Categories: Fluid dynamics, Fluid mechanics, Physics.

# 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.

## 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 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.