diff options
Diffstat (limited to 'content/know/concept/curvature/index.pdc')
| -rw-r--r-- | content/know/concept/curvature/index.pdc | 395 |
1 files changed, 395 insertions, 0 deletions
diff --git a/content/know/concept/curvature/index.pdc b/content/know/concept/curvature/index.pdc new file mode 100644 index 0000000..50c61b2 --- /dev/null +++ b/content/know/concept/curvature/index.pdc @@ -0,0 +1,395 @@ +--- +title: "Curvature" +firstLetter: "C" +publishDate: 2021-03-07 +categories: +- Mathematics + +date: 2021-03-04T09:46:31+01:00 +draft: false +markup: pandoc +--- + +# 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](/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 $x_s(s) = \dv*{x}{s}$. +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: + +$$\begin{aligned} + \kappa(\varphi) + \approx \pdv[2]{h}{r} + = a \cos^2\varphi + b \sin^2\varphi + c \sin(2 \varphi) +\end{aligned}$$ + +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. |