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Suppose I am working in $\Bbb R^3$. Suppose I have a pond and I drop some dust on the surface. If the materials spread out, I have positive divergence, usually.

Let $\mathbf{v}(\mathbf{x})$ denote the velocity of a dust particle. I assume my divergence is a measure of the magnitude of this particle's tendency to move away. So if I want to find the magnitude the particle will move, why wouldn't $||\mathbf{v}(\mathbf{x})||$ denote my divergence?

hmakholm left over Monica
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Stan Shunpike
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    One of the problems is that if you had a steady flow - same $\mathbf{v}$ everywhere - then $||v(x)||$ may be large, but there is no divergence! In each tiny volume you would have as much dust entering as leaving. Divergence seeks to measure just that. If outflow is bigger than inflow, then you have positive divergence. – Jyrki Lahtonen Dec 21 '15 at 21:30
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    See an earlier answer by yours truly. IMHO this is close to being a duplicate of that question. But as I answered that potential duplicate target, I should not vote to close. Furthermore, this is such a natural question that it may have occurred even earlier. Such as here. Read MJD's answer carefully. – Jyrki Lahtonen Dec 21 '15 at 21:31
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    You might also benefit from reading this answer: http://math.stackexchange.com/a/150880. – David K Dec 21 '15 at 21:43
  • Just a note on the question in the title: $\nabla \cdot \mathbf v$ is just a notation meaning the divergence of the vector field $\mathbf v$. It's not a definition. –  Dec 21 '15 at 22:41

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The divergence is not just a single particle's tendency to move, but the tendency of a small ball of test particles to move away from each other. If the test particles stay together, they can move as fast as you like, but they don't diverge from each other.

hmakholm left over Monica
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  • +1 This is very interesting. I've never heard it described that way. Can you give a more rigorous definition of divergence then using a 3-ball (in this case)? And an $n$-ball more generally? – Stan Shunpike Dec 21 '15 at 21:36
  • Like I want it defined in terms of densities then not points. – Stan Shunpike Dec 21 '15 at 21:36
  • @Stan: Density is a scalar field. Only vector fields have divergence. Are you thinking about something like gradient? – Jyrki Lahtonen Dec 21 '15 at 21:38
  • @StanShunpike Here's a rigorous definition of divergence: http://mathworld.wolfram.com/Divergence.html. (The definition is equation (1) and a little of the text before and after it on that page.) – David K Dec 21 '15 at 21:50
  • @Stan: If you have a small ball (or cube) of test particles with velocities given by the field, then the divergence at the ball's location is the instantaneous rate of change of the volume of the ball, divided by its current volume. – hmakholm left over Monica Dec 21 '15 at 22:24