How to show $\displaystyle \kappa=\frac{\langle \alpha^{''}, J(\alpha^{'})\rangle}{\langle \alpha^{'}, \alpha^{'}\rangle^{3/2}}$?

Question

Let $\alpha(t)=(x(t), y(t))$ be a regular curve (not necessarily an unit speed curve) in $\mathbb R^2$. How to show the curvature of $\alpha$ is given by $$\displaystyle \kappa=\frac{\langle \alpha^{''}, J(\alpha^{'})\rangle}{\langle \alpha^{'}, \alpha^{'}\rangle^{3/2}},$$ where $J$ is the operator given by $J(a, b)=(-b, a)$? Any help will be valuable.

Answer 1

By definition, curvature $\kappa$ is given by the equation $T' = \kappa N$ for an arclength parametrization (where $T$, $N$ are, respectively, the unit tangent and unit normal). Note that $N=J(T)$.

In general, $\alpha' = \upsilon T$, where $\upsilon = \|\alpha'\|$ is the speed of the curve. Differentiating, $\alpha'' = \upsilon' T + \kappa\upsilon^2 N$, and so $$\kappa = \frac{\alpha''\cdot N}{\upsilon^2} = \frac{\alpha''\cdot J(T)}{\|\alpha'\|^2}.$$ Your formula is actually wrong. Since $\alpha'$ has length $\upsilon$, the correct formula should be $$\kappa = \frac{\alpha''\cdot J(\alpha')}{\|\alpha'\|^3}.$$

EDIT: OK, the formula is no longer wrong ... lest we confuse readers in the future.