---
tag: 1399T2Q2ZS
kind: definition
name: Geodesic curvature
msc: [53A05]
defines: 550P8PFMVC
uses: [6080X3FAD4, 7JQS6VKFEZ, 0BJDQ2HBH1, 0SEDGXG9S3, 00R1JDDV5X]
sources:
  - url: https://en.wikipedia.org/wiki/Geodesic_curvature
---

## Definition

Let $S \subseteq \R^3$ be a [[regular surface|6080X3FAD4]] with a smooth unit [[normal vector|0BJDQ2HBH1]] field $N$.
Let $\alpha \colon I \to S$ be a curve of class $C^2$ parametrised by [[arc length|00R1JDDV5X]] on an interval $I$.
The *geodesic curvature* of $\alpha$ at $s \in I$ is
$$\kappa_g(s) = \ip{\alpha''(s)}{N(\alpha(s)) \times \alpha'(s)},$$
where $\times$ is the [[cross product|0SEDGXG9S3]].
The *normal curvature* of $\alpha$ at $s$ is $\kappa_n(s) = \ip{\alpha''(s)}{N(\alpha(s))}$.

## Remarks

Since $\alpha'$, $N \times \alpha'$ and $N$ form an orthonormal basis and $\ip{\alpha''}{\alpha'} = 0$, one has $\alpha'' = \kappa_g \, N \times \alpha' + \kappa_n N$.
So $\kappa^2 = \kappa_g^2 + \kappa_n^2$, where $\kappa$ is the [[curvature|7JQS6VKFEZ]] of $\alpha$ as a space curve.
Example: on the unit sphere with the outer normal $N(p) = p$, a great circle $\alpha$ satisfies $\alpha'' = -\alpha$, which is normal to the sphere, so $\kappa_g = 0$.
A circle of latitude at polar angle $\varphi_0 \in (0, \pi)$ has $\kappa = 1 / \sin \varphi_0$ and $\abs{\kappa_g} = \abs{\cot \varphi_0}$.
Non-example: for a curve that is not of unit speed, the formula above is not the geodesic curvature; one must first reparametrise by arc length.
For a curve in the $xy$-plane with $N = (0, 0, 1)$, the geodesic curvature is the signed curvature of the plane curve.