1
$\begingroup$


Prove that any geodesic curve on $S^2$ is a great circle

I want to prove that a geodesic curve ($\gamma$ with $k_g=0$) is in fact a great circle. I know that any curve on a sphere has konstant normal curvature $k_n$. Futheremore we got the Definition of a great circle as the intersection of a plane that passes through the midpoint of the sphere

I already proved the $\tau=0$ and since $k^2=k_n^2+k_g^2 \;$ is constant $\gamma$ is a circle but I dont know how I show that $\gamma$ is a great circle.

Any help would be appreciated.

$\endgroup$
  • $\begingroup$ You can see it geometrically: If you take a circle that is not a great circle, then $k_g\ne 0$ (the principal normal vector of the curve is not normal to the sphere). $\endgroup$ – Ted Shifrin Nov 21 '16 at 22:30
  • $\begingroup$ Doesn't the fact that $k_g=0$ imply that the center of curvature of the geodesic lies on the normal to the surface? Then you just need to show that it must be at the center of the sphere. $\endgroup$ – David K Nov 21 '16 at 22:34

Your Answer

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

Browse other questions tagged or ask your own question.