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+---
+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.
-- 
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