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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.