Add lecture notes for coordinates

src/coordinates.md

@@ -2,21 +2,37 @@
 title: Coordinates
 ---
 
-# Coordinates
-
-
-## Cartesian coordinates
-
+# Cartesian coordinates
 
 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 $j=1, \ldots, n$. The distance $\Delta s$ between
-two points $(x_1, x_2, \ldots, x_n)$ and $(x'_1, x'_2, \ldots, x'_n)$ is
-given by
+axes are $x_j$ where $j=1, \ldots, n$.
+
+Cartesian coordinates are simple, as the coordinate axis are simply
+straight lines and perpendicular to each other. Due to this, it is
+very easy to do calculations in Cartesian coordinates. For example,
+the distance $\Delta s$ between two points $(x_1, x_2, \ldots, x_n)$
+and $(x'_1, x'_2, \ldots, x'_n)$ is easily computed as
+
 $$\Delta s^2 = (x'_1 - x_1)^2 + (x'_2 - x_2)^2 + \ldots + (x'_n - x_n)^2.$$
-The expression on the right hand side can be written using a summation
-sign $\sum$: $$\Delta s^2 = \sum_{i=1}^n (x'_i - x_i)^2.$$ A space with
-such a distance definition is called a *Euclidean space*.
+
+(A space with such a distance definition is called a *Euclidean
+space*.)
+
+In mathematics, we are often dealing with so-called *infinitesimally* small
+distances, for example in the definition of derivatives and integrals.
+In Cartesion coordinates the expressions for infinitesimal distances $ds$ and
+infinitesimal volumes $dV$ are given as
+
+$$ds = \sqrt{dx_1^2 + dx_2^2 + \ldots + dx_n^2}$$
+
+and
+
+$$dV = dx_1 dx_2 \ldots dx_N.$$
+
+The formula for $dV$ also indicates that in Cartesian coordinates, the integral
+over a volume can be expressed as individual integrals over all coordinate directions:
+$\int dV = \idotsint dx_1 dx_2 \ldots dx_N$.
 
 Cartesian coordinates are used a lot. They are particularly suitable for
 infinite spaces or for rectangular volumes.
@@ -24,8 +40,9 @@ infinite spaces or for rectangular volumes.
 ![image](figures/Coordinates_5_1.svg)
 
 
-## Polar coordinates
+# Polar coordinates
 
+## Definition
 
 It often turns out useful to change to a different type of coordinate
 system. For example, if you want to describe the vibrations of a
@@ -48,13 +65,32 @@ from the angular one ($0^\circ$, $45^\circ$ etc.).
 The plot below shows a point on a curve with the polar coordinates
 $(r,\varphi)$ indicated. From this, we can see that the *Cartesian*
 coordinates $(x,y)$ of the point are related to the polar ones as
-follows: $$x = r \cos\varphi;$$ $$y = r \sin \varphi.$$
+follows:
+
+$$x = r \cos\varphi;$$
+$$y = r \sin \varphi.$$
 
 ![image](figures/Coordinates_9_0.svg)
 
+The inverse relation is given as
+
+$$r=\sqrt{x^2 + y^2};$$
+$$\varphi=\begin{cases}
+  \arctan(y/x) & \text{$x>0$,}\\
+  \pi + \arctan(y/x) & \text{$x<0$ and $y>0$,}\\
+  -\pi + \arctan(y/x) & \text{$x<0$ and $y<0$.}
+  \end{cases}$$
+
+The last formula for $\varphi$ warrants a closer explanation: It is easy
+to see that $\tan(\varphi)=y/x$ - but this is not a unique relation, due to
+the fact that the $\tan$ has different branches. Convince yourself that
+the expression above is correct for all the four sectors!
+
+## Distances and areas
+
 Now suppose we want to calculate the distance between two points, one
 with polar coordinates $(r_1, \varphi_1)$, and the other with
-$(r_2, \varphi_2)$. This looks like a difficult exercise. A convenient
+$(r_2, \varphi_2)$. This looks like a difficult exercise. One possible
 way to perform this is by translating the polar coordinates into
 Cartesian coordinates and using the expression given above for this
 distance: $$\Delta s^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2,$$ so
@@ -72,32 +108,36 @@ arc of radius $r_1$ and then we move a small distance radially outward
 from $r_1$ to $r_2$. Provided the difference between the angles
 $\varphi_1$ and $\varphi_2$ is (very) small, these paths are
 approximately perpendicular and we can use Pythagoras’ theorem to find
-the distance $\Delta s$. Note that the arc is approximately straight –
-it has a length $r_1 \Delta \varphi$, where
-$\Delta \varphi = \varphi_2-\varphi_1$. So we have:
-$$\Delta s^2 = (\Delta r)^2 + (arc~length)^2   = (\Delta r)^2 + r_1^2 (\Delta \varphi)^2 .$$
-For $r\equiv r_1 \approx r_2$, replacing the $\Delta$’s by $d$’s to
-emphasise the infinitesimal character of the differences, we finally
-have: $$ds^2 = dr^2+ r^2 d\varphi^2 .$$ You should recall this formula!
-
-If we traverse a curved path in the plane, we must sum up all the $ds$,
-which actually boils down to calculating an integral:
-$$Path~length = \int_{start}^{end} ds = \int_{start}^{end} \sqrt{dr^2 + r^2 d\varphi^2}.$$
-Here, the boundaries $start$ and $end$ stand for the starting and end
-point of the path.
-
-Note that the path can often be written in the form $r(\varphi)$,
-i.e. we write $r$ as a function of $\varphi$. We can then write the
-integral as:
-$$Path~length = \int_{start}^{end} \sqrt{dr^2 + r^2 d\varphi^2} = \int_{\varphi_{start}}^{\varphi_{end}} \sqrt{\left(\frac{dr}{d\varphi}\right)^2 + r^2} \; d\varphi.$$
-(Note: this parametrisation is strictly speaking only possible when $r$
-is a uni-valued function of $\varphi$; a meandering path generally does
-not have this property.)
-
-**Exercise:** write the same expression as an integral over $dr$ instead
-of $d\varphi$. Please sketch two paths, one which is more suitable for
-the expression above (the integral over $\varphi$) and the other more
-convenient for the expression you obtained in this problem.
+the distance $d s$. Note that the arc is approximately straight –
+it has a length $r_1 d \varphi$, where
+$d \varphi = \varphi_2-\varphi_1$. So we have:
+$$d s^2 = (d r)^2 + (arc~length)^2   = (d r)^2 + r_1^2 (d \varphi)^2 .$$
+
+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:
+
+$$dA = r dr d\varphi.$$
+
+This is an important formula to remember for integrating in polar
+coordinates!  The extra $r$ that appears here can be intuitively
+understood: the area swept by an angle difference $d\varphi$
+*increases* as we move further away from the origin.
+
+!!! check "Example: Integrating over a circular area"
+    To check the area element we just derived, let us compute a simple
+    integral. We compute the integral over a circle with radius $r_0$
+    with a very simple function that equals to $1$. In this case,
+    we expect to get as a result the are of the region we integrate over.
+
+    We find:
+
+    $$
+    \int_0^{2\pi} d\varphi \int_0^r_0 r dr = \\
+    2\pi \int_0^r_0 r dr = 2 \pi \frac{1}{2} r_0^2 = \pi r_0^2,
+    $$
+
+    which is indeed the area of a circle with radius 0.
 
 ## Cylindrical coordinates
 
@@ -111,10 +151,10 @@ choose the coordinate system ourselves - this is not imposed by the
 problem.
 
 Cylindrical coordinates are defined straightforwardly: we use polar
-coordinates $\rho$ and $\varphi$ in the $xy$ plane, and the distance $z$
+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
 chosen in physical space, we have two coordinates which have the
-dimension of a distance: $\rho$ and $z$. The other coordinate,
+dimension of a distance: $r$ and $z$. The other coordinate,
 $\varphi$, is of course dimensionless.
 
 What is the distance travelled along a path when we express this in
@@ -122,15 +162,15 @@ cylindrical coordinates? Let’s consider an example (Figure).
 
 ![image](figures/Coordinates_13_0.svg)
 
-We want to find the length of the (small) red segment $\Delta s$. By
+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 $\Delta l$ is perpendicular to the vertical segment
-$dz$. Using for $\Delta l$ the length we obtained before for a line
+$xy$-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
 immediately find:
-$$\Delta s^2 = \Delta l^2 + \Delta z^2 = \Delta \rho^2 + \rho^2 \Delta \varphi^2 + \Delta z^2.$$
-Using $d$ instead of $\Delta$ for infinitesimal changes, this reads:
-$$ds^2 = dl^2 + dz^2 = d\rho^2 + \rho^2 d\varphi^2 + dz^2.$$
+$$d s^2 = d l^2 + d z^2 = d r^2 + r^2 d \varphi^2 + d z^2.$$
+The volume element is consequently given as
+$$dV = r dr d\varphi dz.$$
 
 ## Spherical coordinates
 
@@ -147,12 +187,11 @@ The position of a point on the sphere is specified using the two angles
 $\theta$ and $\phi$ indicated in the figure.
 
 !!! warning
-    Note that in mathematics, usually the angles are labelled the other way
+    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
     the angle of the projection of that line with the $x$-axis. The
-    convention used here is custom in physics. IT IS IMPORTANT TO BE AWARE
-    OF THIS!!
+    convention used here is custom in physics. 
 
 The relation between Cartesian and coordinates is defined by
 $$x = r \cos \varphi \sin \vartheta$$
@@ -184,6 +223,10 @@ displacement.
 
 ![image](figures/Coordinates_19_0.svg)
 
+From these arguments we can again also find the volume element, it is
+here given as
+$$dV = r^2 \sin\theta dr d\theta d\varphi.$$
+
 ### summary
 
 We have discussed four different coordinate systems:
@@ -201,12 +244,14 @@ We have discussed four different coordinate systems:
     given in terms of these coordinates.
 
     Infinitesimal distance: $$ds^2 = dr^2 + r^2 d\phi^2.$$
+    Infinitesimal area: $$dA = r dr d\varphi.$$
 
-3.  *Cylindrical coordinates*: $${\bf r} = (\rho, \phi, z).$$ Can be
+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.
 
-    Infinitesimal distance: $$ds^2 = d\rho^2 + \rho^2 d\phi^2 + dz^2.$$
+    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
@@ -214,6 +259,8 @@ We have discussed four different coordinate systems:
 
     Infinitesimal distance:
     $$ds^2 =r^2 (\sin^2 \theta d\phi^2 + d\theta^2) +  dr^2 .$$
+    Infinitesimal volume:
+    $$dV = r^2 \sin(\theta) dr d\theta d\varphi.$$ 
 
 Problems
 ========
@@ -248,11 +295,3 @@ Problems
     \frac{\partial}{\partial \vartheta}\left( \sin\vartheta \frac{\partial\psi(r,\vartheta, \varphi)}{\partial \vartheta}\right).$$
     This is however even more tedious (you do not have to show this).
 
-4.  [:smirk:] A particle moves over a cone which is characterized
-    by the condition $rho = z$. Calculate the kinetic energy expressed
-    in terms of the cylindrical coordinates $z$ and $\varphi$ and their
-    first order time derivatives.
-
-5.  [:smirk:] The same question for a particle moving on a
-    spherical shell of radius $R$. (Use coordinates $\varphi$ and
-    $\vartheta$.)