I am seeking a motivation for the definition of the curvature of a plane curve. How did people come with the idea of the definition of the curvature? Below I am more specific.

The fundamental theorem for plane curves states the following. Giving two curves $\alpha$ and $\beta$ there exists a rigid motion $M:\mathbb R^2\to \mathbb R^2$ such that $\beta(t)=M(\alpha(t))$ if, and only if, the curves $\alpha$ and $\beta$ have the same curvature function.

Assume for a moment that you do not know about the curvature and that you would like to classify plane curves up to rigid motions. How could one come up with the idea of a geometric invariant that is enough to distinguish curves?


4 Answers 4


The derivative tells you the tangent line, which is the best linear approximation to a curve near a point.

With the second derivative you can find the osculating circle - the circle that best approximates the curve near a point. The curvature is the inverse of its radius. That's a pretty natural geometric invariant.

If these together did not suffice to determine the curve up to a rigid motion you'd look for more invariants.

  • $\begingroup$ Thanks for your answer. This answer showed me that the radius of a circle is a natural geometric invariant: two circles are equal up to rigid motions if and only if they have the same radius. It is of course true that every line is equal up to rigid motions. But the radius of a line is infinity. Therefore, the best geometric invariant would be one over the radius. $\endgroup$
    – ghreis
    Commented Feb 28 at 14:48

The key here is to consider an arclength parametrisation of the curve, i.e., a parameter $s$ such that $v(s)=\alpha'(s)$ is a unit vector for all $s$. Then it turns out that you can choose a single-branched angle $\theta(s)$ (roughly, angle with the positive direction of the $x$-axis) such that $v(s)=e^{i\theta}$. Then it turns out that the curvature of the curve is precisely $\theta'(s)$. If you think the derivative is natural, you will be naturally led to think that the curvature $\theta'(s)$ is also natural.


$\newcommand\R{\mathbb{R}}$ Suppose you have a curve in the plane (think of a road). Suppose you parameterize it: $$ c: I \rightarrow \R^2, $$ where $I \subset \R$ is a connected interval (think of a car driving along the road, where $t$ is time and $c(t)$ is where the car is at time $t$).

Then $v(t) = c'(t)$ is the velocity vector (which tells you the direction of travel and speed of the car at time $t$). The magnitude of the velocity vector is the speed of the parameterized curve (speed the car is traveling at). Let's call this $$ \sigma(t) = |v(t)|. $$ The unit vector $$ u(t) = \frac{v(t)}{|v(t)|} $$ tells you the direction of the velocity vector (direction in which the car is traveling). You can therefore write the velocity vector as $$ c'(t) = v(t) = \sigma(t)u(t). $$

The acceleration vector is the rate of change of the velocity vector, i.e, the second derivative of the position vector, $$ a(t) = v'(t) = c''(t). $$ There are two sources of acceleration: Change in speed and change in direction. This can be seen by differentiating the equation for $v(t)$ above: $$ a(t) = v'(t) = \sigma'(t)u(t) + \sigma(t)u'(t). $$ The first term is the acceleration due to change in speed and the second term is the acceleration due to change in direction.

Now observe that if the speed of the curve (car) is always $1$, the acceleration due to change in direction is due entirely to the shape of the curve (the car changes direction only if the road changes direction).

If the speed is not always $1$, the curvature is (up to sign) the magnitude of the rate of change of direction, normalized for the speed, $$ \kappa = \frac{|u'(t)|}{\sigma(t)}. $$


One way to reinvent curvature of a plane curve is as follows: Given a smooth, oriented curve $C \subset \Bbb E^2$---here $\Bbb E^2$ denotes $\Bbb R^2$ equipped with its usual Euclidean structure---pick an arc length parametrization $\gamma : I \to \Bbb R^2$ of $C$.

We can immediately define a preferred oriented, orthonormal frame $\mathcal E := ({\bf e}_1(s), {\bf e}_2(s))$ determined by the intrinsic geometry of the curve: Set ${\bf e}_1(s) := \gamma'(s)$ and take ${\bf e}_2(s)$ to be the vector given by rotating ${\bf e}_1(s)$ anticlockwise by $\frac\pi2$.

Critically $\mathcal E$ is equivariant under the group $\operatorname{SE}(2)$ of Euclidean motions of $\Bbb R^2$: If we pick $g \in \operatorname{SE}(2)$ and replace $\gamma$ with $g \cdot \gamma$, then ${\bf e}_i(s) \in T_{\gamma(t)} \Bbb R^2$ is replaced by $g \cdot {\bf e}_i(s) \in T_{g \gamma(s)} \Bbb R^2$, $i = 1, 2$. Since ${\bf e}_1(s)$ has unit length, ${\bf e}_1'(s) \perp {\bf e}_1(s)$, hence $${\bf e}_1'(s) = \kappa(s) {\bf e}_2(s)$$ for some function $\kappa$, the (signed) curvature of $C$, and since $\mathcal E$ is equivariant, so is ${\bf e_1}'(s)$ and hence so is $\kappa(s)$, which is thus an invariant under Euclidean motions.

To illustrate the general approach, it might be useful to see how you'd construct a curvature invariant for a curve $C$ in some other (homogeneous) geometry on the plane, e.g., special affine geometry. In this context we denoye the plane by $\Bbb A^2$. In this setting, the analog of an orthonormal frame is a unimodular frame, that is a frame $\mathcal A := ({\bf e}_1(t), {\bf e}_2(t))$ such that $$\det \pmatrix{{\bf e}_1(t) & {\bf e}_2(t)} = 1,$$ and the replacement for the group of Euclidean motions is the special affine group, $G := \Bbb R^2 \rtimes \operatorname{SL}_2(\Bbb R)$.

A natural first attempt at construction $\mathcal A$ would be to choose a parametrization $\gamma$ of $C$ and choose the frame $(\gamma'(t), \gamma''(t))$. (Notice that unlike in the Euclidean case, the first element of the frame does not determine the second, essentially because the group of motions is larger in this setting.) In order for this pair to comprise a frame along $C$, $\gamma'(t), \gamma''(t)$ must be linearly independent---this nondegeneracy condition is the affine analogue of the regularity condition for curves in Euclidean space, and we'll henceforth assume that the curve satisfies it. In particular, that implies that $\det \pmatrix{\gamma' & \gamma''}$ has the same sign everywhere, and for expository convenience we'll take it to be positive; the negative case can be handled similarly. This choice of frame is not necessarily unimodular, but notice that if we reparameterize $\gamma$ (via a standard abuse of notation) $\gamma(s) := \gamma(t(s))$ for some invertible $t$, then $$ \frac{d \gamma}{ds } = t'(s) \frac{d \gamma}{dt }, \qquad \frac{d^2\gamma}{ds^2} = t'(s)^2 \frac{d^2 \gamma}{dt^2} \,\,\,\,\left(\!\!\!\!\!\!\mod {\frac{d\gamma}{ds}}\right) , $$ hence $\det \pmatrix{\gamma'(s) & \gamma''(s)} = t'(s)^3 \det \pmatrix{\gamma'(t) & \gamma''(t)}$. This in turn suggests that we define the (special) affine arc length of $\gamma(t)$ to be $$s(t) := \int_{t_0}^t \sqrt[3]{\det \pmatrix{\gamma'(t) & \gamma''(t)}} \,dt ,$$ an affine analogue of Euclidean arc length. We can parametrize any curve by special affine arc length, and by construction, a curve $\gamma(s)$ is parameterized by special affine arc length iff the frame $(\gamma'(s), \gamma''(s))$ is unimodular. This frame is canonical in the sense that it is equivariant under the action of $G$. It then follows from unimodularity that $\gamma'''(s) = \kappa(s) \gamma'(s)$ for some function $\kappa(s)$, the (special) affine curvature, and $\kappa$ turns out to be a complete invariant for nondegenerate curves in $\Bbb A^2$. (N.b. I think the formula at the link uses the opposite sign convention for $\kappa(s)$.) The curves with $\kappa = 0$ are the parabolas, and the curves with constant $\kappa$ are the conic sections.

For (much) more, see, e.g., Clelland's excellent From Frenet to Cartan: The Method of Moving Frames; the above discussion of invariants of curves in $\Bbb A^2$ is essentially an outline of solutions to Exercises 6.2.12–13 there.


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