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 +-
static/main.css | 5 +-
8 files changed, 460 insertions(+), 230 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
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}
diff --git a/static/main.css b/static/main.css
index 542790e..f3778e5 100644
--- a/static/main.css
+++ b/static/main.css
@@ -6,13 +6,12 @@ body {
padding:1em 0;
font-family:sans-serif;
}
-a {text-decoration:none;color:#00f;}
h1,h2,h3 {text-align:center}
h1 {font-size:200%;}
h2 {font-size:160%;}
h3 {font-size:120%;}
-.noha a:link,a:visited {color:#222;}
-.noha a:hover,a:focus,a:active {color:#00f;}
+a {text-decoration:none;color:#00f;}
+.noha a:link, .noha a:visited {color:#222;}
.nav {height:3rem;font-size:250%;}
.navl {float:left;text-align:left;}
.navr {float:right;text-align:right;}
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