# Curvature and Circumference of Circle

Theorem Let $\gamma\colon [a,b]\rightarrow \mathbb{R}^2$ be a unit speed simple closed curve, with $\gamma'(a)=\gamma'(b)$ and $N$ is the inward-pointing normal. Then $$\int_{a}^b \kappa_N(s)ds=2\pi.$$

Question: If a circle has curvature $\kappa$ and circumference $C$ then from above theorem, how does it follow that $\kappa C=2\pi$?

Ref: Riemannian Manifolds - John M. Lee, Theorem 1.6 and comments, p.4.

• I edited your post to add the "differential geometry" and "plane curves" tags. Cheers! – Robert Lewis Jun 9 '15 at 18:09
• If $\gamma$ is unit speed, then $t$ is the arc-length parametrization and $\kappa$ is constant on circle! – achille hui Jun 9 '15 at 18:13
• Changed the variable to $s$ from $t$ as it is not "any" independent variable.. Also for the simplest case of the circle, $\kappa\ = 1/R$ – Narasimham Jun 9 '15 at 18:22

I started preparing this answer before Narasimham made his edits, wherein the original independent variable name $t$ was replaced by $s$. Since I was nearly done when his edits occurred, I chose to respect the original formulation in my answer; nevertheless, the relationship 'twixt $s$ and $t$ should be clear in what follows.

A unit-speed circle in $\Bbb R^2$, centered at $(x_0, y_0)$ and of radius $\rho > 0$, may be parametrically represented as

$\gamma(t) = (x_0 + \rho \cos (\dfrac{t}{\rho}), y_0 + \rho \sin (\dfrac{t}{\rho})); \tag{1}$

taking $a = 0$ and $b = 2\pi$, we see that

$\gamma(a) = \gamma(0) = \gamma(2\pi) = \gamma(b). \tag{2}$

Also,

$\gamma'(t) = (-\sin(\dfrac{t}{\rho}), \cos(\dfrac{t}{\rho})), \tag{3}$

whence

$\dfrac{ds}{dt} = \Vert \gamma'(t) \Vert = \sqrt{\sin^2(\dfrac{t}{\rho}) + \cos^2(\dfrac{t}{\rho})} = 1, \tag{4}$

where $s$ is the arc-length along $\gamma(t)$; from (4),

$s = t + s_0, \tag{5}$

which shows that both $s$ and $t$ effectively measure distance traversed along $\gamma(t)$, i.e. that $\gamma(t)$ is in fact a unit speed curve; the offset $s_0$ simply indicates a possible difference in parametrized points on $\gamma(t)$ from which the measurement of distance is taken. (By paremetrized point I mean a point $\gamma(t')$ on the curve together with a specific parameter value $t'$; this usage to disambiguate situations in which the same $\gamma(t')$ is in fact the image under $\gamma$ of more than one value $t'$, as in the present case with $\gamma(t' + 2\pi) = \gamma(t')$ for all $t'$.) If we agree, as we will, that $s$ is measured from $\gamma(0)$ at $t = 0$, then $s_0 = 0$ and $s$ and $t$ agree:

$s = t. \tag{6}$

We see from (4) that $\gamma'(t)$ is in fact the unit tangent vector field along $\gamma(t)$; in the notation of Frenet-Serret, we write

$T(t) = \gamma'(t); \tag{7}$

since $t = s$ we have the Frenet-Serret equation

$\dfrac{dT}{ds} = \dfrac{dT}{dt} = \kappa(t) N(t), \tag{8}$

where $\kappa(t)$ is the curvature, and $N(t)$ is the unit normal vector, to $\gamma(t)$ at the parameter value $t$; that is, at the point $\gamma(t)$ itself. From (3), (7) and (8) we have

$\dfrac{dT}{dt} = \gamma''(t) = (-\dfrac{1}{\rho} \cos (\dfrac{t}{\rho}), -\dfrac{1}{\rho} \sin (\dfrac{t}{\rho})) = \dfrac{1}{\rho}(-\cos(\dfrac{t}{\rho}), -\sin (\dfrac{t}{\rho}))$ $= \dfrac{1}{\rho}N(t) = \kappa(t) N(t). \tag{9}$

Inspecting (9), it is clear that (i.)

$\kappa(t) = \dfrac{1}{\rho} \tag{10}$\

is a constant; and (ii.)

$N(t) = (-\cos(\dfrac{t}{\rho}), -\sin (\dfrac{t}{\rho})) = -(\cos(\dfrac{t}{\rho}), \sin (\dfrac{t}{\rho})) = \dfrac{1}{\rho}((x_0, y_0) - \gamma(t)), \tag{11}$

which shows that $N(t)$ is directed from $\gamma(t)$ towards $(x_0, y_0)$, the center of the circle; thus it is an inward-pointing normal vector. Last but not least, we observe that

$\gamma'(a) = \gamma'(0) = \gamma'(2\pi) = \gamma'(b); \tag{12}$

we see that the criterion of the stated theorem are all met by the circle $\gamma(t)$; furthermore, since $\kappa(t) = \kappa(s) = 1/\rho$ is constant, we have

$\int_a^b \kappa(t) dt = \int_a^b \kappa(s) ds = \kappa \int_a^b ds = \kappa C; \tag{13}$

then by the theorem, the integral

$\int_a^b \kappa(t) dt = \int_a^b \kappa(s) ds = 2\pi; \tag{14}$

thus it follows that

$\kappa C = 2\pi, \tag{15}$

as was requested.

Nota Bene: Establishing the formula (15) by calling on the much more general theorem given in the text of the problem seems a little bit like going after flies with a shotgun, since once we see that the curvature of the circle is a constant $\kappa = 1/\rho$, (15) follows immediately from the formula

$C = 2\pi \rho, \tag{16}$

with which I presume most readers are familiar; nevertheless, the relation (15) does provide perhaps the simplest possible case/application of the stated theorem, and so it is worthwhile to work it out; a collection of concrete, simple examples being in many ways the true grist which is ground by the mathematical mill. End of Note.