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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. |