---
title: "Curvature"
sort_title: "Curvature"
date: 2021-03-07
categories:
- Mathematics
layout: "concept"
---

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](/know/concept/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](/know/concept/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](/know/concept/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.