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----
-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.