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I was exploring the concept of curvature and circles of curvature in particular, and it occurred to me that I could design an integral that appears to describe the total fraction of the circles of curvatures' circumferences along a given curve that would be required to construct that curve (I apologize for the poorly worded description). The integral is as follows:

$$\int_0^L\frac{\kappa(s)}{2\pi}\,ds$$

where L stands for the total length of the curve, s is a variable that denotes the arc length of the curve that has been traversed since the starting point of the curve, and $\kappa(s)$ is the instantaneous curvature at the point where an arc length of s has been traversed.

To better explain my understanding of the meaning of this integral, I will start by saying that for any circle, the value of this integral is 1 if one integrates along the entire circumference of the circle (L = 2$\pi$r):

$$\int_0^L\frac{\kappa(s)}{2\pi}\,ds = \int_0^{2{\pi}r}\frac{1}{2{\pi}r}\,ds = \frac{2{\pi}r}{2{\pi}r} = 1$$

For any given circle, only one circle of curvature can be used to describe it, and that is the circle itself. To describe the full circle, the entire circumference of that circle of curvature has to be utilized, so the fraction of the circle of curvature's circumference that would be required to construct the circle is 1.

For other curves that aren't circular, there are instantaneous circles of curvature that can describe the curve at all points along it. The idea is to take the infinitesimally small arc lengths (ds) that correspond with each instantaneous circle of curvature, divide those infinitesimally small arc lengths by the circumferences of their corresponding circles of curvature, and add up the values, hence describing the total fraction of the instantaneous circles of curvatures' circumferences that would be required to construct the curve. That is what my integral is accomplishing. (Incidentally, I also calculated the value of this integral for an ellipse, and it also turns out to be 1 regardless of the eccentricity or size of the ellipse).

My question is, what practical uses are there for this integral?

One possible use that came to my mind was using this integral to characterize curves, especially smooth, closed curves. My current hypothesis is that the value of this integral is always a whole number value greater than or equal to one (i.e. 1,2,3...) if you have a simple smooth closed curve, but the value of this integral can be a decimal value greater than one if you have a non-simple smooth closed curve. It would be awesome if someone can verify/refute this for me.

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  • $\begingroup$ The curvature of a circle at any point along the circle is 1/r, so I substituted $\kappa(s)$ with 1/r. $\endgroup$
    – mathTrials
    Commented Jul 19, 2019 at 5:28
  • $\begingroup$ The $ds$ is there because I am integrating over the arc length. $\endgroup$
    – mathTrials
    Commented Jul 19, 2019 at 5:30
  • $\begingroup$ Oh yeah, i see. $\endgroup$
    – Kinheadpump
    Commented Jul 19, 2019 at 6:16
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    $\begingroup$ For plane curves something like this indeed holds, see e.g. en.wikipedia.org/wiki/Total_curvature $\endgroup$
    – humanStampedist
    Commented Jul 19, 2019 at 6:54

1 Answer 1

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Unfortunately the basic premise of your integral, namely that the integral of a whole circle is equal to $1$ does not work in general. One simple example would be the flat torus. Its scalar curvature is zero at all points, so no matter what curve you pick, the integral will be zero. Another example would be integrating a curve around a saddle point (https://en.wikipedia.org/wiki/Saddle_point) . The scalar curvature is negative at all points, so you would never get one in the integral either.

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