If a curve $\alpha(s)$ on a surface $S$ is parametrized by arc length the geodesic curvature can easily be found

$$k_g(s)=\alpha''(s)\cdot (N(\alpha(s))\times \alpha'(s))$$

where $N$ is the unit normal.

The problem is that there are sometimes when it is extremely hard to reparametrize a curve by arclenght. Sometimes it involves eliptic integrals and so on.

In that case, how could one find the geodesic curvature? How can I find one expression for the geodesic curvature without needing to reparametrize by arc length?


1 Answer 1


If you have a general parametrization $\alpha(t)$, then, letting $\upsilon(t) = \|\alpha'(t)\| = ds/dt$, we have \begin{align*} \alpha'(t) &= \upsilon(t)T(t) \quad\text{and} \\ \alpha''(t) &= \upsilon'(t)T(t) + \kappa(t)\upsilon(t)^2 N(t) \end{align*} (check that the extra $\upsilon$ in the last term comes from the chain rule). Here $T,N$ are the unit tangent and principal normal of the curve. I will write the surface normal as $n$. Remember that $\kappa N = k_n n + k_g (n\times T)$. From this you easily get that $k_g = \kappa N\cdot (n\times T)$ and so $$k_g(t) =\frac 1{\upsilon(t)^2}\alpha''(t)\cdot \big(n(\alpha(t))\times T(t)\big).$$ (The tangential term in the acceleration disappears in the dot product.)

  • $\begingroup$ What does $n(\alpha(t))$ mean? $\endgroup$
    – Divide1918
    Oct 9, 2021 at 10:25
  • $\begingroup$ @Divide1918 $n$ is the surface normal (as I said). $n(P)$ is the surface normal at $P$, so $n(\alpha(t))$ is the surface normal at $\alpha(t)$. $\endgroup$ Oct 9, 2021 at 14:33

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