Does the definition of compressibility depend on the frame of reference?

Question

According to many authors, a fluid is defined to be incompressible if the material derivative of the density $\frac{D\rho}{Dt}$ is zero, that is to say, that in an frame of reference following the motion of an air parcel, density doesn't change. This in turn means, according to the continuity equation,

$$\frac{D\rho}{Dt}+\rho\nabla\cdot{\vec V}= 0,$$

so that $\nabla\cdot{\vec V}= 0$. So far so good.

However, let us consider a simple case in 1D in which the density is of the form $\rho(x,t)=x-t$ and $\vec V=u_x\hat{\imath}$. Both fields satisfy the continuity equation. This is more evident if we use the other form of the continuity equation,

$$ \frac{\partial \rho}{\partial t}+\nabla\cdot{(\rho\vec V)}= 0$$

Clearly, for the velocity field that I gave, $\nabla\cdot{\vec V}= 0$, and the fluid is incompressible, but as we can see, density changes with time and space. Moreover, at a fixed position (i.e. in a stationary frame of reference), density would change with time.

So, does density depend on the frame of reference? What's the real definition of compressibility in fluid mechanics?

Answer 1

The definition of incompressibility is that the density of a fluid parcel (a volume element) does not change (i.e., is constant); this is your first equation: $$ \frac{D}{Dt}\rho(\mathbf x_0,t)=0 $$ which leads to the solenoidal constraint, $\nabla\cdot\mathbf u=0$.

As for your "counter example," there is no issues because the density field can actually vary in space and time in both Lagrangian and Eulerian frames. It is just that in the former, you track the evolution a constant-density fluid parcel ($\rho(\mathbf x_0,t)$) rather than tracking the evolution of the grid in the latter ($\rho(\mathbf x,t)$).

Answer 2

In all of classical physics it's generally taken that the observer is is in an inertial frame. This frame is the frame where objects with no force acting on them move in straight lines (or are at rest). If one uses a non-inertial frame, say with constant acceleration, then the laws of physics will take on a different form.

Einstein argued that the laws of physics should be in general covariant form so that it ought not to matter what frame one is in.

Answer 3

I think your problem is that,according to what you said on April 26 at 11:15, you really don't understand what incompressible means. If you have a certein initial non uniform distribution of $\rho$ in space it doesn't mean that the distribution will stay constant with respect to time, on the contrary, for the value of density of a certein particle(small portion of fluid) to remain constant in time you must follow it along its motion to realize that it remains constant and in doing so you realize that the "constant" value of $\rho$ is assigned to different places at different times. In short, an initial non uniform distribution in space will not remain invariable with time because the different particles carry with them their values of $\rho$ as they move. For your counter example to be general you must put $\rho(x,t)=x-t u_x$ and this is really a good example not a "counter example"

I also want to add a remark to the answer of Kyle kanos, I think that in the Lagrangian formulation incompresibilty does not allow $\rho$ to vary with time