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+---
+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.