# How can we argue $\nabla\cdot v = 0$ when ${\mathscr r} \ne 0$ on this vector function?

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:

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

• 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. – Hanting Zhang Feb 12 at 16:07
• 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. – NickD Feb 12 at 16:30
• Note: the accepted answer has been significantly changed due to being initially somewhat incorrect. Feel free to take whatever action you think this merits. – probably_someone Feb 12 at 18:07
• The plot does not show $\vec r/r^2$ but something like $\vec r$. As the plot is wrong your question is ill posed. – my2cts Feb 12 at 19:11
• @my2cts note that what is plotted is the divergence. More details on the problem: imgur.com/a/SgurYA1 – JD_PM Feb 12 at 19:34

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.