---
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.