Merge branch 'maciejedits' into 'master'

Addressing issues #13 and #14

Closes #14 and #13

See merge request !19

src/2_coordinates.md

@@ -62,7 +62,7 @@ system makes mathematics easier. For example, if you want to describe vibrations
 circular drum, polar coordinates become very convenient. These are
 defined for a two-dimensional space (a plane). The position on this plane is characterised by two
 coordinates: the *distance* $r$ between the point and the origin, and by the
-angle ($\varphi$) between the line connecting the point to the origin and the $x$-axis.
+angle ($\varphi$) between the line connecting the point to the origin and the $x$-axis. The radius is therefore always a non-negative number $r \geq 0$, and the range for the polar angle is $\varphi \in \left< 0,2\pi \right)$
 
 Note that each Cartesian coordinate has a *dimension* of length. 
 In polar coordinates, the radius $r$ has a dimension of *length*, but
@@ -247,7 +247,7 @@ problem.
 
 Cylindrical coordinates are defined straightforwardly: we use polar
 coordinates $r$ and $\varphi$ in the $xy$ plane, and the distance $z$
-along the symmetry-axis as the third coordinate. If the axis system is
+along the symmetry-axis as the third coordinate. The radius is therefore again always defined as a non-negative number $r \in \left<0, \infty \right)$, and the range for the azimuthal angle is analogically $\varphi \in \left< 0,2\pi \right)$. The *height* $z$ along the cylinder axis can take any real value, hence $z \in \mathbb{R}$ . If the axis system is
 chosen in physical space, we have two coordinates which have the
 dimension of a distance: $r$ and $z$. The other coordinate,
 $\varphi$ is of course dimensionless.
@@ -283,30 +283,35 @@ sphere which is centered at the origin:
 
 <figure markdown>
   ![image](figures/Coordinates_15_0.svg)
-  <figcaption>The position of a point on the sphere is specified using the radius $r$ and two angles
-$\theta$ and $\phi$</figcaption>
+  <figcaption>The position of a point on the sphere is specified using the radius $r$ and two angles (azimuthal)
+$\varphi$ and (polar) $\theta$ in the given order </figcaption>
 </figure>
 
+!!! warning "Parameter ranges in spherical coordinates"
+    - The radius ($r$) is defined for $r \in \left<0, \infty \right)$ </br>
+    - The azimuthal angle ($\varphi$) has the range of $\varphi \in \left< 0, 2\pi \right)$  </br>
+    - The polar angle ($\theta$) has the range $\theta \in \left<0, \pi \right>$ 
+
 !!! warning
     In mathematics, the angles are often labeled the other way
-    around: there, $\phi$ is used for the angle between a line running from
+    around: there, $\varphi$ is used for the angle between a line running from
     the origin to the point of interest and the $z$-axis, and $\theta$ for
     the angle of the projection of that line with the $x$-axis. The
     convention used here is customary in physics. 
 
 The relation between Cartesian and spherical coordinates is defined by:
 !!! info "The relation between Cartesian and spherical coordinates"
-    $$x = r \cos \varphi \sin \vartheta$$
-    $$y = r \sin\varphi \sin \vartheta$$ $$z = r \cos\vartheta$$ 
+    $$x = r \cos \varphi \sin \theta$$
+    $$y = r \sin\varphi \sin \theta$$ $$z = r \cos\theta$$ 
 
 The inverse transformation is easy to find: 
 !!! info "The inverse relation between Cartesian and spherical coordinates"
-    $$r = \sqrt{x^2+y^2+z^2}$$
-    $$\theta = \arccos(z/\sqrt{x^2+y^2+z^2})$$
-    $$\phi = \begin{cases} \arctan(y/x) &{\rm for ~} x>0; \\
+    $$r = \sqrt{x^2+y^2+z^2}, \qquad r \in \left<0, \infty \right)$$
+    $$\varphi = \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.
-    \end{cases}$$ 
+    \end{cases}, \qquad \varphi \in \left< 0,2\pi \right)$$ 
+    $$\theta = \arccos(z/\sqrt{x^2+y^2+z^2}), \qquad \theta \in \left< 0,\pi \right> $$
     
 These relations can be derived from the following figure:
 
@@ -318,15 +323,15 @@ 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 the changes in the polar
-angles of $d\vartheta$ and $d\varphi$ is given as
-$$dl^2 = r^2 \left(\sin^2 \vartheta d\varphi^2 + d\vartheta^2\right).$$
+angles of $d\theta$ and $d\varphi$ is given as
+$$dl^2 = r^2 \left(\sin^2 \theta d\varphi^2 + d\theta^2\right).$$
 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.
+*the same* coordinate $\theta$ span a circle in a horizontal plane
+with a radius $r\sin\theta$ 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 surface of the sphere, the combined displacement is:
-$$ds^2 = r^2 \left(\sin^2 \vartheta d\varphi^2 + d\vartheta^2\right) + dr^2.$$
+$$ds^2 = r^2 \left(\sin^2 \theta d\varphi^2 + d\theta^2\right) + dr^2.$$
 
 The picture below shows the geometry behind the calculation of this
 displacement.
@@ -348,35 +353,39 @@ here given as
 We have discussed four different coordinate systems:
 
 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
+    $${\bf r} = (x_1, \ldots, x_n)$$
+    $$ x_{n} \in \mathbb{R}$$
+    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.  !!! 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
+    $${\bf r} = (r, \varphi)$$
+    $$ r \in \left<0, \infty \right), \quad \varphi \in \left< 0,2\pi \right) $$
+    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 distance: $$ds^2 = dr^2 + r^2 d\varphi^2.$$
     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
+    $${\bf r} = (r, \varphi, z)$$
+    $$ r \in \left<0, \infty \right), \quad \varphi \in \left< 0,2\pi \right), \quad z \in \mathbb{R} $$ 
+    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 distance: $$ds^2 = dr^2 + r^2 d\varphi^2 + dz^2.$$
     Infinitesimal volume: $$dV = r dr d\varphi dz.$$
 
 4.  !!! tip "Spherical coordinates" 
-    $${\bf r} = (r, \theta, \phi).$$ This system can be
-    used in three dimensions. It is particularly suitable for systems with spherical
+    $${\bf r} = (r, \varphi, \theta)$$
+    $$ r \in \left<0, \infty \right), \quad \varphi \in \left< 0,2\pi \right), \quad \theta \in \left< 0,\pi \right> $$
+    This system 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: 
-    $$ds^2 =r^2 (\sin^2 \theta d\phi^2 + d\theta^2) +  dr^2 .$$
+    $$ds^2 =r^2 (\sin^2 \theta d\varphi^2 + d\theta^2) +  dr^2 .$$
     Infinitesimal volume:
-    $$dV = r^2 \sin(\theta) dr d\theta d\varphi.$$ 
+    $$dV = r^2 \sin(\theta) dr d\varphi d\theta.$$ 
 
 ## 2.5. Problems
 
@@ -431,9 +440,9 @@ We have discussed four different coordinate systems:
 
     In a similar fashion it can be shown that for spherical coordinates,
     the Laplace operator acting on a function $\psi({\bf r})$ becomes:
-    $$\begin{align} \nabla^2 \psi (r,\vartheta,\varphi) &= 
-    \frac{1}{r^2} \frac{\partial}{\partial r^2} \left( r^2 \frac{\partial \psi(r,\vartheta,\varphi)}{\partial r} \right) \\ &+ \frac{1}{r^2\sin^2\vartheta} \frac{\partial^2 \psi(r,\vartheta, \varphi)}{\partial \varphi^2} \\ &+ \frac{1}{r^2\sin\vartheta} 
-    \frac{\partial}{\partial \vartheta}\left( \sin\vartheta \frac{\partial\psi(r,\vartheta, \varphi)}{\partial \vartheta}\right).
+    $$\begin{align} \nabla^2 \psi (r,\varphi, \theta) &= 
+    \frac{1}{r^2} \frac{\partial}{\partial r^2} \left( r^2 \frac{\partial \psi(r,\varphi,\theta)}{\partial r} \right) \\ &+ \frac{1}{r^2\sin^2\theta} \frac{\partial^2 \psi(r,\varphi, \theta)}{\partial \varphi^2} \\ &+ \frac{1}{r^2\sin\theta} 
+    \frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial\psi(r,\varphi, \theta)}{\partial \theta}\right).
     \end{align}$$
     This is however even more tedious (you do not have to show this).