I have some difficulties trying to understand the viscosity term in the Rayleigh-Plesset equation:

$$R\frac{d^2R}{dt^2} +\frac{3}{2}(\frac{dR}{dt})^2 + \frac{4\nu_L}{R}\frac{dR}{dt} +\frac{2\gamma}{\rho_LR} + \frac{\Delta P(t)}{R} = 0$$

This equation describes the change in the radius of a spherical bubble. The main thing i don't understand is the inclusion of a viscous term - a far as i understand, viscosity is relevant to physical situations where there is shear flow, or equivalently when the gradient of pressure has component orthogonal to the gradient of flow velocity (for example, a laminar flow between two stationary planar plates). In the case of expanding bubble, the flow has spherical symmetry, and that means that both pressure and flow speed are only radius-dependent, hence viscous effects are irelevant here. So what am i misunderstanding here? obviously i have some mistakes in my conception of the radial fluid flow.


The tensorially correct version of Newton's law of viscosity in 3 dimensions includes contribution of viscosity to extensional flows. Extension in one principal strain direction in conjunction with equal compression at 90 degrees translates kinematically into shear on the 45 degree planes. Even in the shear flow you describe translates kinematically into extension at 45 degrees to the shear plane. Fr more details on this, see Chapter 1 of Transport Phenomena by Bird et al.

  • $\begingroup$ Thanks! I'm now trying to imagine the viscous interaction in the case of extension in one strain direction in conjuction with equal compression in 90 degrees... when i'll feel i understand that, i'll accept your answer. Also, if you can add some details from the book you refered to, it will be blessed, as chapter 1 of this book isn't completely available to me on Google books. Anyways, thanks! $\endgroup$
    – user2554
    Oct 12 '20 at 15:00
  • $\begingroup$ I now saw that you already answered another question of me, which was on a very close topic - physics.stackexchange.com/questions/279254/… . This question dealt with Stokes's law of sound attenuation... the answer you gave me now is much better and more intuitive than the previous answer, which was too formal. $\endgroup$
    – user2554
    Oct 12 '20 at 15:10

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