TITLE: Calculate the expression of divergence in spherical coordinates $r, \theta, \varphi$
QUESTION [1 upvotes]: Hi this is my first question in [Physics.SE] I saw a lot of posts and I liked them. I hope that my question will be answered too. 
While I'm solving a problem in vector calculus. I recognized that I need a proof to answer it. 

The problem is the following: Calculate the expression of divergence in spherical coordinates $r, \theta, \varphi$ for a vector field $\boldsymbol{A}$ such that its contravariant components $A^i$

Here's my attempts:
We know that the divergence of a vector field is :
$$\mathbf{div\ V}=\nabla_i v^i$$
Notice that $\mathbf{V}$ is the vector field and $\nabla_k v^i$ its covariant derivative, contracting it we obtain the scalar $\nabla_i v^i$. 
My questions are how I can apply this to solve the main problem ? 
Can I use the developed expression of the covariant derivative? which is : $$\nabla_k v^i=\partial_k v^i+v^j\Gamma_{kj}^i$$

REPLY [1 votes]: and welcome to [Physics.SE], I tried to solve your problem and here is what I found:
As you said the divergence can be written :
$$\mathbf{div \ V}=\nabla_i v^i$$
And the expression of the covariant derivative is :
$$\nabla_k v^i=\partial_k v^i+v^j\Gamma_{kj}^i$$
Using it we obtain :
$$\mathbf{div \ V}=\partial_i v^i +v^j\Gamma_{ij}^i$$
Using Ricci theorem :
$$\nabla_k g_{ij}=\partial_kg_{ij}-\Gamma_{ik}^l g_{lj}-\Gamma_{jk}^l g_{il}=0$$
Multiplying by $g^{ij}$ :

Recall: $g^{ij}g_{jl}=\delta_i^l$

$$g^{ij}\partial_k g_{ij}-\Gamma_{ik}^l \delta_i^l-\Gamma_{jk}^l\delta_l^j =0$$
Thus:
$$g^{ij}\partial_k\ g_{ij}-\Gamma_{ik}^l-\Gamma_{jk}^l=0$$
Since $\Gamma_{ik}^i=\Gamma_{jk}^j$ we have :
$$ g^{ij}\ \partial_k\ g_{ij}=2\Gamma_{ik}^i$$
Let $g$ be the determinant of $g_{ij}$ we obtain :
$$\partial_k g=g\ g_{ij}\ \partial_k\ g_{ij}$$
Thus :
$$\Gamma_{ik}^l=\frac{1}{2g} \partial_k \ g=\frac{1}{\sqrt{|g|}}\partial_k \sqrt{|g|}$$
Applying it we obtain:
$$\mathbf{div \ V}=\partial_iv^i+\frac{v^i}{\sqrt{|g|}}\partial_i \sqrt{|g|}$$

Recall : $$\frac{1}{a} d(ba)=db+\frac{b}{a} da$$
  Let $a=\sqrt{|g|}$ , $b=v^i$ 

finally we have :
$$\fbox{$\mathbf{div \ V}=\frac{1}{\sqrt{|g|}} \partial_i\biggr( v^i \sqrt{|g|}\biggl)$}$$
Using this result in your main problem we get :
$$\mathbf{div \ A}=\partial_i A^i +\frac{A^i}{\sqrt{|g|}}\partial_i \sqrt{|g|}$$
I think I would let you continue. Good luck !
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