First major update of src/2_coordinates.md

src/2_coordinates.md

@@ -264,10 +264,10 @@ We want to find the length of the (small) red segment $d s$. By
 inspecting the figure, we see that the horizontal (i.e. parallel to the
 $xy$-plane) segment $d l$ is perpendicular to the vertical segment
 $dz$. Using for $d l$ the length we obtained before for a line
-segment in the $xy$ plane, expressed in polar coordinates, we
+segment in the $xy$ plane expressed in polar coordinates, we
 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
+The volume element is consequently given as:
 
 !!! info "Volume element in cylindrical coordinates"
     $$dV = r dr d\varphi dz.$$
@@ -281,27 +281,34 @@ 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.svg)
-
-The position of a point on the sphere is specified using the two angles
-$\theta$ and $\phi$ indicated in the figure.
+<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>
+</figure>
 
 !!! warning
-    Note that in mathematics, often the angles are labelled the other way
-    round: there, $\phi$ is used for the angle between a line running from
-    the origin o the point of interest and the $z$-axis, and $\theta$ for
+    In mathematics, the angles are often labelled the other way
+    around: there, $\phi$ 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 custom in physics. 
-
-The relation between Cartesian and coordinates is defined by
-$$x = r \cos \varphi \sin \vartheta$$
-$$y = r \sin\varphi \sin \vartheta$$ $$z = r \cos\vartheta$$ The inverse
-transformation is easy to find: $$r = \sqrt{x^2+y^2+z^2}$$
-$$\theta = \arccos(z/\sqrt{x^2+y^2+z^2})$$
-$$\phi = \begin{cases} \arctan(y/x) &{\rm for ~} x>0; \\
- \pi + \arctan(y/x) & {\rm for ~} x<0 {\rm ~ and ~} y>0;\\
- -\pi + \arctan(y/x) &{\rm ~ for ~} x<0 {\rm ~ and ~} y<0.
- \end{cases}$$ These relations can be derived from the following figure:
+    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"
+    $$x = r \cos \varphi \sin \vartheta$$
+    $$y = r \sin\varphi \sin \vartheta$$ $$z = r \cos\vartheta$$ 
+
+The inverse transformation is easy to find: 
+!!! info "The inverse relatuion between Cartesian and spherical coordinates"
+    $$r = \sqrt{x^2+y^2+z^2}$$
+    $$\theta = \arccos(z/\sqrt{x^2+y^2+z^2})$$
+    $$\phi = \begin{cases} \arctan(y/x) &{\rm for ~} x>0; \\
+    \pi + \arctan(y/x) & {\rm for ~} x<0 {\rm ~ and ~} y>0;\\
+    -\pi + \arctan(y/x) &{\rm ~ for ~} x<0 {\rm ~ and ~} y<0.
+    \end{cases}$$ 
+    
+These relations can be derived from the following figure:
 
 ![image](figures/Coordinates_17_0.svg)