Editing Great circle (section)

==Derivation of shortest paths==
{{see also|Great-circle distance}}
To prove that the minor arc of a great circle is the shortest path connecting two points on the surface of a sphere, one can apply [[calculus of variations]] to it.

Consider the class of all regular paths from a point <math>p</math> to another point <math>q</math>. Introduce [[spherical coordinates]] so that <math>p</math> coincides with the north pole. Any curve on the sphere that does not intersect either pole, except possibly at the endpoints, can be parametrized by

:<math>\theta = \theta(t),\quad \phi = \phi(t),\quad a\le t\le b</math>

provided <math>\phi</math> is allowed to take on arbitrary real values. The infinitesimal arc length in these coordinates is

: <math>
ds=r\sqrt{\theta'^2+\phi'^{2}\sin^{2}\theta}\, dt.
</math>

So the length of a curve <math>\gamma</math> from <math>p</math> to <math>q</math> is a [[functional (mathematics)|functional]] of the curve given by

: <math>
S[\gamma]=r\int_a^b\sqrt{\theta'^2+\phi'^{2}\sin^{2}\theta}\, dt.
</math>

According to the [[Euler–Lagrange equation]], <math>S[\gamma]</math> is minimized if and only if
:<math> \frac{\sin^2\theta\phi'}{\sqrt{\theta'^2+\phi'^2\sin^2\theta}}=C</math>,
where <math>C</math> is a <math>t</math>-independent constant, and
:<math> \frac{\sin\theta\cos\theta\phi'^2}{\sqrt{\theta'^2+\phi'^2\sin^2\theta}}=\frac{d}{dt}\frac{\theta'}{\sqrt{\theta'^2+\phi'^2\sin^2\theta}}.</math>
From the first equation of these two, it can be obtained that
:<math> \phi'=\frac{C\theta'}{\sin\theta\sqrt{\sin^2\theta-C^2}}</math>.
Integrating both sides and considering the boundary condition, the real solution of <math>C</math> is zero. Thus, <math>\phi'=0</math> and <math>\theta</math> can be any value between 0 and <math>\theta_0</math>, indicating that the curve must lie on a meridian of the sphere. In a [[Cartesian coordinate system]], this is
:<math>x\sin\phi_0 - y\cos\phi_0 = 0</math>
which is a plane through the origin, i.e., the center of the sphere.