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First major update of src/2_coordinates.md

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@@ -351,35 +351,33 @@ We have discussed four different coordinate systems:
$${\bf r} = (x_1, \ldots, x_n).$$ 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*: $${\bf r} = (r, \phi).$$ This system can be used in two
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.
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
1. [:grinning:] *Warm-up*
@@ -404,7 +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 $\frac{\partial}{\partial z} f(r, \varphi, \theta)$.
@@ -414,26 +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
@@ -445,7 +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)
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