From 2d15bb2b329b825df262c0a128d391b0dcbede58 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Fri, 12 Mar 2021 17:51:35 +0100
Subject: Expand knowledge base

---
 .../know/concept/hydrostatic-pressure/index.pdc    | 209 +++++++++++++++
 content/know/concept/meniscus/index.pdc            | 192 ++++++++++++++
 .../know/concept/quantum-teleportation/index.pdc   |   2 +-
 .../concept/rayleigh-plateau-instability/index.pdc | 286 +++++++++++++++++++++
 4 files changed, 688 insertions(+), 1 deletion(-)
 create mode 100644 content/know/concept/hydrostatic-pressure/index.pdc
 create mode 100644 content/know/concept/meniscus/index.pdc
 create mode 100644 content/know/concept/rayleigh-plateau-instability/index.pdc

(limited to 'content/know/concept')

diff --git a/content/know/concept/hydrostatic-pressure/index.pdc b/content/know/concept/hydrostatic-pressure/index.pdc
new file mode 100644
index 0000000..3bbb0af
--- /dev/null
+++ b/content/know/concept/hydrostatic-pressure/index.pdc
@@ -0,0 +1,209 @@
+---
+title: "Hydrostatic pressure"
+firstLetter: "H"
+publishDate: 2021-03-12
+categories:
+- Physics
+- Fluid mechanics
+
+date: 2021-03-12T14:37:56+01:00
+draft: false
+markup: pandoc
+---
+
+# Hydrostatic pressure
+
+The pressure $p$ inside a fluid at rest,
+the so-called **hydrostatic pressure**,
+is an important quantity.
+Here we will properly define it,
+and derive the equilibrium condition for the fluid to be at rest,
+both with and without an arbitrary gravity field.
+
+
+## Without gravity
+
+Inside the fluid, we can imagine small arbitrary partition surfaces,
+with normal vector $\vu{n}$ and area $\dd{S}$,
+yielding the following vector element $\dd{\va{S}}$:
+
+$$\begin{aligned}
+    \dd{\va{S}}
+    = \vu{n} \dd{S}
+\end{aligned}$$
+
+The orientation of these surfaces does not matter.
+The **pressure** $p(\va{r})$ is defined as the force-per-area
+of these tiny surface elements:
+
+$$\begin{aligned}
+    \dd{\va{F}}
+    = - p(\va{r}) \dd{\va{S}}
+\end{aligned}$$
+
+The negative sign is there because a positive pressure is conventionally defined
+to push from the positive (normal) side of $\dd{\va{S}}$ to the negative side.
+The total force $\va{F}$ on a larger surface inside the fluid is
+then given by the surface integral over many adjacent $\dd{\va{S}}$:
+
+$$\begin{aligned}
+    \va{F}
+    = - \int_S p(\va{r}) \dd{\va{S}}
+\end{aligned}$$
+
+If we now consider a *closed* surface,
+which encloses a "blob" of the fluid,
+then we can use Gauss' theorem to get a volume integral:
+
+$$\begin{aligned}
+    \va{F}
+    = - \oint_S p \dd{\va{S}}
+    = - \int_V \nabla p \dd{V}
+\end{aligned}$$
+
+Since the total force on the blob is simply the sum of the forces $\dd{\va{F}}$
+on all its constituent volume elements $\dd{V}$,
+we arrive at the following relation:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\va{F}}
+        = - \nabla p \dd{V}
+    }
+\end{aligned}$$
+
+If the fluid is at rest, then all forces on the blob cancel out
+(otherwise it would move).
+Since we are currently neglecting all forces other than pressure,
+this is equivalent to demanding that $\dd{\va{F}} = 0$,
+which implies that $\nabla p = 0$, i.e. the pressure is constant.
+
+$$\begin{aligned}
+    \boxed{
+        \nabla p = 0
+    }
+\end{aligned}$$
+
+
+## With gravity
+
+If we include gravity, then,
+in addition to the pressure's *contact force* $\va{F}_p$ from earlier,
+there is also a *body force* $\va{F}_g$ acting on
+the arbitrary blob $V$ of fluid enclosed by $S$:
+
+$$\begin{aligned}
+    \va{F}_g
+    = \int_V \rho \va{g} \dd{V}
+\end{aligned}$$
+
+Where $\rho$ is the fluid's density (which need not be constant)
+and $\va{g}$ is the gravity field given in units of force-per-mass.
+For a fluid at rest, these forces must cancel out:
+
+$$\begin{aligned}
+    \va{F}
+    = \va{F}_g + \va{F}_p
+    = \int_V \rho \va{g} - \nabla p \dd{V}
+    = 0
+\end{aligned}$$
+
+Since this a single integral over an arbitrary volume,
+it implies that every point of the fluid must
+locally satisfy the following equilibrium condition:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla p
+        = \rho \va{g}
+    }
+\end{aligned}$$
+
+On Earth (or another body with strong gravity),
+it is reasonable to treat $\va{g}$ as only pointing in the downward $z$-direction,
+in which case the above condition turns into:
+
+$$\begin{aligned}
+    p
+    = \rho g_0 z
+\end{aligned}$$
+
+Where $g_0$ is the magnitude of the $z$-component of $\va{g}$.
+We can generalize the equilibrium condition by treating
+the gravity field as the gradient of the gravitational potential $\Phi$:
+
+$$\begin{aligned}
+    \va{g}(\va{r})
+    = - \nabla \Phi(\va{r})
+\end{aligned}$$
+
+With this, the equilibrium condition is turned into the following equation:
+
+$$\begin{aligned}
+    \nabla \Phi + \frac{\nabla p}{\rho}
+    = 0
+\end{aligned}$$
+
+In practice, the density $\rho$ of the fluid
+may be a function of the pressure $p$ (compressibility)
+and/or temperature $T$ (thermal expansion).
+We will tackle the first complication, but neglect the second,
+i.e. we assume that the temperature is equal across the fluid.
+
+We then define the **pressure potential** $w(p)$ as
+the indefinite integral of the density:
+
+$$\begin{aligned}
+    w(p)
+    = \int \frac{1}{\rho(p)} \dd{p}
+\end{aligned}$$
+
+Using this, we can rewrite the equilibrium condition as a single gradient like so:
+
+$$\begin{aligned}
+    0
+    = \nabla \Phi + \frac{\nabla p}{\rho}
+    = \nabla \Phi + \dv{w}{p} \nabla p
+    = \nabla \Big( \Phi + w(p) \Big)
+\end{aligned}$$
+
+By defining the **effective gravitational potential** $\Phi^* = \Phi + w(p)$,
+we get the cleanest form yet of the equilibrium condition,
+which states that $\Phi^*$ must be a constant:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \Phi^*
+        = 0
+    }
+\end{aligned}$$
+
+At every point in the fluid, despite $p$ being variable,
+the force that is applied by the pressure must have the same magnitude in all directions at that point.
+This statement is known as **Pascal's law**,
+and is due to the fact that all forces must cancel out
+for an arbitrary blob:
+
+$$\begin{aligned}
+    \va{F}
+    = \va{F}_g + \va{F}_p
+    = 0
+\end{aligned}$$
+
+Let the blob be a cube with side $a$.
+Now, $\va{F}_p$ is a contact force,
+meaning it acts on the surface, and is thus proportional to $a^2$,
+however, $\va{F}_g$ is a body force,
+meaning it acts on the volume, and is thus proportional to $a^3$.
+Since we are considering a *point* in the fluid,
+$a$ is infinitesimally small,
+so that $\va{F}_p$ dominates $\va{F}_g$.
+Consequently, at equilibrium, $\va{F}_p$ must cancel out by itself,
+which means that the pressure is the same in all directions.
+
+
+
+## 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/meniscus/index.pdc b/content/know/concept/meniscus/index.pdc
new file mode 100644
index 0000000..d11c2d8
--- /dev/null
+++ b/content/know/concept/meniscus/index.pdc
@@ -0,0 +1,192 @@
+---
+title: "Meniscus"
+firstLetter: "M"
+publishDate: 2021-03-11
+categories:
+- Physics
+- Fluid mechanics
+
+date: 2021-03-11T14:39:56+01:00
+draft: false
+markup: pandoc
+---
+
+# Meniscus
+
+When a fluid interface, e.g. the surface of a liquid,
+touches a flat solid wall, it will curve to meet it.
+This small rise or fall is called a **meniscus**,
+and is caused by surface tension and gravity.
+
+In 2D, let the vertical $y$-axis be a flat wall,
+and the fluid tend to $y = 0$ when $x \to \infty$.
+Close to the wall, i.e. for small $x$, the liquid curves up or down
+to touch the wall at a height $y = d$.
+
+Three forces are at work here:
+the first two are the surface tension $\alpha$ of the fluid surface,
+and the counter-pull $\alpha \sin\phi$ of the wall against the tension,
+where $\phi$ is the contact angle.
+The third is the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) gradient
+inside the small portion of the fluid above/below the ambient level,
+which exerts a total force on the wall given by
+(for $\phi < \pi/2$ so that $d > 0$):
+
+$$\begin{aligned}
+    \int_0^d \rho g y \dd{y}
+    = \frac{1}{2} \rho g d^2
+\end{aligned}$$
+
+If you were wondering about the units,
+keep in mind that there is an implicit $z$-direction here too.
+This results in the following balance equation for the forces at the wall:
+
+$$\begin{aligned}
+    \alpha
+    = \alpha \sin\phi + \frac{1}{2} \rho g d^2
+\end{aligned}$$
+
+We isolate this relation for $d$
+and use some trigonometric magic to rewrite it:
+
+$$\begin{aligned}
+    d
+    = \sqrt{\frac{\alpha}{\rho g}} \sqrt{2 (1 - \sin\phi)}
+    = \sqrt{\frac{\alpha}{\rho g}} \sqrt{4 \sin^2\!\Big(\frac{\pi}{4} - \frac{\phi}{2}\Big)}
+\end{aligned}$$
+
+Here, we recognize the definition of the capillary length $L_c = \sqrt{\alpha / (\rho g)}$,
+yielding an expression for $d$
+that is valid both for $\phi < \pi/2$ (where $d > 0$)
+and $\phi > \pi/2$ (where $d < 0$):
+
+$$\begin{aligned}
+    \boxed{
+        d
+        = 2 L_c \sin\!\Big(\frac{\pi/2 - \phi}{2}\Big)
+    }
+\end{aligned}$$
+
+Next, we would like to know the exact shape of the meniscus.
+To do this, we need to describe the liquid surface differently,
+using the elevation angle $\theta$ relative to the $y = 0$ plane.
+The curve $\theta(s)$ is a function of the arc length $s$,
+where $\dd{s}^2 = \dd{x}^2 + \dd{y}^2$,
+and is governed by:
+
+$$\begin{aligned}
+    \dv{x}{s}
+    = \cos\theta
+    \qquad
+    \dv{y}{s}
+    = \sin\theta
+    \qquad
+    \dv{\theta}{s}
+    = \frac{1}{R}
+\end{aligned}$$
+
+The last equation describes the curvature radius $R$
+of the surface along the $x$-axis.
+Since we are considering a flat wall,
+there is no curvature in the orthogonal principal direction.
+
+Just below the liquid surface in the meniscus,
+we expect the hydrostatic pressure
+and the Young-Laplace law agree about the pressure $p$,
+where $p_0$ is the external air pressure:
+
+$$\begin{aligned}
+    p_0 - \rho g y
+    = p_0 - \frac{\alpha}{R}
+\end{aligned}$$
+
+Rearranging this yields that $R = L_c^2 / y$.
+Inserting this into the curvature equation gives us:
+
+$$\begin{aligned}
+    \dv{\theta}{s}
+    = \frac{y}{L_c^2}
+\end{aligned}$$
+
+By differentiating this equation with respect to $s$
+and using $\dv*{y}{s} = \sin\theta$, we arrive at:
+
+$$\begin{aligned}
+    \boxed{
+        L_c^2 \dv[2]{\theta}{s} = \sin\theta
+    }
+\end{aligned}$$
+
+To solve this equation, we multiply it by $\dv*{\theta}{s}$,
+which is nonzero close to the wall:
+
+$$\begin{aligned}
+    L_c^2 \dv[2]{\theta}{s} \dv{\theta}{s}
+    = \dv{\theta}{s} \sin\theta
+\end{aligned}$$
+
+We integrate both sides with respect to $s$
+and set the integration constant to $1$,
+such that we get zero when $\theta \to 0$ away from the wall:
+
+$$\begin{aligned}
+    \frac{L_c^2}{2} \Big( \dv{\theta}{s} \Big)^2
+    = 1 - \cos\theta
+\end{aligned}$$
+
+Isolating this for $\dv*{\theta}{s}$ and using a trigonometric identity then yields:
+
+$$\begin{aligned}
+    \dv{\theta}{s}
+    = \frac{1}{L_c} \sqrt{2 (1 - \cos\theta)}
+    = \frac{1}{L_c} \sqrt{4 \sin^2\!\Big( \frac{\theta}{2} \Big)}
+    = - \frac{2}{L_c} \sin\!\Big( \frac{\theta}{2} \Big)
+\end{aligned}$$
+
+We use trigonometric relations on the equations
+for $\dv*{x}{s}$ and $\dv*{y}{s}$ to get $\theta$-derivatives:
+
+$$\begin{aligned}
+    \dv{x}{\theta}
+    &= \dv{x}{s} \dv{s}{\theta}
+    = \bigg( 1 - 2 \sin^2\!\Big( \frac{\theta}{2} \Big) \bigg) \bigg( \dv{\theta}{s} \bigg)^{-1}
+    = L_c \sin\!\Big( \frac{\theta}{2} \Big) - \frac{L_c}{2 \sin(\theta/2)}
+    \\
+    \dv{y}{\theta}
+    &= \dv{y}{s} \dv{s}{\theta}
+    = \bigg( 2 \sin\!\Big(\frac{\theta}{2}\Big) \cos\!\Big(\frac{\theta}{2}\Big) \bigg) \bigg( \dv{\theta}{s} \bigg)^{-1}
+    = - L_c \cos\!\Big( \frac{\theta}{2} \Big)
+\end{aligned}$$
+
+Let $\theta_0 = \phi - \pi/2$ be the initial elevation angle $\theta(0)$ at the wall.
+Then, by integrating the above equations, we get the following solutions:
+
+$$\begin{gathered}
+    \boxed{
+        \frac{x}{L_c}
+        = 2 \cos\!\Big(\frac{\theta_0}{2}\Big) + \log\!\bigg| \tan\!\Big(\frac{\theta_0}{4}\Big) \bigg|
+        - 2 \cos\!\Big(\frac{\theta}{2}\Big) + \log\!\bigg| \tan\!\Big(\frac{\theta}{4}\Big) \bigg|
+    }
+    \\
+    \boxed{
+        \frac{y}{L_c}
+        = - 2 \sin\!\Big(\frac{\theta}{2}\Big)
+    }
+\end{gathered}$$
+
+Where the integration constant has been chosen such that $y \to 0$ for $\theta \to 0$ away from the wall,
+and $x = 0$ for $\theta = \theta_0$.
+This result is consistent with our earlier expression for $d$:
+
+$$\begin{aligned}
+    d
+    = y(\theta_0)
+    = - 2 L_c \sin\!\Big(\frac{\theta_0}{2}\Big)
+    = 2 L_c \sin\!\Big( \frac{\pi/2 - \phi}{2} \Big)
+\end{aligned}$$
+
+
+## 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/quantum-teleportation/index.pdc b/content/know/concept/quantum-teleportation/index.pdc
index b4394c9..bf33a11 100644
--- a/content/know/concept/quantum-teleportation/index.pdc
+++ b/content/know/concept/quantum-teleportation/index.pdc
@@ -30,7 +30,7 @@ This is not enough: even if Alice did know $\alpha$ and $\beta$ exactly
 (which would need her having infinitely many copies to measure),
 sending an arbitrary real number requires an infinite amount of classical data.
 
-However, between them, she and Bob also have an entangled Bell state,
+However, between them, she and Bob also have an entangled [Bell state](/know/concept/bell-state/),
 e.g. $\ket*{\Phi^+}_{AB}$ (it does not matter which Bell state it is)
 The state of the composite system is then as follows,
 with $A'$ being Alice' qubit, $A$ her side of the Bell state, and $B$ Bob's side:
diff --git a/content/know/concept/rayleigh-plateau-instability/index.pdc b/content/know/concept/rayleigh-plateau-instability/index.pdc
new file mode 100644
index 0000000..f7e10d4
--- /dev/null
+++ b/content/know/concept/rayleigh-plateau-instability/index.pdc
@@ -0,0 +1,286 @@
+---
+title: "Rayleigh-Plateau instability"
+firstLetter: "R"
+publishDate: 2021-03-10
+categories:
+- Physics
+- Fluid mechanics
+- Perturbation
+
+date: 2021-03-10T09:13:22+01:00
+draft: false
+markup: pandoc
+---
+
+# Rayleigh-Plateau instability
+
+In fluid mechanics, the **Rayleigh-Plateau instability** causes
+a column of liquid to break up due to surface tension.
+It is the reason why a smooth stream of water (e.g. from a tap)
+eventually breaks into droplets as it falls.
+
+Consider an infinitely long cylinder of liquid
+with radius $R_0$ and surface tension $\alpha$.
+In this case, the Young-Laplace equation states that its internal pressure
+is a constant $p_i$ expressed as follows,
+where $p_o$ is the exterior air pressure:
+
+$$\begin{aligned}
+    p_i
+    = p_o + \frac{\alpha}{R_0}
+\end{aligned}$$
+
+We assume that the liquid is at rest.
+Alternatively, if it is moving in the $z$-direction,
+we can also let our coordinate system travel at the same speed.
+Anyway, for convenience,
+we neglect any motion or acceleration of the liquid column.
+
+Next, we add a perturbation $p_\epsilon$, assumed to be small,
+to the internal pressure, which we allow to vary with time and space.
+We use cylindrical coordinates:
+
+$$\begin{aligned}
+    p(r, \phi, z, t) = p_i + p_\epsilon(r, \phi, z, t)
+\end{aligned}$$
+
+This internal pressure difference will cause the liquid to start to flow.
+We express the flow velocity as a vector $\vec{u} = (u_r, u_\phi, u_z)$,
+which obeys the following Euler equations:
+
+$$\begin{aligned}
+    \pdv{\vec{u}}{t} + (\vec{u} \cdot \nabla) \vec{u}
+    = - \frac{1}{\rho} \nabla p
+    \qquad \qquad
+    \nabla \cdot \vec{u} = 0
+\end{aligned}$$
+
+The latter equation states that the fluid is incompressible.
+We assume that $\vec{u}$ is so small that we can ignore
+the quadratic term in the former equation, leaving:
+
+$$\begin{aligned}
+    \pdv{\vec{u}}{t}
+    = - \frac{1}{\rho} \nabla p_\epsilon
+\end{aligned}$$
+
+Taking the divergence and using incompressibility
+yields the Laplace equation for $p_\epsilon$:
+
+$$\begin{aligned}
+    - \frac{1}{\rho} \nabla^2 p_\epsilon
+    = \pdv{t} (\nabla \cdot \vec{u})
+    = 0
+    \qquad \implies \qquad
+    \nabla^2 p_\epsilon = 0
+\end{aligned}$$
+
+We write out the Laplacian in cylindrical coordinates
+to get the following problem:
+
+$$\begin{aligned}
+    \nabla^2 p_\epsilon
+    = \pdv[2]{p_\epsilon}{r} + \frac{1}{r} \pdv{p_\epsilon}{r} + \pdv[2]{p_\epsilon}{z} + \frac{1}{r^2} \pdv[2]{p_\epsilon}{\phi}
+    = 0
+\end{aligned}$$
+
+Finally, we add a perturbation $R_\epsilon \ll R_0$
+to the radius of the surface of the liquid column:
+
+$$\begin{aligned}
+    R(z, t)
+    = R_0 + R_\epsilon(z, t)
+\end{aligned}$$
+
+Note that there is no dependence on the angle $\phi$;
+the deformation is assumed to be symmetric.
+Imagine the cross-section of the cylinder,
+and convince yourself that all asymmetric deformations
+will be removed by surface tension, which prefers a circular shape.
+We thus assume that $R_\epsilon$, $p_\epsilon$ and $\vec{u}$
+do not depend on $\phi$.
+The Laplace equation then reduces to:
+
+$$\begin{aligned}
+    \nabla^2 p_\epsilon
+    = \pdv[2]{p_\epsilon}{r} + \frac{1}{r} \pdv{p_\epsilon}{r} + \pdv[2]{p_\epsilon}{z}
+    = 0
+\end{aligned}$$
+
+Before solving this, we need boundary conditions.
+The radial fluid velocity $u_r$ (the $r$-component of $\vec{u}$)
+at the column surface $r\!=\!R$ is the *material derivative* of $R_\epsilon$:
+
+$$\begin{aligned}
+    u_r(r\!=\!R)
+    = \frac{\mathrm{D} R_\epsilon}{\mathrm{D} t}
+    = \pdv{R_\epsilon}{t} + u_z(r\!=\!R) \pdv{R_\epsilon}{z}
+\end{aligned}$$
+
+We linearize this by assuming that the deformation $R_\epsilon$
+varies slowly with respect to $z$:
+
+$$\begin{aligned}
+    u_r(r\!=\!R)
+    \approx \pdv{R_\epsilon}{t}
+\end{aligned}$$
+
+Meanwhile, we can write the boundary condition of the pressure $p$
+in two ways, respectively from the Young-Laplace equation
+and the definition of the perturbation $p_\epsilon$:
+
+$$\begin{aligned}
+    p(r\!=\!R)
+    = p_o + \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big)
+    \qquad \quad
+    p(r\!=\!R)
+    = p_i + p_\epsilon(r\!=\!R)
+\end{aligned}$$
+
+Where $R_1$ and $R_2$ are the principal curvature radii of the column surface.
+These two expressions must be equivalent,
+so, by inserting the definition of $p_i = p_o + \alpha / R_0$:
+
+$$\begin{aligned}
+    p_o + \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big)
+    %= p_i + p_\epsilon(r\!=\!R)
+    = p_o + \frac{\alpha}{R_0} + p_\epsilon(r\!=\!R)
+\end{aligned}$$
+
+Isolating this equation for $p_\epsilon$ yields the desired boundary condition:
+
+$$\begin{aligned}
+    p_\epsilon(r\!=\!R)
+    = \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big) - \frac{\alpha}{R_0}
+\end{aligned}$$
+
+The principal radius around the circumference is $R_0 + R_\epsilon$,
+while the curvature along the length can be approximated
+using the second $z$-derivative of $R_\epsilon$:
+
+$$\begin{aligned}
+    p_\epsilon(r\!=\!R)
+    \approx \alpha \Big( \frac{1}{R_0 + R_\epsilon} - \pdv[2]{R_\epsilon}{z} \Big) - \frac{\alpha}{R_0}
+\end{aligned}$$
+
+This can be simplified a bit by using the assumption that $R_\epsilon$ is small:
+
+$$\begin{aligned}
+    p_\epsilon(r\!=\!R)
+    \approx - \alpha \Big( \frac{R_\epsilon}{R_0^2 + R_\epsilon} + \pdv[2]{R_\epsilon}{z} \Big)
+    \approx - \alpha \Big( \frac{R_\epsilon}{R_0^2} + \pdv[2]{R_\epsilon}{z} \Big)
+\end{aligned}$$
+
+At last, we have all the necessary boundary condition.
+We now make the following ansatz,
+where $k$ is the wavenumber
+and $\sigma$ describes exponential growth or decay:
+
+$$\begin{aligned}
+    \vec{u}(r, z, t)
+    &= \vec{u}(r) \exp(\sigma t) \cos(k z)
+    \\
+    p_\epsilon(r, z, t)
+    &= p_\epsilon(r) \exp(\sigma t) \cos(k z)
+    \\
+    R_\epsilon(z, t)
+    &= R_\epsilon \exp(\sigma t) \cos(k z)
+\end{aligned}$$
+
+This is justified by the fact that we can Fourier-expand any perturbation;
+this ansatz is simply the dominant term of the resulting series.
+
+Inserting this into the Laplace equation for $p_\epsilon$ yields
+Bessel's modified equation of order zero:
+
+$$\begin{aligned}
+    \dv[2]{p_\epsilon}{r} + \frac{1}{r} \dv{p_\epsilon}{r} - k^2 p_\epsilon
+    = 0
+\end{aligned}$$
+
+This has well-known solutions: the modified Bessel functions $I_0$ and $K_0$.
+However, because $K_0$ diverges at $r = 0$, we must set the constant $B = 0$:
+
+$$\begin{aligned}
+    p_\epsilon(r)
+    = A I_0(kr) + B K_0(kr)
+    = A I_0(kr)
+\end{aligned}$$
+
+Inserting the ansatz into the boundary condition for $p_\epsilon$
+gives us the following relation:
+
+$$\begin{aligned}
+    p_\epsilon(r\!=\!R)
+    = - \alpha R_\epsilon \Big( \frac{1}{R_0^2} + k^2 \Big)
+    = A I_0(k R)
+\end{aligned}$$
+
+Meanwhile, the linearized Euler equation governing $\vec{u}$
+states that $u_r$ is given by:
+
+$$\begin{aligned}
+    \sigma u_r
+    = - \frac{1}{\rho} \dv{p_\epsilon}{r}
+    = - \frac{A k}{\rho} I_0'(kr)
+\end{aligned}$$
+
+Now that we have an expression for $u_r$,
+we can revisit its boundary condition:
+
+$$\begin{aligned}
+    u_r(r\!=\!R)
+    = - \frac{A k}{\rho \sigma} I_0'(k R)
+    = \sigma R_\epsilon
+\end{aligned}$$
+
+Isolating this for $R_\epsilon$ and inserting it
+into the boundary condition for $p_\epsilon$ yields:
+
+$$\begin{aligned}
+    p_\epsilon(r\!=\!R)
+    = A I_0(kR)
+    = \alpha \Big( \frac{1}{R_0^2} + k^2 \Big) \Big( \frac{A k}{\rho \sigma^2} I_0'(k R) \Big)
+\end{aligned}$$
+
+Isolating this for the exponential growth/decay parameter $\sigma$
+gives us the desired result,
+where we have also used the fact that $R \approx R_0$:
+
+$$\begin{aligned}
+    \sigma^2
+    = \frac{\alpha k}{\rho R_0^2} (1 - k^2 R_0^2) \frac{I_0'(kR_0)}{I_0(kR_0)}
+\end{aligned}$$
+
+To get exponential growth (i.e. instability), we need $\sigma^2 > 0$.
+Since $(1 - k^2 R_0^2)$ is the only factor that can be negative,
+we need $k R_0 < 1$, leading us to the **critical wavelength** $\lambda_c$:
+
+$$\begin{aligned}
+    \boxed{
+        \lambda_c
+        = \frac{2 \pi}{k}
+        = 2 \pi R_0
+    }
+\end{aligned}$$
+
+If the perturbation wavelength $\lambda$ is larger than $\lambda_c$,
+surface tension creates a higher pressure in the narrower sections
+compared to the wider ones, thereby pumping the liquid into the bulges,
+further increasing their size until they become droplets.
+
+Else, if $\lambda < \lambda_c$, the tighter curvatures
+dominate the action of surface tension,
+which will then try to smoothen the surface by shrinking the bulges
+and widening the constrictions.
+In other words, the liquid column is stable in this case.
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
+2.  T. Bohr, A. Anderson,
+    *The Rayleigh-Plateau instability of a liquid column*, 2020,
+    unpublished.
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