Stokes flow is reversible because it is linear and instantaneous. Instantaneous means that is entirely the boundary conditions that define the movement at any given time.

What does the definition of "instantaneity" really means?

And why is this flow linear?

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    $\begingroup$ The flow is governed by a set of differential equations. Do you know what they are? Do you know what it means for differential equations to be linear or nonlinear? What form are the equations? Regarding the instantaneous -- again, it comes back to the equations. What is the mathematical classification of the equations? Are they elliptic, parabolic, hyperbolic? $\endgroup$ – tpg2114 Mar 4 '17 at 17:26
  • $\begingroup$ @tpg2114 When the force is removed, instantaneously the flow stops. Where is the inertia? Does this make sense to you? This flow is very viscous and at the same time reversible. People think viscosity as dissipation. How could that be? $\endgroup$ – veronika Mar 4 '17 at 17:49
  • $\begingroup$ @tpg2114 You gave me an amazing example of stokes flow yesterday: hydrodynamic lubrication.( through the book that you suggested) Oil coming out of bearings is really hot. $\endgroup$ – veronika Mar 4 '17 at 17:56
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    $\begingroup$ @veronika I'm glad you agree, but I highly recommend answering (even if you don't do it here) the questions I posed in my earlier comments. If you can figure out the mathematics of the differential equations, then you'll begin to understand what "instantaneous" means etc.. It all comes back to the mathematical classifications of the equations and once you understand how to figure those out, you will understand the behavior of the equation immediately. Including how to solve it and what other, similar, equations might exist to help you understand better. $\endgroup$ – tpg2114 Mar 5 '17 at 3:21
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    $\begingroup$ As a good example of what I mean, think about this -- the solution of potential flow around a cylinder looks identical to the Stokes flow around a cylinder. Potential flow assumes no viscosity while Stokes flow assumes viscosity dominates everything. So why are the solutions the same? If you go to the equations and understand the nature of them, it becomes clear why this is the case. $\endgroup$ – tpg2114 Mar 5 '17 at 3:25

I think the statement is false: Stokes flow is irreversible.

Viscosity is essential in understanding the Stokes problem. Without viscosity there is no force on the sphere. This is sometimes called d'Alembert's paradox. But viscosity will lead to irreversible dissipation of heat in the fluid. Another way to say the same thing: Pushing the sphere through the fluid (or pushing the fluid past a stationary fluid) does work, and that work goes into heating the fluid.

The only thing I can imagine that the "reversible" statement refers to is the following: The Navier Stokes equation is $$ \partial_t v + v\cdot \nabla v = -\frac{1}{\rho}\nabla P + \frac{\eta}{\rho} \nabla^2 v . $$ Under $T$ the velocity and $\partial_t$ are odd. So the LHS is even, the first terms on the RHS is even, and the viscosity terms is odd (as expected). In Stokes flow, the first term on the LHS is zero (stationary flow) and the second term is neglected. That still leaves the RHS which does not transform simply under $T$. But now we can take the curl and get an equation that only involves $v$ $$ \nabla^2 \nabla\times v =0 $$ which does not contain $\eta$ and transform homogeneoulsy under $T$.

  • $\begingroup$ I also think that viscous flows are irreversible. Thanks for the answer. $\endgroup$ – veronika Mar 4 '17 at 17:59
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    $\begingroup$ As I mentioned on the question -- the classic model of Stokes flow is reversible. There's a reason for it. It's a mathematical model. The energy equation is not part of the model and so there is no dissipation or losses. Model and "reality" need to be separated when talking about these types of simplified problems. $\endgroup$ – tpg2114 Mar 4 '17 at 18:01
  • $\begingroup$ @tpg2114 The force in the classic treatment of the Stokes problem is $F=\gamma u$, where $u$ is the relative velocity of the sphere and the fluid. This is $T$-odd, like any other force of friction. So I would argue that the standard treatment is indeed irreversible. $\endgroup$ – Thomas Mar 4 '17 at 19:15

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