I want to calculate the covariant divergence of a vector field in spherical coordinates $(x^1,x^2,x^3)=(r,\theta, \phi)$, i.e. write out $E^i_{\ ||i}$.
I know that
$$E^i_{\ ||i}=\partial_iE^i+\Gamma^i_{im} E^m=\partial_iE^i+\frac{\partial_i\sqrt{|g|}}{\sqrt{|g|}}E^i.$$
The line element is $ds^2=dr^2+r^2d\theta^2+r^2\sin^2\phi d\phi^2$. Therefore, the Gram determinant is $g=r^4\sin^2\theta$.
But how do i get from the $E^1,E^2,E^3, \partial_1,\partial_2,\partial_3$ to the $E_r, E_\theta, E_\phi, \partial_r, \partial_\theta, \partial_\phi$?
Best Answer
I have figured it out!
The derivatives in this formula are with respect to unnormalised unit vectors.
We have the contravariant base $$dx^1=h_rdr, dx^2=rd\theta, dx^3=r\sin\theta d\theta,$$ and therefore $$\partial_1=\partial_r, \partial_2=\frac{\partial_\theta}{r}, \partial_3=\frac{\partial_\phi}{r\sin\theta}.$$
The only non-vanishing connection coefficients are $\Gamma^2_{12}, \Gamma^3_{13}, \Gamma^3_{23}$. For demonstration, we have $$\Gamma^3_{23}=\frac{g^{33}}{2}\partial_2g_{33} =\frac{g^{33}}{2r}\partial_\theta g_{33}=\frac{\cot\theta}{r}.$$
Now we have $$\nabla_iE^i=\partial_iE^i+\Gamma^i_{mi}E^m =\partial_rE_r+\frac{1}{r}\partial_\theta E_\theta+\frac{1}{r\sin\theta}\partial_\phi E_\phi+\frac{2}{r}E_r+\frac{\cot\theta}{r}E_\theta.$$
This is the same result one would obtain, if one were to calculate the divergence in spherical coordinates using the formula
$$\nabla\cdot\mathbf{E}=\frac{1}{h_rh_\theta h_\phi}\sum\limits_{i=r, \theta, \phi}\partial_i\frac{h_rh_\theta h_\phi}{h_i}E_i. $$
Note that in the last formula the index takes on the (Greek) letters and not any numbers.
Note also that in my first post, I assumed $\partial_1=\partial_r, \partial_2=\partial_\theta, \partial_3=\partial_\phi. $