# Divergence

• January 31st 2008, 07:40 AM
Pinsky
Divergence
Prove that $div \vec{e}=0$ if $e=\frac{\vec{r}}{r^3}$

1. $div\vec{e}=div\frac{\vec{r}}{r^3}$
2. $=(grad\frac{1}{r^3})\vec{r} + \frac{1}{r^3}div\vec{r}$
3. $=-3\vec{r}\frac{1}{r^4}grad{\vec{r}}+\frac{1}{r^3}3$

Here are the first three steps of the solution i have in the book, i know how to go after step three, here is what i can't figure out:

What is done in step 1 that leads to step 2?
Why is $div\vec{r}=3$?
• January 31st 2008, 07:57 AM
ThePerfectHacker
What is $r$?
• January 31st 2008, 08:15 AM
Pinsky
it's a radivector. Wasn't able to find a right translation of the word. If you don't understand what it is i'll try to explain.
• January 31st 2008, 08:32 AM
Plato
Here is one way to rewrite the notation to avoid confusion.
$R = xi + yj + zk\,\& \,r = \left\| R \right\|$.

Now the definition $e = \frac{R}{{r^3 }}$ is a bit easier to understand.
Clearly $\nabla \cdot R = \frac{\partial }{{\partial x}}(x) + \frac{\partial }{{\partial y}}(y) + \frac{\partial }{{\partial z}}(z) = 3$.