_2$$ $\beta(s) = \gamma(\bar s)$ then rewrites to $s = \bar s + (c_2 - c_1)$ as needed. The proof clearly extends to the other cases.