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+---
+title: "Rayleigh-Plesset equation"
+date: 2021-04-06
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+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 equation](/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 $\ipdv{p}{r}$, 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](/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:
+
+$$\begin{aligned}
+ \boxed{
+ \frac{\Delta p}{\rho}
+ = R \dvn{2}{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2 + \nu \frac{4}{R} \dv{R}{t} + \frac{\alpha}{\rho} \frac{2}{R}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.