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 $(x_1, x_2, \ldots, x_n)$
@@ -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
@@ -317,15 +317,15 @@ These relations can be derived from the following 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
@@ -339,47 +339,48 @@ displacement.
 
 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:: $$dV = r dr d\varphi dz.$$
-
-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 volume: $$dV = r dr d\varphi dz.$$
 
-    Infinitesimal distance:
+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. <br/>
+    Infinitesimal distance: 
     $$ds^2 =r^2 (\sin^2 \theta d\phi^2 + d\theta^2) +  dr^2 .$$
     Infinitesimal volume:
     $$dV = r^2 \sin(\theta) dr d\theta d\varphi.$$ 
 
+## 2.5. Problems
 
-## Problems
-
-1.  [:grinning:]
+1.  [:grinning:] *Warm-up*
 
     1.  Find the polar coordinates of the point with Cartesian
         coordinates $${\bf r} = \sqrt{2} (1,1).$$
@@ -401,7 +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 $\frac{\partial}{\partial z} f(r, \varphi, \theta)$.
@@ -411,26 +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
@@ -442,7 +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)