Let $\alpha:I\subset\mathbb{R}^{3}$ be a regular curve (not nec. parametrized by arc length), $\beta: J\subset\mathbb{R}\longrightarrow\mathbb{R}^{3}$ a reparametrization of $\alpha(I)$ by arc length $s=s(t)$, from $t_{0}\in I$. Let $t=t(s)$ the inverse function of $s$ and $d\alpha/dt=\alpha^{\prime}$, $d^{2}\alpha/dt^{2}=\alpha^{\prime\prime}$, etc.
Knowing that $dt/ds=1/\Vert\alpha^{\prime}\Vert$, I want to prove that $d^{2}t/ds^{2}=-\frac{\langle\alpha^{\prime},\alpha^{\prime\prime}\rangle}{\Vert\alpha^{\prime}\Vert^{4}}$, but I obtain $d^{2}t/ds^{2}=-\frac{\langle\alpha^{\prime},\alpha^{\prime\prime}\rangle}{\Vert\alpha^{\prime}\Vert^{3}}$ thru the following method:
We have $\frac{d^2t}{ds^2} = \frac{d}{ds}\frac{1}{|\alpha'|} = \frac{d}{ds}\left<\alpha',\alpha'\right>^{-1/2} = -\frac{1}{2}\left<\alpha',\alpha'\right>^{-1/2-1}(\left<\alpha'',\alpha'\right> + \left<\alpha',\alpha''\right>) =-\frac{1}{2|\alpha|^3}{(2\left<\alpha',\alpha''\right>)}{} = -\frac{\langle\alpha^{\prime},\alpha^{\prime\prime}\rangle}{|\alpha'|^3}$.
Who is wrong here? Where can I read more?
Cheers!
It's a chain rule mistake:
$$\frac{d}{ds}\langle \alpha', \alpha' \rangle = \frac{dt}{ds}\frac{d}{dt}\langle \alpha', \alpha' \rangle$$ $$~~~~~~~~~~~~ = \frac{2\langle \alpha', \alpha''\rangle}{||\alpha'||} $$
The problem is you are taking the derivative with respect to $s$, but you want to write the result it in terms of things defined by $t$. So you need to think of $\alpha'$ as $\alpha'(t(s))$. This is where the missing $\frac{1}{||\alpha'||}$ shows up.