1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
|
---
title: "Parabolic cylindrical coordinates"
firstLetter: "P"
publishDate: 2021-03-04
categories:
- Mathematics
- Physics
date: 2021-03-04T15:07:46+01:00
draft: false
markup: pandoc
---
# Parabolic cylindrical coordinates
**Parabolic cylindrical coordinates** are a coordinate system
that describes a point in space using three coordinates $(\sigma, \tau, z)$.
The $z$-axis is unchanged from the Cartesian system,
hence it is called a *cylindrical* system.
In the $z$-isoplane, however, confocal parabolas are used.
These coordinates can be converted to the Cartesian $(x, y, z)$ as follows:
$$\begin{aligned}
\boxed{
x = \frac{1}{2} (\tau^2 - \sigma^2 )
\qquad
y = \sigma \tau
\qquad
z = z
}
\end{aligned}$$
Converting the other way is a bit trickier.
It can be done by solving the following equations,
and potentially involves some fiddling with signs:
$$\begin{aligned}
2 x
= \frac{y^2}{\sigma^2} - \sigma^2
\qquad \quad
2 x
= - \frac{y^2}{\tau^2} + \tau^2
\end{aligned}$$
Parabolic cylindrical coordinates form an orthogonal
[curvilinear](/know/concept/curvilinear-coordinates/) system,
so we would like to find its scale factors $h_\sigma$, $h_\tau$ and $h_z$.
The differentials of the Cartesian coordinates are as follows:
$$\begin{aligned}
\dd{x} = - \sigma \dd{\sigma} + \tau \dd{\tau}
\qquad
\dd{y} = \tau \dd{\sigma} + \sigma \dd{\tau}
\qquad
\dd{z} = \dd{z}
\end{aligned}$$
We calculate the line segment $\dd{\ell}^2$,
skipping many terms thanks to orthogonality:
$$\begin{aligned}
\dd{\ell}^2
&= (\sigma^2 + \tau^2) \:\dd{\sigma}^2 + (\tau^2 + \sigma^2) \:\dd{\tau}^2 + \dd{z}^2
\end{aligned}$$
From this, we can directly read off the scale factors $h_\sigma^2$, $h_\tau^2$ and $h_z^2$,
which turn out to be:
$$\begin{aligned}
\boxed{
h_\sigma = \sqrt{\sigma^2 + \tau^2}
\qquad
h_\tau = \sqrt{\sigma^2 + \tau^2}
\qquad
h_z = 1
}
\end{aligned}$$
With these scale factors, we can use
the general formulae for orthogonal curvilinear coordinates
to easily to convert things from the Cartesian system.
The basis vectors are:
$$\begin{aligned}
\boxed{
\begin{aligned}
\vu{e}_\sigma
&= \frac{- \sigma}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_x + \frac{\tau}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_y
\\
\vu{e}_\tau
&= \frac{\tau}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_x + \frac{\sigma}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_y
\\
\vu{e}_z
&= \vu{e}_z
\end{aligned}
}
\end{aligned}$$
The basic vector operations (gradient, divergence, Laplacian and curl) are given by:
$$\begin{aligned}
\boxed{
\nabla f
= \frac{\mathbf{e}_\sigma}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\sigma}
+ \frac{\mathbf{e}_\tau}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\tau}
+ \mathbf{e}_z \pdv{f}{z}
}
\end{aligned}$$
$$\begin{aligned}
\boxed{
\nabla \cdot \mathbf{V}
= \frac{1}{\sigma^2 + \tau^2}
\Big( \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{d\sigma} + \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{d\tau} \Big) + \pdv{V_z}{z}
}
\end{aligned}$$
$$\begin{aligned}
\boxed{
\nabla^2 f
= \frac{1}{\sigma^2 + \tau^2} \Big( \pdv[2]{f}{\sigma} + \pdv[2]{f}{\tau} \Big) + \pdv[2]{f}{z}
}
\end{aligned}$$
$$\begin{aligned}
\boxed{
\begin{aligned}
\nabla \times \mathbf{V}
&= \mathbf{e}_\sigma \Big( \frac{\mathbf{e}_1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\tau} - \pdv{V_\tau}{z} \Big)
\\
&+ \mathbf{e}_\tau \Big( \pdv{V_\sigma}{z} - \frac{1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\sigma} \Big)
\\
&+ \frac{\mathbf{e}_z}{\sigma^2 + \tau^2}
\Big( \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{\sigma} - \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{\tau} \Big)
\end{aligned}
}
\end{aligned}$$
The differential element of volume $\dd{V}$
in parabolic cylindrical coordinates is given by:
$$\begin{aligned}
\boxed{
\dd{V} = (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau} \dd{z}
}
\end{aligned}$$
The differential elements of the isosurfaces are as follows,
where $\dd{S_\sigma}$ is the $\sigma$-isosurface, etc.:
$$\begin{aligned}
\boxed{
\begin{aligned}
\dd{S_\sigma} &= \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z}
\\
\dd{S_\tau} &= \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z}
\\
\dd{S_z} &= (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau}
\end{aligned}
}
\end{aligned}$$
The normal element $\dd{\vu{S}}$ of a surface and
the tangent element $\dd{\vu{\ell}}$ of a curve are respectively:
$$\begin{aligned}
\boxed{
\dd{\vu{S}}
= \mathbf{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z}
+ \mathbf{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z}
+ \mathbf{e}_z (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau}
}
\end{aligned}$$
$$\begin{aligned}
\boxed{
\dd{\vu{\ell}}
= \mathbf{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\sigma}
+ \mathbf{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\tau}
+ \mathbf{e}_z \dd{z}
}
\end{aligned}$$
## References
1. M.L. Boas,
*Mathematical methods in the physical sciences*, 2nd edition,
Wiley.
|
