Question about divergence of $\vec{F} = \frac{1}{r^2} \hat{r}$

Question

I am looking at the divergence of this famous expression:

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

i saw this calculation which looks reasonable ...

$\hat{r} = (x,y,z)/\sqrt{x^2 + y^2 + z^2}$ $$ F(x,y,z) = \frac{1}{(x^2 + y^2 + z^2)^{3/2}} (x,y,z) \\ \frac{\partial}{\partial x} F_x = \frac{\partial}{\partial x} \frac{x}{(x^2 + y^2 + z^2)^{3/2}} = \frac{-2x^2 + y^2 + z^2}{(x^2 + y^2 + z^2)^{5/2}} \\ \frac{\partial}{\partial y} F_y = \frac{\partial}{\partial y} \frac{y}{(x^2 + y^2 + z^2)^{3/2}} = \frac{x^2 -2y^2 + z^2}{(x^2 + y^2 + z^2)^{5/2}} \\ \frac{\partial}{\partial z} F_z = \frac{\partial}{\partial z} \frac{z}{(x^2 + y^2 + z^2)^{3/2}} = \frac{x^2 + y^2 -2z^2}{(x^2 + y^2 + z^2)^{5/2}} \\ $$ Putting together : $$ \nabla\cdot F = \frac{\partial}{\partial x} F + \frac{\partial}{\partial y} F + \frac{\partial}{\partial z} F = \frac{0}{(x^2 + y^2 + z^2)^{5/2}} = 0 $$

However I am confused about one thing.

1. Suppose vector field $\vec{F}$ has a value $F_1$ at (1,1,1).

2. Now if x changes by dx, that is x2 = 1+dx, y and z remaining unchanged,

3. Then the change in x component vector F will be 0 since partial derivative of $\vec{F_x}$ w.r.t to x evaluates to 0. Similarly the change in y and z component will also be 0, since their partial derivatives also evaluate to 0 at (1,1,1).

This seems confusing to me. How can there be no change in the vector when x changes by dx, at (1,1,1) ?

What am I doing wrong ?

Answer 1

$ \def\l{\lambda} \def\n{\nabla} \def\L#1{\l^{-#1}} \def\qiq{\quad\implies\quad} \def\tr{\,{\rm Trace}} \def\c#1{\color{red}{#1}} $Differentiate the length $\l$ of the position vector $r$ $$\eqalign{ \l^2 &= r\cdot r \\ 2\l\:d\l &= 2r\cdot dr \\ d\l &= \L1r\cdot dr \\ }$$ Then calculate the gradient of $F$ $$\eqalign{ F &= \L3r \\ dF &= \L3dr - r(3\L4d\l) \\ &= \L3I\cdot dr - r(3\L5\,r\cdot dr) \\ &= (\L3I - 3\L5rr)\cdot dr \\ \n F &= \L3I - 3\L5rr \\ }$$ Finally, calculate the divergence $$\eqalign{ \n\cdot F &= \tr(\n F) \\ &= \L3\tr(I) - 3\L5\tr(rr) \\ &= \L3(n) - 3\L5(\l^2) \\ &= (n-3)\L3 \\ }$$ Therefore in three dimensions $(n=3)$ the divergence is equal to zero, but in any other dimension it is non-zero.

Note however, that the Taylor series at $r=r_1$ uses the gradient of $F\,$ not its divergence $$\eqalign{ F(r_2) - F(r_1) &= (r_2 - r_1)\cdot\c{\n F(r_1)} \;+\; {\cal O}\left(\|r_2-r_1\|^2\right) \\ dF &\approx dr\cdot\c{(\L3I - 3\L5r_1r_1)} \\ dF &\approx \L3dr - 3\L5(dr\cdot r_1)\,r_1 \\ }$$ So the change in the function value is $dF$ which (as you suspected) is not equal to zero.