# divergence of unit vector

• May 29th 2011, 06:06 PM
ulysses123
divergence of unit vector
show that the divergence of the unit vecotor r is 2/r

I have done this using cartesian coordinates which is quite tedious,i was hoping someone could show me a way to do it with tensors
• May 29th 2011, 08:30 PM
TheEmptySet
Quote:

Originally Posted by ulysses123
show that the divergence of the unit vecotor r is 2/r

I have done this using cartesian coordinates which is quite tedious,i was hoping someone could show me a way to do it with tensors

First we need to know that

$\hat{r}=dr \quad \hat{\theta}=rd\theta \quad \hat{\phi}=r\sin(\theta)d\phi$

is a basis and

$*1=r^2\sin(\theta)dr \wedge d\theta \wedge d\phi$

we have the 1-form

$dr$

The divergence is given by

$*d*f=\nabla \cdot \mathbf{f}$

where * is the Hodge Dual

$*dr=*\hat{r}=r^2\sin(\theta)d\theta \wedge d\phi$

This gives a two form, now taking the derivative

$d*dr=2r\sin(\theta)dr \wedge d\theta \wedge d\phi=\frac{2}{r}(dr) \wedge (rd\theta) \wedge (r\sin(\theta)d\phi)$

This gives a three form and now taking the dual gives a 0-form

$*d*dr=\frac{2}{r}$

So this is the divergence in spherical coordinates.