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+---
+title: "Cylindrical polar coordinates"
+date: 2021-07-26
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+**Cylindrical polar coordinates** are an extension of polar coordinates to 3D,
+which describes the location of a point in space
+using the coordinates $(r, \varphi, z)$.
+The $z$-axis is unchanged from Cartesian coordinates,
+hence it is called a *cylindrical* system.
+
+Cartesian coordinates $(x, y, z)$
+and the cylindrical system $(r, \varphi, z)$ are related by:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            x &= r \cos\varphi \\
+            y &= r \sin\varphi \\
+            z &= z
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Conversely, a point given in $(x, y, z)$
+can be converted to $(r, \varphi, z)$
+using these formulae:
+
+$$\begin{aligned}
+    \boxed{
+        r = \sqrt{x^2 + y^2}
+        \qquad
+        \varphi = \mathtt{atan2}(y, x)
+        \qquad
+        z = z
+    }
+\end{aligned}$$
+
+The cylindrical polar coordinates form an orthogonal
+[curvilinear system](/know/concept/curvilinear-coordinates/),
+whose scale factors $h_r$, $h_\varphi$ and $h_z$ we want to find.
+To do so, we calculate the differentials of the Cartesian coordinates:
+
+$$\begin{aligned}
+    \dd{x} = \dd{r} \cos\varphi - \dd{\varphi} r \sin\varphi
+    \qquad
+    \dd{y} = \dd{r} \sin\varphi + \dd{\varphi} r \cos\varphi
+    \qquad
+    \dd{z} = \dd{z}
+\end{aligned}$$
+
+And then we calculate the line element $\dd{\ell}^2$,
+skipping many terms thanks to orthogonality,
+
+$$\begin{aligned}
+    \dd{\ell}^2
+    &= \dd{r}^2 \big( \cos^2(\varphi) + \sin^2(\varphi) \big)
+    + \dd{\varphi}^2 \big( r^2 \sin^2(\varphi) + r^2 \cos^2(\varphi) \big)
+    + \dd{z}^2
+    \\
+    &= \dd{r}^2 + r^2 \: \dd{\varphi}^2 + \dd{z}^2
+\end{aligned}$$
+
+Finally, we can simply read off
+the squares of the desired scale factors
+$h_r^2$, $h_\varphi^2$ and $h_z^2$:
+
+$$\begin{aligned}
+    \boxed{
+        h_r = 1
+        \qquad
+        h_\varphi = r
+        \qquad
+        h_z = 1
+    }
+\end{aligned}$$
+
+With these factors, we can easily convert things from the Cartesian system
+using the standard formulae for orthogonal curvilinear coordinates.
+The basis vectors are:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \vu{e}_r
+            &= \cos\varphi \:\vu{e}_x + \sin\varphi \:\vu{e}_y
+            \\
+            \vu{e}_\varphi
+            &= - \sin\varphi \:\vu{e}_x + \cos\varphi \:\vu{e}_y
+            \\
+            \vu{e}_z
+            &= \vu{e}_z
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The basic vector operations (gradient, divergence, Laplacian and curl) are given by:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla f
+        = \vu{e}_r \pdv{f}{r}
+        + \vu{e}_\varphi \frac{1}{r} \pdv{f}{\varphi}
+        + \mathbf{e}_z \pdv{f}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{V}
+        = \frac{1}{r} \pdv{(r V_r)}{r}
+        + \frac{1}{r} \pdv{V_\varphi}{\varphi}
+        + \pdv{V_z}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \nabla^2 f
+        = \frac{1}{r} \pdv{}{r}\Big( r \pdv{f}{r} \Big)
+        + \frac{1}{r^2} \pdvn{2}{f}{\varphi}
+        + \pdvn{2}{f}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \nabla \times \vb{V}
+            &= \vu{e}_r \Big( \frac{1}{r} \pdv{V_z}{\varphi} - \pdv{V_\varphi}{z} \Big)
+            \\
+            &+ \vu{e}_\varphi \Big( \pdv{V_r}{z} - \pdv{V_z}{r} \Big)
+            \\
+            &+ \frac{\vu{e}_z}{r} \Big( \pdv{(r V_\varphi)}{r} - \pdv{V_r}{\varphi} \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The differential element of volume $\dd{V}$
+takes the following form:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{V}
+        = r \dd{r} \dd{\varphi} \dd{z}
+    }
+\end{aligned}$$
+
+So, for example, an integral over all of space is converted like so:
+
+$$\begin{aligned}
+    \iiint_{-\infty}^\infty f(x, y, z) \dd{V}
+    = \int_{-\infty}^{\infty} \int_0^{2\pi} \int_0^\infty f(r, \varphi, z) \: r \dd{r} \dd{\varphi} \dd{z}
+\end{aligned}$$
+
+The isosurface elements are as follows, where $S_r$ is a surface at constant $r$, etc.:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \dd{S}_r = r \dd{\varphi} \dd{z}
+            \qquad
+            \dd{S}_\varphi = \dd{r} \dd{z}
+            \qquad
+            \dd{S}_z = r \dd{r} \dd{\varphi}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Similarly, the normal vector element $\dd{\vu{S}}$ for an arbitrary surface is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{S}}
+        = \vu{e}_r \: r \dd{\varphi} \dd{z}
+        + \vu{e}_\varphi \dd{r} \dd{z}
+        + \vu{e}_z \: r \dd{r} \dd{\varphi}
+    }
+\end{aligned}$$
+
+And finally, the tangent vector element $\dd{\vu{\ell}}$ of a given curve is as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{\ell}}
+        = \vu{e}_r \dd{r}
+        + \vu{e}_\varphi \: r \dd{\varphi}
+        + \vu{e}_z \dd{z}
+    }
+\end{aligned}$$
+
+
+## References
+1.  M.L. Boas,
+    *Mathematical methods in the physical sciences*, 2nd edition,
+    Wiley.