From where does irreversibility arise in the Navier-Stokes momentum equation?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

There are different notions of irreversibility. The relevant one is the third one, but first, it may be useful to start from a more microscopic notion.

• A deterministic or quantum process is time-reversible if the time-reversed process satisfies the same dynamic equations as the original process: the equations are invariant or symmetrical under a change in the sign of $t$. This is not the notion that applies to the Couette experiment at low Re.

  For example, take a system of particles labelled by $i$ undergoing elastic binary collisions with no external fields: at every instant, we have a collection of variables $C_t=(x_i(t),v_i(t))$, the position and velocity of each particle. Take a movie of the evolution from $C_{t_1}$ to $C_{t_2}$ (movie F). Now do a different experiment: prepare the system in the configuration $C_{t_0} = (x_i(t_2),-v_i(t_2))$, namely take $C_{t_2}$ as initial configuration but invert all the velocities. Follow the evolution and take a movie (movie B) till configuration $C_{t_0+(t_2-t_1)}$. You will have that $C_{t_0+(t_2-t_1)}$ is exactly $C_{t_1}$, but with the velocities reversed: movie F is just movie B played backwards. Note that there is no external field or force, this is a consequence of the reversibility of the microscopic dynamics (we are considering the totality of the degrees of freedom $C_t=(x_i(t),v_i(t))$ at every instant of time).

• Reversibility of a thermodynamic process: basically what happens for quasi-stationary transformations that allow going through a sequence of equilibrium states. Here we have no access to all the microscopic degrees of freedom, we only have collective variables (a few thermodynamic quantities). To vary the system we have to act via the slow change of an external field (e.g. the potential of the walls of a piston). There is no explicit time notion (the process is infinitely slow, so the system has time to find a new thermodynamic equilibrium at each instant), and if you move the external field "in the other direction" you do not reverse the velocities of the microscopic components like in the case of time-inversion. However, the macroscopic, collective variables can be driven back to their initial state.

• Reversibility of the Stokes flow: this is the case relevant to the question. At a low Re number, you can neglect the non-linear advection term in the Navier-Stokes (NS) equation. This makes NS time-reversible (in the sense given in the first point). However, there is a difference with respect to the first point: now the velocity is a collective (macroscopic) variable, given by the average of all the many molecular velocities at the level of the fluid element. Only this macroscopic variable (and not all the microscopic velocities) is reversed. Moreover, to obtain a cyclic transformation you have to act on the system with an external field (like in the thermodynamic case: you turn a cylinder boundary in the Couette experiment). In fact, the Stokes flow can have a dependence on time only through time-dependent boundary conditions or time-varying external fields: given the boundary conditions of a Stokes flow, the flow can be found without knowledge of the flow at any other time. Not surprisingly, this kind of reversibility looks much more similar to the thermodynamic one: it involves only a few macroscopic variables and it is obtained via the "inversion" of an external agent that slowly drives the system. Of course, entropy increases (see this answer) because of the viscous term in NS, but the entropy production rate is extremely small because the process is slow (it is typically ignored).