Add lecture notes for coordinates

src/coordinates.md

@@ -21,7 +21,7 @@ such a distance definition is called a *Euclidean space*.
 Cartesian coordinates are used a lot. They are particularly suitable for
 infinite spaces or for rectangular volumes.
 
-![image](figures/Coordinates_5_1.pdf)
+![image](figures/Coordinates_5_1.svg)
 
 
 ## Polar coordinates
@@ -40,7 +40,7 @@ Note that each Cartesian coordinate has a *dimension* of length; in
 polar coordinates, the radius $r$ has a dimension of *length*, whereas
 the angular coordinate $\varphi$ is dimensionless.
 
-![image](figures/Coordinates_7_0.pdf)
+![image](figures/Coordinates_7_0.svg)
 
 In this plot you can distinguish the radial coordinate (0.2, 0.4 etc.)
 from the angular one ($0^\circ$, $45^\circ$ etc.).
@@ -50,7 +50,7 @@ $(r,\varphi)$ indicated. From this, we can see that the *Cartesian*
 coordinates $(x,y)$ of the point are related to the polar ones as
 follows: $$x = r \cos\varphi;$$ $$y = r \sin \varphi.$$
 
-![image](figures/Coordinates_9_0.pdf)
+![image](figures/Coordinates_9_0.svg)
 
 Now suppose we want to calculate the distance between two points, one
 with polar coordinates $(r_1, \varphi_1)$, and the other with
@@ -65,7 +65,7 @@ If we consider two points which are *very close*, the analysis
 simplifies however. We can use the geometry of the problem to find the
 distance (see the figure below).
 
-![image](figures/Coordinates_11_0.pdf)
+![image](figures/Coordinates_11_0.svg)
 
 When going from point 1 to point 2, we first traverse a small circular
 arc of radius $r_1$ and then we move a small distance radially outward
@@ -120,7 +120,7 @@ $\varphi$, is of course dimensionless.
 What is the distance travelled along a path when we express this in
 cylindrical coordinates? Let’s consider an example (Figure).
 
-![image](figures/Coordinates_13_0.pdf)
+![image](figures/Coordinates_13_0.svg)
 
 We want to find the length of the (small) red segment $\Delta s$. By
 inspecting the figure, we see that the horizontal (i.e. parallel to the
@@ -141,7 +141,7 @@ and (2) the direction of the line connecting the origin to our point.
 The specification of this direction can be identified with a point on a
 sphere which is centered at the origin:
 
-![image](figures/Coordinates_15_0.pdf)
+![image](figures/Coordinates_15_0.svg)
 
 The position of a point on the sphere is specified using the two angles
 $\theta$ and $\phi$ indicated in the figure.
@@ -164,7 +164,7 @@ $$\phi = \begin{cases} \arctan(y/x) &{\rm for ~} x>0; \\
  -\pi + \arctan(y/x) &{\rm ~ for ~} x<0 {\rm ~ and ~} y<0.
  \end{cases}$$ These relations can be derived from the following figure:
 
-![image](figures/Coordinates_17_0.pdf)
+![image](figures/Coordinates_17_0.svg)
 
 The distance related to a change in the spherical coordinates is
 calculated using Pythagoras’ theorem. The length $ds$ of a short segment
@@ -182,35 +182,35 @@ $$ds^2 = r^2 \left(\sin^2 \vartheta d\varphi^2 + d\vartheta^2\right) + dr^2.$$
 The picture below shows the geometry behind the calculation of this
 displacement.
 
-![image](figures/Coordinates_19_0.pdf)
+![image](figures/Coordinates_19_0.svg)
 
 ### summary
 
 We have discussed four different coordinate systems:
 
-1.  *Cartesian coordinates*: $$\bfr = (x_1, \ldots, x_n).$$ Can be used
-    for any dimension $n$. Convenient for: infinite spaces, systems with
-    rectangular symmatry.
+1.  *Cartesian coordinates*: $${\bf r} = (x_1, \ldots, x_n).$$ Can be
+    used for any dimension $n$. Convenient for: infinite spaces, systems
+    with rectangular symmatry.
 
-    Distance between two points $\bfr = (x_1, \ldots, x_n)$ and
-    $\bfr' = (x'_1, \ldots, x'_n)$:
+    Distance between two points ${\bf r} = (x_1, \ldots, x_n)$ and
+    ${\bf r}' = (x'_1, \ldots, x'_n)$:
     $$\Delta s^2 = (x'_1 - x_1)^2 + (x'_2 - x_2)^2 + \ldots + (x'_n - x_n)^2.$$
 
-2.  *Polar coordinates*: $$\bfr = (r, \phi).$$ Can be used in two
+2.  *Polar coordinates*: $${\bf r} = (r, \phi).$$ Can be used in two
     dimensions. Suitable for systems with circular symmetry or functions
     given in terms of these coordinates.
 
     Infinitesimal distance: $$ds^2 = dr^2 + r^2 d\phi^2.$$
 
-3.  *Cylindrical coordinates*: $$\bfr = (\rho, \phi, z).$$ Can be used
-    in three dimensions. Suitable for systems with axial symmetry or
-    functions given in terms of these coordinates.
+3.  *Cylindrical coordinates*: $${\bf r} = (\rho, \phi, z).$$ Can be
+    used in three dimensions. Suitable for systems with axial symmetry
+    or functions given in terms of these coordinates.
 
     Infinitesimal distance: $$ds^2 = d\rho^2 + \rho^2 d\phi^2 + dz^2.$$
 
-4.  *Spherical coordinates*: $$\bfr = (r, \theta, \phi).$$ Can be used
-    in three dimensions. Suitable for systems with spherical symmetry or
-    functions given in terms of these coordinates.
+4.  *Spherical coordinates*: $${\bf r} = (r, \theta, \phi).$$ Can be
+    used in three dimensions. Suitable for systems with spherical
+    symmetry or functions given in terms of these coordinates.
 
     Infinitesimal distance:
     $$ds^2 =r^2 (\sin^2 \theta d\phi^2 + d\theta^2) +  dr^2 .$$
@@ -221,13 +221,13 @@ Problems
 1.  [:grinning:]
 
     1.  Find the polar coordinates of the point with Cartesian
-        coordinates $$\bfr = \sqrt{2} (1,1).$$
+        coordinates $${\bf r} = \sqrt{2} (1,1).$$
 
     2.  Find the cylindrical coordinates of the point with Cartesian
-        coordinates $$\bfr = \frac{3}{2} (\sqrt{3}, 1, 1).$$
+        coordinates $${\bf r} = \frac{3}{2} (\sqrt{3}, 1, 1).$$
 
     3.  Find the spherical coordinates of the points
-        $$\bfr = (3/2, \sqrt{3}/2, 1).$$
+        $${\bf r} = (3/2, \sqrt{3}/2, 1).$$
 
 2.  [:smirk:] From the transformation from polar to Cartesian
     coordinates, show that
@@ -237,12 +237,12 @@ Problems
     (Use the chain rule for differentiation).
 
 3.  [:sweat:] Using the result of problem 2, show that the Laplace
-    operator acting on a function $\psi(\bfr)$ in polar coordinates
+    operator acting on a function $\psi({\bf r})$ in polar coordinates
     takes the form
-    $$\nabla^2 \psi(\bfr) =\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) \psi(\bfr) = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \psi(r,\varphi)}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \psi(r,\varphi)}{\partial \varphi^2}.$$
+    $$\nabla^2 \psi({\bf r}) =\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) \psi({\bf r}) = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \psi(r,\varphi)}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \psi(r,\varphi)}{\partial \varphi^2}.$$
 
     In a similar fashion it can be shown that for spherical coordinates,
-    the Laplace operator acting on a function $\psi(\bfr)$ becomes:
+    the Laplace operator acting on a function $\psi({\bf r})$ becomes:
     $$\nabla^2 \psi (r,\vartheta,\varphi) = 
     \frac{1}{r^2} \frac{\partial}{\partial r^2} \left( r^2 \frac{\partial \psi(r,\vartheta,\varphi)}{\partial r} \right) + \frac{1}{r^2\sin^2\vartheta} \frac{\partial^2 \psi(r,\vartheta, \varphi)}{\partial \varphi^2} + \frac{1}{r^2\sin\vartheta} 
     \frac{\partial}{\partial \vartheta}\left( \sin\vartheta \frac{\partial\psi(r,\vartheta, \varphi)}{\partial \vartheta}\right).$$