3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ It's not clear to me what you mean by «shortest». $\endgroup$ – Loïc Teyssier Aug 11 '16 at 14:39
  • $\begingroup$ @LoïcTeyssier, I tried to clarify. $\endgroup$ – Mikhail Katz Aug 11 '16 at 14:43
7
$\begingroup$

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 $\cot(\phi) = \tan(\gamma) \cos(\theta - \theta_0)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.