Categories: Mathematics.

# Curvature

Given a curve or surface, its curvature $\kappa$ describes how sharply it is bending at a given point. It is defined as the inverse of the radius of curvature $R$, which is the radius of the tangent circle that osculates (i.e. best approximates) the curve/surface at that point:

\begin{aligned} \kappa = \frac{1}{R} \end{aligned}

Typically, $\kappa$ is positive for convex curves/surfaces, and negative for concave ones, although this distinction is somewhat arbitrary. Below, we calculate the curvature in several general cases.

## 2D height functions

We start with a specialized case: height functions, where one coordinate is a function of the other one (2D) or two (3D). In this case, we can use the calculus of variations to find the curvature.

This approach relies on the fact that a circle has the highest area-perimeter ratio of any 2D shape, and a sphere has the highest volume-surface ratio of any 3D body. By the definition of curvature, these shapes have constant $\kappa$.

We will thus minimize the perimeter/surface while keeping the area/volume fixed, which will give us a shape with constant curvature, and from that we can extrapolate an expression for $\kappa$.

In 2D, for a single-variable height function $h(x)$, the length of a small segment of the curve is:

\begin{aligned} \sqrt{\dd{x}^2 + \dd{h}^2} = \dd{x} \sqrt{\Big( \dv{x}{x} \Big)^2 + \Big( \dv{h}{x} \Big)^2} = \dd{x} \sqrt{1 + h_x^2} \end{aligned}

Which leads us to define the following Lagrangian $\mathcal{L}$ describing the “energy cost” of the curve:

\begin{aligned} \mathcal{L} = \sqrt{1 + h_x^2} \end{aligned}

Furthermore, we demand that the area under the curve (i.e. the “volume”) is constant:

\begin{aligned} V = \int_{x_0}^{x_1} h(x) \dd{x} \end{aligned}

By putting these things together, we arrive at the following energy functional $E[h]$, where $\kappa$ is an ominously-named Lagrange multiplier:

\begin{aligned} E[h] = \int (\mathcal{L} + \kappa h) \dd{x} \end{aligned}

Minimizing this functional leads to the following Lagrange equation of the first kind:

\begin{aligned} 0 = \pdv{\mathcal{L}}{h} - \dv{}{x}\Big( \pdv{\mathcal{L}}{h_x} \Big) + \kappa \end{aligned}

We evaluate the terms of this equation to arrive at an expression for the curvature $\kappa$:

\begin{aligned} \boxed{ \kappa = \frac{h_{xx}}{\big(1 + h_x^2\big)^{3/2}} } \end{aligned}

In this optimization problem, $\kappa$ is a constant, but in fact the statement above is valid for variable curvatures too, in which case $\kappa$ is a function of $x$.

## 2D in general

We can parametrically describe an arbitrary plane curve as a function of the arc length $s$:

\begin{aligned} \big( x(s), y(s) \big) \qquad \mathrm{where} \qquad \dd{s}^2 = \dd{x}^2 + \dd{y}^2 \end{aligned}

If we choose the horizontal $x$-axis as a reference, we can furthermore define the elevation angle $\theta(s)$ as the angle between the reference and the curve’s tangent vector $\vu{t}$:

\begin{aligned} \vu{t} = \big( x_s(s), y_s(s) \big) = \big( \cos\theta(s), \sin\theta(s) \big) \end{aligned}

Where $x_s(s) = \idv{x}{s}$. The curvature $\kappa$ is defined as the $s$-derivative of this elevation angle:

\begin{aligned} \kappa = \dv{\theta}{s} = \theta_s(s) \end{aligned}

We have two ways of writing $\vu{t}$: using the derivatives $x_s$ and $y_s$, or the elevation angle $\theta$. Now, let us take the $s$-derivative of both expressions, and equate them:

\begin{aligned} \big( x_{ss}, y_{ss} \big) = \dv{\vu{t}}{s} = \theta_s \: \big( \!-\!\sin\theta, \cos\theta \big) = \kappa \big( \!-\!y_s, x_s \big) \end{aligned} \begin{aligned} x_{ss} = - \kappa y_s \qquad y_{ss} = \kappa x_s \end{aligned}

We multiply these equation by $y_s$ and $x_s$, respectively, and subtract the first from the last:

\begin{aligned} y_{ss} x_s - x_{ss} y_s = \kappa x_s^2 + \kappa y_s^2 \end{aligned}

Isolating this for $\kappa$ and using the fact that $x_s^2 + y_s^2 = 1$ thanks to $s$ being the arc length:

\begin{aligned} \kappa = \frac{y_{ss} x_s - x_{ss} y_s}{x_s^2 + y_s^2} = y_{ss} x_s - x_{ss} y_s \end{aligned}

While this result is correct, we would like to generalize it to cases where the curve is parametrized by some other $t$, not necessarily the arc length. Let prime denote the $t$-derivative:

\begin{aligned} x_s = x' t_s \qquad x_{ss} = x'' t_s^2 + x' t_{ss} \\ y_s = y' t_s \qquad \: y_{ss} = y'' t_s^2 + x' t_{ss} \end{aligned}

By inserting these expression into the earlier formula for $\kappa$, we find:

\begin{aligned} \kappa = y_{ss} x_s - x_{ss} y_s &= x' t_s (y'' t_s^2 + y' t_{ss}) - y' t_s (x'' t_s^2 + x' t_{ss}) \\ &= t_s t_{ss} (x' y' - y' x') + t_s^3 (x' y'' - y' x'') \\ &= t_s^3 (x' y'' - y' x'') \end{aligned}

Since $x_s^2 + y_s^2 = 1$, we know that $(x')^2 + (y')^2 = 1 / t_s^2$, which leads us to the following general expression for the curvature $\kappa$ of a plane curve:

\begin{aligned} \boxed{ \kappa = \frac{y'' x' - x'' y'}{\big((x')^2 + (y')^2\big)^{3/2}} } \end{aligned}

If the curve happens to be a height function, i.e. $y(x)$, then $x' = 1$ and $x'' = 0$, and we arrive at our previous result again.

## 3D height functions

The generalization to a 3D height function $h(x, y)$ is straightforward: the cost of an infinitesimal portion of the surface is as follows, using the same reasoning as before:

\begin{aligned} \mathcal{L} = \sqrt{1 + h_x^2 + h_y^2} \end{aligned}

Keeping the volume $V$ constant, we get the following energy functional $E$ to minimize:

\begin{aligned} E[h] = \iint (\mathcal{L} + \lambda h) \dd{x} \dd{y} \end{aligned}

Which gives us an Euler-Lagrange equation involving the Lagrange multiplier $\lambda$:

\begin{aligned} 0 = \pdv{\mathcal{L}}{h} - \dv{}{x}\Big( \pdv{\mathcal{L}}{h_x} \Big) - \dv{}{y}\Big( \pdv{\mathcal{L}}{h_y} \Big) + \lambda \end{aligned}

Inserting $\mathcal{L}$ into this and evaluating all the derivatives yields a result for the (variable) curvature:

\begin{aligned} \boxed{ \lambda = \kappa_1 + \kappa_2 = \frac{(1 + h_y^2) h_{xx} - 2 h_x h_y h_{xy} + (1 + h_x^2) h_{yy}}{\big(1 + h_x^2 + h_y^2\big)^{3/2}} } \end{aligned}

What are $\kappa_1$ and $\kappa_2$? Well, the problem in 3D is that the curvature of an osculating circle depends on the orientation of that circle. The principal curvatures $\kappa_1$ and $\kappa_2$ are the largest and smallest curvatures at a given point, but finding their values and the corresponding principal directions is not so easy. Fortunately, in practice, we are often only interested in their sum:

\begin{aligned} \lambda = \kappa_1 + \kappa_2 = \frac{1}{R_1} + \frac{1}{R_2} \end{aligned}

These principal radii $R_1$ and $R_2$ are important for e.g. the Young-Laplace law.

## 3D in general

To find a general expression for the mean curvature of an arbitrary surface, we “cut off” a small part of the surface that we can regard as a height function. We call the “cutting” reference plane $(x, y)$, and the surface it describes $h(x, y)$. We then define the unit tangent vectors $\vu{t}_x$ and $\vu{t}_y$ to be parallel to the $x$-axis and $y$-axis, respectively:

\begin{aligned} \vu{t}_x = \frac{1}{\sqrt{1 + (h_x)^2}} \begin{bmatrix} 1 \\ 0 \\ h_x \end{bmatrix} \qquad \vu{t}_y = \frac{1}{\sqrt{1 + (h_y)^2}} \begin{bmatrix} 0 \\ 1 \\ h_y \end{bmatrix} \end{aligned}

Since they were chosen to lie along the axes, these vectors are not necessarily orthogonal, so we need to normalize the resulting normal vector $\vu{n}$:

\begin{aligned} \vu{n} = \vu{t}_x \cross \vu{t}_y = \frac{1}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \begin{bmatrix} - h_x \\ - h_y \\ 1 \end{bmatrix} \end{aligned}

Let us take a look at the divergence of $\vu{n}$, or to be precise, its projection onto the reference plane (although this distinction is not really important for our purposes):

\begin{aligned} \nabla \cdot \vu{n} = - \dv{}{x}\bigg( \frac{h_x}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg) - \dv{}{y}\bigg( \frac{h_y}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg) \end{aligned}

Compare this with the expression for $\lambda$ we found earlier, with the help of variational calculus:

\begin{aligned} \lambda &= \dv{}{x}\Big( \pdv{\mathcal{L}}{h_x} \Big) + \dv{}{y}\Big( \pdv{\mathcal{L}}{h_y} \Big) \\ &= \dv{}{x}\bigg( \frac{h_x}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg) + \dv{}{y}\bigg( \frac{h_y}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg) \end{aligned}

The similarity is clearly visible. This leads us to the following general expression:

\begin{aligned} \boxed{ \kappa_1 + \kappa_2 = - \nabla \cdot \vu{n} } \end{aligned}

A useful property is that the principal directions of curvature are always orthogonal. To show this, consider the most general second-order approximating surface, in polar coordinates:

\begin{aligned} h(x, y) &= \frac{1}{2} a x^2 + \frac{1}{2} b y^2 + c x y \\ &= \frac{1}{2} a r^2 \cos^2\varphi + \frac{1}{2} b r^2 \sin^2\varphi + c r^2 \cos\varphi \sin\varphi \end{aligned}

Sufficiently close to the extremum, where $h_x$ and $h_y$ are negligible, the curvature along a certain direction $\varphi$ is given by our earlier formula for a 2D height function:

\begin{aligned} \kappa(\varphi) \approx \pdvn{2}{h}{r} = a \cos^2\varphi + b \sin^2\varphi + c \sin(2 \varphi) \end{aligned}

To find the extremes of $\kappa$, we differentiate with respect to $\varphi$ and demand that it is zero:

\begin{aligned} 0 &= - 2 a \cos\varphi \sin\varphi + 2 b \sin\varphi \cos\varphi + 2 c \cos(2 \varphi) \\ &= - a \sin(2 \varphi) + b \sin(2 \varphi) + 2 c \cos(2 \varphi) \end{aligned}

After rearranging this a bit, we arrive at the following transcendental equation:

\begin{aligned} \frac{2 c}{a - b} = \frac{\sin(2 \varphi)}{\cos(2 \varphi)} = \tan(2 \varphi) \end{aligned}

Since the $\tan$ function is $\pi$-periodic, this has two solutions, $\varphi_0$ and $\varphi_0 + \pi/2$, which are clearly orthogonal, hence the principal directions are at an angle of $\pi/2$.

Finally, it is also worth mentioning that the principal directions always lie in planes containing the normal of the surface.

## References

1. T. Bohr, Curvature of plane curves and surfaces, 2020, unpublished.
2. B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.