Short version:

Apparently it is possible for a $C^1$-continuous parametric surface to (locally) have infinite (Gaussian) curvature. I find this quite counter-intuitive, because I always thought that only a $C^0$-continuous surface (i.e. containing a kink) can have infinite curvature. What am I missing?

Long version:

Since the curvature of a curve is defined as $1/r$, where $r$ is the radius of the circle that best describes the curve at the point of interest, infinite curvature is only possible when the circle has zero radius (it reduces to a single point). In case of a $C^0$-connection between two parametric curves, it is obvious that the curvature goes to infinity. But apparently, this can also happen when two parametric curves meet with $C^1$-continuity. Any examples (e.g. using parametric polynomial curves)?

With regard to surfaces — the two principal curvatures $\kappa_1$ and $\kappa_2$ at the point $(u,v)$ of a parametric surface $X(u,v)$ are respectively the minimum and maximum normal curvature at this point $(u,v)$. The normal curvature in the direction $\phi$ is obtained when a plane containing the normal at $(u,v)$ is oriented in the direction $\phi$. The intersection of this plane and the surface results in a curve — the curvature of this curve in our point $(u,v)$ is called the normal curvature in the direction of $\phi$.

The Gaussian curvature $\kappa_G$ is simply the product of the two principal curvatures, the mean curvature $\kappa_M$ is the sum of the two principal curvatures.

If the Gaussian curvature is infinite, then at least one of the principal curvatures has to be infinite as well. I always thought that the curve (i.e. the intersection of the plane — oriented in the corresponding principal direction — and the surface) should therefore contain a kink, but apparently this does not have to be true. What am I missing?


Short answer: the curvature is a property of the second derviative, and it's perfectly possible for a $C^1$ function to have a badly behaved second derivative.

In one variable, think about the function $f(t) = |t|^{3/2}$; it is $C^1$ but $f''(0) = +\infty$.

To turn this into a surface with infinite curvature, try $$X(u,v) = (u, |u|^{3/2}, v)$$ i.e. just extending this curve in the $z$ direction, forming a sheet. You'll find the normal curvature at the origin in the direction normal to the $xy$-plane is infinite.

  • $\begingroup$ Thanks for the example! However, I'm still thinking about curvature being equal to $1/r$, where $r$ is the radius of the circle that best describes the curve at the point of interest. For infinite curvature, $r$ has to go to zero. How is that possible for anything else than a kink or cusp? $\endgroup$ – Ailurus Mar 21 '13 at 11:09
  • $\begingroup$ @Ailurus: I don't quite know what you're looking for. Presumably you know the proof that these two notions of curvature are the same. And you could check explicitly that for this curve, smaller and smaller circles give you better and better approximations, so there is no circle of positive radius that "best" describes it. You could also try zooming in on a graph of $f$ near $t=0$; this may give some insight. $\endgroup$ – Nate Eldredge Mar 21 '13 at 12:28

Consider the parametric curve given by $x(t) = t^3$, $y(t)=t^2$. Clearly this curve is $C_\infty$. However, at the point where $t=0$, there is a sharp corner in the curve, and its curvature is infinite (radius of curvature is zero).

The flaw in your thinking is related to parametrization. The derivatives of a curve depend on how it's parameterized, but radius of curvature is an intrinsic geometric property that is independent of parametrization. Therefore, you can't expect any hard-and-fast relationships between derivatives and radius of curvature. Everything breaks down at points where several derivatives are zero (which is what happens at $t=0$ on the example curve mentioned above).


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