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I am preparing a presentation in which I need to calculate the divergence of the vector field $$\mathbf{F}(x,y,z) = \frac{(x,y,z)}{ \lVert (x,y,z) \lVert ^3}$$ The thing is I don't want to make the straight calculation because is very time consuming, nor just give the result without any argument.

Can you think of any way to get the result easily? I know it is 0, so maybe there is some geometric interpretation or something alike... ?

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    Use the divergence theorem to show that the integral of divergence over a sphere of radius $r$ is independent of $r$. – 2017-02-23

2 Answers 2

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It has zero divergence. Take one derivative, $$\partial_x F = \frac{-2x^2 + y^2 + z^2}{||(x,y,z)||^5},$$ Then by symmetry, the other derivatives will be the same and hence upon adding them up, you get zero.

Perhaps it is interesting to note that $F = -\nabla (1/||(x,y,z)||) = -\nabla (\frac{1}{r}),$ meaning that the divergence of $F$ is really the question if $1/r$ is a harmonic function in $\mathbb{R^3\setminus{\{0\}}}.$ Since it is in fact, $F$ has zero divergence. But again, only the calculation will show either of this fact that given a harmonic function as a potential $\phi$, and letting the force field be defined by $F = -\nabla \phi,$ you get a divergenceless force field.

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In spherical coordinates,

$$\mathrm v (r, \theta, \phi) := \frac{1}{r^2} \hat{r} + 0 \cdot \hat{\theta} + 0 \cdot \hat{\phi} = \frac{1}{r^2} \hat{r}$$

Hence, the divergence of vector field $\mathrm v$ is given by

$$\boxed{\nabla \cdot \mathrm v = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{1}{r^2} \right) = 0}$$

The geometric interpretation is that the magnitude of the vector field decays inversely proportional to $r^2$, whereas the surface area of a sphere centered at the origin grows directly proportional to $r^2$. Hence, the double integral of the vector field over the surface of a sphere centered at the origin is independent of the radius $r$.

One can also use symbolic computation. In SymPy:

>>> from sympy import *
>>> x1, x2, x3 = symbols('x1 x2 x3')
>>> f1 = x1 / (sqrt(x1**2 + x2**2 + x3**2))**3
>>> f2 = x2 / (sqrt(x1**2 + x2**2 + x3**2))**3
>>> f3 = x3 / (sqrt(x1**2 + x2**2 + x3**2))**3
>>> divergence = diff(f1,x1) + diff(f2,x2) + diff(f3,x3)
>>> divergence
-3*x1**2/(x1**2 + x2**2 + x3**2)**(5/2) - 3*x2**2/(x1**2 + x2**2 + x3**2)**(5/2) - 3*x3**2/(x1**2 + x2**2 + x3**2)**(5/2) + 3/(x1**2 + x2**2 + x3**2)**(3/2)
>>> simplify(divergence)
0