From 45af77f068aaa57c052cd861412d53beecbe5e3b Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Fri, 9 Apr 2021 20:44:44 +0200
Subject: Expand knowledge base

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+---
+title: "Rayleigh-Plesset equation"
+firstLetter: "R"
+publishDate: 2021-04-06
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+
+date: 2021-04-06T19:03:36+02:00
+draft: false
+markup: pandoc
+---
+
+# Rayleigh-Plesset equation
+
+In fluid dynamics, the **Rayleigh-Plesset equation**
+describes how the radius of a spherical bubble evolves in time
+inside an incompressible liquid.
+Notably, it leads to [cavitation](/know/concept/cavitation/).
+
+
+## Simple form
+
+The simplest version of the Rayleigh-Plesset equation is found
+in the limiting case of a liquid with zero viscosity zero surface tension.
+
+Consider one of the [Euler equations](/know/concept/euler-equations/)
+for the velocity field $\va{v}$,
+where $\rho$ is the (constant) density:
+
+$$\begin{aligned}
+    \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+    = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v}
+    = - \frac{\nabla p}{\rho}
+\end{aligned}$$
+
+We make the ansatz $\va{v} = v(r, t) \vu{e}_r$,
+where $\vu{e}_r$ is the basis vector;
+in other words, we demand that the only spatial variation of the flow is in $r$.
+The above Euler equation then becomes:
+
+$$\begin{aligned}
+    \pdv{v}{t} + v \pdv{v}{r}
+    = - \frac{1}{\rho} \pdv{p}{r}
+\end{aligned}$$
+
+Meanwhile, the incompressibility condition
+is as follows in this situation:
+
+$$\begin{aligned}
+    \nabla \cdot \va{v}
+    = \frac{1}{r^2} \pdv{(r^2 v)}{r}
+    = 0
+\end{aligned}$$
+
+This is only satisfied if $r^2 v$ is constant with respect to $r$,
+leading us to a solution $v(r)$ given by:
+
+$$\begin{aligned}
+    v(r)
+    = \frac{C(t)}{r^2}
+\end{aligned}$$
+
+Where $C(t)$ is an unknown function that does not depend on $r$.
+We then insert this result in the earlier Euler equation,
+and isolate it for $\pdv*{p}{r}$, yielding:
+
+$$\begin{aligned}
+    \pdv{p}{r}
+    = - \rho \bigg( \pdv{v}{t} + v \pdv{v}{r} \bigg)
+    = - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2 \bigg)
+\end{aligned}$$
+
+Integrating this with respect to $r$ yields the following expression for $p$,
+where $p_\infty$ is the (possibly time-dependent) pressure at $r = \infty$:
+
+$$\begin{aligned}
+    p(r, t)
+    = p_\infty(t) + \rho \bigg( \frac{1}{r} C'(t) - \frac{1}{2 r^4} C^2(t) \bigg)
+\end{aligned}$$
+
+We now consider a spherical bubble with radius $R(t)$ and pressure $P(t)$ along the liquid surface.
+To study the liquid boundary's movement, we set $r = R$ and $p = P$,
+and see that $R'(t) = v(t)$, such that $C = r^2 V = R^2 R'$.
+We thus arrive at:
+
+$$\begin{aligned}
+    P
+    &= p_\infty + \rho \bigg( \frac{1}{R} \dv{(R^2 R')}{t} - \frac{1}{2 R^4} (R^2 R')^2 \bigg)
+    \\
+    &= p_\infty + \rho \bigg( 2 (R')^2 + R R'' - \frac{1}{2} (R')^2 \bigg)
+\end{aligned}$$
+
+Rearranging this and defining $\Delta p = P - p_\infty$
+leads to the simple Rayleigh-Plesset equation:
+
+$$\begin{aligned}
+    \boxed{
+        R \dv[2]{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2
+        = \frac{\Delta p}{\rho}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
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