---

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 $\va{v}$:

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

We make the ansatz $\va{v} = v(r, t) \vu{e}_r$, 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 equation then becomes:

$$\begin{aligned} \pdv{v}{t} + v \pdv{v}{r} = - \frac{1}{\rho} \pdv{p}{r} + \nu \bigg( \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{v}{r} \Big) - \frac{2}{r^2} v \bigg) \end{aligned}$$

Meanwhile, the incompressibility condition in spherical coordinates yields:

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

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 main Navier-Stokes equation, and isolate it for $\ipdv{p}{r}$, yielding:

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

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

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

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

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

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

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

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

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

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

$$\begin{aligned} P &= p_\infty + \rho \bigg( \frac{1}{R} \dv{(R^2 R')}{t} - \frac{1}{2 R^4} (R^2 R')^2 + \nu \frac{4}{R^3} (R^2 R') \bigg) + \alpha \frac{2}{R} \\ &= p_\infty + \rho \bigg( 2 (R')^2 + R R'' - \frac{1}{2} (R')^2 + \nu \frac{4}{R} R' \bigg) + \alpha \frac{2}{R} \end{aligned}$$

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

$$\begin{aligned} \boxed{ \frac{\Delta p}{\rho} = R \dvn{2}{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2 + \nu \frac{4}{R} \dv{R}{t} + \frac{\alpha}{\rho} \frac{2}{R} } \end{aligned}$$

References

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

---