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First major update of src/2_coordinates.md

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@@ -29,7 +29,7 @@ and $(x'_1, x'_2, \ldots, x'_n)$ can be quickly computed using a general formula
$$\Delta s^2 = (x'_1 - x_1)^2 + (x'_2 - x_2)^2 + \ldots + (x'_n - x_n)^2.$$
(A space with such a distance definition is called a *Euclidean
(A space with such a distance definition is called an *Euclidean
space*.)
In mathematics, we are often dealing with so-called *infinitesimally* small
@@ -70,7 +70,7 @@ the angular coordinate $\varphi$ is dimensionless.
<figure markdown>
![image](figures/Coordinates_7_0.svg)
<figcaption>In this example of a polar plot, you can distinguish the radial coordinate (0.2, 0.4 etc.) \\from the angular one expressed in degrees ($0^\circ$, $45^\circ$ etc.).</figcaption>
<figcaption>In this example of a polar plot, you can distinguish the radial coordinate (0.2, 0.4 etc.) from the angular one expressed in degrees ($0^\circ$, $45^\circ$ etc.).</figcaption>
</figure>
@@ -87,10 +87,10 @@ $$\begin{equation} y = r \sin \varphi.\end{equation}$$
<figcaption></figcaption>
</figure>
The inverse relation is given as:
#### The inverse relation
!!! info "Inverse relation between polar and Cartesian coordinate systems"
$$\begin{equation} r=\sqrt{x^2 + y^2}; \label{rxy}\end{equation}$$
\begin{equation} r=\sqrt{x^2 + y^2}; \label{rxy}\end{equation}
\begin{equation} \varphi=\begin{cases}
\arctan(y/x) & \text{$x>0$,}\\
\pi + \arctan(y/x) & \text{$x<0$ and $y>0$,}\\
@@ -117,7 +117,10 @@ If we consider two points which are *very close*, the analysis
simplifies however. We can use the geometry of the problem to find the
distance (see the figure below).
![image](figures/Coordinates_11_0.svg)
<figure markdown>
![image](figures/Coordinates_11_0.svg)
<figcaption></figcaption>
</figure>
When going from point 1 to point 2, we first traverse a small circular
arc of radius $r_1$ and then we move a small distance radially outward
@@ -161,7 +164,7 @@ understood: the area swept by an angle difference $d\varphi$
<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/NGQWGx71w98?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
Often, in physics important equations involve derivatives given in terms
Important equations in physics often involve derivatives given in terms
of Cartesian coordinates. One prominent example are equations of the form
$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)
f(x, y) = \ldots.$$
@@ -169,7 +172,7 @@ The derivative operator $\left(\frac{\partial^2}{\partial x^2} +
\frac{\partial^2}{\partial y^2}\right)$ is so common it has its own name:
the Laplacian (here for two-dimensional space).
Such an equation is universal, but for particular situations it might be
This equation is universal, but for particular situations it might be
advantageous to use a different coordinate system, such as polar coordinates
for a system with rotational symmetry. The question then is: How does the
corresponding equation look like in a different coordinate system?
@@ -178,7 +181,7 @@ There are different ways to find the answer. Here, we will focus on
directly deriving the transformed equation through an explicit calculation
involving the chain rule for a function of several variables.
!!! info Chain rule for a multi-variable function
!!! info "Chain rule for a multi-variable function"
Let $f$ be a function of $n$ variables: $f(y_1, y_2, \ldots, y_n)$,
as well as $g_i(x_1, x_2, \ldots, x_n)$ for $i=1,2,\ldots, n$. Then
@@ -192,7 +195,7 @@ of a function of $r, \varphi$ in terms of $x$?
For this, we need to realize that there are relations between the coordinate systems.
In particular, $r = r(x,y)$ and $\varphi = \varphi(x, y)$ as defined in equations
\ref{rxy} and \ref{phixy}. In fact, we have been rather sloppy in our notation above,
of [the inverse relations](#the-inverse-relation). In fact, we have been rather sloppy in our notation above,
as the functions $f(x,y)$ and $f(r, \varphi)$ do not mean that I substitute $x=r$
and $y=\varphi$! It is more precise to state that there are two diferent
functions $f_\text{cart}(x,y)$ and $f_\text{polar}(r, \varphi)$ that are equivalent,
@@ -211,22 +214,22 @@ $$ \frac{\partial}{\partial x} f(r, \varphi) =
$$
and it is now a matter of (tedious) calculus to arrive at the right result.
This is the task of exercises 3 and 4, which finally compute the Laplacian
This is the task of exercises 3 and 4, which lead you to compute the Laplacian
in polar coordinates.
!!! warning Inverse function theorem
In this calculation one might be tempted to use the inverse
!!! warning "Inverse function theorem"
In this calculation, one might be tempted to use the inverse
function theorem to compute derivatives like
$\frac{\partial \varphi}{\partial x}$ from the much simpler
$\frac{\partial x}{\partial \varphi}$. Note though that here we
are dealing with functions depending on several variables, so the
*Jacobian* has to be used (see [Wikipedia](https://en.wikipedia.org/wiki/Inverse_function_theorem)). A direct calculation is in this particular case more easy.
$\frac{\partial x}{\partial \varphi}$. However, note that here we
are dealing with functions depending on several variables, so an appropriate
*Jacobian* has to be used (see [Wikipedia](https://en.wikipedia.org/wiki/Inverse_function_theorem)). A direct calculation is in this particular case considerably easier.
Note that this procedure also carries over to other coordinate systems,
although the calculations can become quite tedious. In these cases,
it's usually best to look up the correct form.
Note that this procedure also applies to transformations to other coordinate systems,
although the calculations can become quite tedious. In conventional cases,
it is usually advised to look up the correct form.
## Coordinate systems in 3D
## 2.3. Coordinate systems in 3D
<iframe width="100%" height=315 src="https://www.youtube-nocookie.com/embed/VjUbnZN1BvA?rel=0" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
@@ -236,10 +239,10 @@ it's usually best to look up the correct form.
Three dimensional systems may have axial symmetry. An example is an
electrically charged wire of which we would like to calculate the
electric field, or a current-carrying wire for which we would like to
calculate the magnetic field. For such problems the most convenient
coordinates are *cylindrical coordinates*. For convenience, we choose
the symmetry-axis as the $z$-axis. Note that this can be done as we can
choose the coordinate system ourselves - this is not imposed by the
calculate the magnetic field. For such problems, the most convenient
coordinates are *cylindrical coordinates*. For a further convenience, we choose
the symmetry-axis as the $z$-axis. Note that this allowed, because we may
choose the coordinate system ourselves - it is not imposed by the
problem.
Cylindrical coordinates are defined straightforwardly: we use polar
@@ -247,22 +250,27 @@ 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: $r$ and $z$. The other coordinate,
$\varphi$, is of course dimensionless.
$\varphi$ is of course dimensionless.
What is the distance travelled along a path when we express this in
cylindrical coordinates? Let’s consider an example (Figure).
cylindrical coordinates? Let’s consider an example shown in the figure below.
![image](figures/Coordinates_13_0.svg)
<figure markdown>
![image](figures/Coordinates_13_0.svg)
<figcaption></figcaption>
</figure>
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
$$dV = r dr d\varphi dz.$$
The volume element is consequently given as:
!!! info "Volume element in cylindrical coordinates"
$$dV = r dr d\varphi dz.$$
### Spherical coordinates
@@ -273,29 +281,39 @@ 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)
<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
@@ -313,7 +331,11 @@ $$ds^2 = r^2 \left(\sin^2 \vartheta d\varphi^2 + d\vartheta^2\right) + dr^2.$$
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