# Clairaut's relation and the equation of great circle in spherical coordinates

Clairaut's relation for a great circle parametrized by $t$ is $r(t)\cos\gamma(t)=\text{Const}$ where $r$ is the distance to the $z$-axis and $\gamma$ is the angle with the latitude. The implicit equation of great circle in spherical coordinates $(\theta,\phi)$ is $\cot \phi= a\cos(\theta-\theta_0)$ where $\phi$ is the angle with the positive $z$-axis and $\theta$ is the usual angle of polar coordinates $(r,\theta)$ of the projection of the point to the $(x,y)$ plane.

What would be a reasonably short derivation of the latter from the former? The proof I am familiar with is somewhat roundabout and involves first developing a differential equation equivalent to Clairaut's relation, and then performing a clever trigonometric substitution to get the answer. Given the simplicity of the two relations there should be a more direct route to get from one to the other.

• It's not clear to me what you mean by «shortest». – Loïc Teyssier Aug 11 '16 at 14:39
• @LoïcTeyssier, I tried to clarify. – Mikhail Katz Aug 11 '16 at 14:43

A great circle is the intersection of the sphere with a plane through the origin. Let a unit normal to that plane be ${\bf u} = [-\sin(\gamma), 0,\cos(\gamma)]$, where for convenience we choose our $x$ and $y$ axes so that $u_2 = 0$. Then in spherical coordinates, the equation ${\bf u} \cdot [\sin(\phi) \cos\theta), \sin(\phi)\sin(\theta), \cos(\phi)] = 0$ becomes $\cot(\phi) = \tan(\gamma) \cos(\theta)$. Rotating around the $z$ axis, this becomes $\cot(\phi) = \tan(\gamma) \cos(\theta - \theta_0)$.