prove: $\nabla \cdot \frac{\vec{r}}{r^3} = 0$

Question

prove:

$\nabla \cdot \frac{\vec{r}}{r^3} = 0$

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a few useful identities for proof:

$\vec{r} \cdot \vec{r} = 1$

$\nabla r^n = n r^{n-2} \vec{r}$

$\nabla \cdot (\vec{v} + \vec{w}) = \nabla \cdot \vec{v} + \nabla \cdot \vec{w}$

$\nabla \cdot (A \vec{v}) = (\nabla A) \cdot \vec{v} + A (\nabla \cdot \vec{v})$

$r = (x^2 + y^2 + z^2)^{-\frac{1}{2}}$

$\vec{r} = x\hat{\text{i}} + y\hat{\text{j}} + z\hat{\text{k}}$

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ok...so i attempt this proof as follows:

$\nabla \cdot \frac{\vec{r}}{r^3} = \nabla ( r^{-3}~ \vec{r})$

by product rule:

$\nabla \cdot \frac{\vec{r}}{r^3} = (\nabla r^{-3}) \cdot \vec{r} + r^{-3}(\nabla \cdot \vec{r})$

by one of the identities that kind of looks like a derivative rule:

$\nabla \cdot \frac{\vec{r}}{r^3} = -3r^{-5}\vec{r} \cdot \vec{r} + r^{-3}(\nabla \cdot \vec{r})$

then because $\vec{r} \cdot \vec{r} = 1$:

$\nabla \cdot \frac{\vec{r}}{r^3} = -3r^{-5} + r^{-3}(\nabla \cdot \vec{r})$

At this point i get stuck trying to get this expression to equal zero...

Answer 1

Your error is the assumption: $$ \vec{r}\cdot\vec{r}=1 $$ Correct value is $$ \vec{r}\cdot\vec{r}=r^2 $$ Another intresting fact is $$ \nabla\cdot\vec r=3. $$

Answer 2

Writing

$\nabla \cdot \left (\dfrac{\vec r}{r^3} \right ) = \nabla \cdot (r^{-3} \vec r), \tag 1$

and using the general rule from this wikipedia page

$\nabla \cdot (f \vec X) = \nabla f \cdot \vec X + f\nabla \cdot \vec X, \tag 2$

we find

$\nabla \cdot (r^{-3} \vec r) = \nabla (r^{-3}) \cdot \vec r + r^{-3} \nabla \cdot \vec r; \tag 3$

for functions $f:I \to \Bbb R$ and $g:\Bbb R^n \to I$, $I$ open in $\Bbb R$, we also have the identity

$\nabla f(g) = f'(g) \nabla g, \tag{3.5}$

whence

$\nabla (r^{-3}) = -3r^{-4} \nabla r = -3r^{-4}\dfrac{\vec r}{r} = -3r^{-5}\vec r, \tag 4$

and so

$\nabla (r^{-3}) \cdot \vec r = -3r^{-5} \vec r \cdot \vec r = -3r^{-5}r^2 = -3r^{-3}; \tag 5$

also

$\nabla \cdot \vec r = 3, \tag 6$

so

$r^{-3} \nabla \cdot \vec r = 3r^{-3}; \tag 7$

combining (3), (5) and (7) yields

$\nabla \cdot (r^{-3} \vec r) = -3r^{-3} + 3r^{-3} = 0, \tag 8$

which was to be shown.

Answer 3

Alternatively we can use: $$\nabla\cdot \left(\frac{\vec r}{r^3}\right)=?$$

$$\vec r=r\hat r$$

$$\nabla\cdot \left(\frac{\vec r}{r^3}\right)=\nabla\cdot \left(\frac{r\hat r}{r^3}\right)$$

$$=\nabla\cdot \left(\frac{\hat r}{r^2}\right)$$

$$=4\pi \delta^{(3)}(\vec r)$$

where $\delta^{(3)}(\vec r)$ is three-dimensional $\textbf{Dirac-Delta function}$

[Reference:Appendix of Quantum Mechanics By Zetilli and D.J. Griffith of Electrodynamics]

Answer 4

Solution : $$\nabla \cdot\frac{\vec r}{r^3}$$ $$=(\nabla \cdot \vec r)\frac{1}{r^3} + \vec r \cdot \left( \nabla \frac{1}{r^3} \right)$$ where, $\nabla \cdot(\phi \vec A)=(\nabla \cdot \vec A) \phi + \vec A \cdot (\nabla \phi)$ $$=\frac{3}{r^3}+\vec r \cdot [ (-3)r^{-3-2}\vec r ] $$ where , $ \nabla r^n=nr^{n-2} \vec r$ $$=\frac{3}{r^3}+\frac{-3}{r^3}=0$$