--- 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/). Consider the main [Navier-Stokes equations](/know/concept/navier-stokes-equations/) 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](/know/concept/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 $\pdv*{p}{r}$, 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](/know/concept/viscosity/), we know that the normal [stress](/know/concept/cauchy-stress-tensor/) $\sigma_{rr}$ 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](/know/concept/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: $$\begin{aligned} \boxed{ \frac{\Delta p}{\rho} = R \dv[2]{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2 + \nu \frac{4}{R} \dv{R}{t} + \frac{\alpha}{\rho} \frac{2}{R} } \end{aligned}$$ ## References 1. B. Lautrup, *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, CRC Press.