I'm struggling with some differential geometry's homework, I've read the Shifrin's lecture notes but I can't attack this:
assume that the curves are regular and their curvature isn't zero.
Two curves $\alpha(s)$ and $\beta(s)$ ($s$ might not be the arclength parameter for both) are called Bertrand
mates if for each $s_0$, the normal lines of $\alpha$ and $\beta$ at $s_0$ are the same lines.
Show that $s$ is the arclength parameter for both $α$ and its Bertrand mate $\beta$, if and only if
$\alpha(s) = \beta(s)$ for all $s$ (i.e. they are the same curves).
new edit
I think it goes like this:
right to left side is obvious, for left to right side:
we know that $\beta(s)-\alpha(s)=rN(s)$ where $r = constant$ so we have
$(\beta(s)-\alpha(s))'=rN'(s)$ and as we are disgussing the left to right side we assume that both $\alpha$ and $\beta$ are arc-length parametrized so we conclude that $T_\beta – T_\alpha = r (-\kappa T_\alpha+\tau B_\alpha)$ now we multiply both sides with $T_\alpha$ then we will have
$T_\beta T_\alpha = 1 – r\kappa_\alpha$.
$T'_\beta T_\alpha + T_\beta T'_\alpha = -r\kappa'_\alpha \Rightarrow -r\kappa'_\alpha = 0 \Rightarrow r=0 \vee \kappa_\alpha=\kappa$
in the case that $r = 0$ we conclude that $\alpha = \beta$ but in the case that $\kappa_\alpha = constant$ our curves are lines or circles or $\alpha(t)= (a cos t, a sin t, bt)$ in cases of circles and lines as they're arc-length parametrized they're equal but we have:
$\alpha(s) = (\frac{a}{c}cos \frac{s}{c},\frac{a}{c}sin \frac{s}{c},\frac{bs}{c} )$ and $\beta(s) = (-\frac{a}{c}cos \frac{s}{c},-\frac{a}{c}sin \frac{s}{c},\frac{bs}{c} )$ while $c = \sqrt {a^2+b^2}$ which are arc-length parametrized, with a same normal line but not equal, so the proposition is right for just plane curves.
Best Answer
Interesting that my name ended up here, even though I never even thought to ask that question :)
When you start with an arclength parametrization of $\alpha$ and parametrize $\beta(s) = \alpha(s) + r N(s)$, what do you get for $\|\beta'(s)\|$? How can this always be equal to $1$? You can check that the necessary and sufficient condition is $$(1-r\kappa)^2 + (r\tau)^2 = 1.$$
I actually do not believe the result is correct. I believe that you can start with any circular helix and find one (different) Bertrand mate of that circular helix that is also arclength-parametrized. You can use the result of part c. of Exercise 1.2.20 to check that this works out.