Add lecture notes for coordinates

src/coordinates.md

@@ -67,19 +67,19 @@ $(r,\varphi)$ indicated. From this, we can see that the *Cartesian*
 coordinates $(x,y)$ of the point are related to the polar ones as
 follows:
 
-$$x = r \cos\varphi;$$
-$$y = r \sin \varphi.$$
+$$\begin{equation} x = r \cos\varphi; \end{equation}$$
+$$\begin{equation} y = r \sin \varphi.\end{equation}$$
 
 ![image](figures/Coordinates_9_0.svg)
 
 The inverse relation is given as
 
-$$r=\sqrt{x^2 + y^2};$$
-$$\varphi=\begin{cases}
+$$\begin{equation} r=\sqrt{x^2 + y^2}; \label{rxy}\end{equation}$$
+$$\begin{equation} \varphi=\begin{cases}
   \arctan(y/x) & \text{$x>0$,}\\
   \pi + \arctan(y/x) & \text{$x<0$ and $y>0$,}\\
   -\pi + \arctan(y/x) & \text{$x<0$ and $y<0$.}
-  \end{cases}$$
+  \end{cases} \label{phixy}\end{equation}$$
 
 The last formula for $\varphi$ warrants a closer explanation: It is easy
 to see that $\tan(\varphi)=y/x$ - but this is not a unique relation, due to
@@ -132,12 +132,49 @@ understood: the area swept by an angle difference $d\varphi$
 
     We find:
 
-    $$ \int_0^{2\pi} d\varphi \int_0^r_0 r dr = \\
-    2\pi \int_0^r_0 r dr = 2 \pi \frac{1}{2} r_0^2 = \pi r_0^2$$
+    $$
+    \int_0^{2\pi} d\varphi \int_0^r_0 r dr =
+    2\pi \int_0^r_0 r dr = 2 \pi \frac{1}{2} r_0^2 = \pi r_0^2,
+    $$
 
     which is indeed the area of a circle with radius 0.
 
-## Cylindrical coordinates
+## Converting derivatives between coordinate systems
+
+Often, in physics important equations involve derivatives given in terms
+of Cartesian coordinates. One prominent example are equations of the form
+$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)
+f(x, y) = \ldots.$$
+The derivative operator $\left(\frac{\partial^2}{\partial x^2} +
+\frac{\partial^2}{\partial y^2}\right)$ is so common it has its own name:
+the Laplacian (here for two-dimensional space).
+
+Such an equation is universal, but for particular situations it might be
+advantageous to use a different coordinate system, such as polar coordinates
+for a system with rotational symmetry. The question then is: How does the
+corresponding equation look like in a different coordinate system?
+
+There are different ways to find the answer. Here, we will focus on
+directly deriving the transformed equation through an explicit calculation
+involving the chain rule for a function of several variables.
+
+!!! info Chain rule for a multi-variable function
+    Let $f$ be a function of $n$ variables: $f(y_1, y_2, \ldots, y_n)$,
+    as well as $g_i(x_1, x_2, \ldots, x_n)$ for $i=1,2,\ldots, n$. Then
+
+    $$\frac{\partial}{\partial x_i} = \sum_{j=1}^n
+    \fac{partial f}{\partial y_j} \frac{\partial g_j}{\partial x_i}$$
+
+We start by replacing the function $f(x, y)$ by a function in polar coordinates
+$f(r, \varphi)$, and ask what is $\frac{\partial}{\partial x} f(r, \varphi)$. When
+we look at this expression, we need to understand what it *means* to take the derivative
+of a function of $r, \varphi$ in terms of $x$?
+
+For this, we need to realize that there are relations between the coordinate systems.
+In particular, $r = r(x,y)$ and $\varphi = \varphi(x, y)$ as defined in equations
+\ref{rxy} and \ref{phixy}.
+
+# Cylindrical coordinates
 
 Three dimensional systems may have axial symmetry. An example is an
 electrically charged wire of which we would like to calculate the
@@ -170,7 +207,7 @@ $$d s^2 = d l^2 + d z^2 = d r^2 + r^2 d \varphi^2 + d z^2.$$
 The volume element is consequently given as
 $$dV = r dr d\varphi dz.$$
 
-## Spherical coordinates
+# Spherical coordinates
 
 For problems with spherical symmetry, we use *spherical coordinates*.
 These work as follows. For a point $\bf r$ in 3D space, we can specify