Clairaut's relation (differential geometry)

In classical differential geometry, Clairaut's relation, named after Alexis Claude de Clairaut, is a formula that characterizes the great circle paths on the unit sphere. The formula states that if ${\displaystyle \gamma }$ is a parametrization of a great circle then

${\displaystyle \rho (\gamma (t))\sin \psi (\gamma (t))={\text{constant}},\,}$

where ${\displaystyle \rho (P)}$ is the distance from a point ${\displaystyle P}$ on the great circle to the ${\displaystyle z}$-axis, and ${\displaystyle \psi (P)}$ is the angle between the great circle and the meridian through the point ${\displaystyle P}$.

The relation remains valid for a geodesic on an arbitrary surface of revolution.

A statement of the general version of Clairaut's relation is:¹

Let ${\displaystyle \gamma }$ be a geodesic on a surface of revolution ${\displaystyle S}$, let ${\displaystyle \rho }$ be the distance of a point of ${\displaystyle S}$ from the axis of rotation, and let ${\displaystyle \psi }$ be the angle between ${\displaystyle \gamma }$ and the meridian of ${\displaystyle S}$. Then ${\displaystyle \rho \sin \psi }$ is constant along ${\displaystyle \gamma }$. Conversely, if ${\displaystyle \rho \sin \psi }$ is constant along some curve ${\displaystyle \gamma }$ in the surface, and if no part of ${\displaystyle \gamma }$ is part of some parallel of ${\displaystyle S}$, then ${\displaystyle \gamma }$ is a geodesic.

Pressley (p. 185) explains this theorem as an expression of conservation of angular momentum about the axis of revolution when a particle moves along a geodesic under no forces other than those that keep it on the surface.

Now imagine a particle constrained to move on a surface of revolution, without external torque around the axis. By conservation of angular momentum:

${\displaystyle r\,v_{\theta }=L,}$

where

• ${\displaystyle r}$ = distance to the axis,
• ${\displaystyle v_{\theta }}$ = component of velocity orthogonal to the meridian,
• ${\displaystyle L}$ = conserved angular momentum around the axis.

But geometrically,

${\displaystyle v_{\theta }=|v|\sin \psi ,}$

If we normalize so the speed ${\displaystyle |v|=1}$ (unit speed geodesics), we get:

${\displaystyle r\sin \psi ={\frac {L}{|v|}}={\text{constant}}.}$