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First major update of src/2_coordinates.md

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@@ -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
@@ -264,12 +264,12 @@ 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"
!!! info "Infinitesimal volume element in cylindrical coordinates"
$$dV = r dr d\varphi dz.$$
### Spherical coordinates
@@ -281,91 +281,106 @@ 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
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
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
!!! 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:: $$dV = r dr d\varphi dz.$$
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
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.
Infinitesimal distance:
Infinitesimal distance: <br/>
$$ds^2 =r^2 (\sin^2 \theta d\phi^2 + d\theta^2) + dr^2 .$$
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 $${\bf r} = \sqrt{2} (1,1).$$
@@ -387,7 +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 $\frac{\partial}{\partial z} f(r, \varphi, \theta)$.
@@ -397,26 +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
@@ -428,7 +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)