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Geodesics on z = F(x,y)

Here is a question about geodesics on the surface of the graph of

z = x^2 + y^2:

Excluding the case of geodesics that pass through x=y=0, do geodesics spiral towards or away from the point x=y=0 ? Or do they have a zero winding number about the z axis - remaining to one side?

Answer: Any geodesic on this surface spirals away from zero but not towards it.There is a lemma: For any unit tangent and normal at a point on a surface, the plane through the point containing the tangent and normal cuts the surface in an arc through the point with curvature that matches the curvature of any geodesic with that same tangent vector through the point. In other words the arc cut by this plane is representative of the geodesic.

On the surface z = x^2 + y^2 , as we get near 0, the radius of curvature of all such plane-cut-arcs approaches a constant. But a spiral getting closer to x=y=0 has curvature approaching infinity, so it could never be a geodesic without violating the lemma.

I still have no answer for the winding number question. Do all geodesics not through x=y=0 have a self crossing?

The latter question has me stumped but is lots of fun to think about. It is a growth rate question.

Update: Good news! The geodesics spiral outward on the parabaloid because of another simple lemma: On a geodesic, the tangent vector always turns more in the direction of greatest curvature. So for a horizontal tangent vector on z = x^2 + y^2, it will be turned in the horizontal more than in the vertical so the moving point moves horizontally faster than vertically. So you run out of angular dimension (<=2pi) long before you run out of vertical dimension (infinite).

Google tells me a fellow Ling writes about multiple intersections of geodesics on parabaloids.

ReplyDeleteMore fun aspects: the geodesic cannot remain trapped in any finite part of the parabolic surface because of the Archmedean principle - once the tangent gets some component positive in the z direction, it must only get more positive. Already with a small positive amount, it must continue to accumulate to larger and larger z values of the moving point.

ReplyDeleteAlso a geodesic cannot become tangent to another, preventing the geodesic from turning past the vertical or past the horizontal [except at its minimum].

You would make more progress, sir, if you stated things clearly. It is a RELATIVE growth rate problem. Maybe try showing growth in theta is always a "substantial" fraction of growth in z. (z=r^2)

ReplyDeleteActually it is simpler. The growth rate in theta is greater than in z - which I had misunderstood. So the point moves faster in theta than in z and 2PI is exceeded over and over again as z becomes infinite.

ReplyDelete