Explanation for why an ideal fluid doesn't have increasing entropy?

Question

The equations of motion for a very simple ideal fluid (specifically a calorically perfect, monatomic, ideal gas) are \begin{align*}\dot{\rho}+\nabla \cdot (\rho u)=0 \;&\text{(mass conservation)} \\ \dot{(\rho u)}+\nabla \cdot (\rho u u) + \nabla p=0 \;&\text{(momentum conservation)} \\ \dot{(\rho e)} +\nabla \cdot (\rho ue+\rho p)=0 \;&\text{(energy conservation)} \\ e=\frac{1}{2}u^2+\frac{3}{2}p \;&\text{(equation of state)}\end{align*} where $\rho$ is the density, $u$ the velocity, $p$ the pressure, and $e$ the total energy (including the internal energy).

I've noticed that these equations are time reversible, i.e. if we have a solution on the time interval $[0,T]$, then by simply sending $u \to -u$, $t \to -t$ we get a solution on $[-T,0]$. From the point of view of thermodynamics, specifically the fact that total entropy is (weakly) increasing, this only makes sense if entropy is constant.

EDIT: In response to some of the comments I've deleted the example of gas expansion, since as pointed out, this wasn't strange. However I'd like to mention that a very simple model of particle collisions in a gas gives rise to the above equations:

Assume particles are interacting through collisions only (i.e. not through 'long range' forces), and there are sufficient collisions occuring that to a good degree of accuracy, the distribution of velocities of particles at any point is isotropic (after subtracting the mean velocity). For example if the velocity distribution of all particles at a given point is always a spherical gaussian this would be the case. Under just this assumption, the above equations follow.

I'm not denying that they are time reversible, and so must have constant entropy. It's just that I have no intuitive explanation for this, other than simply computing the equations. The statistical model mentioned is not time reversible, in fact it relies heavily on frequent collisions, and so time reversibility is a rather suprising fact.

Is there some other explanation, perhaps more intuitive than simply computing the equations, that explains time reversibility?

Answer 1

This is a very important issue, to which an answer must be made in mathematical terms, rather than by waving hands.

Yes, the Euler system (conservation of mass, momentum and energy) is time-reversible. So where is the error when we say Entropy is non-decreasing, but the system is time-reversible, therefore the entropy must be constant along trajectories ?

The point is that the Cauchy problem (i.e. solving the PDEs together with imposing an initial data) is not uniquely solvable. It is so when the initial data is smooth enough, but only for some finite time interval. For rather general smooth initial data, the smooth solution exists only for a finite time interval $(0,T_{\max})$. As $t\rightarrow T_\max$, some first derivative becomes infinite somewhere (generic behaviour). Beyond $T_\max$, the solution is not any more smooth ; it is at best piecewise smooth, with $\rho,u,e,p$ being discontinuous accross hypersurfaces. These discontinuities are known as shock waves and contact discontinuities.

It turns out that once shock waves develop, the Euler system is not any more sufficient to select a unique solution. There are actually infinity many, among which only one has a physical sense. The way to recognize that one, and to select it from a mathematical perspective, is to add a so-called entropy criterion. This is nothing more than saying that when a particle crosses a shock, then its entropy increases.

This entropy criterion is expected to guaranty the uniqueness of the solution (this is still an open problem). But, being an inequality, is not compatible with the time reversal. This is why the Cauchy problem, at far as physicall meaningful solutions are concerned, is irreversible, despite the apparent time-reversibility of the Euler system.

Edit. To answer Michael's concerns, the constancy of entropy along smooth solutions is just the well accepted fact in thermodynamics that smooth flows are time-reversible. Of course, if you have in mind a finer description, the situation will be different. The mesoscopic level is described by a kinetic equation, say that of Boltzmann, which is irreversible: the entropy does increase whenever the local distribution of the gas deviates from the Gaussians (= Maxwell's equilibria). Thus the only reversible model is at the microscopic level, where particles obey to Newton's law and interact through short range forces (or hard spheres dynamics).

To come back to the Euler system, the boundary conditions at discontinuities are not ad hoc. They just express that the mass, momentum and energy are conserved.

Answer 2

Q: Explanation for why an ideal fluid doesn't have increasing entropy?
A: The entropy will in fact increase for the most probable initial conditions.

---

The question in the OP refers to the socalled irreversibility paradox (or Loschmidt paradox): The statistical evolution of the ideal gas is irreversible even though the equations of motion are reversible. The evolution of the statistical distribution $f(r,v,t)$ is governed by the Boltzmann equation, $$\frac{\partial f}{\partial t}+v\cdot\frac{\partial f}{\partial r}=\int dv_2 dv'_1 dv'_2 w(v'_1,v'_2;v,v_2)[f(v'_1,r,t)f(v'_2,r,t)-f(v,r,t)f(v_2,r,t)],$$ for some collision rate function $w$. This equation describes the relaxation of $f$ towards the equilibrium Maxwell-Boltzmann distribution, in accordance with the second law of thermodynamics.

From a modern perspective, a resolution of the irreversibility paradox is discussed by Freddy Bouchet in arXiv:2002.10398: The Boltzmann equation only holds for a subset of microscopic initial conditions compatible with a certain $f(r,v,0)$. Not every microscopic state of a macroscopic system will evolve in accordance with the second law, but only the majority of states, a majority which however becomes so overwhelming when the number of atoms in the system becomes very large that irreversible behavior becomes a near certainty.