We may define the Frenet Frame $(T, N, B)$ of a regular curve $\alpha$ as follows:
$$T:=\frac{\alpha'}{| \alpha'|}$$
$$N:=\frac{T'}{|T'|}$$
$$B:=T\times N$$
and it can be shown that if $\beta$ is an arc-length parametrization of $\alpha$ by $s$, then $T_{\alpha}(t)=T_{\beta}(s(t))$,
$N_{\alpha}(t)=N_{\beta}(s(t))$, and
$B_{\alpha}(t)=B_{\beta}(s(t))$.
In this post both answers point to another way of expressing the Frenet frame of the curve $\alpha$, namely by defining $T$ as above and letting
$$N=\frac{\alpha'\times(\alpha''\times \alpha')}{|\alpha'|\ |\alpha'' \times \alpha'|}$$
$$B=\frac{\alpha'\times \alpha''}{|\alpha' \times \alpha''|}.$$
How can these last two formulas be proven equivalent to their former definitions?
Best Answer
$\newcommand{\R}{\mathbb{R}}$ My preferred approach is to derive the Frenet-Serret equations from scratch without assuming arclength parameterization. The equations you want are then an easy consequence. Let $\alpha: I \rightarrow \R^3$ be a smooth curve such that $\alpha'(t), \alpha''(t)$ are linearly independent for all $t \in I$. For each $t$, $\alpha'(t)$ is tangent to $\alpha$ and therefore, $$ T = \frac{\alpha'}{|\alpha'|} $$ is a unit tangent vector that points in the direction of the curve. If we view $\alpha'$ as velocity, then $$ \sigma = |\alpha'| $$ is the speed. Therefore, the length of the curve is $$ \ell = \int_{t\in I} \sigma(t)\,dt. $$
Since $$ \alpha' = \sigma T, $$ the acceleration of $\alpha$ is given by \begin{align} \alpha'' &= \sigma' T + \sigma T'. \end{align} The assumption that $\alpha', \alpha''$ are linearly independent implies that $T'\ne 0$. Since $T\cdot T = 1$, $$ 0 = (T\cdot T)' = 2T\cdot T'. $$ It follows that the equation above is an orthogonal decomposition of $\alpha'$. In particular, it decomposes $\alpha''$ into a tangential term whose magnitude is how quickly the speed is changing and a normal vector that indicates how the directionof $\alpha$ is changing. In particular, if we let $$ N = \frac{T'}{|T'|}, $$ then $$ T' = \sigma\kappa N, $$ where $\sigma\kappa$ measures how quickly the direction of $\alpha$ is changing and $N$ indicates the direction of the change in the direction of $\alpha$. Finally, there is a unique unit vector $B$ such that $(T, N, B)$ is an oriented basis of $\R^3$. We can now derive the equations satisfied by these vectors: \begin{align*} 0 &= (T\cdot T)' = (N\cdot N)' = (B\cdot B)'\\ \implies\ T'\cdot T &= 0\\ 0 &= (T\cdot N)'\\ &= T'\cdot N + T\cdot N'\\ &= \sigma\kappa N\cdot N + T\cdot N'\\ \implies\ N'\cdot T &= -\sigma\kappa\\ 0 &= (N\cdot B)'\\ &= N'\cdot B + N\cdot B'\\ \implies\ B'\cdot N &= -N'\cdot B.\\ 0 &= (T\cdot B)'\\ &= T'\cdot B + T\cdot B'\\ &= T\cdot B'\\ \implies\ B'\cdot T &= 0. \end{align*} Therefore, if we let $$ N' = \sigma\tau B, $$ it follows that $$ \begin{bmatrix} T & N & B\end{bmatrix}' = \sigma \begin{bmatrix} T & N & B\end{bmatrix}' \begin{bmatrix} 0 & -\kappa & 0\\ \kappa & 0 & \tau\\ 0 & -\tau & 0 \end{bmatrix}. $$ These are the Frenet-Serret equations. Observe that if $c$ is parameterized by arclength, then they become the more familiar form of the equations.
Now observe that \begin{align*} \alpha'\times\alpha'' &= \sigma T\times (\sigma T + \kappa N)\\ &= \sigma \kappa T\times N\\ &= \sigma \kappa B\\ \implies\ B &= \frac{\alpha'\times \alpha''}{|\alpha'\times \alpha''|}\\ \text{and }|\alpha'\times \alpha''| &= \sigma\kappa\\ \alpha'\times (\alpha'\times \alpha'') &= \sigma T\times(\sigma\kappa B)\\ &= \sigma^2\kappa(T\times B)\\ &= \sigma^2\kappa N\\ &= |\alpha'||\alpha'\times\alpha''| N\\ \implies\ N &= \frac{\alpha'\times (\alpha'\times \alpha'')}{|\alpha'||\alpha'\times\alpha''|}. \end{align*}