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Curl of a vector field A is non-zero. So that means that the vector which has curl or rotates does not diverge does not spread. So if we take A as the velocity. Then the curl would be circulation per unit area. Then does it mean that when a fluid rotates it does not spread?

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  • $\begingroup$ Why do you think that a vector field with a non-zero curl must have $0$ divergence? $\endgroup$ Commented Aug 9, 2018 at 10:15
  • $\begingroup$ The divergence of the curl of a vector is not the same as the divergence of the vector. $\endgroup$
    – user197851
    Commented Aug 9, 2018 at 10:16
  • $\begingroup$ @AaronStevens I am saying a divergence of curl of a vector . $\endgroup$ Commented Aug 9, 2018 at 10:49
  • $\begingroup$ @LonelyProf can you interpret the problem physically in my scenario. It would be helpful. $\endgroup$ Commented Aug 9, 2018 at 10:50

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As pointed out in the comments, you are mixing up the divergence of the vector and the divergence of the curl of the vector.

Let's say we have a vector $\boldsymbol{A}$. Then, as you have noted, $$\boldsymbol{\nabla} \cdot(\boldsymbol{\nabla} \times \boldsymbol{A})=0$$

But this is looking at the divergence of the curl of the vector.

If you want to talk about how the vector field "spreads out" we want to look at the divergence of the vector itself $$\boldsymbol{\nabla} \cdot \boldsymbol{A}$$

This quantity does not necessarily have to be $0$ even when the curl $\boldsymbol{\nabla} \times \boldsymbol{A}$ is non-zero.

In other words, it is possible to have a vector field with, as you put it, both "circulation" and "spreading out". To determine this all you need to do is calculate the divergence and curl separately, not the divergence of the curl (which is always $0$).

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  • $\begingroup$ I always thought the idealized (2-d) flow around a whirlpool (or tornado/hurricane) was irrotational, and all the vorticity was fixed in the core (funnel/eye). $\endgroup$
    – JEB
    Commented Aug 9, 2018 at 11:40
  • $\begingroup$ Aaron, I think this formatting is more elegant $$ \boldsymbol{\nabla\cdot}\left(\boldsymbol{\nabla \times} \mathbf{A}\right)=0 $$ $\endgroup$
    – Frobenius
    Commented Aug 9, 2018 at 12:31
  • $\begingroup$ @Frobenius Always helping with the formatting tips :) thanks haha $\endgroup$ Commented Aug 9, 2018 at 12:35
  • $\begingroup$ Now I must up-vote it, haha !!! $\endgroup$
    – Frobenius
    Commented Aug 9, 2018 at 12:37
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    $\begingroup$ Giving nice answers must be our priority, then well-formatting is the next priority. $\endgroup$
    – Frobenius
    Commented Aug 9, 2018 at 12:42

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