What is the geodesic curvature of the sphere? The great circle on the sphere is curved. Why is the geodesic curvature of the great circle equal to zero? I don't understand.

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    $\begingroup$ If this is something you encountered during a course of study, the first thing to do when you have such a question is to look up the definition of the thing you are asking about. Looking in your textbook or course notes is a much better way to look that up than asking a lot of strangers on the internet. $\endgroup$ – Toby Mak Aug 4 at 2:38
  • $\begingroup$ Geodesic curvature is the curvature a small creature observes living entirely in the surface. Geodesics (being the analogues of straight lines) have geodesic curvature $0$. You might learn a great deal by reading my differential geometry text and working exercises. For geodesic curvature, see section 2.4, but you probably should start at the beginning. $\endgroup$ – Ted Shifrin Aug 4 at 16:57
  • $\begingroup$ @TedShifrin Thank you for the link. I will read it carefully. In addition, if this tiny biological measurement of the distance from the equator (great circle) to the latitude line (small circle), then it will find that the distance is unchanged, then it will conclude that the latitude line (small circle) is also a straight line, right? $\endgroup$ – z.qmpx Aug 6 at 2:27
  • $\begingroup$ What do you mean by “the distance is unchanged”? $\endgroup$ – Ted Shifrin Aug 6 at 4:13
  • $\begingroup$ The distance from any point on the equator to a definite latitude line is constant. $\endgroup$ – z.qmpx Aug 6 at 6:38

Part of the confusion is that "curvature" is a highly overloaded term. Some kinds of curvature (such as mean or Gaussian curvature) are properties of the surface and measure, in different ways, how curved a piece of the surface is "on average." Other kinds of curvature, like geodesic curvature, are properties of curves. So it doesn't make sense to ask what the "geodesic curvature of the sphere" is.

Suppose you have a curve on the sphere. It has to bend, as viewed by a 3D observer, because the underlying surface is bending. But you can ask whether the curve is "as straight as possible" given that it must bend to stay on the sphere; the failure to be as straight as possible is geodesic curvature. You can quantify the amount of in-plane bending by looking at the acceleration vector $\gamma''(s)$ of a curve $\gamma(s)$; this vector will have some component along the normal of the sphere and some component that is tangent. The magnitude of the part that's tangent (also often written as the covariant derivative of the curve in its own direction, $\nabla_{\gamma'} \gamma'$) is the geodesic curvature. Notice that for a great arc, the derivative of the tangent $\gamma'$ is purely in the direction normal to the sphere, and so the geodesic curvature is zero. A great arc is "as straight as possible" with no bending in the tangent plane of the sphere.

Note for nitpickers: I'm assuming here that the sphere is equipped with the metric pulled back from ambient space. You can of course equip the sphere with other metrics or connections so that the geodesics are no longer great arcs.

  • $\begingroup$ The great circle is curved. Why is its curvature zero? Geodetic curvature is meant to measure the straightness of a surface? $\endgroup$ – z.qmpx Aug 4 at 11:29
  • $\begingroup$ From a four-dimensional point of view, does the sphere also have geodesic curvature? $\endgroup$ – z.qmpx Aug 4 at 11:33
  • $\begingroup$ @z.qmpx No. Geodesic curvature measures curvature of curves on surfaces, even in higher dimensions. There are other types of curvature for manifolds (eg sectional curvature). $\endgroup$ – user7530 Aug 4 at 15:53
  • $\begingroup$ Looking at the sphere from the four-dimensional space, what is the sphere? $\endgroup$ – z.qmpx Aug 6 at 6:39
  • $\begingroup$ @z.qmpx I'm not sure what you're asking; the standard (round) sphere in 4D is the set of points $x^2+y^2+z^2+w^2=1.$ $\endgroup$ – user7530 Aug 6 at 6:40

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