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5 files changed, 769 insertions, 8 deletions

diff --git a/content/know/concept/curvilinear-coordinates/index.pdc b/content/know/concept/curvilinear-coordinates/index.pdc
new file mode 100644
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@@ -0,0 +1,350 @@
+---
+title: "Curvilinear coordinates"
+firstLetter: "C"
+publishDate: 2021-03-03
+categories:
+- Mathematics
+- Physics
+
+date: 2021-03-03T19:47:34+01:00
+draft: false
+markup: pandoc
+---
+
+# Curvilinear coordinates
+
+In a 3D coordinate system, the isosurface of a coordinate
+(i.e. the surface where that coordinate is constant while the others vary)
+is known as a **coordinate surface**, and the intersections of
+the surfaces of different coordinates are called **coordinate lines**.
+
+A **curvilinear** coordinate system is one where at least one of the coordinate surfaces is curved,
+e.g. in cylindrical coordinates the line between $r$ and $z$ is a circle.
+If the coordinate surfaces are mutually perpendicular,
+it is an **orthogonal** system, which is generally desirable.
+
+A useful attribute of a coordinate system is its **line element** $\dd{\ell}$,
+which represents the differential element of a line in any direction.
+For an orthogonal system, its square $\dd{\ell}^2$ is calculated
+by taking the differential elements of the old Cartesian $(x, y, z)$ system
+and writing them out in the new $(x_1, x_2, x_3)$ system.
+The resulting expression will be of the form:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\ell}^2
+        = \dd{x}^2 + \dd{y}^2 + \dd{z}^2
+        = h_1^2 \dd{x_1}^2 + h_2^2 \dd{x_2}^2 + h_3^2 \dd{x_3}^2
+    }
+\end{aligned}$$
+
+Where $h_1$, $h_2$, and $h_3$ are called **scale factors**,
+and need not be constants.
+The equation above only contains quadratic terms
+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/).
+
+In the following subsections,
+we derive general formulae to convert expressions
+from Cartesian coordinates in the new orthogonal system $(x_1, x_2, x_3)$.
+
+
+## Basis vectors
+
+Consider the the vector form of the line element $\dd{\ell}$,
+denoted by $\dd{\vu{\ell}}$ and expressed as:
+
+$$\begin{aligned}
+    \dd{\vu{\ell}}
+    = \vu{e}_x \dd{x} + \vu{e}_y \dd{y} + \vu{e}_z \dd{z}
+\end{aligned}$$
+
+We can expand the Cartesian differential elements, e.g. $\dd{y}$,
+in the new basis as follows:
+
+$$\begin{aligned}
+    \dd{y}
+    = \pdv{y}{x_1} \dd{x_1} + \pdv{y}{x_2} \dd{x_2} + \pdv{y}{x_3} \dd{x_3}
+\end{aligned}$$
+
+If we write this out for $\dd{x}$, $\dd{y}$ and $\dd{z}$,
+and group the terms according to $\dd{x}_1$, $\dd{x}_2$ and $\dd{x}_3$,
+we can compare it the alternative form of $\dd{\vu{\ell}}$:
+
+$$\begin{aligned}
+    \dd{\vu{\ell}}
+    = \vu{e}_1 \:h_1 \dd{x_1} + \vu{e}_2 \:h_2 \dd{x_2} + \vu{e}_3 \:h_3 \dd{x_4}
+\end{aligned}$$
+
+From this, we can read off $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$.
+Here we only give $\vu{e}_1$, since $\vu{e}_2$ and $\vu{e}_3$ are analogous:
+
+$$\begin{aligned}
+    \boxed{
+        h_1 \vu{e}_1
+        = \vu{e}_x \pdv{x}{x_1} + \vu{e}_y \pdv{y}{x_1} + \vu{e}_z \pdv{y}{x_1}
+    }
+\end{aligned}$$
+
+
+## Gradient
+
+For a given direction $\dd{\ell}$, we know that
+$\dv*{f}{\ell}$ is the component of $\nabla f$ in that direction:
+
+$$\begin{aligned}
+    \dv{f}{\ell}
+    = \pdv{f}{x} \dv{x}{\ell} + \pdv{f}{y} \dv{y}{\ell} + \pdv{f}{z} \dv{z}{\ell}
+    = \nabla f \cdot \bigg( \dv{x}{\ell}, \dv{y}{\ell}, \dv{z}{\ell} \bigg)
+    = \nabla f \cdot \vu{u}
+\end{aligned}$$
+
+Where $\vu{u}$ is simply a unit vector in the direction of $\dd{\ell}$.
+We can thus find an expression for the gradient $\nabla f$
+by choosing $\dd{\ell}$ to be $h_1 \dd{x_1}$, $h_2 \dd{x_2}$ and $h_3 \dd{x_3}$ in turn:
+
+$$\begin{gathered}
+    \nabla f
+    = \vu{e}_1 \dv{x_1}{\ell} \pdv{f}{x_1}
+    + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2}
+    + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3}
+    \\
+    \boxed{
+        \nabla f
+        = \vu{e}_1 \frac{1}{h_1} \pdv{f}{x_1}
+        + \vu{e}_2 \frac{1}{h_2} \pdv{f}{x_2}
+        + \vu{e}_3 \frac{1}{h_3} \pdv{f}{x_3}
+    }
+\end{gathered}$$
+
+Where $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$
+are the basis unit vectors respectively corresponding to $x_1$, $x_2$ and $x_3$.
+
+
+## Divergence
+
+Consider a vector $\vb{V}$ in the target coordinate system
+with components $V_1$, $V_2$ and $V_3$:
+
+$$\begin{aligned}
+    \vb{V}
+    &= \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3
+    \\
+    &= \vu{e}_1 \frac{1}{h_2 h_3} (h_2 h_3 V_1)
+    + \vu{e}_2 \frac{1}{h_1 h_3} (h_1 h_3 V_2)
+    + \vu{e}_3 \frac{1}{h_1 h_2} (h_1 h_2 V_3)
+\end{aligned}$$
+
+We take only the $\vu{e}_1$-component of this vector,
+and expand its divergence using a vector identity,
+where $f = h_2 h_3 V_1$ is a scalar
+and $\vb{U} = \vu{e}_1 / (h_2 h_3)$ is a vector:
+
+$$\begin{gathered}
+    \nabla \cdot (\vb{U} \: f)
+    = \vb{U} \cdot (\nabla f) + (\nabla \cdot \vb{U}) \: f
+    \\
+    \nabla \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} (h_2 h_3 V_1) \Big)
+    = \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big)
+    + \Big( \nabla \cdot \frac{\vu{e}_1}{h_2 h_3} \Big) (h_2 h_3 V_1)
+\end{gathered}$$
+
+The first term is straightforward to calculate
+thanks to our preceding expression for the gradient.
+Only the $\vu{e}_1$-component survives due to the dot product:
+
+$$\begin{aligned}
+    \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big)
+    = \frac{\vu{e}_1}{h_1 h_2 h_3} \pdv{(h_2 h_3 V_1)}{x_1}
+\end{aligned}$$
+
+The second term is a bit more involved.
+To begin with, we use the gradient formula to note that:
+
+$$\begin{aligned}
+    \nabla x_1
+    = \frac{\vu{e}_1}{h_1}
+    \qquad \quad
+    \nabla x_2
+    = \frac{\vu{e}_2}{h_2}
+    \qquad \quad
+    \nabla x_3
+    = \frac{\vu{e}_3}{h_3}
+\end{aligned}$$
+
+Because $\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$ in an orthogonal basis,
+we can get the vector whose divergence we want:
+
+$$\begin{aligned}
+    \nabla x_2 \cross \nabla x_3
+    = \frac{\vu{e}_2}{h_2} \cross \frac{\vu{e}_3}{h_3}
+    = \frac{\vu{e}_1}{h_2 h_3}
+\end{aligned}$$
+
+We then apply the divergence and expand the expression using a vector identity.
+In all cases, the curl of a gradient $\nabla \cross \nabla f$ is zero, so:
+
+$$\begin{aligned}
+    \nabla \cdot \frac{\vu{e}_1}{h_2 h_3}
+    = \nabla \cdot \big( \nabla x_2 \cross \nabla x_3 \big)
+    = \nabla x_3 \cdot (\nabla \cross \nabla x_2) - \nabla x_2 \cdot (\nabla \cross \nabla x_3)
+    = 0
+\end{aligned}$$
+
+After repeating this procedure for the other components of $\vb{V}$,
+we arrive at the following general expression for the divergence $\nabla \cdot \vb{V}$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{V}
+        = \frac{1}{h_1 h_2 h_3}
+        \Big( \pdv{(h_2 h_3 V_1)}{x_1} + \pdv{(h_1 h_3 V_2)}{x_2} + \pdv{(h_1 h_2 V_3)}{x_3} \Big)
+    }
+\end{aligned}$$
+
+
+## Laplacian
+
+The Laplacian $\nabla^2 f$ is simply $\nabla \cdot \nabla f$,
+so we can find the general formula
+by combining the two preceding results
+for the gradient and the divergence:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla^2 f
+        = \frac{1}{h_1 h_2 h_3}
+        \bigg(
+            \pdv{x_1} \Big(\! \frac{h_2 h_3}{h_1} \pdv{f}{x_1} \!\Big)
+            + \pdv{x_2} \Big(\! \frac{h_1 h_3}{h_2} \pdv{f}{x_2} \!\Big)
+            + \pdv{x_3} \Big(\! \frac{h_1 h_2}{h_3} \pdv{f}{x_3} \!\Big)
+        \bigg)
+    }
+\end{aligned}$$
+
+
+## Curl
+
+We find the curl in a similar way as the divergence.
+Consider an arbitrary vector $\vb{V}$:
+
+$$\begin{aligned}
+    \vb{V}
+    = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3
+    = \frac{\vu{e}_1}{h_1} (h_1 V_1) + \frac{\vu{e}_2}{h_2} (h_2 V_2) + \frac{\vu{e}_3}{h_3} (h_3 V_3)
+\end{aligned}$$
+
+We expand the curl of its $\vu{e}_1$-component using a vector identity,
+where $f = h_1 V_1$ is a scalar and $\vb{U} = \vu{e}_1 / h_1$ is a vector:
+
+$$\begin{gathered}
+    \nabla \cross (\vb{U} \: f)
+    = (\nabla \cross \vb{U}) \: f - \vb{U} \cross (\nabla f)
+    \\
+    \nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big)
+    = \Big( \nabla \cross \frac{\vu{e}_1}{h_1} \Big) (h_1 V_1) - \frac{\vu{e}_1}{h_1} \cross \Big( \nabla (h_1 V_1) \Big)
+\end{gathered}$$
+
+Previously, when calculating the divergence,
+we already showed that $\vu{e}_1 / h_1 = \nabla x_1$.
+Because the curl of a gradient is zero,
+the first term thus disappears, leaving only the second,
+which contains a gradient turning out to be:
+
+$$\begin{aligned}
+    \nabla (h_1 V_1)
+    = \vu{e}_1 \frac{1}{h_1} \pdv{(h_1 V_1)}{x_1}
+    + \vu{e}_2 \frac{1}{h_2} \pdv{(h_1 V_1)}{x_2}
+    + \vu{e}_3 \frac{1}{h_3} \pdv{(h_1 V_1)}{x_3}
+\end{aligned}$$
+
+Consequently, the curl of the first component of $\vb{V}$ is as follows,
+using the fact that $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$
+are related to each other by cross products:
+
+$$\begin{aligned}
+    \nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big)
+    = - \frac{\vu{e}_1}{h_1} \cross \Big( \nabla (h_1 V_1) \Big)
+    = - \frac{\vu{e}_3}{h_1 h_2} \pdv{(h_1 V_1)}{x_2} + \frac{\vu{e}_2}{h_1 h_3} \pdv{(h_1 V_1)}{x_3}
+\end{aligned}$$
+
+If we go through the same process for the other components of $\vb{V}$
+and add the results together, we get the following expression for the curl $\nabla \cross \vb{V}$:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \nabla \times \vb{V}
+            &= \frac{\vu{e}_1}{h_2 h_3} \Big( \pdv{(h_3 V_3)}{x_2} - \pdv{(h_2 V_2)}{x_3} \Big)
+            \\
+            &+ \frac{\vu{e}_2}{h_1 h_3} \Big( \pdv{(h_1 V_1)}{x_3} - \pdv{(h_3 V_3)}{x_1} \Big)
+            \\
+            &+ \frac{\vu{e}_3}{h_1 h_2} \Big( \pdv{(h_2 V_2)}{x_1} - \pdv{(h_1 V_1)}{x_2} \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+
+## Differential elements
+
+The point of the scale factors $h_1$, $h_2$ and $h_3$, as can seen from their derivation,
+is to correct for "distortions" of the coordinates compared to the Cartesian system,
+such that the line element $\dd{\ell}$ retains its length.
+This property extends to the surface $\dd{S}$ and volume $\dd{V}$.
+
+When handling a differential volume in curvilinear coordinates,
+e.g. for a volume integral,
+the size of the box $\dd{V}$ must be corrected by the scale factors:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{V}
+        = \dd{x}\dd{y}\dd{z}
+        = h_1 h_2 h_3 \dd{x_1} \dd{x_2} \dd{x_3}
+    }
+\end{aligned}$$
+
+The same is true for the isosurfaces $\dd{S_1}$, $\dd{S_2}$ and $\dd{S_3}$
+where the coordinates $x_1$, $x_2$ and $x_3$ are respectively kept constant:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \dd{S_1} &= h_2 h_3 \dd{x_2} \dd{x_3}
+            \\
+            \dd{S_2} &= h_1 h_3 \dd{x_1} \dd{x_3}
+            \\
+            \dd{S_3} &= h_1 h_2 \dd{x_1} \dd{x_2}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Using the same logic, the normal vector element $\dd{\vu{S}}$
+of an arbitrary surface is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{S}}
+        = \vu{e}_1 h_2 h_3 \dd{x_2} \dd{x_3} + \vu{e}_2 h_1 h_3 \dd{x_1} \dd{x_3} + \vu{e}_3 h_1 h_2 \dd{x_1} \dd{x_2}
+    }
+\end{aligned}$$
+
+Finally, the tangent vector element $\dd{\vu{\ell}}$ takes the following form:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{\ell}}
+        = \vu{e}_1 h_1 \dd{x_1} + \vu{e}_2 h_2 \dd{x_2} + \vu{e}_3 h_3 \dd{x_3}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  M.L. Boas,
+    *Mathematical methods in the physical sciences*, 2nd edition,
+    Wiley.
diff --git a/content/know/concept/lagrange-multiplier/index.pdc b/content/know/concept/lagrange-multiplier/index.pdc
index 2b14897..fffe85f 100644
--- a/content/know/concept/lagrange-multiplier/index.pdc
+++ b/content/know/concept/lagrange-multiplier/index.pdc
@@ -48,7 +48,9 @@ Solving this directly would be a delicate balancing act
 of all the partial derivatives.
 
 To help us solve this, we introduce a "dummy" parameter $\lambda$,
-the so-called **Lagrange multiplier**, and contruct a new function $L$ given by:
+the so-called **Lagrange multiplier**,
+which need not be constant,
+and contruct a new function $L$ given by:
 
 $$\begin{aligned}
     L(x, y, z) = f(x, y, z) + \lambda \phi(x, y, z)
diff --git a/content/know/concept/parabolic-cylindrical-coordinates/index.pdc b/content/know/concept/parabolic-cylindrical-coordinates/index.pdc
new file mode 100644
index 0000000..56544ae
--- /dev/null
+++ b/content/know/concept/parabolic-cylindrical-coordinates/index.pdc
@@ -0,0 +1,188 @@
+---
+title: "Parabolic cylindrical coordinates"
+firstLetter: "P"
+publishDate: 2021-03-04
+categories:
+- Mathematics
+- Physics
+
+date: 2021-03-04T15:07:46+01:00
+draft: false
+markup: pandoc
+---
+
+# Parabolic cylindrical coordinates
+
+**Parabolic cylindrical coordinates** are a coordinate system
+that describes a point in space using three coordinates $(\sigma, \tau, z)$.
+The $z$-axis is unchanged from the Cartesian system,
+hence it is called a *cylindrical* system.
+In the $z$-isoplane, however, confocal parabolas are used.
+These coordinates can be converted to the Cartesian $(x, y, z)$ as follows:
+
+$$\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/pulay-mixing/index.pdc b/content/know/concept/pulay-mixing/index.pdc
index 9102c0e..8daa54f 100644
--- a/content/know/concept/pulay-mixing/index.pdc
+++ b/content/know/concept/pulay-mixing/index.pdc
@@ -83,13 +83,14 @@ We thus want to minimize the following quantity,
 where $\lambda$ is a [Lagrange multiplier](/know/concept/lagrange-multiplier/):
 
 $$\begin{aligned}
-	\braket{R_{n+1}}{R_{n+1}} + \lambda \sum_{m = 1}^n \alpha_m
-	= \sum_{m=1}^n \alpha_m \Big( \sum_{k=1}^n \alpha_k \braket{R_m}{R_k} + \lambda \Big)
+	\braket{R_{n+1}}{R_{n+1}} + \lambda \sum_{m = 1}^n \alpha_m^*
+	= \sum_{m=1}^n \alpha_m^* \Big( \sum_{k=1}^n \alpha_k \braket{R_m}{R_k} + \lambda \Big)
 \end{aligned}$$
 
-By differentiating the right-hand side with respect to $\alpha_m$,
+By differentiating the right-hand side with respect to $\alpha_m^*$
+and demanding that the result is zero,
 we get a system of equations that we can write in matrix form,
-which is relatively cheap to solve numerically:
+which is cheap to solve:
 
 $$\begin{aligned}
 	\begin{bmatrix}
@@ -107,6 +108,11 @@ $$\begin{aligned}
 	\end{bmatrix}
 \end{aligned}$$
 
+From this, we can also see that the Lagrange multiplier
+$\lambda = - \braket{R_{n+1}}{R_{n+1}}$,
+where $R_{n+1}$ is the *predicted* residual of the next iteration,
+subject to the two assumptions.
+
 This method is very effective.
 However, in practice, the earlier inputs $\rho_1$, $\rho_2$, etc.
 are much further from $\rho_*$ than $\rho_n$,
@@ -121,7 +127,7 @@ You might be confused by the absence of all $\rho_m^\mathrm{new}$
 in the creation of $\rho_{n+1}$, as if the iteration's outputs are being ignored.
 This is due to the first assumption,
 which states that $\rho_n^\mathrm{new}$ are $\rho_n$ are already similar,
-such that they are interchangeable.
+such that they are basically interchangeable.
 
 Speaking of which, about those assumptions:
 while they will clearly become more accurate as $\rho_n$ approaches $\rho_*$,
@@ -147,8 +153,9 @@ $$\begin{aligned}
 In other words, we end up introducing a small amount of the raw outputs $\rho_m^\mathrm{new}$,
 while still giving more weight to iterations with smaller residuals.
 
-Pulay mixing is very effective:
-it can accelerate convergence by up to one order of magnitude!
+Pulay mixing is very effective for certain types of problems,
+e.g. density functional theory,
+where it can accelerate convergence by up to one order of magnitude!
 
 
 
diff --git a/content/know/concept/spherical-coordinates/index.pdc b/content/know/concept/spherical-coordinates/index.pdc
new file mode 100644
index 0000000..4338ab4
--- /dev/null
+++ b/content/know/concept/spherical-coordinates/index.pdc
@@ -0,0 +1,214 @@
+---
+title: "Spherical coordinates"
+firstLetter: "S"
+publishDate: 2021-03-04
+categories:
+- Mathematics
+- Physics
+
+date: 2021-03-04T15:05:21+01:00
+draft: false
+markup: pandoc
+---
+
+# Spherical coordinates
+
+**Spherical coordinates** are an extension of polar coordinates to 3D.
+The position of a given point in space is described by
+three coordinates $(r, \theta, \varphi)$, defined as:
+
+*   $r$: the **radius** or **radial distance**: distance to the origin.
+*   $\theta$: the **elevation**, **polar angle** or **colatitude**:
+    angle to the positive $z$-axis, or **zenith**, i.e. the "north pole".
+*   $\varphi$: the **azimuth**, **azimuthal angle** or **longitude**:
+    angle from the positive $x$-axis, typically in the counter-clockwise sense.
+
+Cartesian coordinates $(x, y, z)$ and the spherical system
+$(r, \theta, \varphi)$ are related by:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            x &= r \sin\theta \cos\varphi \\
+            y &= r \sin\theta \sin\varphi \\
+            z &= r \cos\theta
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Conversely, a point given in $(x, y, z)$
+can be converted to $(r, \theta, \varphi)$
+using these formulae:
+
+$$\begin{aligned}
+    \boxed{
+        r = \sqrt{x^2 + y^2 + z^2}
+        \qquad
+        \theta = \arccos(z / r)
+        \qquad
+        \varphi = \mathtt{atan2}(y, x)
+    }
+\end{aligned}$$
+
+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.
+To do so, we calculate the differentials of the Cartesian coordinates:
+
+$$\begin{aligned}
+    \dd{x} &= \dd{r} \sin\theta \cos\varphi + \dd{\theta} r \cos\theta \cos\varphi - \dd{\varphi} r \sin\theta \sin\varphi
+    \\
+    \dd{y} &= \dd{r} \sin\theta \sin\varphi + \dd{\theta} r \cos\theta \sin\varphi + \dd{\varphi} r \sin\theta \cos\varphi
+    \\
+    \dd{z} &= \dd{r} \cos\theta - \dd{\theta} r \sin\theta
+\end{aligned}$$
+
+And then we calculate the line element $\dd{\ell}^2$,
+skipping many terms thanks to orthogonality,
+
+$$\begin{aligned}
+    \dd{\ell}^2
+    &= \:\:\:\: \dd{r}^2 \big( \sin^2(\theta) \cos^2(\varphi) + \sin^2(\theta) \sin^2(\varphi) + \cos^2(\theta) \big)
+    \\
+    &\quad + \dd{\theta}^2 \big( r^2 \cos^2(\theta) \cos^2(\varphi) + r^2 \cos^2(\theta) \sin^2(\varphi) + r^2 \sin^2(\theta) \big)
+    \\
+    &\quad + \dd{\varphi}^2 \big( r^2 \sin^2(\theta) \sin^2(\varphi) + r^2 \sin^2(\theta) \cos^2(\varphi) \big)
+    \\
+    &= \dd{r}^2 + r^2 \: \dd{\theta}^2 + r^2 \sin^2(\theta) \: \dd{\varphi}^2
+\end{aligned}$$
+
+Finally, we can simply read off
+the squares of the desired scale factors
+$h_r^2$, $h_\theta^2$ and $h_\varphi^2$:
+
+$$\begin{aligned}
+    \boxed{
+        h_r = 1
+        \qquad
+        h_\theta = r
+        \qquad
+        h_\varphi = r \sin\theta
+    }
+\end{aligned}$$
+
+With to these factors, we can easily convert things from the Cartesian system
+using the standard formulae for orthogonal curvilinear coordinates.
+The basis vectors are:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \vu{e}_r
+            &= \sin\theta \cos\varphi \:\vu{e}_x + \sin\theta \sin\varphi \:\vu{e}_y + \cos\theta \:\vu{e}_z
+            \\
+            \vu{e}_\theta
+            &= \cos\theta \cos\varphi \:\vu{e}_x + \cos\theta \sin\varphi \:\vu{e}_y - \sin\theta \:\vu{e}_z
+            \\
+            \vu{e}_\varphi
+            &= - \sin\varphi \:\vu{e}_x + \cos\varphi \:\vu{e}_y
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The basic vector operations (gradient, divergence, Laplacian and curl) are given by:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla f
+        = \vu{e}_r \pdv{f}{r}
+        + \vu{e}_\theta \frac{1}{r} \pdv{f}{\theta} + \mathbf{e}_\varphi \frac{1}{r \sin\theta} \pdv{f}{\varphi}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{V}
+        = \frac{1}{r^2} \pdv{(r^2 V_r)}{r}
+        + \frac{1}{r \sin\theta} \pdv{(\sin\theta V_\theta)}{\theta}
+        + \frac{1}{r \sin\theta} \pdv{V_\varphi}{\varphi}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \nabla^2 f
+        = \frac{1}{r^2} \pdv{r} \Big( r^2 \pdv{f}{r} \Big)
+        + \frac{1}{r^2 \sin\theta} \pdv{\theta} \Big( \sin\theta \pdv{f}{\theta} \Big)
+        + \frac{1}{r^2 \sin^2(\theta)} \pdv[2]{f}{\varphi}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \nabla \times \vb{V}
+            &= \frac{\vu{e}_r}{r \sin\theta} \Big( \pdv{(\sin\theta V_\varphi)}{\theta} - \pdv{V_\theta}{\varphi} \Big)
+            \\
+            &+ \frac{\vu{e}_\theta}{r} \Big( \frac{1}{\sin\theta} \pdv{V_r}{\varphi} - \pdv{(r V_\varphi)}{r} \Big)
+            \\
+            &+ \frac{\vu{e}_\varphi}{r} \Big( \pdv{(r V_\theta)}{r} - \pdv{V_r}{\theta} \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The differential element of volume $\dd{V}$
+takes the following form:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{V}
+        = r^2 \sin\theta \dd{r} \dd{\theta} \dd{\varphi}
+    }
+\end{aligned}$$
+
+So, for example, an integral over all of space in Cartesian is converted like so:
+
+$$\begin{aligned}
+    \iiint_{-\infty}^\infty f(x, y, z) \dd{V}
+    = \int_0^{2\pi} \int_0^\pi \int_0^\infty f(r, \theta, \varphi) \: r^2 \sin\theta \dd{r} \dd{\theta} \dd{\varphi}
+\end{aligned}$$
+
+The isosurface elements are as follows, where $S_r$ is a surface at constant $r$, etc.:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \dd{S}_r = r^2 \sin\theta \dd{\theta} \dd{\varphi}
+            \qquad
+            \dd{S}_\theta = r \sin\theta \dd{r} \dd{\varphi}
+            \qquad
+            \dd{S}_\varphi = r \dd{r} \dd{\theta}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Similarly, the normal vector element $\dd{\vu{S}}$ for an arbitrary surface is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{S}}
+        = \vu{e}_r \: r^2 \sin\theta \dd{\theta} \dd{\varphi}
+        + \vu{e}_\theta \: r \sin\theta \dd{r} \dd{\varphi}
+        + \vu{e}_\varphi \: r \dd{r} \dd{\theta}
+    }
+\end{aligned}$$
+
+And finally, the tangent vector element $\dd{\vu{\ell}}$ of a given curve is as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{\ell}}
+        = \vu{e}_r \: \dd{r}
+        + \vu{e}_\theta \: r \dd{\theta}
+        + \vu{e}_\varphi \: r \sin\theta \dd{\varphi}
+    }
+\end{aligned}$$
+
+
+## References
+1.  M.L. Boas,
+    *Mathematical methods in the physical sciences*, 2nd edition,
+    Wiley.