I need to find the divergence in spherical co-ordinates using the expression $$ \nabla \cdot \vec{v} = \frac{1}{\sqrt{g}} \frac{\partial}{\partial u^{j}} (\sqrt{g} v^{j})$$ I've figured out the diagonal metric tensor elements: $g_{rr} = 1, g_{\theta \theta} = r^2, g_{\phi \phi} = r^2 \sin^2 \theta$. All other elements are 0. The inverse matrix $g^{ij}$ has components: $g^{rr} = 1, g^{\theta \theta} = \frac{1}{r^2}, g^{\phi \phi} = \frac{1}{r^2 \sin^2 \theta}$. The square root of the determinant, $\sqrt g = r^2 \sin \theta.$ Hence we have $$v^{r} = g^{rr}v_{r} + g^{r \theta} v_{\theta} + g^{r \phi} v_{\phi} = v_{r} $$ $$v^{\theta} = g^{\theta r}v_{r} + g^{\theta \theta} v_{\theta} + g^{\theta \phi} v_{\phi} = \frac{1}{r^2}v_{\theta} $$ $$v^{\phi} = g^{\phi r}v_{r} + g^{\phi \theta} v_{\theta} + g^{\phi \phi} v_{\phi} = \frac{1}{r^2 \sin^2 \theta}v_{\phi} $$ This leads to a divergence of $$ \nabla \cdot \vec{v} = \frac{1}{r^2}\frac{\partial}{\partial r} (r^2 v_{r}) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} (\sin \theta v_{\theta}) + \frac{1}{r^2 \sin^2 \theta} \frac{\partial v_{\phi}}{\partial \phi}$$
I probably went wrong in converting the $v^{i}$s to $v_{j}$s. Is my lowering of indices here incorrect? Is this why my divergence is wrong?
