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diff --git a/content/know/category/fluid-mechanics.md b/content/know/category/fluid-mechanics.md new file mode 100644 index 0000000..727f214 --- /dev/null +++ b/content/know/category/fluid-mechanics.md @@ -0,0 +1,9 @@ +--- +title: "Fluid mechanics" +firstLetter: "F" +date: 2021-02-26T20:30:24+01:00 +draft: false +layout: "category" +--- + +This page will fill itself. 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. |