Spherically symmetric incompressible radial flow

For radial flow outside a sphere, incompressible flow makesindependent of radius. Its velocity potential is $\varphi =-Q/(4\pi r)$.

In an unbounded incompressible flow with spherical symmetry, the kinematic boundary condition on a sphere of radius $R(t)$ gives $u_r=R^2\dot{R}/r^2$ and velocity potential $\varphi =G(t)-R^2\dot{R}/r$. The Unsteady Bernoulli equation at infinity has $F(t)=\dot{G}+p_{\infty }/\rho$. Subtracting it from the finite-radius equation gives the displayed pressure, independently of $G$. Time differentiation is at fixed $r$. Evaluating at $r=R(t)$ yields the Rayleigh equation for an inviscid spherical bubble when the boundary pressure is prescribed.

In radial potential flow outside a spherical bubble, incompressibility and the moving-boundary condition give $u_r=R^2\dot{R}/r^2$ and $\varphi =-R^2\dot{R}/r$. The Unsteady Bernoulli equation, with its gauge fixed at infinity, yields the displayed radius equation when surface tension and viscosity are neglected. Pressure continuity at the interface identifies the liquid pressure with the gas pressure $p_g$. The vacuum special case is Rayleigh collapse of a spherical cavity.

For spherical expansion in an unbounded incompressible flow, continuity gives $u_r=R^2\dot{R}/r^2$. Integrating the kinetic energy outside the bubble gives $K=2\pi \rho R^3{\dot{R}}^2$. The Rayleigh equation for an inviscid spherical bubble then yields $\dot{K}=(p_b-p_{\infty })\dot{V}$, where $V=4\pi R^3/3$. With $p_b=p_{\infty }{V}_0/V$ and constant $p_{\infty }$, integration gives $K-K_0=p_{\infty }[V_0\log (V/V_0)-V+V_0]$. This neglects viscosity, surface tension and external body-force differences, which contribute additional terms in other bubble models.

For gas pressure depending only on radius and initial speed zero, multiplying the Rayleigh equation for an inviscid spherical bubble by $2R^2\dot{R}$ integrates to the displayed identity. A polytropic pressure $p_g=K/R^{3\kappa }$ gives elementary power integrals when $\kappa \ne 1$, and a logarithm when $\kappa =1$. Turning radii are the positive zeros of this first integral. At $\kappa =2$ and $p_g(R_0)=p_{\infty }/2$, they are $R_0$ and $R_0/2^{1/3}$; the inward and outward accelerations at them have opposite signs, giving bounded undamped oscillation in the ideal model.

The Rayleigh-Plesset equation balances the radial inertia of an incompressible liquid around a spherical bubble against internal and ambient pressure, surface tension and viscous stress. Removing the last two effects and setting the bubble pressure to zero gives the following vacuum-collapse case.

A vacuum cavity of radius $a(t)$ in an infinite inviscid incompressible fluid of density $\rho$ and far-field pressure $p_0$ obeysIf $a(0)=a_0$ and $\dot{a}(0)=0$, its collapsing branch satisfies

For a spherical vacuum cavity initially at rest with radius $a$, in an inviscid liquid of constant mass density $\rho$ and positive constant far-field pressure $p_{\infty }$, the Rayleigh equation for an inviscid spherical bubble has the first integral $R^3{\dot{R}}^2=2p_{\infty }(a^3-R^3)/(3\rho )$. To obtain it, multiply the radius equation by $2R^2\dot{R}$ and integrate the resulting exact derivative. Choosing the negative square root and integrating $dt=dR/\dot{R}$ givesThis improper integral converges at both endpoints, proving finite-time Rayleigh collapse of a spherical cavity in the ideal model. It has numerical value approximately $0.914681$. The prediction neglects viscosity, surface tension and liquid compressibility.