1
$\begingroup$

Please pardon, I wasn't sure how to phrase this to get an answer on Google. I'm familiar with the geodesic equation for a Riemannian Manifold which generalizes the concept of a straight line to this space. I was wondering if such an equation exists for circles?

I had some ideas.

The length between two points is $$\delta L=\int_a^b\sqrt{g_{\alpha\beta}\,dx^\alpha\,dx^\beta}$$ given the constraint, $$\ddot{x}^k=-\Gamma_{ij}^k\dot{x}^i\dot{x}^j.$$ .

Leave a constant and adjust b, preserving the length, we should draw out a circle. The Fundamental Theorem of Calculus might come into play.

Alternatively, the gradient of $$\nabla\delta L\cdot{d\vec{p}}=0$$ might yield a useful path, but I'm not quite sure how to properly differentiate this under the integral.

My intuition suggests I can assume any direction perpendicular to the tangent of the curve at the point on the circle will suffice, but I have some gaps in a vague proof in my head.

Given an equation for a a surface, the cross product of the normal with the tangent of the least distance curve might give a valid direction to goo in.

$$\frac{d\vec{p}}{d\lambda}=\frac{d\vec{s}}{d\lambda}\times\nabla{f}$$ where f is a constraint defining the surface, and $$\frac{d\vec{s}}{d\lambda}$$ is the tangent of the geodesic at b?

$\endgroup$
  • $\begingroup$ What is your definition of circle for a general Riemannian manifold? $\endgroup$ – Travis Willse Sep 26 '18 at 19:37
  • $\begingroup$ @Travis, the set of all points having the property that the length of a geodesic from a point designated as "the center" to the given point is a constant. I believe this will produce a Euclidean circle on the surface of a sphere, but break down on the surface of an ellipsoid. $\endgroup$ – TurlocTheRed Sep 26 '18 at 19:40
  • 1
    $\begingroup$ There cannot be such an equation for general $M$, as the differential equation on the curve is local but the definition is nonlocal: Given such a circle, one could deform the metric in a suitable open set in the inside of the curve. In general, the curve with not be a circle for the new metric. One /could/ expect an equation that captures whether a curve is locally like a circle in Euclidean space. Circular arcs are characterized by having constant curvature (and torsion and its analogues in dimension $\geq 3$), leading to the suggestion in Seub's answer. $\endgroup$ – Travis Willse Sep 27 '18 at 3:45
  • $\begingroup$ @Travis. Thanks. Has me curious what manifolds do admit of such solutions and what criteria metrics must satisfy. Had an old prof insist there exists a map projection that can at least preserve circles as circles when mapped from the globe to the plane, if distorting their radius or position. $\endgroup$ – TurlocTheRed Sep 27 '18 at 13:39
  • $\begingroup$ Stereographic projection has that property (if one allows "circles" to include lines). Both the sphere and plane are homogeneous Riemannian manifolds, so informally they look at same from every point, and this in a way fills in the gap in information I mentioned above. In particular, on the sphere and the plane the constant-curvature curves are the circles (and lines). $\endgroup$ – Travis Willse Sep 27 '18 at 15:04
2
$\begingroup$

That's an interesting question. I don't think you could properly talk about a "circle" in a Riemannian manifold, but you could certainly consider curves with constant curvature as a natural generalization. I'll let you figure what the equation for that is in coordinates, if that's what you want. Intrinsically, you can define the curvature of a curve $\gamma(t)$ as:

$$\kappa(t) = \left\Vert \frac{D}{dt} T(t)\right\Vert$$ where $T(t) = \frac{\gamma'(t)}{\Vert \gamma'(t)\Vert}$ is the unit tangent vector to the curve and $\frac{D}{dt}$ is the covariant derivative along $\gamma$.

On the other hand, another generalization of "circle" would be a curve all of whose points are at a constant geodesic distance from some point (the "center").

In general, these two notions of "generalized circle" do not agree.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think I've sorted out constant curvature. I believe it would be just like the geodesic equation, only with a positive number replacing zero. Going to write that out in detail when I have more time. I've also taken a crack at the constant geodesic length approach above. $\endgroup$ – TurlocTheRed Sep 26 '18 at 21:33

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.