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A great circle is geodesic

  1. Dec 12, 2008 #1
    I've been reading a few proofs showing that a great circle is geodesic. Most of these proofs start with a parametrization and then show that it satisfies the differential equations of geodesics. The book that I have doesn't even give a proof. It just tells me that the great circles on the sphere are geodesic. Is there a way to find a particular curve on the sphere and then show that curve is geodesic?
  2. jcsd
  3. Dec 12, 2008 #2


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    There's a much cooler way for sphere's in particular, and other symmetrical objects in general. Intersect the sphere with a plane through the center (so the intersection is a great circle). You should know geodesics are mapped to geodesics under an isometry. Pick a point on the intersection. There's a unique geodesic going through the point that is tangent to the plane at this point. Under reflection across the sphere, the point and tangent vector are unchanged, and hence the geodesic is too. The only curve that is invariant under this reflection is the great circle. Therefore all great circles are geodesics
  4. Dec 12, 2008 #3
    The problem with this argument is that geodesics are not the only curves preserved by isometries. E.g. lines of latitude are also preserved (but not fixed) by certain rotations.

    A correct proof would need to show directly or indirectly that the great circles are the only curves on the sphere that satisfy the definition of geodesic. There is a way to show this using the isometry group and without necessarily getting into differential equations, but I can't recall it right now.
    Last edited: Dec 12, 2008
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