Categories: Fluid dynamics, Fluid mechanics, Physics.

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.

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:

\[\begin{aligned} \boxed{ R \dv[2]{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2 = \frac{\Delta p}{\rho} } \end{aligned}\]

- B. Lautrup,
*Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, CRC Press.

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