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---
title: "Cylindrical parabolic coordinates"
date: 2021-03-04
categories:
- Mathematics
- Physics
layout: "concept"
---
**Cylindrical parabolic 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:
$$\begin{aligned}
2 x
= \frac{y^2}{\sigma^2} - \sigma^2
\qquad \qquad
2 x
= - \frac{y^2}{\tau^2} + \tau^2
\end{aligned}$$
Cylindrical parabolic coordinates form an orthogonal
[curvilinear system](/know/concept/curvilinear-coordinates/),
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:
$$\begin{aligned}
\boxed{
\nabla f
= \frac{\vu{e}_\sigma}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\sigma}
+ \frac{\vu{e}_\tau}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\tau}
+ \vu{e}_z \pdv{f}{z}
}
\end{aligned}$$
$$\begin{aligned}
\boxed{
\nabla \cdot \vb{V}
= \frac{1}{\sigma^2 + \tau^2}
\Big( \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{\sigma} + \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{\tau} \Big) + \pdv{V_z}{z}
}
\end{aligned}$$
$$\begin{aligned}
\boxed{
\nabla^2 f
= \frac{1}{\sigma^2 + \tau^2} \Big( \pdvn{2}{f}{\sigma} + \pdvn{2}{f}{\tau} \Big) + \pdvn{2}{f}{z}
}
\end{aligned}$$
$$\begin{aligned}
\boxed{
\begin{aligned}
\nabla \times \vb{V}
&= \vu{e}_\sigma \Big( \frac{\vu{e}_1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\tau} - \pdv{V_\tau}{z} \Big)
\\
&+ \vu{e}_\tau \Big( \pdv{V_\sigma}{z} - \frac{1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\sigma} \Big)
\\
&+ \frac{\vu{e}_z}{\sigma^2 + \tau^2}
\Big( \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{\sigma} - \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{\tau} \Big)
\end{aligned}
}
\end{aligned}$$
The differential element of volume $\dd{V}$
in cylindrical parabolic 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:
$$\begin{aligned}
\boxed{
\dd{\vu{S}}
= \vu{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z}
+ \vu{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z}
+ \vu{e}_z (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau}
}
\end{aligned}$$
$$\begin{aligned}
\boxed{
\dd{\vu{\ell}}
= \vu{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\sigma}
+ \vu{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\tau}
+ \vu{e}_z \dd{z}
}
\end{aligned}$$
## References
1. M.L. Boas,
*Mathematical methods in the physical sciences*, 2nd edition,
Wiley.
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