From e12c7ce372ecaa042d85d9fb76371a75ff518d1a Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Wed, 28 Jul 2021 14:27:37 +0200
Subject: Expand knowledge base, fix a:visited CSS

---
 .../know/concept/curvilinear-coordinates/index.pdc |   4 +-
 .../cylindrical-parabolic-coordinates/index.pdc    | 188 +++++++++++++++++++
 .../cylindrical-polar-coordinates/index.pdc        | 206 +++++++++++++++++++++
 .../parabolic-cylindrical-coordinates/index.pdc    | 188 -------------------
 .../concept/path-integral-formulation/index.pdc    |   8 +-
 .../concept/rayleigh-plesset-equation/index.pdc    |  82 +++++---
 .../know/concept/spherical-coordinates/index.pdc   |   9 +-
 7 files changed, 458 insertions(+), 227 deletions(-)
 create mode 100644 content/know/concept/cylindrical-parabolic-coordinates/index.pdc
 create mode 100644 content/know/concept/cylindrical-polar-coordinates/index.pdc
 delete mode 100644 content/know/concept/parabolic-cylindrical-coordinates/index.pdc

(limited to 'content/know')

diff --git a/content/know/concept/curvilinear-coordinates/index.pdc b/content/know/concept/curvilinear-coordinates/index.pdc
index 925eda3..62efe39 100644
--- a/content/know/concept/curvilinear-coordinates/index.pdc
+++ b/content/know/concept/curvilinear-coordinates/index.pdc
@@ -45,8 +45,8 @@ because the coordinate system is orthogonal by assumption.
 
 Examples of orthogonal curvilinear coordinate systems include
 [spherical coordinates](/know/concept/spherical-coordinates/),
-cylindrical coordinates,
-and [parabolic cylindrical coordinates](/know/concept/parabolic-cylindrical-coordinates/).
+[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/),
+and [cylindrical parabolic coordinates](/know/concept/cylindrical-parabolic-coordinates/).
 
 In the following subsections,
 we derive general formulae to convert expressions
diff --git a/content/know/concept/cylindrical-parabolic-coordinates/index.pdc b/content/know/concept/cylindrical-parabolic-coordinates/index.pdc
new file mode 100644
index 0000000..3460de4
--- /dev/null
+++ b/content/know/concept/cylindrical-parabolic-coordinates/index.pdc
@@ -0,0 +1,188 @@
+---
+title: "Cylindrical parabolic coordinates"
+firstLetter: "C"
+publishDate: 2021-03-04
+categories:
+- Mathematics
+- Physics
+
+date: 2021-03-04T15:07:46+01:00
+draft: false
+markup: pandoc
+---
+
+# Cylindrical parabolic coordinates
+
+**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:
+
+$$\begin{aligned}
+    2 x
+    = \frac{y^2}{\sigma^2} - \sigma^2
+    \qquad \quad
+    2 x
+    = - \frac{y^2}{\tau^2} + \tau^2
+\end{aligned}$$
+
+Cylindrical parabolic 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 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:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{S}}
+        = \mathbf{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z}
+        + \mathbf{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z}
+        + \mathbf{e}_z (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{\ell}}
+        = \mathbf{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\sigma}
+        + \mathbf{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\tau}
+        + \mathbf{e}_z \dd{z}
+    }
+\end{aligned}$$
+
+
+## References
+1.  M.L. Boas,
+    *Mathematical methods in the physical sciences*, 2nd edition,
+    Wiley.
diff --git a/content/know/concept/cylindrical-polar-coordinates/index.pdc b/content/know/concept/cylindrical-polar-coordinates/index.pdc
new file mode 100644
index 0000000..6242b9f
--- /dev/null
+++ b/content/know/concept/cylindrical-polar-coordinates/index.pdc
@@ -0,0 +1,206 @@
+---
+title: "Cylindrical polar coordinates"
+firstLetter: "C"
+publishDate: 2021-07-26
+categories:
+- Mathematics
+- Physics
+
+date: 2021-07-26T16:08:46+02:00
+draft: false
+markup: pandoc
+---
+
+# Cylindrical polar coordinates
+
+**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](/know/concept/curvilinear-coordinates/) system,
+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:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla f
+        = \vu{e}_r \pdv{f}{r}
+        + \vu{e}_\varphi \frac{1}{r} \pdv{f}{\varphi}
+        + \mathbf{e}_z \pdv{f}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{V}
+        = \frac{1}{r} \pdv{(r V_r)}{r}
+        + \frac{1}{r} \pdv{V_\varphi}{\varphi}
+        + \pdv{V_z}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \nabla^2 f
+        = \frac{1}{r} \pdv{r} \Big( r \pdv{f}{r} \Big)
+        + \frac{1}{r^2} \pdv[2]{f}{\varphi}
+        + \pdv[2]{f}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \nabla \times \vb{V}
+            &= \vu{e}_r \Big( \frac{1}{r} \pdv{V_z}{\varphi} - \pdv{V_\varphi}{z} \Big)
+            \\
+            &+ \vu{e}_\varphi \Big( \pdv{V_r}{z} - \pdv{V_z}{r} \Big)
+            \\
+            &+ \frac{\vu{e}_\varphi}{r} \Big( \pdv{(r V_\varphi)}{r} - \pdv{V_r}{\varphi} \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+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.
diff --git a/content/know/concept/parabolic-cylindrical-coordinates/index.pdc b/content/know/concept/parabolic-cylindrical-coordinates/index.pdc
deleted file mode 100644
index 56544ae..0000000
--- a/content/know/concept/parabolic-cylindrical-coordinates/index.pdc
+++ /dev/null
@@ -1,188 +0,0 @@
----
-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.
diff --git a/content/know/concept/path-integral-formulation/index.pdc b/content/know/concept/path-integral-formulation/index.pdc
index c66aed8..1698e27 100644
--- a/content/know/concept/path-integral-formulation/index.pdc
+++ b/content/know/concept/path-integral-formulation/index.pdc
@@ -27,7 +27,7 @@ to the destination $x_N$ at time $t_N$:
 $$\begin{aligned}
     \boxed{
         K(x_N, t_N; x_0, t_0)
-        = A \sum_{\mathrm{all}\:x(t)} \exp(i S[x] / \hbar)
+        = A \sum_{\mathrm{all}\:x(t)} \exp\!(i S[x] / \hbar)
     }
 \end{aligned}$$
 
@@ -156,7 +156,7 @@ The definition of the propagator $K$ is then further reduced to the following:
 $$\begin{aligned}
     K
     = \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2}
-    \idotsint \exp(i S[x] / \hbar) \dd{x_1} \cdots \dd{x_{N-1}}
+    \idotsint \exp\!(i S[x] / \hbar) \dd{x_1} \cdots \dd{x_{N-1}}
 \end{aligned}$$
 
 Finally, for the purpose of normalization,
@@ -173,8 +173,8 @@ which sums over all possible paths $x(t)$:
 
 $$\begin{aligned}
     K
-    = \int \exp(i S[x] / \hbar) \:D[x]
-    = A \sum_{\mathrm{all}\:x(t)} \exp(i S[x] / \hbar)
+    = \int \exp\!(i S[x] / \hbar) \:D[x]
+    = A \sum_{\mathrm{all}\:x(t)} \exp\!(i S[x] / \hbar)
 \end{aligned}$$
 
 
diff --git a/content/know/concept/rayleigh-plesset-equation/index.pdc b/content/know/concept/rayleigh-plesset-equation/index.pdc
index ee8622b..9325f3f 100644
--- a/content/know/concept/rayleigh-plesset-equation/index.pdc
+++ b/content/know/concept/rayleigh-plesset-equation/index.pdc
@@ -19,34 +19,29 @@ describes how the radius of a spherical bubble evolves in time
 inside an incompressible liquid.
 Notably, it leads to [cavitation](/know/concept/cavitation/).
 
-
-## Simple form
-
-The simplest version of the Rayleigh-Plesset equation is found
-in the limiting case of a liquid with zero viscosity zero surface tension.
-
-Consider one of the [Euler equations](/know/concept/euler-equations/)
-for the velocity field $\va{v}$,
-where $\rho$ is the (constant) density:
+Consider the main
+[Navier-Stokes equations](/know/concept/navier-stokes-equations/)
+for the velocity field $\va{v}$:
 
 $$\begin{aligned}
     \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
     = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v}
-    = - \frac{\nabla p}{\rho}
+    = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
 \end{aligned}$$
 
 We make the ansatz $\va{v} = v(r, t) \vu{e}_r$,
 where $\vu{e}_r$ is the basis vector;
 in other words, we demand that the only spatial variation of the flow is in $r$.
-The above Euler equation then becomes:
+The above equation then becomes:
 
 $$\begin{aligned}
     \pdv{v}{t} + v \pdv{v}{r}
     = - \frac{1}{\rho} \pdv{p}{r}
+    + \nu \bigg( \frac{1}{r^2} \pdv{r} \Big( r^2 \pdv{v}{r} \Big) - \frac{2}{r^2} v \bigg)
 \end{aligned}$$
 
 Meanwhile, the incompressibility condition
-is as follows in this situation:
+in [spherical coordinates](/know/concept/spherical-coordinates/) yields:
 
 $$\begin{aligned}
     \nabla \cdot \va{v}
@@ -63,42 +58,75 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where $C(t)$ is an unknown function that does not depend on $r$.
-We then insert this result in the earlier Euler equation,
+We then insert this result in the main Navier-Stokes equation,
 and isolate it for $\pdv*{p}{r}$, yielding:
 
 $$\begin{aligned}
     \pdv{p}{r}
-    = - \rho \bigg( \pdv{v}{t} + v \pdv{v}{r} \bigg)
+    = - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2
+    - \nu \Big( \frac{2}{r^4} C - \frac{2}{r^4} C \Big) \bigg)
     = - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2 \bigg)
 \end{aligned}$$
 
 Integrating this with respect to $r$ yields the following expression for $p$,
-where $p_\infty$ is the (possibly time-dependent) pressure at $r = \infty$:
+where $p_\infty(t)$ is the (possibly time-dependent) pressure at $r = \infty$:
+
+$$\begin{aligned}
+    p(r)
+    = p_\infty + \rho \bigg( \frac{1}{r} C' - \frac{1}{2 r^4} C^2 \bigg)
+\end{aligned}$$
+
+From the definition of [viscosity](/know/concept/viscosity/),
+we know that the normal [stress](/know/concept/cauchy-stress-tensor/)
+$\sigma_{rr}$ in the liquid is given by:
+
+$$\begin{aligned}
+    \sigma_{rr}(r)
+    = - p(r) + 2 \rho \nu \pdv{v(r)}{r}
+\end{aligned}$$
+
+We now consider a spherical bubble
+with radius $R(t)$ and interior pressure $P(t)$ along its surface.
+Since we know the liquid pressure $p(r)$,
+we can find $P$ from $\sigma_{rr}(r)$.
+Furthermore, to include the effects of surface tension, we simply add
+the [Young-Laplace law](/know/concept/young-laplace-law/) to $P$:
+
+$$\begin{aligned}
+    P
+    = - \sigma_{rr}(R) + \alpha \frac{2}{R}
+    = p(R) - 2 \rho \nu \Big( \frac{-2}{R^3} C \Big) + \alpha \frac{2}{R}
+\end{aligned}$$
+
+We isolate this for $p(R)$, and equate it to
+our expression for $p(r)$
+at the surface $r\!=\!R$:
 
 $$\begin{aligned}
-    p(r, t)
-    = p_\infty(t) + \rho \bigg( \frac{1}{r} C'(t) - \frac{1}{2 r^4} C^2(t) \bigg)
+    P - \rho \nu \frac{4}{R^3} C - \alpha \frac{2}{R}
+    = p_\infty + \rho \bigg( \frac{1}{R} C' - \frac{1}{2 R^4} C^2 \bigg)
 \end{aligned}$$
 
-We now consider a spherical bubble with radius $R(t)$ and pressure $P(t)$ along the liquid surface.
-To study the liquid boundary's movement, we set $r = R$ and $p = P$,
-and see that $R'(t) = v(t)$, such that $C = r^2 V = R^2 R'$.
-We thus arrive at:
+Isolating for $P$,
+and inserting the fact that $R'(t) = v(t)$,
+such that $C = r^2 v = R^2 R'$,
+yields:
 
 $$\begin{aligned}
     P
-    &= p_\infty + \rho \bigg( \frac{1}{R} \dv{(R^2 R')}{t} - \frac{1}{2 R^4} (R^2 R')^2 \bigg)
+    &= p_\infty + \rho \bigg( \frac{1}{R} \dv{(R^2 R')}{t} - \frac{1}{2 R^4} (R^2 R')^2
+    + \nu \frac{4}{R^3} (R^2 R') \bigg) + \alpha \frac{2}{R}
     \\
-    &= p_\infty + \rho \bigg( 2 (R')^2 + R R'' - \frac{1}{2} (R')^2 \bigg)
+    &= p_\infty + \rho \bigg( 2 (R')^2 + R R'' - \frac{1}{2} (R')^2 + \nu \frac{4}{R} R' \bigg) + \alpha \frac{2}{R}
 \end{aligned}$$
 
-Rearranging this and defining $\Delta p = P - p_\infty$
-leads to the simple Rayleigh-Plesset equation:
+Rearranging this and defining $\Delta p \equiv P - p_\infty$
+leads to the Rayleigh-Plesset equation:
 
 $$\begin{aligned}
     \boxed{
-        R \dv[2]{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2
-        = \frac{\Delta p}{\rho}
+        \frac{\Delta p}{\rho}
+        = R \dv[2]{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2 + \nu \frac{4}{R} \dv{R}{t} + \frac{\alpha}{\rho} \frac{2}{R}
     }
 \end{aligned}$$
 
diff --git a/content/know/concept/spherical-coordinates/index.pdc b/content/know/concept/spherical-coordinates/index.pdc
index 4338ab4..4768110 100644
--- a/content/know/concept/spherical-coordinates/index.pdc
+++ b/content/know/concept/spherical-coordinates/index.pdc
@@ -50,9 +50,6 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The spherical basis vectors $\vu{e}_r$, $\vu{e}_\theta$ and $\vu{e}_\varphi$
-are expressed in the Cartesian basis like so:
-
 The spherical coordinate system is an orthogonal
 [curvilinear](/know/concept/curvilinear-coordinates/) system,
 whose scale factors $h_r$, $h_\theta$ and $h_\varphi$ we want to find.
@@ -67,7 +64,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 And then we calculate the line element $\dd{\ell}^2$,
-skipping many terms thanks to orthogonality,
+skipping many terms thanks to orthogonality:
 
 $$\begin{aligned}
     \dd{\ell}^2
@@ -94,7 +91,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-With to these factors, we can easily convert things from the Cartesian system
+With these factors, we can easily convert things from the Cartesian system
 using the standard formulae for orthogonal curvilinear coordinates.
 The basis vectors are:
 
@@ -164,7 +161,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-So, for example, an integral over all of space in Cartesian is converted like so:
+So, for example, an integral over all of space is converted like so:
 
 $$\begin{aligned}
     \iiint_{-\infty}^\infty f(x, y, z) \dd{V}
-- 
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