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,

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

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

But let's consider a simple case in 1D in which the density was of the form $\rho(x,t)=x-t$ and $\vec V=\vec u_x$. 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?


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 ($\rho(\mathbf x,t)$).

  • $\begingroup$ I understand your arguments, but I don't see how they refute the validity of my counter-example. I get the whole point of what compressibility means, but I would redefine it this way: a fluid is incompressible when given a initial distribution of density even for which density may vary in space (this is what puzzles me, because then the fluid would be compressible), this distribution doesn't change (i.e., the density of a fluid parcel doesn't change in time). I think that would be a better way to understand it. $\endgroup$ – alfdc80 Apr 26 '15 at 11:15
  • $\begingroup$ Your problem is conflating the two frames. The Lagrangian frame considers the evolution of a deformable volume of constant density. The Eulerian frame considers the evolution of a fixed grid. Your definition is essentially already employed, when properly understood, in the form of $D_t\rho=0$ and as defined in my answer. $\endgroup$ – Kyle Kanos Apr 26 '15 at 13:43
  • $\begingroup$ Note also that your "counter-example" only works if $V=1$, otherwise it neither satisfies continuity nor incompressibility. $\endgroup$ – Kyle Kanos Apr 26 '15 at 13:45
  • $\begingroup$ I think I understand both approaches: in a Lagrangian frame, you move with fluid and in a Eulerian frame, you stay fixed in space and watch the fluid move pass you. What I really mean is that it is basically an issue of terminology. I personally believe that incompressible should not be used in the definition, because the idea I have of incompressibility is that of something constant both in space and in time, and I gave an example of a fluid that is incompressible by the definition, yet it changes in space at a fixed time. $\endgroup$ – alfdc80 Apr 28 '15 at 23:14
  • $\begingroup$ By the way, could you give an example where it wouldn't work if $ V \neq 1 $? $\endgroup$ – alfdc80 Apr 28 '15 at 23:18

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

  • $\begingroup$ I think the phrase "this distribution doesn't change (i.e., the density of a fluid parcel doesn't change in time)" on my comment on April 26 at 11:15 is not clear enough. What I meant by that is that, in a frame of reference moving at the same speed as that of the fluid, density may change in space, but not in time, hence $\frac{D\rho}{Dt}=0$, and the fluid is incompressible according to the definition. $\endgroup$ – alfdc80 May 7 '15 at 13:33
  • $\begingroup$ But like you said, if you have a certain initial non uniform distribution of ρ in space it doesn't mean that the distribution will stay constant with respect to time, and my example (like yours too) is a good example of that. Therefore, I think it should not be called incompressible, although it is by the definition. $\endgroup$ – alfdc80 May 7 '15 at 13:34
  • $\begingroup$ Moreover, it seems like the definition depends on the frame of reference. If you move with the fluid, it doesn't change with time, if stay fixed at some point, it does. $\endgroup$ – alfdc80 May 7 '15 at 13:35

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