I must admit that the differences between arc-length and a general parameterization of a curve are confusing.

Consider a curve $\gamma$ parameterized not by arc-length. Let $n$ be the unit normal to the curve. Is it true that the curvature is $\ddot \gamma \cdot n$? This is not homework, I'm just trying to understand this concept.

If we take $\gamma (t)=\gamma (s(t))$ then we get $\ddot \gamma \cdot n=\frac {\gamma''(s(t))\cdot n}{dt/ds}$. Is that the curvature?

I'm trying to get a definition of a curvature using the projection on the unit normal to the curve that will be true for all parameterizations. Is that even possible? I know that it is true for the normal curvature (which involves a surface). Is it also true in this case?


1 Answer 1


Curvature is defined by $\kappa=\langle\mathbf t',\mathbf n\rangle$, where $\bf t$ is the unit tangent vector. This definition is independent of the parametrization. Curvature is a geoemtric property.

For arc length parametrized curves holds $\bf t'=\gamma''$ , hence the curvature can be expressed as $\kappa=\langle \mathbf t',\mathbf n\rangle=\langle\gamma'',\mathbf n\rangle$.

But since $\gamma''$ is not a geometric property and might change with the used parametrization, $\langle\gamma'',\bf n\rangle$ can not give the curvature in general. The curvature must be independent of the parametrization.


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