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
$\displaystyle \hat{r}=dr \quad \hat{\theta}=rd\theta \quad \hat{\phi}=r\sin(\theta)d\phi$
is a basis and
$\displaystyle *1=r^2\sin(\theta)dr \wedge d\theta \wedge d\phi$
we have the 1-form
$\displaystyle dr$
The divergence is given by
$\displaystyle *d*f=\nabla \cdot \mathbf{f}$
where * is the Hodge Dual
$\displaystyle *dr=*\hat{r}=r^2\sin(\theta)d\theta \wedge d\phi$
This gives a two form, now taking the derivative
$\displaystyle d*dr=2r\sin(\theta)dr \wedge d\theta \wedge d\phi=\frac{2}{r}(dr) \wedge (rd\theta) \wedge (r\sin(\theta)d\phi)$
This gives a three form and now taking the dual gives a 0-form
$\displaystyle *d*dr=\frac{2}{r}$
So this is the divergence in spherical coordinates.