Time inversion for Euler equation in fluid dynamics

Question

Consider Euler equation for continuum body:

$$\frac{\partial u^i}{\partial t}+\mathbf u\cdot \nabla u^i=- \frac{1}{\rho} \frac{\partial p}{\partial x^i} $$

where $\rho$ is the mass density, $p$ is the pressure and $\mathbf u (t, \mathbf x)$ is the velocity field. Under the effect of a boost $(t=t',R=1)$:

$$\mathbf u'(t,\mathbf x')= \mathbf u (t,\mathbf x' + \mathbf v t)- \mathbf v=\mathbf u (t,\mathbf x)- \mathbf v$$

(here $\mathbf v$ is the velocity of reference frame)

we note that if we put in a natural way $\rho '(t,\mathbf x')=\rho (t, \mathbf x)$ and $p'(t,\mathbf x')=p (t, \mathbf x)$ we obtain that Euler equation is Galilean covariant BUT is not invariant for a time inversion (i.e. $t \rightarrow -t$).

My question is: which transformations of $\rho$ and $p$ do we have to use in order to obtain Euler equation that is invariant for time-inversion, without losing Galilean covariance?

Answer 1

I can not answer you question mathematically, but my experience with Burgers equation tells me that there is no such transformation.

If you think of Euler equation as Navier-Stokes in the limit where the viscosity vanishes $\nu \to 0$, then time reversal symmetry is simply spontaneously broken. As long as you have a finite viscosity the system is dissipative and there should be no time reversal symmetry. I don't think that such a symmetry can just pop up as $\nu \to 0$ because it does not contain any continuously tunable parameter.

In the case of Burgers equation (no pressure term),

$\partial_t \mathbf{u} + (\mathbf{u} \cdot \nabla) \mathbf{u} = \nu \Delta \mathbf{u} \, ,$

when the viscosity is small, most of the dissipation happens at shocks. I.e. small regions of space with strong gradients of the velocity field. The velocity field is almost everywhere smooth but jumps almost discontinuously at the position of the shock. You can read more at http://arxiv.org/abs/0704.1611v1 and see pictures in wikipedia. The shocks are true singularities in the limit $\nu \to 0$. When there is a non-vanishing viscosity they are smoothed out.

This explains why, even in the limit $\nu \to 0$, the solutions of Burgers equation are not symmetric under time reversal. There still is dissipation. It is just happening at places that have a vanishing volume.

Answer 2

Maybe it's too late to answer this question more properly but I will post it just for future references. I believe Steven's answer is incorrect. At least its part, which is about Euler equation is not accurate. Euler equation comes directly from second law of Newton without any non-conservative force and as a result of no viscous term everything should be reversible. In fact, it could be shown mathematically that under T transformation when $t \rightarrow -t$, the acceleration does not change its sign so $\frac{\partial \mathbf{u} (\mathbf{r},t)}{\partial t} = \frac{\partial \mathbf{u}(\mathbf{r},-t)}{\partial (-t)}$. Also nonlinear convective term, which comes from convective acceleration mechanism does not change its sign, so: $\mathbf{u} (\mathbf{r},t) \cdot \nabla \mathbf{u}(\mathbf{r},t) = \mathbf{u}(\mathbf{u},-t) \cdot \nabla \mathbf{u}(\mathbf{r},-t)$. Also pressure will not change its sign, so: $P(\mathbf{r},t) = P(\mathbf{r},-t)$. So finally the Euler equation is invariant under T transformation as:

$$\rho (\mathbf{r},t) \frac{\partial \mathbf{u}(\mathbf{r},t)}{\partial t} + \rho(\mathbf{r},t) \mathbf{u}(\mathbf{r},t) \cdot \nabla \mathbf{u}(\mathbf{r},t) + \nabla P(\mathbf{r},t) = \rho(\mathbf{r},-t) \frac{\partial \mathbf{u}(\mathbf{r},-t)}{\partial (-t)} + \rho(\mathbf{r},-t) \mathbf{u}(\mathbf{r},-t) \cdot \nabla \mathbf{u}(\mathbf{r},-t) + \nabla P(\mathbf{r},-t) = 0$$

But Navier-Stokes equation because of viscous term always shows an increase of entropy and as a result it will not be time reversal under T transformation. I believe it's a general rule that non-conservative forces of viscous dissipative mechanisms are responsible for irreversibility under T transformation.