# Geometric interpretation of Divergence of $\vec{f} = \frac{1}{r^2} \hat{r}$

I know that the mathematics tells me that the divergence is zero for the below vector field:

$$\vec{f} = \frac{1}{r^2} \hat{r}$$

But I am more interested in the geometric intuition of it. Here is what I am looking at. The vector length is decreasing as I am increasing the radii of the sphere around origin in 3D space. Now divergence is defined as $$\partial {v_x}/\partial x +\partial {v_y}/\partial y+\partial {v_z}/\partial z$$ in cartesian coordinates.

Now lets think about a point other than the origin. Lets take $$\partial {v_x}/\partial x$$. Now as the vector is decreasing in length as we are increasing the radii, this slope must be less than zero, i.e., $$\partial {v_x}/\partial x < 0$$ as the value is decreasing as we are increasing the $$x$$. The same logic can be applied to other dimensions, i.e.,

$$\partial {v_y}/\partial y < 0$$

$$\partial {v_z}/\partial z < 0$$

Now given all these inequalities, how can $$\partial {v_x}/\partial x +\partial {v_y}/\partial y+\partial {v_z}/\partial z=0$$ ?

• Maybe this can help : youtu.be/rB83DpBJQsE Commented Aug 28, 2020 at 15:45
• still, what is wrong in my argument?. Commented Aug 28, 2020 at 15:51
• Consider the $y$ axis, along it, the $x$ component is zero, but at any point on this axis, the value of the field increases for and infinitesimal increment parallel to x. Commented Aug 28, 2020 at 17:25

Note that we have

$$\frac{\hat r}{r^2}=\frac{\vec r}{r^3}$$

So, the Cartesian components are $$\displaystyle \frac{x}{(x^2+y^2+z^2)^{3/2}}$$, $$\displaystyle \frac{y}{(x^2+y^2+z^2)^{3/2}}$$, and $$\displaystyle \frac{z}{(x^2+y^2+z^2)^{3/2}}$$.

Therefore, the partial derivative with respect to the $$i$$'th Cartesian coordinate of the $$i$$'th component of $$\displaystyle \frac{\hat r}{r^2}$$ is

$$\frac{\partial }{\partial x_i}\frac{\hat x_i\cdot \vec r}{r^3}=\frac{r^2-3x_i^2 }{r^5}\tag1$$

for $$r\ne 0$$. Clearly these partial derivatives are not negative for all $$(x,y,z)$$.

However, summing $$(1)$$ over $$i$$ reveals for $$r\ne0$$

$$\nabla\cdot \left(\frac{\vec r}{r^3}\right)=\frac1{r^5}\sum_{j=1}^3 ( r^2-3x_i^2)=0$$

as expected!

• @user3001408 Please let me know how I can improve my answer. I really want to give you the best answer I can. Commented Aug 29, 2020 at 3:13
• And feel free to up vote and accept an answer as you see fit of course. ;-) Commented Aug 29, 2020 at 3:13

Actually, it's $$\vec{\nabla}\cdot\vec{f}=4\pi\delta^{(3)}(\vec{r})$$. The geometric-cum-physical intuition is that$$\int_{r\le R}\vec{\nabla}\cdot\vec{f}\mathrm{d}^3\vec{x}=\int_{r=R}\vec{f}\cdot\mathrm{d}\vec{S}$$equates a "charge" enclosed within a ball to a field's surface integral at its edge. Here the charge density $$\vec{\nabla}\cdot\vec{f}$$ is $$0$$ except at $$r=0$$, because of a point charge $$4\pi$$. Meanwhile, since $$f=1/R^2$$ at the surface $$r=R$$ of area $$4\pi R^2$$, the surface integral on the right-hand side is the enclosed charge of density $$4\pi\delta^{(3)}(\vec{r})$$.