Divergence of a vector field, going through the math [closed]

Question

The example I'm working on has this given identity: $\bigtriangledown \cdot \mathbf{\bar{r}}=3$.

The question is: find the divergence of a vector field $\bar{\mathbf{E}}=\frac{\mathbf{r}}{r^{3}}$.

So, using the product rule relationship, i arrange it as such:

$\bigtriangledown \cdot (\psi \bar{\mathbf{E}})=(\bigtriangledown \psi) \cdot E + \psi (\bigtriangledown \cdot \bar{\mathbf{E}})$

as then substitute in:

$\bigtriangledown \cdot (\psi \frac{\mathbf{\bar{r}}}{r^{3}})=(\bigtriangledown \psi) \cdot \frac{\mathbf{\bar{r}}}{r^{3}} + \psi (\bigtriangledown \cdot \frac{\mathbf{\bar{r}}}{r^{3}})$

I can then do $\bigtriangledown \cdot \frac{\mathbf{\bar{r}}}{r^{3}}$ to $\frac{-3}{r^{4}}$ by differentiation, (i think this is wrong)

And using the given relationship $\bigtriangledown \mathbf{\bar{r}}=3$

I can re-write as: $\bigtriangledown \cdot ( \frac{\mathbf{\bar{r}}}{r^{3}})=\frac{3}{r^{3}}-\frac{3}{r^{4}}$

This is where i'm a little lost, I have the list of solutions, and the solution is 0

I can see its there is something mathematically wrong I have done to get $\frac{1}{r^4}$ instead of what i presume should be $\frac{1}{r^3}$, but really cant find out what it is.

Answer 1

In your decompositon you have

$$ \psi = 1/r^3$$

and

$$ \vec{E} = \vec{r} \, .$$

Then, as you wrote,

$$ \vec{\nabla} \cdot \left(\vec{r} \frac{1}{r^3}\right) = \vec{\nabla}\psi \cdot \vec{r} + \psi \vec{\nabla}\cdot \vec{r} \, .$$

Insert

$$ \vec{\nabla} \psi = -\frac{3}{r^5} \vec{r}$$

and

$$\vec{\nabla} \cdot \vec{r} = 3$$

and you are done.

Note that this is only true for $r\neq 0$. The full answer is

$$\vec{\nabla} \left(\frac{r}{r^3}\right) = 4\pi\delta(\vec{r}) \, .$$

Indeed, if you ignore the fact that $\vec{\nabla} \left(\frac{r}{r^3}\right) =0$ and take the volume integral (which contains the origin $\vec{r}=0$!) over a sphere of volume $R$:

$$ \int \text{dr}^3 \vec{\nabla} \left(\frac{r}{r^3}\right) \, ,$$

you can use gauss's law and convert this to an intereal over the surface of the sphere

$$ \int \frac{r}{r^3} \cdot \vec{dS} = 4 \pi R^2 \, \frac{R}{R^3} = 4 \pi \neq 0 \, .$$

We see that the integral over a finite volume of a function that is zero everywhere except at $\vec{r}=0$ gives a finite answer. This means that $\vec{\nabla} \left(\frac{r}{r^3}\right)$ is infinite at the origin.