Conservation of mass in Navier-Stokes equation I just started looking into Navier-Stokes Equations and one of the two equation of Navier-Stokes is:
$$\nabla \cdot \vec u = 0.$$
This equation is said to be the conservation of mass. The concept of conservation of mass in fluid dynamics makes sense. My understanding is since we are considering a given system, say a river, it is neither losing mass nor gaining any.
I cannot understand how the equation ${\rm div}\,\vec{u} = 0$ implies that since it is basically saying that the divergence of velocity vector field is zero?
 A: $\nabla \cdot \mathbf{u} = 0$ is a kinematic constraint on the velocity field that is equivalent to $\frac{D \rho}{D t} = 0$ (using mass equation)
It can be interpreted in this way: each material particle keeps a constant value of density. Let's call $\mathbf{x}_p (t)$ the position of the particle in space as a function of time, and $\rho_p(t) = \rho(\mathbf{x}_p(t),t)$ its density, that is constant if the incompressibility constraint holds, $\rho_p(t) = \overline{\rho}_p$
Approximating a material particle with an elementary material volume (that is a closed system, since it doesn't exchange mass with through its boundary, and thus has a constant mass, $\Delta m_p(t) = \Delta \overline{m}_p$), if the density inside the material volume is constant in time, thus the volume of the elementary material volume is constant as well, since $\Delta m_p(t) = \rho_p (t) \Delta V_p (t)$ and thus
$\Delta V_p(t) = \dfrac{\Delta m_p(t)}{\rho_p(t)} = \dfrac{\Delta \overline{m}_p}{\overline{\rho}_p} = \Delta \overline{V}_p = \text{const.}$
A: Generally speaking, the "intuitive" meaning of $\nabla \cdot$ is that it measures how much each particular point of a vector field locally resembles a material source or a sink (for comparison, $\nabla \times$ represents how much it locally fails to conserve energy, were it a force field). That is, were you to zoom in extremely (ideally: infinitely) far into a given point $X$, you would find that the field lines were tending away from (or toward) each other, moreso the larger that $[\nabla \cdot (...)](X)$ is. Now think about what that would have to mean if you consider matter moving along them, as in the velocity lines for the fluid. Unfortunately, I don't have a neat drawing program readily on hand to draw you a picture of a constant-divergence field, which is what the microscopic shape of a field of variable divergence will look like. In any case, you should agree it would have to look like this.
If the fluid is compressible, then that would mean that the fluid had to be rarefying (if tending away, i.e. the divergence is positive) or compressing (if tending together, i.e. the divergence is negative) at that point.
If the fluid is incompressible, then that means the density has to remain constant yet somehow fluid has to still move away or toward, which means it must move away or toward without leaving an absence of mass or a surfeit of mass. That would mean mass would be having to be created or destroyed at that point.
FWIW, an infinite divergence corresponds to a radiating point, i.e. where that all the lines are leaving or entering like spokes. An interesting example is the field $\mathbf{u}(X) = \frac{1}{|X - O|^2}$, where $O$ is the origin (or if you like, any particular point). $(\nabla \cdot \mathbf{u})(O)$ "is infinite", but it is zero everywhere else - actually, it's a delta "function".
Hence, $\nabla \cdot \mathbf{u} = (X \mapsto 0)$ - i.e. $(\nabla \cdot \mathbf{u})(X) = 0$ everywhere - expresses conservation of mass, provided the fluid is incompressible.
A: The conservation of mass in the Navier-Stokes equations is,
$$\frac{\partial\rho}{\partial t}+\nabla\cdot\left(\rho\mathbf v\right)=0.\tag{1}$$
Using the material derivative, Equation (1) can be written as,
$$\frac{D\rho}{Dt}+\rho\left(\nabla\cdot\mathbf v\right)=0.$$
Then by using the condition of incompressibility $D_t\rho=0$ and we are left with,
$$\rho\left(\nabla\cdot\mathbf v\right)=0\implies\nabla\cdot\mathbf v=0.$$
So what you have written is the result of applying incompressibility to the conservation of mass (that the velocity field is divergence-free), rather than the conservation of mass.
