Let $\alpha:\, I \to \mathbb{R}^3$ be a curve parametrized by arc
length $s$, with curvature $k(s) \neq 0$, for all $s \in I$.Show that the torsion $\tau$ of $\alpha$ is given by:
$$ \tau(s) = -\frac{\alpha'(s) \times \alpha''(s) \cdot \alpha'''(s)}{|k(s)|^2} $$
The definition of $\tau$ is the unique scalar such that $b'(s) = \tau(s)n(s)$, where $b$ is the binormal vector and $n$ is the normal vector. So my plan is to show that this forumla for $\tau$ satisfies the definition. We can do this by making showing
$\tau(s)n(s) =b'(s)=t(s) \times n'(s)$.
\begin{align}
\tau(s)n(s) &= -\left(\frac{\alpha'(s) \times \alpha''(s) \cdot \alpha'''(s)}{|k(s)|^2}\right)n(s) \\
&= -\left(\frac{t(s) \times k(s)n(s) \cdot \alpha'''(s)}{|k(s)|^2}\right)n(s) \\
&= -\frac{k(s)}{|k(s)|^2}\left[t(s) \times n(s) \cdot \alpha'''(s)\right]\,n(s) \\
&= -\frac{k(s)}{|k(s)|^2}\left[b(s) \cdot \alpha'''(s)\right]\,n(s) \\
\end{align}
where do I go from here?
The book also mentions that $a''' = kn' + k'n = -k^2t + k'n -k \tau b$, but I don't know how that helps.
Best Answer
Note: I worked through this with a slightly different convention on torsion (which I believe is more standard) before I realized the definition given above. This should account for the minus sign.
The standard Frenet-Serret Formula for a curve parametrized by arc length are
\begin{array} \vec{T}^{\prime} =& \kappa(s) \vec{N}\\ \vec{N}^{\prime} =& -\kappa(s)\vec{T} + \tau(s) \vec{B} \\ \vec{B}^{\prime} =& -\tau(s)\vec{N},\\ \end{array}
where the primes indicate derivatives with respect to arc length and $\vec{T} = \alpha^{\prime}$, $\vec{N} = \frac{\vec{T}^\prime}{\lvert \lvert \vec{T}^{\prime}\rvert \rvert}$ and $\vec{B} = \vec{T} \times \vec{N}$.
Note that the ultimate goal is to express torsion in terms of the parametrization. With this in mind, we will start with an expression that expresses $\vec{B}$ in terms of the parametrization, differentiate and apply the Frenet-Serret equations.
Starting with the identity $\alpha^{\prime} \times \alpha^{\prime\prime} = \vec{T} \times \vec{T}^{\prime} = \kappa \vec{B}$ (this streamlines the differentiation slightly) and differentiating with respect to $s$ we obtain
$$ \alpha^{\prime} \times {\alpha}^{\prime\prime\prime} = \kappa^{\prime}\vec{B} + \kappa\vec{B}^{\prime}.$$
Using the Frenet-Serret formula, substitution yields the following equation
$$\alpha^{\prime} \times \alpha^{\prime\prime\prime} = \kappa^{\prime}\vec{B} - \kappa\tau \vec{N}.$$
"Dotting" both sides of the above equation with $\vec{N}$ and using the defining relations of $\vec{T}$, $\vec{N}$ and $\vec{B}$ and we obtain
$$ \left(\alpha^{\prime} \times \alpha^{\prime\prime\prime}\right)\cdot \vec{N} = -\kappa \tau,$$
which yields
$$ -\frac{1}{\kappa^2}\left(\alpha^{\prime} \times \alpha^{\prime\prime\prime}\right)\cdot \alpha^{\prime\prime} = \tau.$$
Applying the scalar triple product to the above gives
$$ \frac{1}{\kappa^2}\left(\alpha^{\prime} \times \alpha^{\prime\prime}\right)\cdot \alpha^{\prime\prime\prime} = \tau.$$