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I have always been interested in generating discrete approximations that tend to mimic the properties of their continuum counterparts. Usually, one approximation that is desired is that the discrete solution is divergence free. Take Maxwell equations in the time domain as an example: $$ -\frac{\partial\mathbf{b}(\mathbf{r},t)}{\partial t} = \nabla\times\mathbf{e}(\mathbf{r},t) \\ \mathbf{j}(\mathbf{r},t) = \nabla\times\mathbf{h}(\mathbf{r},t) \\ \nabla\cdot\mathbf{b}(\mathbf{r},t) = 0 \\ \nabla\cdot\mathbf{j}(\mathbf{r},t) = 0 $$ The third equation implies that only two out of the three components of $\mathbf{b}$ are independent of each other.

However, I would like a more intuitive explanation of what it means to be divergent free. Even if the context is other than EM.

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    $\begingroup$ The usual E&M intuition is pretty good: simply that there are no sources or sinks. So consider any closed surface. With no sources, there can't be any net flux coming out of that surface. And with no sinks, there can't be any net flux going in. So all flux that goes in must come out, and vice versa. So guess what the integral over any closed surface adds up to. $\endgroup$ May 30 '17 at 10:49
  • $\begingroup$ Any divergence-free function can be expressed as the curl of some other vector function, implying that div-free functions have the general property of circuitous streamlines. Think of the classic picture of a bar magnet's field lines. Piggy-backing on John's good answer above, I just wanted to point out that the surface integral of the field components normal to that closed surface vanishes. But that may have already been clear from the context of the discussion of flux. $\endgroup$ May 31 '17 at 12:34
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I find the following two notions helpful.

  1. The continuity equation $$ \frac{\partial\rho}{\partial t} + \nabla \cdot (\rho\boldsymbol{u}) = \frac{\mathrm{d}\rho}{\mathrm{d}t} + \rho (\nabla \cdot \boldsymbol{u}) = 0, $$ implies that a flow with divergence-free velocity $\boldsymbol{u}$ is incompressible ($\mathrm{d}\rho/\mathrm{d}t=0$: the density $\rho$ remains unchanged along the flow). The matter must neither diverge (when $\rho$ would drop) nor converge (when $\rho$ would grow), but the motions of adjacent fluid elements must conspire to keep the density constant along the flow.

  2. Gauß's theorem $$ \int_{\partial V}\boldsymbol{f}\cdot\mathrm{d}\boldsymbol{\Omega}= \int_V\,(\boldsymbol{\nabla}\cdot\boldsymbol{f})\,\mathrm{d}V $$ implies that the total flux of a divergence-free field $\boldsymbol{f}$ through any closed surface vanishes. (Applying this to $\boldsymbol{f}=\rho\boldsymbol{u}$ shows that the continuity equation expresses matter conservation.)

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  • $\begingroup$ @RichardZhang thanks for the edit. I simply forgot the $\cdot$. $\endgroup$
    – Walter
    May 31 '17 at 15:58
  • $\begingroup$ That's not completely correct. First of all, you seem to assume that the density is constant in space (which is not true in many applications). Secondly, incompressibility is not defined as $d\rho/dt = 0$ but rather as any control volume in the flow keeps its volume over time ("it's not getting compressed"). $\endgroup$
    – Jan
    Jun 1 '17 at 8:20
  • $\begingroup$ @Jan 1. What (do you think) is not completely correct? I don't assume that $\nabla\rho=0$ (why do think that? – note the change from partial to total time derivative) 2. Your definition of incompressible is sloppy (it's correct only if the control volume is deforming with the flow, but not for a fixed volume) and equivalent to $\mathrm{d}\rho/\mathrm{d}t=0$ if $\rho$ is the volume density of a conserved quantity such as mass in (most) fluids. Note also that Wikipedia defines incompressibility in terms of density, not volume. $\endgroup$
    – Walter
    Jun 1 '17 at 14:02
  • $\begingroup$ Alright, I see. As for 1. you are correct. As for the definition, though written down sloppy, defining incompressibililty through volumes is more general. (And, because it is required for any control volume, equivalent to $d\rho/dt$ in the case that there exists a smooth mass density function $\rho$.) $\endgroup$
    – Jan
    Jun 3 '17 at 8:44
  • $\begingroup$ OK. Now it came to me what I thought was wrong with your answer... You basically say that a flow that is div free and satisfies the continuity equation is incompressible. I somehow read that div free implies incompressibility because of the continuity equation (as if the latter was a generally valid identity). $\endgroup$
    – Jan
    Jun 4 '17 at 16:49

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