First major update of src/2_coordinates.md

src/2_coordinates.md

@@ -21,7 +21,7 @@ The most common coordinates are *Cartesian coordinates*, where we use a
 number $n$ of perpendicular axes. The coordinates corresponding to these
 axes are $x_j$ where $j=1, \ldots, n$.
 
-Cartesian coordinates are simple to describe and operate in. The coordinate axes are 
+Cartesian coordinates are simple to describe and operate on. The coordinate axes are 
 straight lines perpendicular to each other. It is therefore
 very easy to do calculations in Cartesian coordinates. For example,
 the distance $\Delta s$ between two points $(x_1, x_2, \ldots, x_n)$
@@ -358,22 +358,22 @@ We have discussed four different coordinate systems:
 2.  !!! tip "Polar coordinates" 
     $${\bf r} = (r, \phi).$$ This system can be used in two
     dimensions. It is particularly suitable for systems with circular symmetry or functions
-    given in terms of these coordinates.
+    given in terms of these coordinates. <br/>
     Infinitesimal distance: $$ds^2 = dr^2 + r^2 d\phi^2.$$
     Infinitesimal area: $$dA = r dr d\varphi.$$
 
 3.  !!! tip "Cylindrical coordinates" 
     $${\bf r} = (r, \phi, z).$$ This system can be
     used in three dimensions. It is particularly suitable for systems with axial symmetry
-    or functions given in terms of these coordinates.
+    or functions given in terms of these coordinates. <br/>
     Infinitesimal distance: $$ds^2 = dr^2 + r^2 d\phi^2 + dz^2.$$
-    Infinitesimal volume:: $$dV = r dr d\varphi dz.$$
+    Infinitesimal volume: $$dV = r dr d\varphi dz.$$
 
 4.  !!! tip "Spherical coordinates" 
     $${\bf r} = (r, \theta, \phi).$$ This sysytem can be
     used in three dimensions. It is particularly suitable for systems with spherical
-    symmetry or functions given in terms of these coordinates.
-    Infinitesimal distance:
+    symmetry or functions given in terms of these coordinates. <br/>
+    Infinitesimal distance: 
     $$ds^2 =r^2 (\sin^2 \theta d\phi^2 + d\theta^2) +  dr^2 .$$
     Infinitesimal volume:
     $$dV = r^2 \sin(\theta) dr d\theta d\varphi.$$ 
@@ -402,7 +402,8 @@ We have discussed four different coordinate systems:
     5. $r=1$ and $\theta=\pi/4$ in spherical coordinates,
     6. $\varphi=\pi/2$ and $\theta=\pi/2$ in spherical coordinates.
 
-3.  [:smirk:]
+3.  [:smirk:] *Partial derivatives*
+
     (a) Consider the function $f(r,\varphi,\theta)=\frac{1}{r^2}$ defined
         using spherical coordinates.
         Compute $\frac{\partial}{\partial z} f(r, \varphi, \theta)$.
@@ -412,26 +413,32 @@ We have discussed four different coordinate systems:
         question).
         Compute again $\frac{\partial}{\partial z} f(r, \varphi, z)$.
  
-4.  [:smirk:] From the transformation from polar to Cartesian
+4.  [:smirk:] *Chain rule practice* 
+
+    From the transformation from polar to Cartesian
     coordinates, show that
     $$\frac{\partial}{\partial x} = \cos\varphi \frac{\partial}{\partial r} - \frac{\sin\varphi}{r} \frac{\partial}{\partial \varphi}$$
     and
     $$\frac{\partial}{\partial y} = \sin\varphi \frac{\partial}{\partial r} + \frac{\cos\varphi}{r} \frac{\partial}{\partial \varphi}.$$
     (Use the chain rule for differentiation).
 
-5.  [:sweat:] Using the result of problem 4, show that the Laplace
+5.  [:sweat:] *Laplace operator in spherical coordinates*
+
+    Using the result of problem 4, show that the Laplace
     operator acting on a function $\psi({\bf r})$ in polar coordinates
     takes the form
     $$\nabla^2 \psi({\bf r}) =\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) \psi({\bf r}) = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \psi(r,\varphi)}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \psi(r,\varphi)}{\partial \varphi^2}.$$
 
     In a similar fashion it can be shown that for spherical coordinates,
     the Laplace operator acting on a function $\psi({\bf r})$ becomes:
-    \begin{align} \nabla^2 \psi (r,\vartheta,\varphi) &= 
+    $$\begin{align} \nabla^2 \psi (r,\vartheta,\varphi) &= 
     \frac{1}{r^2} \frac{\partial}{\partial r^2} \left( r^2 \frac{\partial \psi(r,\vartheta,\varphi)}{\partial r} \right) \\ &+ \frac{1}{r^2\sin^2\vartheta} \frac{\partial^2 \psi(r,\vartheta, \varphi)}{\partial \varphi^2} \\ &+ \frac{1}{r^2\sin\vartheta} 
-    \frac{\partial}{\partial \vartheta}\left( \sin\vartheta \frac{\partial\psi(r,\vartheta, \varphi)}{\partial \vartheta}\right).\end{align}
+    \frac{\partial}{\partial \vartheta}\left( \sin\vartheta \frac{\partial\psi(r,\vartheta, \varphi)}{\partial \vartheta}\right).
+    \end{align}$$
     This is however even more tedious (you do not have to show this).
 
 6.  [:grinning:] *Integration and coordinates I*
+
     We define $f(r, \varphi) = \frac{1}{r}$ in polar coordinates. Explain how
     a circular region, centered at the origin and with radius $r_0$, can be described
     using polar coordinates. Then compute the integral of $f(r,\varphi)$ over
@@ -443,7 +450,7 @@ We have discussed four different coordinate systems:
 
 8.  [:smirk:] *Integration and coordinates III*
 
-    In 2D we can define a shape by specifying a function $r(\varphi)$:
+    In 2D, we can define a shape by specifying a function $r(\varphi)$:
 
     ![image](figures/shape_polar.svg)