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I am dealing with the vector field:

$$v = \dfrac{\hat{\mathscr r} }{{\mathscr r}^2}$$

And I am studying its divergence. If we compute it we get:

$$\nabla\cdot\left(\dfrac{\hat{\mathscr r} }{{\mathscr r}^2}\right) = 0, \qquad {\mathscr r} \ne 0.$$

I understand we are dealing with a delta function, which explains why we get $\nabla\cdot v = 0$ everywhere but in the origin, where it blows up.

But the fact that $\nabla\cdot v = 0$ does not make sense to me looking at the graph of the vector function:

enter image description here

Where we can see how the vector field spreads out. The only reasoning I see is that at the origin, the vector field spreads out so much that once we look out of it the field cannot spread any more, thus we get zero divergence. But I insist, this cannot be seen in the plot of the function

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    $\begingroup$ Note that your graph is not to scale. The magnitude drops off as $1/r$, and the divergence will be much more accurately represented with a better visual. $\endgroup$
    – cxx
    Commented Feb 12, 2019 at 16:07
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    $\begingroup$ Look at a diagram for the volume element in spherical coords: it's a box with four sides in the radial direction (no flux through those since the field is radial) and two spherical caps as the top and bottom of the box. The "bottom" cap which is nearer to the origin is smaller than the top cap, but the field is stronger at the bottom cap than it is at the top cap: the net effect is that the flux into the box through the bottom cap is the same as the flux out of the box through the top cap, i.e. the divergence is 0. $\endgroup$
    – NickD
    Commented Feb 12, 2019 at 16:30
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    $\begingroup$ Note: the accepted answer has been significantly changed due to being initially somewhat incorrect. Feel free to take whatever action you think this merits. $\endgroup$
    – probably_someone
    Commented Feb 12, 2019 at 18:07
  • $\begingroup$ The plot does not show $\vec r/r^2$ but something like $\vec r$. As the plot is wrong your question is ill posed. $\endgroup$
    – my2cts
    Commented Feb 12, 2019 at 19:11
  • $\begingroup$ @my2cts note that what is plotted is the divergence. More details on the problem: imgur.com/a/SgurYA1 $\endgroup$
    – JD_PM
    Commented Feb 12, 2019 at 19:34

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Despite the name, divergence does not say whether the field diverges in the meaning "spreads". A vector field can have zero divergence and be spreading, or it can have non-zero divergence and not be spreading. An example of the former, you have in your post. An example of the latter is $\vec{F} = x \hat{x}$ where $\hat{x}$ is the unit vector in the $x$ direction.

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  • $\begingroup$ Mm I see. I took Griffiths' words too literally: 'is a measure of how much the vector field spreads out from the point in question.' What would you say divergence is then? $\endgroup$
    – JD_PM
    Commented Feb 15, 2019 at 17:28
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    $\begingroup$ I'd rather say that it's how much of the vector field is produced in the point. Consider a small volume element. The divergence is how much more flows out of this small volume than flows into it, divided by the volume of the element. $\endgroup$
    – md2perpe
    Commented Feb 15, 2019 at 17:33
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Your intuition about what divergence means is incorrect. Divergence measures how much field enters vs. leaves a very small volume about a particular point. At all points in this diagram except for the very center, if you were to zoom in enough, you would find a sufficiently small volume where you would see the following pattern: on the side of the volume closer to the center, the field would be slightly larger, but slightly more spread apart, while the field on the other side is smaller, but more uniform in direction. The effect of the spreading-out on the near side causes cancellation that reduces the amount of field poking perpendicularly through the surface on that side, so the field entering and leaving the volume are the same. Hence, there is no divergence at points other than the center (where, no matter how small the volume, more field is leaving the volume than is entering it). At the center, in fact, the more you shrink the volume, the stronger the field is on the border, so the more field is leaving the volume; this goes to infinity, which means the divergence is also infinity.

Important note: The first version of this answer had an incomplete intuitive definition of divergence. This version has been fixed.

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  • $\begingroup$ could you add a plot of the diagram's zoom you explained or am I asking too much? $\endgroup$
    – JD_PM
    Commented Feb 15, 2019 at 17:24

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