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diff --git a/source/know/concept/rayleigh-plateau-instability/index.md b/source/know/concept/rayleigh-plateau-instability/index.md index 2ba060c..5dcf77f 100644 --- a/source/know/concept/rayleigh-plateau-instability/index.md +++ b/source/know/concept/rayleigh-plateau-instability/index.md @@ -16,11 +16,11 @@ 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 $\alpha$. +with radius $$R_0$$ and surface tension $$\alpha$$. In this case, the [Young-Laplace equation](/know/concept/young-laplace-law/) states that its internal pressure -is a constant $p_i$ expressed as follows, -where $p_o$ is the exterior air pressure: +is a constant $$p_i$$ expressed as follows, +where $$p_o$$ is the exterior air pressure: $$\begin{aligned} p_i @@ -28,12 +28,12 @@ $$\begin{aligned} \end{aligned}$$ We assume that the liquid is at rest. -Alternatively, if it is moving in the $z$-direction, +Alternatively, if it is moving in the $$z$$-direction, we can also let our coordinate system travel at the same speed. Anyway, for convenience, we neglect any motion or acceleration of the liquid column. -Next, we add a perturbation $p_\epsilon$, assumed to be small, +Next, we add a perturbation $$p_\epsilon$$, assumed to be small, to the internal pressure, which we allow to vary with time and space. We use cylindrical coordinates: @@ -42,7 +42,7 @@ $$\begin{aligned} \end{aligned}$$ This internal pressure difference will cause the liquid to start to flow. -We express the flow velocity as a vector $\vec{u} = (u_r, u_\phi, u_z)$, +We express the flow velocity as a vector $$\vec{u} = (u_r, u_\phi, u_z)$$, which obeys the following Euler equations: $$\begin{aligned} @@ -53,7 +53,7 @@ $$\begin{aligned} \end{aligned}$$ The latter equation states that the fluid is incompressible. -We assume that $\vec{u}$ is so small that we can ignore +We assume that $$\vec{u}$$ is so small that we can ignore the quadratic term in the former equation, leaving: $$\begin{aligned} @@ -62,7 +62,7 @@ $$\begin{aligned} \end{aligned}$$ Taking the divergence and using incompressibility -yields the Laplace equation for $p_\epsilon$: +yields the Laplace equation for $$p_\epsilon$$: $$\begin{aligned} - \frac{1}{\rho} \nabla^2 p_\epsilon @@ -81,7 +81,7 @@ $$\begin{aligned} = 0 \end{aligned}$$ -Finally, we add a perturbation $R_\epsilon \ll R_0$ +Finally, we add a perturbation $$R_\epsilon \ll R_0$$ to the radius of the surface of the liquid column: $$\begin{aligned} @@ -89,13 +89,13 @@ $$\begin{aligned} = R_0 + R_\epsilon(z, t) \end{aligned}$$ -Note that there is no dependence on the angle $\phi$; +Note that there is no dependence on the angle $$\phi$$; the deformation is assumed to be symmetric. Imagine the cross-section of the cylinder, and convince yourself that all asymmetric deformations will be removed by surface tension, which prefers a circular shape. -We thus assume that $R_\epsilon$, $p_\epsilon$ and $\vec{u}$ -do not depend on $\phi$. +We thus assume that $$R_\epsilon$$, $$p_\epsilon$$ and $$\vec{u}$$ +do not depend on $$\phi$$. The Laplace equation then reduces to: $$\begin{aligned} @@ -105,9 +105,9 @@ $$\begin{aligned} \end{aligned}$$ 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](/know/concept/material-derivative/) of $R_\epsilon$: +The radial fluid velocity $$u_r$$ (the $$r$$-component of $$\vec{u}$$) +at the column surface $$r\!=\!R$$ is the +[material derivative](/know/concept/material-derivative/) of $$R_\epsilon$$: $$\begin{aligned} u_r(r\!=\!R) @@ -115,17 +115,17 @@ $$\begin{aligned} = \pdv{R_\epsilon}{t} + u_z(r\!=\!R) \pdv{R_\epsilon}{z} \end{aligned}$$ -We linearize this by assuming that the deformation $R_\epsilon$ -varies slowly with respect to $z$: +We linearize this by assuming that the deformation $$R_\epsilon$$ +varies slowly with respect to $$z$$: $$\begin{aligned} u_r(r\!=\!R) \approx \pdv{R_\epsilon}{t} \end{aligned}$$ -Meanwhile, we can write the boundary condition of the pressure $p$ +Meanwhile, we can write the boundary condition of the pressure $$p$$ in two ways, respectively from the Young-Laplace equation -and the definition of the perturbation $p_\epsilon$: +and the definition of the perturbation $$p_\epsilon$$: $$\begin{aligned} p(r\!=\!R) @@ -135,32 +135,32 @@ $$\begin{aligned} = p_i + p_\epsilon(r\!=\!R) \end{aligned}$$ -Where $R_1$ and $R_2$ are the principal curvature radii of the column surface. +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 $p_i = p_o + \alpha / R_0$: +so, by inserting the definition of $$p_i = p_o + \alpha / R_0$$: $$\begin{aligned} p_o + \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big) = p_o + \frac{\alpha}{R_0} + p_\epsilon(r\!=\!R) \end{aligned}$$ -Isolating this equation for $p_\epsilon$ yields the desired boundary condition: +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 $R_0 + R_\epsilon$, +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 $R_\epsilon$: +using the second $$z$$-derivative of $$R_\epsilon$$: $$\begin{aligned} p_\epsilon(r\!=\!R) \approx \alpha \Big( \frac{1}{R_0 + R_\epsilon} - \pdvn{2}{R_\epsilon}{z} \Big) - \frac{\alpha}{R_0} \end{aligned}$$ -This can be simplified a bit by using the assumption that $R_\epsilon$ is small: +This can be simplified a bit by using the assumption that $$R_\epsilon$$ is small: $$\begin{aligned} p_\epsilon(r\!=\!R) @@ -170,8 +170,8 @@ $$\begin{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: +where $$k$$ is the wavenumber +and $$\sigma$$ describes exponential growth or decay: $$\begin{aligned} \vec{u}(r, z, t) @@ -187,7 +187,7 @@ $$\begin{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 +Inserting this into the Laplace equation for $$p_\epsilon$$ yields Bessel's modified equation of order zero: $$\begin{aligned} @@ -195,8 +195,8 @@ $$\begin{aligned} = 0 \end{aligned}$$ -This has well-known solutions: the modified Bessel functions $I_0$ and $K_0$. -However, because $K_0$ diverges at $r = 0$, we must set the constant $B = 0$: +This has well-known solutions: the modified Bessel functions $$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) @@ -204,7 +204,7 @@ $$\begin{aligned} = A I_0(kr) \end{aligned}$$ -Inserting the ansatz into the boundary condition for $p_\epsilon$ +Inserting the ansatz into the boundary condition for $$p_\epsilon$$ gives us the following relation: $$\begin{aligned} @@ -213,8 +213,8 @@ $$\begin{aligned} = A I_0(k R) \end{aligned}$$ -Meanwhile, the linearized Euler equation governing $\vec{u}$ -states that $u_r$ is given by: +Meanwhile, the linearized Euler equation governing $$\vec{u}$$ +states that $$u_r$$ is given by: $$\begin{aligned} \sigma u_r @@ -222,7 +222,7 @@ $$\begin{aligned} = - \frac{A k}{\rho} I_0'(kr) \end{aligned}$$ -Now that we have an expression for $u_r$, +Now that we have an expression for $$u_r$$, we can revisit its boundary condition: $$\begin{aligned} @@ -231,8 +231,8 @@ $$\begin{aligned} = \sigma R_\epsilon \end{aligned}$$ -Isolating this for $R_\epsilon$ and inserting it -into the boundary condition for $p_\epsilon$ yields: +Isolating this for $$R_\epsilon$$ and inserting it +into the boundary condition for $$p_\epsilon$$ yields: $$\begin{aligned} p_\epsilon(r\!=\!R) @@ -240,18 +240,18 @@ $$\begin{aligned} = \alpha \Big( \frac{1}{R_0^2} + k^2 \Big) \Big( \frac{A k}{\rho \sigma^2} I_0'(k R) \Big) \end{aligned}$$ -Isolating this for the exponential growth/decay parameter $\sigma$ +Isolating this for the exponential growth/decay parameter $$\sigma$$ gives us the desired result, -where we have also used the fact that $R \approx R_0$: +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** $\lambda_c$: +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** $$\lambda_c$$: $$\begin{aligned} \boxed{ @@ -261,12 +261,12 @@ $$\begin{aligned} } \end{aligned}$$ -If the perturbation wavelength $\lambda$ is larger than $\lambda_c$, +If the perturbation wavelength $$\lambda$$ is larger than $$\lambda_c$$, surface tension creates a higher pressure in the narrower sections compared to the wider ones, thereby pumping the liquid into the bulges, further increasing their size until they become droplets. -Else, if $\lambda < \lambda_c$, the tighter curvatures +Else, if $$\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. |