divergence of $\frac{\hat{r}}{r}$ is positive, $\frac{\hat{r}}{r^2}$ is zero and $\frac{\hat{r}}{r^3}$ is negative at points other than origin.
As I have studied, divergence in 3D space tells whether there are sources or sinks present.My initial thinking was that since the magnitude of the vectors are decreasing, so the divergence should be negative for all the three cases but that's not the case.It is not like more vector field is being generated, the same number of vector field passes through the spherical cross section of radius $r+dr$ that is being passed through the cross section of radius $r$.
My Question is can anyone provide the physical intuition of why the divergence is positive for the first vector field, zero for the second and negative for the third?
Best Answer
First let build up some intuition by looking at the divergence in 1D. Imagine you have a river which has a very uniform flow so you can approximate the flow as 1 dimensionsal. The flow is also static in time. This particular river flows in the positive x direction so $v(x)$ is positive and assume at a particular location we have that $\frac{\partial v}{\partial x}>0$. This means the divergence can be approximated by $$\nabla\cdot\vec v=\frac{\partial v}{\partial x}\approx\frac{1}{\Delta x}(v(x+\Delta x)-v(x))>0$$ We can conclude that $v(x+\Delta x)> v(x)$. The quantity $v(x)\times\text{cross section}$ gives how much volume is passing through the plane at $x$ at each second. We must conclude that more volume is passing through $v(x+\Delta x)$ than through $v(x)$. (we assume constant cross section). The density of the water in the region $[x,x+\Delta x]$ must be decreasing as the water flows through this section of river given that no water is destroyed/created. So if you imagine a vector field $\vec v$ as being the flow of water/particles then $\nabla\cdot \vec v>0$ means the density decreases locally and conversely when $\nabla\cdot \vec v<0$ the density increases locally.
So now that we have built up some intuition I want to show it visually. It's better to draw this in 2D so let's consider the divergence of these functions in 2D: \begin{align} \nabla\cdot\left(\hat r r\right)&=2>0\\ \nabla\cdot\left(\frac{\hat r}{r}\right)&=0\\ \nabla\cdot\left(\frac{\hat r}{r^3}\right)&=-\frac{2}{r^4}<0 \end{align} I picked these functions to give the best visual results.
In the following animations you see particles following a vector field as velocity. The vector fields are, from top to bottom, given by $\vec v(\vec r)\propto\vec r, \frac{\hat r}{r}, \frac{\hat r}{r^3}$. They are also scaled a bit because otherwise some of these animations would be too quick or too slow.
It may be a bit hard to see but you can tell that in the top picture (positive divergence) the density goes down as the particles move outwards. This effect is a combination of the particles becoming more spread out and the particles speeding up, both of which increase divergence. In the bottom picture you see the particles bunching up. The effect of slowing down is so strong that it completely overpowers the spreading out effect. In the middle picture the spreading out and slowing down are perfectly balanced which results in a nice uniform density.
These effects can be translated to 3D but the particular dimensions change, the effect of spreading out is stronger in higher dimensions.