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.

Consider the main Navier-Stokes equation for the velocity field v\va{v}:

DvDt=vt+(v)v=pρ+ν2v\begin{aligned} \frac{\mathrm{D} \va{v}}{\mathrm{D} t} = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v} = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} \end{aligned}

We make the ansatz v=v(r,t)e^r\va{v} = v(r, t) \vu{e}_r, where e^r\vu{e}_r is the basis vector; in other words, we demand that the only spatial variation of the flow is in rr. The above equation then becomes:

vt+vvr=1ρpr+ν(1r2r(r2vr)2r2v)\begin{aligned} \pdv{v}{t} + v \pdv{v}{r} = - \frac{1}{\rho} \pdv{p}{r} + \nu \bigg( \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{v}{r} \Big) - \frac{2}{r^2} v \bigg) \end{aligned}

Meanwhile, the incompressibility condition in spherical coordinates yields:

v=1r2(r2v)r=0\begin{aligned} \nabla \cdot \va{v} = \frac{1}{r^2} \pdv{(r^2 v)}{r} = 0 \end{aligned}

This is only satisfied if r2vr^2 v is constant with respect to rr, leading us to a solution v(r)v(r) given by:

v(r)=C(t)r2\begin{aligned} v(r) = \frac{C(t)}{r^2} \end{aligned}

Where C(t)C(t) is an unknown function that does not depend on rr. We then insert this result in the main Navier-Stokes equation, and isolate it for p/r\ipdv{p}{r}, yielding:

pr=ρ(1r2C2r5C2ν(2r4C2r4C))=ρ(1r2C2r5C2)\begin{aligned} \pdv{p}{r} = - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2 - \nu \Big( \frac{2}{r^4} C - \frac{2}{r^4} C \Big) \bigg) = - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2 \bigg) \end{aligned}

Integrating this with respect to rr yields the following expression for pp, where p(t)p_\infty(t) is the (possibly time-dependent) pressure at r=r = \infty:

p(r)=p+ρ(1rC12r4C2)\begin{aligned} p(r) = p_\infty + \rho \bigg( \frac{1}{r} C' - \frac{1}{2 r^4} C^2 \bigg) \end{aligned}

From the definition of viscosity, we know that the normal stress σrr\sigma_{rr} in the liquid is given by:

σrr(r)=p(r)+2ρνv(r)r\begin{aligned} \sigma_{rr}(r) = - p(r) + 2 \rho \nu \pdv{v(r)}{r} \end{aligned}

We now consider a spherical bubble with radius R(t)R(t) and interior pressure P(t)P(t) along its surface. Since we know the liquid pressure p(r)p(r), we can find PP from σrr(r)\sigma_{rr}(r). Furthermore, to include the effects of surface tension, we simply add the Young-Laplace law to PP:

P=σrr(R)+α2R=p(R)2ρν(2R3C)+α2R\begin{aligned} P = - \sigma_{rr}(R) + \alpha \frac{2}{R} = p(R) - 2 \rho \nu \Big( \frac{-2}{R^3} C \Big) + \alpha \frac{2}{R} \end{aligned}

We isolate this for p(R)p(R), and equate it to our expression for p(r)p(r) at the surface r ⁣= ⁣Rr\!=\!R:

Pρν4R3Cα2R=p+ρ(1RC12R4C2)\begin{aligned} P - \rho \nu \frac{4}{R^3} C - \alpha \frac{2}{R} = p_\infty + \rho \bigg( \frac{1}{R} C' - \frac{1}{2 R^4} C^2 \bigg) \end{aligned}

Isolating for PP, and inserting the fact that R(t)=v(t)R'(t) = v(t), such that C=r2v=R2RC = r^2 v = R^2 R', yields:

P=p+ρ(1Rd(R2R)dt12R4(R2R)2+ν4R3(R2R))+α2R=p+ρ(2(R)2+RR12(R)2+ν4RR)+α2R\begin{aligned} P &= p_\infty + \rho \bigg( \frac{1}{R} \dv{(R^2 R')}{t} - \frac{1}{2 R^4} (R^2 R')^2 + \nu \frac{4}{R^3} (R^2 R') \bigg) + \alpha \frac{2}{R} \\ &= p_\infty + \rho \bigg( 2 (R')^2 + R R'' - \frac{1}{2} (R')^2 + \nu \frac{4}{R} R' \bigg) + \alpha \frac{2}{R} \end{aligned}

Rearranging this and defining ΔpPp\Delta p \equiv P - p_\infty leads to the Rayleigh-Plesset equation:

Δpρ=Rd2Rdt2+32(dRdt)2+ν4RdRdt+αρ2R\begin{aligned} \boxed{ \frac{\Delta p}{\rho} = R \dvn{2}{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2 + \nu \frac{4}{R} \dv{R}{t} + \frac{\alpha}{\rho} \frac{2}{R} } \end{aligned}


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