First major update of src/2_coordinates.md

src/2_coordinates.md

@@ -37,7 +37,7 @@ distances, for example in the definition of derivatives and integrals.
 In Cartesian coordinates, the expressions for infinitesimal distances $ds$ and
 infinitesimal volumes $dV$ are given as:
 
-!!! info "Segment and volume element in n-dimensional Cartesian coordinates"
+!!! info "Infinitesimal segment and volume elements in n-dimensional Cartesian coordinates"
     $$ds = \sqrt{dx_1^2 + dx_2^2 + \ldots + dx_n^2}$$
     $$dV = dx_1 dx_2 \ldots dx_N.$$
 
@@ -136,7 +136,7 @@ We can use the same arguments also for the area: since the different
 segments are approximately perpendicular, we find the area by simply
 multiplying them:
 
-!!! info "Surface element in polar coordinates"
+!!! info "Infinitesimal surface element in polar coordinates"
     	$$dA = r dr d\varphi.$$
 
 This is an important formula to remember for integrating in polar
@@ -269,7 +269,7 @@ immediately find:
 $$d s^2 = d l^2 + d z^2 = d r^2 + r^2 d \varphi^2 + d z^2.$$
 The volume element is consequently given as:
 
-!!! info "Volume element in cylindrical coordinates"
+!!! info "Infinitesimal volume element in cylindrical coordinates"
     $$dV = r dr d\varphi dz.$$
 
 ### Spherical coordinates
@@ -305,74 +305,82 @@ The inverse transformation is easy to find:
     $$\theta = \arccos(z/\sqrt{x^2+y^2+z^2})$$
     $$\phi = \begin{cases} \arctan(y/x) &{\rm for ~} x>0; \\
     \pi + \arctan(y/x) & {\rm for ~} x<0 {\rm ~ and ~} y>0;\\
-    -\pi + \arctan(y/x) &{\rm ~ for ~} x<0 {\rm ~ and ~} y<0.
+    -\pi + \arctan(y/x) &{\rm for ~} x<0 {\rm ~ and ~} y<0.
     \end{cases}$$ 
     
 These relations can be derived from the following figure:
 
-![image](figures/Coordinates_17_0.svg)
+<figure markdown>
+  ![image](figures/Coordinates_17_0.svg)
+  <figcaption></figcaption>
+</figure>
 
 The distance related to a change in the spherical coordinates is
 calculated using Pythagoras’ theorem. The length $ds$ of a short segment
-on the sphere with radius $r$ corresponding to changes in the polar
+on the sphere with radius $r$ corresponding to the changes in the polar
 angles of $d\vartheta$ and $d\varphi$ is given as
 $$dl^2 = r^2 \left(\sin^2 \vartheta d\varphi^2 + d\vartheta^2\right).$$
 In order to verify this, it is important to realize that all points with
 *the same* coordinate $\vartheta$ span a circle in a horizontal plane
 with a radius $r\sin\vartheta$ as shown in the figure below.
 
-From this we can also infer that, for a segment with a radial component
-$dr$ in addition to the displacement on the sphere, the displacement is:
+From this, we can also infer that for a segment with a radial component
+$dr$ in addition to the displacement on the surface of the sphere, the combined displacement is:
 $$ds^2 = r^2 \left(\sin^2 \vartheta d\varphi^2 + d\vartheta^2\right) + dr^2.$$
 
 The picture below shows the geometry behind the calculation of this
 displacement.
 
-![image](figures/Coordinates_19_0.svg)
+<figure markdown>
+  ![image](figures/Coordinates_19_0.svg)
+  <figcaption></figcaption>
+</figure>
+
 
 From these arguments we can again also find the volume element, it is
 here given as
+
+!!! info "Infinitesimal volume element in spherical coordinates"
 $$dV = r^2 \sin\theta dr d\theta d\varphi.$$
 
-## Summary
+## 2.4. Summary
 
 We have discussed four different coordinate systems:
 
-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.
-
+1.  !!! tip "Cartesian coordinates" 
+    $${\bf r} = (x_1, \ldots, x_n).$$ This systems can be
+    used for any dimension $n$. It is particularly convenient for: infinite spaces, systems
+    with rectangular symmetry.
     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*: $${\bf r} = (r, \phi).$$ Can be used in two
-    dimensions. Suitable for systems with circular symmetry or functions
-    given in terms of these coordinates.
-
+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. <br/>
     Infinitesimal distance: $$ds^2 = dr^2 + r^2 d\phi^2.$$
     Infinitesimal area: $$dA = r dr d\varphi.$$
 
-3.  *Cylindrical coordinates*: $${\bf r} = (r, \phi, z).$$ Can be
-    used in three dimensions. Suitable for systems with axial symmetry
-    or functions given in terms of these coordinates.
-
+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. <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.  *Spherical coordinates*: $${\bf r} = (r, \theta, \phi).$$ Can be
-    used in three dimensions. Suitable for systems with spherical
+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:
+    Infinitesimal distance: <br/>
     $$ds^2 =r^2 (\sin^2 \theta d\phi^2 + d\theta^2) +  dr^2 .$$
     Infinitesimal volume:
     $$dV = r^2 \sin(\theta) dr d\theta d\varphi.$$ 
 
+## 2.5. Problems
 
-## Problems
-
-1.  [:grinning:]
+1.  [:grinning:] *Warm-up*
 
     1.  Find the polar coordinates of the point with Cartesian
         coordinates $${\bf r} = \sqrt{2} (1,1).$$
@@ -394,7 +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 $\frac{\partial}{\partial z} f(r, \varphi, \theta)$.
@@ -404,26 +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:
-    $$\nabla^2 \psi (r,\vartheta,\varphi) = 
-    \frac{1}{r^2} \frac{\partial}{\partial r^2} \left( r^2 \frac{\partial \psi(r,\vartheta,\varphi)}{\partial r} \right) + \frac{1}{r^2\sin^2\vartheta} \frac{\partial^2 \psi(r,\vartheta, \varphi)}{\partial \varphi^2} + \frac{1}{r^2\sin\vartheta} 
-    \frac{\partial}{\partial \vartheta}\left( \sin\vartheta \frac{\partial\psi(r,\vartheta, \varphi)}{\partial \vartheta}\right).$$
+    $$\begin{align} \nabla^2 \psi (r,\vartheta,\varphi) &= 
+    \frac{1}{r^2} \frac{\partial}{\partial r^2} \left( r^2 \frac{\partial \psi(r,\vartheta,\varphi)}{\partial r} \right) \\ &+ \frac{1}{r^2\sin^2\vartheta} \frac{\partial^2 \psi(r,\vartheta, \varphi)}{\partial \varphi^2} \\ &+ \frac{1}{r^2\sin\vartheta} 
+    \frac{\partial}{\partial \vartheta}\left( \sin\vartheta \frac{\partial\psi(r,\vartheta, \varphi)}{\partial \vartheta}\right).
+    \end{align}$$
     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
@@ -435,7 +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)