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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?

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  • $\begingroup$ Without getting into physics: How is a river defined? Clearly, if it rains, a river can gain mass, if by river we mean water flowing in a certain spot. $\endgroup$ Oct 17, 2022 at 21:27
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    $\begingroup$ Instead of the "divergence of a velocity" it might make more sense to think of it as "the divergence of the velocity vector field" being zero btw. $\endgroup$
    – user541686
    Oct 18, 2022 at 6:26
  • $\begingroup$ @ user541686 I agree with your comment. I have made edited my question. $\endgroup$ Oct 18, 2022 at 18:58
  • $\begingroup$ @ JosephDoggie, I am thinking about it in a more ideal case scenario where we are looking at a certain volume of water such that the amount of water entering the river is equal to the amount of water exiting that certain volume. $\endgroup$ Oct 18, 2022 at 19:01

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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.

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    $\begingroup$ That makes sense, didn't know that Div(u) = 0 was not the generic equation but for the special case of incompressibility. $\endgroup$ Oct 17, 2022 at 12:14
  • $\begingroup$ The true condition of incompressibility is rather the very $\nabla . \mathbf{u} = 0$ saying that the volume of a fluid particle does not change. The relation to the actual density is more problematic especially when using the incompressible N.-S. equations for slow flows of gases or liquids that are actually somewhat compressible. Or even more when using the Boussinesq approximation. The whole mathematical machinery is rather complicated arxiv.org/abs/1307.6030 $\endgroup$ Oct 18, 2022 at 9:08
  • $\begingroup$ @VladimirFГероямслава No, $\nabla\cdot\mathbf u=0$ is the result of incompressibility, not the definition. The definition is that the mass density is constant along a flow in time, which is what $D\rho/Dt=0$ states. $\endgroup$
    – Kyle Kanos
    Oct 20, 2022 at 14:40
  • $\begingroup$ nope. 7 more to go... $\endgroup$ Oct 20, 2022 at 14:55
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$\nabla \cdot \mathbf{u} = 0$ 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

$\Delta V_p(t) = \dfrac{\Delta m_p(t)}{\rho_p(t)} = \dfrac{\Delta \overline{m}_p}{\overline{\rho}_p} = \Delta \overline{V}_p = \text{const.}$

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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.

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  • $\begingroup$ I agree with your intuition for $\nabla \cdot$. I do not understand the remark you made about $\nabla \times$. The intuition I have for $\nabla \times$ is that suppose you place a boat in the water, if it rotates the velocity vector ($\vec v$) has some curl to it, that is $\nabla \times \vec v \neq 0$. I do not understand what you mean when you say "$\nabla \times$ represents how much it locally fails to conserve energy". $\endgroup$ Oct 18, 2022 at 18:55
  • $\begingroup$ @Shivank Chadda: $\nabla \times \mathbf{F}$ can actually be given as an integral, but this isn't necessarily typically mentioned in the usual calculus books. In particular, $(\nabla \times \mathbf{F})(X) \cdot \hat{\mathbf{u}} = \lim_{A \rightarrow 0} \frac{1}{|A|} \oint_\gamma \mathbf{F} \cdot d\mathbf{r}$. Note the last integral is a work integral; $A$ is the area enclosed by the curve $\gamma$ surrounding point $X$ in the plane with surface normal $\hat{\mathbf{u}}$, so in effect the curl measures the normalized work done in moving around an arbitrarily small loop enclosing the point. $\endgroup$ Oct 18, 2022 at 22:55

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