A form of the Navier-Stokes momentum equation can be written as: $$ \rho \left( \frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u} \right) = - \nabla \bar{p} + \mu \, \nabla^2 \mathbf u + \tfrac13 \mu \, \nabla (\nabla\cdot\mathbf{u}) + \rho\mathbf{g}$$

This question feels quite basic, but from where can irreversibility arise in this equation? For example, in this video exhibiting the reversibility of Taylor-Couette flow, I believe $|\rho \left(\mathbf{u} \cdot \nabla \mathbf{u} \right)|/|\mu \nabla^2 \mathbf u| = {\rm Re} \ll 1$ since it is in a regime of laminar flow (i.e. low Reynold's number). But why explicitly is Taylor-Couette flow reversible, while the stirring of coffee is irreversible based upon the mathematical terms present in the momentum equation? Is it caused by the nonlinear term due to a possible interaction between various scales in the system that makes the fluid hard to "unmix" or is it somehow due to the dynamic viscosity, $\mu$, being strictly positive? Or can irreversibility arise from other origins such as the initial/boundary conditions or fluid closure applied? Mathematical insights and physical intuition would be greatly appreciated.

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    $\begingroup$ I am not sure to understand properly your question but irreversibility in the sense of absence of $t\to -t$ symmetry is evident in the form of the equation. The appearance of the first order $t$ derivative together with the velocity $u$ in the left hand side makes time reversal invariant this term, whereas the presence of only $u$ in the right hand side makes not time-reversal invariant that term. $\endgroup$ – Valter Moretti Nov 7 '20 at 19:37
  • $\begingroup$ To clarify, the momentum equation is applicable to both Taylor-Couette flow and the stirring of coffee. But I am wondering, from both a mathematical and physical intuition perspective, why the former flow is reversible while the latter is not. I'm not sure I quite follow your response, namely, $t \rightarrow -t$ would affect both $\mathbf{u} \cdot \nabla \mathbf{u}$ and $\mu \, \nabla^2 \mathbf u$ equivalently (i.e. time-reversal does not affect the sign of these terms whether the flow is turbulent or laminar). $\endgroup$ – Mathews24 Nov 7 '20 at 19:53
  • $\begingroup$ This is valid for the left-hand side, but not for the right-hand side: it is not invariant under time reflection just due to dissipative terms proportional to $\mu$. The latter term you wrote changes sign under time reversal, the former does not. $\endgroup$ – Valter Moretti Nov 7 '20 at 20:01
  • $\begingroup$ @ChiralAnomaly By irreversibility, I mean that the system can go from an initial state A to a consequent state B, but it cannot return to state A. $\endgroup$ – Mathews24 Nov 7 '20 at 23:16
  • $\begingroup$ @ValterMoretti Regarding the left-hand side terms, time-reversal only appears to affect $\frac{\partial \mathbf{u}}{\partial t}$ but not $\mathbf{u} \cdot \nabla \mathbf{u}$, no? Maybe I am missing obvious dimensional analysis, but clarification on why $\mathbf{u} \cdot \nabla \mathbf{u}$ changes sign under $t \rightarrow -t$ would be appreciated. $\endgroup$ – Mathews24 Nov 7 '20 at 23:21

Which "reversibility" are you referring to? If you mean "thermodynamically reversible" (a flow which does not generate entropy) then viscous dissipation ($\mu\nabla^2\mathbf{u}$) always ensures irreversibility, whatever the Reynolds number. But perhaps you are referring to "kinematic reversibility", which implies reversal of flow in every detail upon reversal of external forces acting on it - that is, upon reversal of external forces, every fluid particle would retrace its trajectory backwards.

Low Reynolds number flows, called "creeping flows", indeed display kinematic reversibility as Reynolds number ($Re$) $\to 0$ (see G.I. Taylor's demo). Here, the non-linear advection term ($\mathbf{u}\cdot\nabla\mathbf{u}$) is negligible (because $Re\ll 1$). To see why the advection term being negligible results in kinematic reversibility of the flow, consider the opposite extreme of a turbulent flow in which the advection term is not negligible. As a specific example, consider turbulent mixing between two initially separate fluids. You could imagine it to be a lab experiment or a numerical simulation - we shall imagine the latter. For simplicity, we imagine that the two fluids are identical but the two portions are given different colours. The turbulent mixing flow - in fact turbulence in general - will be chaotic. After mixing has progressed for some time, we stop the simulation and reverse the velocity field everywhere. Will the two fluids now "un-mix" and separate from each other? It will not, because dominance of the non-linear advection term makes the flow depend sensitively on the initial conditions; this sensitivity is the reason why turbulent flows cannot be reproduced in exact detail; and since we can only specify the initial conditions for the reversed flow with finite precision, the flow will not exactly reverse itself and we will still have mixing as before (in other words, no fluid particle will exactly retrace its previous trajectory). When the non-linear advection term is negligible, we have a highly ordered creeping flow which can be reversed because it is more tolerant to small errors in the specification of the initial conditions.

To summarize: Although Navier-Stokes equation governs all flows, the degree of non-linearity as measured by $Re$ is not the same for all flows; thus flows in the extreme limits of $Re$ can exhibit qualitatively different behaviour.

  • $\begingroup$ I think you have precisely identified my confusion, namely, the appropriate distinction/overlap between kinematic and thermodynamic irreversibility. In particular, are all thermodynamically irreversible processes also kinematically irreversible? Or, conversely, are all kinematically irreversible processes also thermodynamically irreversible? For example, is your scenario of kinematic irreversibility (i.e. high $Re$ flow with mixing) not also thermodynamically irreversible (i.e. increasing the fluid's entropy)? $\endgroup$ – Mathews24 Nov 13 '20 at 20:23
  • $\begingroup$ Further, it is stated that viscous dissipation is thermodynamically irreversible (i.e. increases entropy) yet kinematically reversible, correct? But, by definition, if the fluid's entropy is increased, how could it ever be kinematically reversible? Kinematic irreversibility would imply every single particle can be retraced back to its original position and velocity, yet this would also imply the entropy was also decreased back to its original value. This begs the question, where did the entropy go and how is it being reversed if viscous dissipation is truly a kinematically reversible process? $\endgroup$ – Mathews24 Nov 13 '20 at 20:30
  • $\begingroup$ @Mathews24 Stokes flow has kinematic but not thermodynamic reversibility. Flow governed by Euler equation (which is essentially Navier-Stokes equation with viscous dissipation term removed) has thermodynamic but not kinematic reversibility. This applies to my example of turbulent mixing flow as well $\endgroup$ – Deep Nov 14 '20 at 3:27
  • $\begingroup$ @Mathews24 Imagine sliding an object on ground by applying constant force. Due to friction, this is a thermodynamically irreversible process. But by reversing the force, you can always make the object reverse its trajectory. It is therefore kinematically reversible. The reversed motion is also thermodynamically irreversible, again because of friction, which does not care about whether the object is going forward or backward in its trajectory. $\endgroup$ – Deep Nov 14 '20 at 3:31
  • $\begingroup$ The demo by G.I. Taylor and accompanying notes help tremendously. Namely, kinematic reversibility is based upon the direction of motion of the fluid boundary being reversed, but the larger system (i.e. fluid + environment) appears to be thermodynamically irreversibile. In particular, the low $Re$ flow's entropy increases when turned at first by the boundary. And when the boundary's motion is reversed, the fluid's motion is also reversed and its net entropy change is essentially 0 (neglecting molecular diffusion for now). $\endgroup$ – Mathews24 Nov 15 '20 at 15:43

Reversibility is characteristic of the regime of Stokes flow—also known as “creeping flow.” In this case, the velocity is always, in an appropriate sense, small. It is small enough that without external forcing, the viscosity terms damp the fluid to momentum to zero essentially instantaneously. (How small this is in practice obvious depends on how viscous the fluid is. In terms of the Reynolds number, this means ${\rm Re}\ll 1$, so that without an external force, ${\bf u}$ is already small enough that it will be damped out on a length scale much smaller than the size of the system.) This puts this system in a standard overdamped regime, like the motion of a falling object that rapidly reaches a terminal velocity proportional to the external gravitational force.

In the Navier-Stokes equations, the term that impedes the reversibility is the advective term ${\bf u}\cdot\nabla{\bf u}$. This term represents the change in the local fluid velocity as the the fluid is carried along by a bulk motion. In a low-viscosity fluid, if you set part of the fluid in motion, it will keep moving for a fairly long time, and the ${\bf u}\cdot\nabla{\bf u}$ term will create irreversible changes in the spatial distribution of ${\bf u}$. However, if the motion essentially ceases once the applied force ceases, this term is very small; everything more or less stays in place, so if you push back on it the other direction, the fluid will move back to were it started.

In terms of the equation of motion, notice that if you omit the advective term and invert the external forces, there is a time-reversed solution. (In the Couette problem, the external forces come from shear stresses at the boundaries, whereas in the version of the equation in the question, there can be an external pressure head and gravity.) That is, if $u({\bf x},t)$ was a solution with external force ${\bf f}$ over the period $0<t<T$ [taking the velocity field from an initial configuration ${\bf u}_{0}({\bf x})$ to ${\bf u}_{1}({\bf x})$], then $-{\bf u}({\bf x},2T-t)$ is a reversed solution over the period $T<t<2T$ with a similarly reversed force $-{\bf f}$, which carries the configuration from ${\bf u}_{1}({\bf x})$ back to ${\bf u}_{0}({\bf x})$. This doesn't work with the advection term present, because ${\bf u}\cdot\nabla{\bf u}$ is quadratic in ${\bf u}$, and so the minus signs in $-{\bf u}\cdot\nabla(-{\bf u})$ cancel out.

There is actually a Taylor-Couette flow film featuring G. I. Taylor, who first solved the stability problem in the system, himself, in which he explains this fairly well. However, I cannot seem to find the video anywhere on the Web at the moment.

  • $\begingroup$ To clarify: 1) velocity does not have to necessarily be small for the flow to be reversible, does it? What appears mathematically important are parallel (not perpendicular) gradients in the velocity need to be small, correct? 2) Is there a good way to relate the magnitude of ${\bf u}\cdot\nabla{\bf u}$ to the change in entropy of the system? The latter is thermodynamically how one defines irreversible processes, and maybe that will help clarify the heart of this question since I don't see why the advective term necessarily makes "irreversible changes in the spatial distribution of ${\bf u}$." $\endgroup$ – Mathews24 Nov 7 '20 at 23:47
  • $\begingroup$ Also, I agree with everything you've added, but it's still missing the key part, i.e. the negative of the scenario you've described. Namely, if we do not omit the advective term, how can we prove that there is not a time-reversed solution? $\endgroup$ – Mathews24 Nov 8 '20 at 5:47
  • $\begingroup$ Thank you for the clarification and I hope I can find the video with Taylor. This response helps tremendously. But using the argument you've provided, wouldn't velocity diffusion (i.e. $\mu \nabla^2 \mathbf u$ which is still present in the momentum equation, yet by applying $t \rightarrow -t$, we know this term is not time-symmetric) then be considered reversible? To be clear, based upon the answer, it appears irreversibility only arises from the nonlinear advective term present in the equations, but I want to clarify if that interpretation is in fact correct. $\endgroup$ – Mathews24 Nov 9 '20 at 18:25

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