I am getting into differential geometry and came up with the problem which is unexpectedly difficult to solve. Suppose we have a curve $\alpha(t):[0,L]\rightarrow \mathbb{R}^2 $ parametrized by arclength. I would like to find reparametrization such that new map $\beta(t)$ satisfies $||\beta''(t)|| = const$. Let us rewrite the condition as $$ 1=||\beta''(t)||=||\frac{d^2}{dt^2}\alpha(g^{-1}(t))|| $$ I used inverse function to get differential equation on $g(t)$. Expanding all derivatives and setting $g^{-1}(t) = s$ $$ 1= \left|\left|\frac{\alpha''(s)g'(s)-\alpha'(s)g''(s)}{g'(s)^3}\right|\right| = \frac{\sqrt{||\alpha''(s)||^2 g'(s)^2 + g''(s)^2}}{g'(s)^3} $$ Now we get final differential equation. Also setting $\gamma(s) = g'(s)$ $$ ||\alpha''(s)||^2 \gamma(s)^2 + \gamma'(s)^2 = \gamma(s)^6 $$ This equation seems to be unsolvable analytically for general $\alpha(s)$. Am I missing something? Is there a simpler way to find this reparametrization? Or is this just how it is and reparametrization is just described by this equation? Maybe the equation itself is solvable with some nice substitution but I was not able to find one.
Upd Since we are updating reparametrization $||\beta(t)||$ is not curvature. However it does not make a problem simpler. The original question still holds (now I have 2 questions -- reparametrization for mentioned condition and reparametrization for constant curvature).