1

On Berger's 《A Panoramic View of Riemannian Geometry》 I read an explanation of curvature which is new to me.

He writes

'For a geometric curve, we can introduce it(curvature) by looking at the variation of the length of the curves drawn at a constant distance, called the parallel curves.If the absolute value of the infinitesimal change of length of these equidistant curves, close to a point $m = c(t)$, is the same as for a circle of radius r we say that the curve c has radius of curvature r at $m = c(t)$, and that its curvature is $K = 1/r$.'

I'm not familiar with the 'variation' he said.In fact the 'variation' he mentioned in the previous section means two points $p(t)$ and $q(t)$ running along two curves in the Euclidean plane, and we watch how the distance $d (p (t), q (t))$ varies with time t.But I can not understand in the context of curvature.

Any help will be thanked.

Tree23
  • 1,198
  • A very strange introduction to curvature, indeed. – Michael Hoppe Nov 20 '19 at 12:26
  • See https://math.stackexchange.com/questions/2202579/intuitive-definition-for-curvature/2202754#2202754 for an intuitive explanation. – Michael Hoppe Nov 20 '19 at 12:31
  • @MichaelHoppe I can understand the usual explanation of curvature.But what's 'the variation of the length of the parallel curves' here means?Specifically, what's the variation of the length of a curve? – Tree23 Nov 20 '19 at 13:19

2 Answers2

2

This is not so complicated. If we take an arclength-parametrized plane curve $\alpha(s)$, then the family of parallel curves Berger is talking about is given by $\beta_\epsilon(s) = \alpha(s) - \epsilon N(s)$, where $N$ is the principal normal. (I chose the negative sign so that the parallel curves move outward as $\epsilon$ increases.) ($s$ is arclength for $\alpha$, but not for $\beta$.) Then we have $\beta'_\epsilon(s) = (1+\epsilon\kappa(s))T(s)$, and so the arclength of $\beta$ comes from integrating $|1+\epsilon\kappa(s)| = 1+\epsilon\kappa(s)$. Then the variation — or rate of change — of an infinitesimal amount of arclength of $\beta_\epsilon$ is $$\frac d{d\epsilon}\Big|_{\epsilon = 0} (1+\epsilon\kappa(s)) = \kappa(s).$$

Ted Shifrin
  • 125,228
  • Yes I understand.But why he introduces curvature in this way? – Tree23 Nov 21 '19 at 04:54
  • I know Berger, but I don't know this book. He probably was trying to lay groundwork for later work. On the other hand, it's pretty intuitive: Think about thickening up an ellipse and notice that you get more arclength (instantaneously) at points of lower curvature. – Ted Shifrin Nov 21 '19 at 05:33
0

I think it's an interpretation related to Jacobi fields. Taking a variation of geodesics $\alpha (s,t)\colon (-\varepsilon,\varepsilon )\times [0,l]$ of a geodesic $\gamma $ and look at the second variation, then we will find the parallel vector field $Y(t)=\frac{\partial\alpha }{\partial s} $ satisfying $\nabla Y(t)+R_{\dot{\gamma } Y }\dot{\gamma } =0$, where $R$ is the Riemannian curvature tensor. From the Rauch comparison theorem, we could compare $\Vert Y\Vert $ with the corresponding Jacobi field on spheres.

For reference, I recommend reading books on Riemannian geometry, such as do Carmo's book or Peterson's book. This is a long story for to write down all the details and definitions.