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4 files changed, 721 insertions, 12 deletions

diff --git a/content/know/concept/euler-bernoulli-law/index.pdc b/content/know/concept/euler-bernoulli-law/index.pdc
new file mode 100644
index 0000000..e9d3fdd
--- /dev/null
+++ b/content/know/concept/euler-bernoulli-law/index.pdc
@@ -0,0 +1,314 @@
+---
+title: "Euler-Bernoulli law"
+firstLetter: "E"
+publishDate: 2021-06-03
+categories:
+- Physics
+
+date: 2021-05-14T09:15:30+02:00
+draft: false
+markup: pandoc
+---
+
+# Euler-Bernoulli law
+
+**Euler-Bernoulli beam theory** concerns itself with the bending of beams
+(e.g. the metal beams used in large buildings),
+subject to certain simplifying assumptions,
+which are generally valid for beams that are narrow,
+i.e. longitudinally much larger than transversely.
+
+Consider a beam of length $L$, placed upright
+on the $z = 0$ plane, above the origin.
+If we pull the top of this beam in the postive $y$-direction,
+we assume that it bends uniformly,
+i.e. with constant radius of [curvature](/know/concept/curvature/) $R$.
+We also assume that the bending is **shear-free**:
+if we treat the beam as a bundle of elastic strings,
+then there is no friction between them.
+
+The central string has its length unchanged (i.e. still $L$),
+while an arbitrary non-central string is extended or compressed to $L'$.
+The [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) element $u_{zz}$ is then:
+
+$$\begin{aligned}
+    u_{zz}
+    = \frac{L' - L}{L}
+\end{aligned}$$
+
+Because the bending is uniform, the central string
+is an arc with radius $R$ and central angle $\theta$,
+where $L = \theta R$.
+The non-central string has $L' = \theta R'$,
+where $R'$ is geometrically shown to be $R' = R - y$,
+with $y$ being the $y$-coordinate of that string at the beam's base.
+So:
+
+$$\begin{aligned}
+    u_{zz}
+    = \frac{R' - R}{R}
+    = - \frac{y}{R}
+\end{aligned}$$
+
+By assumption, there are no shear stresses
+and no forces acting on the beam's sides,
+so the only nonzero component of the
+[Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$
+is $\sigma_{zz}$, given by [Hooke's law](/know/concept/hookes-law/):
+
+$$\begin{aligned}
+    \sigma_{zz} = E u_{zz}
+\end{aligned}$$
+
+Where $E$ is the elastic modulus of the material.
+By Hooke's inverse law,
+the other nonzero strain components are as follows,
+where $\nu$ is Poisson's ratio:
+
+$$\begin{aligned}
+    u_{xx}
+    = u_{yy}
+    = - \frac{\nu}{E} \sigma_{zz}
+    = - \nu u_{zz}
+    = \nu \frac{y}{R}
+\end{aligned}$$
+
+For completeness, we turn the strain tensor $\hat{u}$
+into a full displacement field $\va{u}$:
+
+$$\begin{aligned}
+    \boxed{
+        u_x = \nu \frac{x y}{R}
+        \qquad
+        u_y = \frac{z^2}{2 R} + \nu \frac{y^2 - x^2}{2 R}
+        \qquad
+        u_z = - \frac{y z}{R}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-field"/>
+<label for="proof-field">Proof</label>
+<div class="hidden">
+<label for="proof-field">Proof.</label>
+By integrating the above strains $u_{ii} = \pdv*{u_i}{i}$,
+we get the components of $\va{u}$:
+
+$$\begin{aligned}
+    u_x
+    = \nu \frac{x y}{R} + f_x(y, z)
+    \qquad
+    u_y
+    = \nu \frac{y^2}{2 R} + f_y(x, z)
+    \qquad
+    u_z
+    = - \frac{y z}{R} + f_z(x, y)
+\end{aligned}$$
+
+Where $f_x$, $f_y$ and $f_z$ are integration constants,
+which we find by demanding that the off-diagonal strains $u_{ij}$ are zero.
+Starting with $u_{xz} = 0$:
+
+$$\begin{aligned}
+    0
+    = u_{xz}
+    = \frac{1}{2} \Big( \pdv{u_x}{z} + \pdv{u_z}{x} \Big)
+    = \frac{1}{2} \Big( \pdv{f_x}{z} + \pdv{f_z}{x} \Big)
+\end{aligned}$$
+
+Here, only $f_x$ may depend on $z$,
+and only $f_z$ may depend on $x$.
+This equation thus tell us:
+
+$$\begin{aligned}
+    f_x(y, z)
+    = z \: g(y)
+    \qquad \quad
+    f_z(x, y)
+    = - x \: g(y)
+\end{aligned}$$
+
+Where $g(y)$ is an unknown integration constant.
+Moving on to $u_{xy} = 0$:
+
+$$\begin{aligned}
+    0
+    = \frac{1}{2} \Big( \pdv{u_x}{y} + \pdv{u_y}{x} \Big)
+    = \frac{1}{2} \Big( \nu \frac{x}{R} + \pdv{f_x}{y} + \pdv{f_y}{x} \Big)
+\end{aligned}$$
+
+Only $f_x$ may contain $y$,
+so its $y$-derivative must be a constant,
+so $g(y) = C y$. Therefore:
+
+$$\begin{aligned}
+    f_x(y, z)
+    = C y z
+    \qquad
+    f_y(x, z)
+    = - \nu \frac{x^2}{2 R} - C x z + h(z)
+    \qquad
+    f_z(x, y)
+    = - C x y
+\end{aligned}$$
+
+Where $h(z)$ is an unknown integration constant.
+Finally, we put everything in $u_{yz} = 0$:
+
+$$\begin{aligned}
+    0
+    = \frac{1}{2} \Big( \pdv{u_y}{z} + \pdv{u_z}{y} \Big)
+    = \frac{1}{2} \Big( \pdv{f_y}{z} - \frac{z}{R} + \pdv{f_z}{y} \Big)
+    = \frac{1}{2} \Big( \!-\! 2 C x + \dv{h}{z} - \frac{z}{R} \Big)
+\end{aligned}$$
+
+Only the first term contains $x$, so to satisfy this equation, we must set $C = 0$.
+The remaining terms then tell us that $h(z) = z^2 / (2 R)$.
+Therefore:
+
+$$\begin{aligned}
+    f_x = 0
+    \qquad
+    f_y = - \nu \frac{x^2}{2 R} + \frac{z^2}{2 R}
+    \qquad
+    f_z = 0
+\end{aligned}$$
+
+Inserting this into the components $u_x$, $u_y$ and $u_z$
+then yields the full displacement field.
+</div>
+</div>
+
+In any case, the beam experiences a bending torque with an $x$-component $T_x$ given by:
+
+$$\begin{aligned}
+    T_x
+    = - \int_A y \sigma_{zz} \dd{A}
+    = - \frac{E}{R} \int_A y^2 \dd{A}
+\end{aligned}$$
+
+Where $A$ is the cross-section.
+Th above integral is known as the **area moment**,
+and is typically abbreviated by $I$.
+This brings us to the **Euler-Bernoulli law**:
+
+$$\begin{aligned}
+    \boxed{
+        T_x
+        = - \frac{E I}{R}
+    }
+    \qquad \quad
+    I
+    \equiv \int_A y^2 \dd{A}
+\end{aligned}$$
+
+The product $E I$ is called the **flexural rigidity**,
+i.e. the beam's "stiffness".
+For a small deformation, i.e. a large radius of curvature $R$,
+the law can be approximated by:
+
+$$\begin{aligned}
+    T_x
+    \approx - E I \dv[2]{y}{z}
+\end{aligned}$$
+
+
+
+## Slender rods
+
+A beam that is very thin in the transverse directions ($x$ and $y$ in this case),
+can be approximated as a single string or rod $y(z)$.
+Each infinitesimal piece $(\dd{y}, \dd{z})$ of the rod
+exerts forces $F_y$ and $F_z$ on the next piece,
+and is feels external forces-per-length $K_y$ and $K_z$, e.g. gravity.
+In order to have equilibrium, the total force must be zero:
+
+$$\begin{aligned}
+    0
+    &= F_y(z + \dd{z}) - F_y(z) + K_y(z) \dd{z}
+    \\
+    0
+    &= F_z(z + \dd{z}) - F_z(z) + K_z(z) \dd{z}
+\end{aligned}$$
+
+Rearranging these relations yields these equations for the internal forces $F_y$ and $F_z$:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{F_y}{z}
+        = - K_y
+    }
+    \qquad \quad
+    \boxed{
+        \dv{F_z}{z}
+        = - K_z
+    }
+\end{aligned}$$
+
+Meanwhile, the rod also feels a torque with $x$-component $T_x$,
+where equilibrium entails:
+
+$$\begin{aligned}
+    0
+    = T_x(z + \dd{z}) - T_x(z) + F_z(z) \dd{y} - F_y(z) \dd{z}
+\end{aligned}$$
+
+This can be rearranged to get a differential equation for $T_x$, namely:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{T_x}{z}
+        = F_y - F_z \dv{y}{z}
+    }
+\end{aligned}$$
+
+If $F_z$ and $\dv*{y}{z}$ are small, the last term can be dropped.
+These equations are widely applicable,
+but there is one especially important application,
+so much so that it is usually what is meant by "Euler-Bernoulli law":
+the shape of a laterally loaded rod.
+
+Consider a beam along the $z$-axis, carrying a lateral load $K_y$,
+e.g. its own weight $A g \rho$ or more.
+Assuming there is no other load $K_z = 0$
+and $F_y \ll F_z$, the above equations become:
+
+$$\begin{aligned}
+    T_x
+    = - E I \dv[2]{y}{z}
+    \qquad
+    \dv{T_x}{z}
+    = F_y
+    \qquad
+    \dv{F_y}{z}
+    = - K_y
+\end{aligned}$$
+
+Which we can simply substitute into each other,
+eventually leading to:
+
+$$\begin{aligned}
+    \boxed{
+        K_y
+        = \dv[2]{z} \Big( E I \dv[2]{y}{z} \Big)
+    }
+\end{aligned}$$
+
+This is often referred to as the **Euler-Bernoulli law** as well.
+Typically the flexural rigidity $EI$ is a constant in $z$,
+in which case we can reduce the equation to:
+
+$$\begin{aligned}
+    K_y
+    = E I \dv[4]{y}{z}
+\end{aligned}$$
+
+Which is clearly solved by a fourth-order polynomial,
+given some boundary conditions.
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
diff --git a/content/know/concept/impulse-response/index.pdc b/content/know/concept/impulse-response/index.pdc
index b055fe7..fa921fa 100644
--- a/content/know/concept/impulse-response/index.pdc
+++ b/content/know/concept/impulse-response/index.pdc
@@ -35,9 +35,14 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-*__Proof.__ Starting from the definition of $u_p(t)$,
+<div class="accordion">
+<input type="checkbox" id="proof-main"/>
+<label for="proof-main">Proof</label>
+<div class="hidden">
+<label for="proof-main">Proof.</label>
+Starting from the definition of $u_p(t)$,
 we shift the argument by some constant $\tau$,
-and multiply both sides by the constant $f(\tau)$:*
+and multiply both sides by the constant $f(\tau)$:
 
 $$\begin{aligned}
     \hat{L} \{ u_p(t - \tau) \} &= \delta(t - \tau)
@@ -45,8 +50,8 @@ $$\begin{aligned}
     \hat{L} \{ f(\tau) \: u_p(t - \tau) \} &= f(\tau) \: \delta(t - \tau)
 \end{aligned}$$
 
-*Where $f(\tau)$ can be moved inside using the
-linearity of $\hat{L}$. Integrating over $\tau$ then gives us:*
+Where $f(\tau)$ can be moved inside using the
+linearity of $\hat{L}$. Integrating over $\tau$ then gives us:
 
 $$\begin{aligned}
     \int_0^\infty \hat{L} \{ f(\tau) \: u_p(t - \tau) \} \dd{\tau}
@@ -54,14 +59,14 @@ $$\begin{aligned}
     = f(t)
 \end{aligned}$$
 
-*The integral and $\hat{L}$ are operators of different variables, so we reorder them:*
+The integral and $\hat{L}$ are operators of different variables, so we reorder them:
 
 $$\begin{aligned}
     \hat{L} \int_0^\infty f(\tau) \: u_p(t - \tau) \dd{\tau}
     &= (f * u_p)(t) = \hat{L}\{ u(t) \} = f(t)
 \end{aligned}$$
-
-*__Q.E.D.__*
+</div>
+</div>
 
 
 
diff --git a/content/know/concept/lubrication-theory/index.pdc b/content/know/concept/lubrication-theory/index.pdc
new file mode 100644
index 0000000..7641440
--- /dev/null
+++ b/content/know/concept/lubrication-theory/index.pdc
@@ -0,0 +1,223 @@
+---
+title: "Lubrication theory"
+firstLetter: "L"
+publishDate: 2021-06-03
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+
+date: 2021-05-11T14:34:07+02:00
+draft: false
+markup: pandoc
+---
+
+# Lubrication theory
+
+**Lubricants** are widely used
+to reduce friction between two moving surfaces.
+In fluid mechanics, **lubrication theory**
+is the study of fluids that are tightly constrained in one dimension,
+especially those in small gaps between moving surfaces.
+
+For simplicity, we limit ourselves to 2D
+by assuming that everything is constant along the $z$-axis.
+Consider a gap of width $d$ (along $y$) and length $L$ (along $x$),
+where $d \ll L$, containing the fluid.
+Outside the gap, the lubricant has a
+[Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \approx U L / \nu$.
+
+Inside the gap, the Reynolds number $\mathrm{Re}_\mathrm{gap}$ is different.
+This is because advection will dominate along the $x$-axis (gap length),
+and viscosity along the $y$-axis (gap width).
+Therefore:
+
+$$\begin{aligned}
+    \mathrm{Re}_\mathrm{gap}
+    \approx \frac{|(\va{v} \cdot \nabla) \va{v}|}{|\nu \nabla^2 \va{v}|}
+    \approx \frac{U^2 / L}{\nu U / d^2}
+    \approx \frac{d^2}{L^2} \mathrm{Re}
+\end{aligned}$$
+
+If $d$ is small enough compared to $L$,
+then $\mathrm{Re}_\mathrm{gap} \ll 1$.
+More formally, we need $d \ll L / \sqrt{\mathrm{Re}}$,
+so we are inside the boundary layer,
+in the realm of the [Prandtl equations](/know/concept/prandtl-equations/).
+
+Let $\mathrm{Re}_\mathrm{gap} \ll 1$.
+We are thus dealing with *Stokes flow*, in which case
+the [Navier-Stokes equations](/know/concept/navier-stokes/equations/)
+can be reduced to the following *Stokes equations*:
+
+$$\begin{aligned}
+    \pdv{p}{x}
+    = \eta \: \Big( \pdv[2]{v_x}{x} + \pdv[2]{v_x}{y} \Big)
+    \qquad \quad
+    \pdv{p}{y}
+    = \eta \: \Big( \pdv[2]{v_y}{x} + \pdv[2]{v_y}{y} \Big)
+\end{aligned}$$
+
+Let the $y = 0$ plane be an infinite flat surface,
+sliding in the positive $x$-direction at a constant velocity $U$.
+On the other side of the gap,
+an arbitrary surface is described by $h(x)$.
+
+Since the gap is so narrow,
+and the surfaces' movements cause large shear stresses inside,
+$v_y$ is negligible compared to $v_x$.
+Furthermore, because the gap is so long,
+we assume that $\pdv*{v_x}{x}$ is negligible compared to $\pdv*{v_x}{y}$.
+This reduces the Stokes equations to:
+
+$$\begin{aligned}
+    \pdv{p}{x}
+    = \eta \pdv[2]{v_x}{y}
+    \qquad \quad
+    \pdv{p}{y}
+    = 0
+\end{aligned}$$
+
+This result could also be derived from the Prandtl equations.
+In any case, it tells us that $p$ only depends on $x$,
+allowing us to integrate the former equation:
+
+$$\begin{aligned}
+    v_x
+    = \frac{p'}{2 \eta} y^2 + C_1 y + C_2
+\end{aligned}$$
+
+Where $C_1$ and $C_2$ are integration constants.
+At $y = 0$, the viscous *no-slip* condition demands that $v_x = U$, so $C_2 = U$.
+Likewise, at $y = h(x)$, we need $v_x = 0$, leading us to:
+
+$$\begin{aligned}
+    v_x
+    = \frac{p'}{2 \eta} y^2 - \Big( \frac{p'}{2 \eta} h + \frac{U}{h} \Big) y + U
+\end{aligned}$$
+
+The moving bottom surface drags fluid in the $x$-direction
+at a volumetric rate $Q$, given by:
+
+$$\begin{aligned}
+    Q
+    = \int_0^{h(x)} v_x(x, y) \dd{y}
+    = \bigg[ \frac{p'}{6 \eta} y^3 - \frac{p'}{4 \eta} h y^2 - \frac{U}{2 h} y^2 + U y \bigg]_0^{h}
+    = - \frac{p'}{12 \eta} h^3 + \frac{U}{2} h
+\end{aligned}$$
+
+Assuming that the lubricant is incompressible,
+meaning that the same volume of fluid must be leaving a point as is entering it.
+In other words, $Q$ is independent of $x$,
+which allows us to write $p'(x)$ in terms of
+measurable constants and the known function $h(x)$:
+
+$$\begin{aligned}
+    \boxed{
+        p'
+        = 6 \eta \: \Big( \frac{U}{h^2} - \frac{2 Q}{h^3} \Big)
+    }
+\end{aligned}$$
+
+Then we insert this into our earlier expression for $v_x$, yielding:
+
+$$\begin{aligned}
+    v_x
+    &= 3 y (y - h) \Big( \frac{U}{h^2} - \frac{2 Q}{h^3} \Big) - \frac{U h}{h^2} y + \frac{U h^2}{h^2}
+    %\\
+    %&= U \frac{3 y (y - h) - h y + h^2}{h^2} - Q \frac{6 y (y - h)}{h^3}
+\end{aligned}$$
+
+Which, after some rearranging, can be written in the following form:
+
+$$\begin{aligned}
+    \boxed{
+        v_x
+        = U \frac{(3 y - h) (y - h)}{h^2} - Q \frac{6 y (y - h)}{h^3}
+    }
+\end{aligned}$$
+
+With this, we can find $v_y$ by exploiting incompressibility,
+i.e. the continuity equation states:
+
+$$\begin{aligned}
+    \pdv{v_y}{y}
+    = - \pdv{v_x}{x}
+    = - 2 h' \frac{U h - 3 Q}{h^4} \big( 2 h y - 3 y^2 \big)
+\end{aligned}$$
+
+Integrating with respect to $y$ thus leads to the following transverse velocity $v_y$:
+
+$$\begin{aligned}
+    \boxed{
+        v_y
+        = - 2 h' \frac{U h - 3 Q}{h^4} y^2 (h - y)
+    }
+\end{aligned}$$
+
+Typically, the lubricant is not in a preexisting pressure differential,
+i.e it is not getting pumped through the system.
+Although the pressure gradient $p'$ need not be zero,
+we therefore expect that its integral vanishes:
+
+$$\begin{aligned}
+    0
+    = \int_L p'(x) \dd{x}
+    = 6 \eta U \int_L \frac{1}{h(x)^2} \dd{x} - 12 \eta Q \int_L \frac{1}{h(x)^3} \dd{x}
+\end{aligned}$$
+
+Isolating this for $Q$, and defining $q$ as below, yields a simple equation:
+
+$$\begin{aligned}
+    Q
+    = \frac{1}{2} U q
+    \qquad \quad
+    q
+    \equiv \frac{\int_L h^{-2} \dd{x}}{\int_L h^{-3} \dd{x}}
+\end{aligned}$$
+
+We substitute this into $v_x$ and rearrange to get an interesting expression:
+
+$$\begin{aligned}
+    v_x
+    &= U \frac{3 y^2 - h y - 3 h y + h^2}{h^2} - U q \frac{3 y^2 - 3 h y}{h^3}
+    \\
+    &= U \Big( 1 - \frac{y}{h} \Big) \Big( 1 - \frac{3 y (h - q)}{h^2} \Big)
+\end{aligned}$$
+
+The first factor is always positive,
+but the second can be negative,
+if for some $y$-values:
+
+$$\begin{aligned}
+    h^2 < 3 y (h - q)
+    \quad \implies \quad
+    y > \frac{h^2}{3 (h - q)}
+\end{aligned}$$
+
+Since $h > y$, such $y$-values will only exist
+if $h$ is larger than some threshold:
+
+$$\begin{aligned}
+    3 (h - q) > h
+    \quad \implies \quad
+    h > \frac{3}{2} q
+\end{aligned}$$
+
+If this condition is satisfied,
+there will be some flow reversal:
+rather than just getting dragged by the shearing motion,
+the lubricant instead "rolls" inside the gap.
+This is confirmed by $v_y$:
+
+$$\begin{aligned}
+    v_y
+    = - U h' \frac{2 h - 3 q}{h^4} y^2 (h - y)
+\end{aligned}$$
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
diff --git a/content/know/concept/selection-rules/index.pdc b/content/know/concept/selection-rules/index.pdc
index 5da97e7..22dfd64 100644
--- a/content/know/concept/selection-rules/index.pdc
+++ b/content/know/concept/selection-rules/index.pdc
@@ -20,7 +20,7 @@ the total angular momentum and its $z$-component:
 
 $$\begin{aligned}
     \matrixel{f}{\hat{O}}{i}
-    = \matrixel{n_f \ell_f m_f}{\hat{O}}{n_i \ell_i m_i}
+    = \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i}
 \end{aligned}$$
 
 Where $\hat{O}$ is an operator, $\ket{i}$ is an initial state, and
@@ -28,8 +28,6 @@ $\ket{f}$ is a final state (usually at least; $\ket{i}$ and $\ket{f}$
 can be any states). **Selection rules** are requirements on the relations
 between $\ell_i$, $\ell_f$, $m_i$ and $m_f$, which, if not met,
 guarantee that the above matrix element is zero.
-Note that $n_f$ and $n_i$ typically do not matter in this context,
-so they will be omitted from now on.
 
 
 ## Parity rules
@@ -478,11 +476,180 @@ $$\begin{aligned}
 </div>
 
 
+## Rotational rules
+
+Given a general (pseudo)scalar operator $\hat{s}$,
+which, by nature, must satisfy the
+following relations with the angular momentum operators:
+
+$$\begin{aligned}
+    \comm*{\hat{L}^2}{\hat{s}} = 0
+    \qquad
+    \comm*{\hat{L}_z}{\hat{s}} = 0
+    \qquad
+    \comm*{\hat{L}_{\pm}}{\hat{s}} = 0
+\end{aligned}$$
+
+Where $\hat{L}_\pm \equiv \hat{L}_x \pm i \hat{L}_y$.
+The inner product of any such $\hat{s}$ must obey these selection rules:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta \ell = 0
+    }
+    \qquad \quad
+    \boxed{
+        \Delta m = 0
+    }
+\end{aligned}$$
+
+It is common to write this in the following more complete way, where
+$\matrixel{\ell_f}{|\hat{s}|}{\ell_i}$ is the **reduced matrix element**,
+which is identical to $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}$, but
+with a different notation to say that it does not depend on $m_f$ or $m_i$:
+
+$$\begin{aligned}
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
+        = \delta_{\ell_f \ell_i} \delta_{m_f m_i} \matrixel{\ell_f}{|\hat{s}|}{\ell_i}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-rot-scalar"/>
+<label for="proof-rot-scalar">Proof</label>
+<div class="hidden">
+<label for="proof-rot-scalar">Proof.</label>
+Firstly, we look at the commutator of $\hat{s}$ with the $z$-component $\hat{L}_z$:
+$$\begin{aligned}
+    0
+    = \matrixel{\ell_f m_f}{\comm*{\hat{L}_z}{\hat{s}}}{\ell_i m_i}
+    &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}_z}{\ell_i m_i}
+    \\
+    &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
+\end{aligned}$$
+
+Which can only be true if $m_f \!-\! m_i = 0$, unless,
+of course, $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} = 0$ by itself.
+
+Secondly, we look at the commutator of $\hat{s}$ with the total angular momentum $\hat{L}^2$:
+
+$$\begin{aligned}
+    0
+    = \matrixel{\ell_f m_f}{\comm*{\hat{L}^2}{\hat{s}}}{\ell_i m_i}
+    &= \matrixel{\ell_f m_f}{\hat{L}^2 \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}^2}{\ell_i m_i}
+    \\
+    &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
+\end{aligned}$$
+
+Assuming $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \neq 0$,
+this can only be satisfied if the following holds:
+
+$$\begin{aligned}
+    0
+    = \ell_f^2 + \ell_f - \ell_i^2 - \ell_i
+    = (\ell_f + \ell_i) (\ell_f - \ell_i) + (\ell_f - \ell_i)
+\end{aligned}$$
+
+If $\ell_f = \ell_i = 0$ this equation is trivially satisfied.
+Otherwise, the only option is $\ell_f \!-\! \ell_i = 0$,
+which is another part of the selection rule.
+
+Thirdly, we look at the commutator of $\hat{s}$ with the ladder operators $\hat{L}_\pm$:
+
+$$\begin{aligned}
+    0
+    = \matrixel{\ell_f m_f}{\comm*{\hat{L}_\pm}{\hat{s}}}{\ell_i m_i}
+    &= \matrixel{\ell_f m_f}{\hat{L}_\pm \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}_\pm}{\ell_i m_i}
+    \\
+    &= C_f \matrixel{\ell_f (m_f\!\mp\!1)}{\hat{s}}{\ell_i m_i} - C_i \matrixel{\ell_f m_f}{\hat{s}}{\ell_i (m_i\!\pm\!1)}
+\end{aligned}$$
+
+Where $C_f$ and $C_i$ are constants given below.
+We already know that $\Delta \ell = 0$ and $\Delta m = 0$,
+so the above matrix elements are only nonzero if $m_f = m_i \pm 1$.
+Therefore:
+
+$$\begin{aligned}
+    C_i
+    &= \hbar \sqrt{\ell_i (\ell_i + 1) - m_i (m_i \pm 1)}
+    \\
+    C_f
+    &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_f (m_f \!\mp\! 1)}
+    \\
+    &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - (m_i \!\pm\! 1) (m_i \!\pm\! 1 \!\mp\! 1)}
+    \\
+    &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_i (m_i \!\pm\! 1)}
+\end{aligned}$$
+
+In other words, $C_f = C_i$. The above equation therefore reduces to:
+
+$$\begin{aligned}
+    \matrixel{\ell_f m_i}{\hat{s}}{\ell_i m_i}
+    &= \matrixel{\ell_f (m_i \!\pm\! 1)}{\hat{s}}{\ell_i (m_i\!\pm\!1)}
+\end{aligned}$$
+
+Which means that the value of the matrix element
+does not depend on $m_i$ (or $m_f$) at all.
+</div>
+</div>
+
+Similarly, given a general (pseudo)vector operator $\vu{V}$,
+which, by nature, must satisfy the following commutation relations,
+where $\hat{V}_\pm \equiv \hat{V}_x \pm i \hat{V}_y$:
+
+$$\begin{gathered}
+    \comm*{\hat{L}_z}{\hat{V}_z} = 0
+    \qquad
+    \comm*{\hat{L}_z}{\hat{V}_{\pm}} = \pm \hbar \hat{V}_{\pm}
+    \qquad
+    \comm*{\hat{L}_{\pm}}{\hat{V}_z} = \mp \hbar \hat{V}_{\pm}
+    \\
+    \comm*{\hat{L}_{\pm}}{\hat{V}_{\pm}} = 0
+    \qquad
+    \comm*{\hat{L}_{\pm}}{\hat{V}_{\mp}} = \pm 2 \hbar \hat{V}_z
+\end{gathered}$$
+
+The inner product of any such $\vu{V}$ must obey the following selection rules:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta \ell
+        = 0 \:\:\mathrm{or}\: \pm 1
+    }
+    \qquad
+    \boxed{
+        \Delta m
+        = 0 \:\:\mathrm{or}\: \pm 1
+    }
+\end{aligned}$$
+
+In fact, the complete result involves the Clebsch-Gordan coefficients (from spin addition):
+
+$$\begin{gathered}
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{V}_{z}}{\ell_i m_i}
+        = C^{\ell_i \: 1 \: \ell_f}_{m_i \: 0 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i}
+    }
+    \\
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{V}_{+}}{\ell_i m_i}
+        = - \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: 1 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i}
+    }
+    \\
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{V}_{-}}{\ell_i m_i}
+        = \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: -1 \:m_f} \matrixel{\ell_f}{|\hat{V}}{|\ell_i}
+    }
+\end{gathered}$$
+
+
 ## Superselection rule
 
-Selection rules need not always be about atomic electron transitions.
+Selection rules are not always about atomic electron transitions, or angular momenta even.
+
 According to the **principle of indistinguishability**,
-permutating identical particles never leads to an observable difference.
+permuting identical particles never leads to an observable difference.
 In other words, the particles are fundamentally indistinguishable,
 so for any observable $\hat{O}$ and multi-particle state $\ket{\Psi}$, we can say: