Rayleigh–Plesset equation

By conservation of mass, the inverse-square law requires that the radially outward velocity ${\displaystyle u(r,t)}$ must be inversely proportional to the square of the distance from the origin (the center of the bubble).[6] Therefore, letting ${\displaystyle F(t)}$ be some function of time,

${\displaystyle u(r,t)={\frac {F(t)}{r^{2}}}}$

In the case of zero mass transport across the bubble surface, the velocity at the interface must be

${\displaystyle u(R,t)={\frac {dR}{dt}}={\frac {F(t)}{R^{2}}}}$

which gives that

${\displaystyle F(t)=R^{2}dR/dt}$

In the case where mass transport occurs, the rate of mass increase inside the bubble is given by

${\displaystyle {\frac {dm_{V}}{dt}}=\rho _{V}{\frac {dV}{dt}}=\rho _{V}{\frac {d(4\pi R^{3}/3)}{dt}}=4\pi \rho _{V}R^{2}{\frac {dR}{dt}}}$

with ${\displaystyle V}$ being the volume of the bubble. If ${\displaystyle u_{L}}$ is the velocity of the liquid relative to the bubble at ${\displaystyle r=R}$, then the mass entering the bubble is given by

${\displaystyle {\frac {dm_{L}}{dt}}=\rho _{L}Au_{L}=\rho _{L}(4\pi R^{2})u_{L}}$

with ${\displaystyle A}$ being the surface area of the bubble. Now by conservation of mass ${\displaystyle dm_{v}/dt=dm_{L}/dt}$, therefore ${\displaystyle u_{L}=(\rho _{V}/\rho _{L})dR/dt}$. Hence

${\displaystyle u(R,t)={\frac {dR}{dt}}-u_{L}={\frac {dR}{dt}}-{\frac {\rho _{V}}{\rho _{L}}}{\frac {dR}{dt}}=\left(1-{\frac {\rho _{V}}{\rho _{L}}}\right){\frac {dR}{dt}}}$

Therefore

${\displaystyle F(t)=\left(1-{\frac {\rho _{V}}{\rho _{L}}}\right)R^{2}{\frac {dR}{dt}}}$

In many cases, the liquid density is much greater than the vapor density, ${\displaystyle \rho _{L}\gg \rho _{V}}$, so that ${\displaystyle F(t)}$ can be approximated by the original zero mass transfer form ${\displaystyle F(t)=R^{2}dR/dt}$, so that[6]

${\displaystyle u(r,t)={\frac {F(t)}{r^{2}}}={\frac {R^{2}}{r^{2}}}{\frac {dR}{dt}}}$

Assuming that the liquid is a Newtonian fluid, the incompressible Navier–Stokes equation in spherical coordinates for motion in the radial direction gives

${\displaystyle \rho _{L}\left({\frac {\partial u}{\partial t}}+u{\frac {\partial u}{\partial r}}\right)=-{\frac {\partial P}{\partial r}}+\mu _{L}\left[{\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial u}{\partial r}}\right)-{\frac {2u}{r^{2}}}\right]}$

Substituting kinematic viscosity ${\displaystyle \nu _{L}=\mu _{L}/\rho _{L}}$ and rearranging gives

${\displaystyle -{\frac {1}{\rho _{L}}}{\frac {\partial P}{\partial r}}={\frac {\partial u}{\partial t}}+u{\frac {\partial u}{\partial r}}-\nu _{L}\left[{\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial u}{\partial r}}\right)-{\frac {2u}{r^{2}}}\right]}$

whereby substituting ${\displaystyle u(r,t)}$ from mass conservation yields

${\displaystyle -{\frac {1}{\rho _{L}}}{\frac {\partial P}{\partial r}}={\frac {2R}{r^{2}}}\left({\frac {dR}{dt}}\right)^{2}+{\frac {R^{2}}{r^{2}}}{\frac {d^{2}R}{dt^{2}}}-{\frac {2R^{4}}{r^{5}}}\left({\frac {dR}{dt}}\right)^{2}={\frac {1}{r^{2}}}\left(2R\left({\frac {dR}{dt}}\right)^{2}+R^{2}{\frac {d^{2}R}{dt^{2}}}\right)-{\frac {2R^{4}}{r^{5}}}\left({\frac {dR}{dt}}\right)^{2}}$

Note that the viscous terms cancel during substitution.[6] Separating variables and integrating from the bubble boundary ${\displaystyle r=R}$ to ${\displaystyle r\rightarrow \infty }$ gives

${\displaystyle -{\frac {1}{\rho _{L}}}\int _{P(R)}^{P(\infty )}dP=\int _{R}^{\infty }\left[{\frac {1}{r^{2}}}\left(2R\left({\frac {dR}{dt}}\right)^{2}+R^{2}{\frac {d^{2}R}{dt^{2}}}\right)-{\frac {2R^{4}}{r^{5}}}\left({\frac {dR}{dt}}\right)^{2}\right]dr}$

${\displaystyle {{\frac {P(R)-P_{\infty }}{\rho _{L}}}=\left[-{\frac {1}{r}}\left(2R\left({\frac {dR}{dt}}\right)^{2}+R^{2}{\frac {d^{2}R}{dt^{2}}}\right)+{\frac {R^{4}}{2r^{4}}}\left({\frac {dR}{dt}}\right)^{2}\right]_{R}^{\infty }=R{\frac {d^{2}R}{dt^{2}}}+{\frac {3}{2}}\left({\frac {dR}{dt}}\right)^{2}}}$

Let ${\displaystyle \sigma _{rr}}$ be the normal stress in the liquid that points radially outward from the center of the bubble. In spherical coordinates, for a fluid with constant density and constant viscosity,

${\displaystyle \sigma _{rr}=-P+2\mu _{L}{\frac {\partial u}{\partial r}}}$

Therefore at some small portion of the bubble surface, the net force per unit area acting on the lamina is

${\displaystyle {\begin{aligned}\sigma _{rr}(R)+P_{B}-{\frac {2\sigma }{R}}&=-P(R)+\left.2\mu _{L}{\frac {\partial u}{\partial r}}\right|_{r=R}+P_{B}-{\frac {2\sigma }{R}}\\&=-P(R)+2\mu _{L}{\frac {\partial }{\partial r}}\left({\frac {R^{2}}{r^{2}}}{\frac {dR}{dt}}\right)_{r=R}+P_{B}-{\frac {2\sigma }{R}}\\&=-P(R)-{\frac {4\mu _{L}}{R}}{\frac {dR}{dt}}+P_{B}-{\frac {2\sigma }{R}}\\\end{aligned}}}$

where ${\displaystyle \sigma }$ is the surface tension.[6] If there is no mass transfer across the boundary, then this force per unit area must be zero, therefore

${\displaystyle P(R)=P_{B}-{\frac {4\mu _{L}}{R}}{\frac {dR}{dt}}-{\frac {2\sigma }{R}}}$

and so the result from momentum conservation becomes

${\displaystyle {\frac {P(R)-P_{\infty }}{\rho _{L}}}={\frac {P_{B}-P_{\infty }}{\rho _{L}}}-{\frac {4\mu _{L}}{\rho _{L}R}}{\frac {dR}{dt}}-{\frac {2\sigma }{\rho _{L}R}}=R{\frac {d^{2}R}{dt^{2}}}+{\frac {3}{2}}\left({\frac {dR}{dt}}\right)^{2}}$

whereby rearranging and letting ${\displaystyle \nu _{L}=\mu _{L}/\rho _{L}}$ gives the Rayleigh–Plesset equation[6]

${\displaystyle {\frac {P_{B}(t)-P_{\infty }(t)}{\rho _{L}}}=R{\frac {d^{2}R}{dt^{2}}}+{\frac {3}{2}}\left({\frac {dR}{dt}}\right)^{2}+{\frac {4\nu _{L}}{R}}{\frac {dR}{dt}}+{\frac {2\sigma }{\rho _{L}R}}}$

Using dot notation to represent derivatives with respect to time, the Rayleigh–Plesset equation can be more succinctly written as

${\displaystyle {\frac {P_{B}(t)-P_{\infty }(t)}{\rho _{L}}}=R{\ddot {R}}+{\frac {3}{2}}({\dot {R}})^{2}+{\frac {4\nu _{L}{\dot {R}}}{R}}+{\frac {2\sigma }{\rho _{L}R}}}$