Calculate the expression of divergence in spherical coordinates $r, \theta, \varphi$

Question

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 : $$\nabla_k v^i=\partial_k v^i+v^j\Gamma_{kj}^i$$

Answer 1

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 !

Answer 2

Minkowski metric$(-+++) (\eta_{\mu\nu} =\eta^{\mu\nu})$ in spherical coordinates is: $$ \begin{bmatrix} -1&0&0&0 \\0&1&0&0 \\0&0&r^2&0 \\0&0&0&r^2\sin^2(\theta) \end{bmatrix} \tag{1} $$ And Christoffel symbols are defined as $$ \Gamma^\alpha_{\beta\gamma} = \frac{1}{2}g^{\delta\alpha}(g_{\beta\delta,\gamma}+g_{\gamma\delta,\beta}-g_{\beta\gamma,\delta}) \tag{2} $$ This is much easier in minkowski space as only the diagonals of the metric are non-zero.

This should allow you enough information to calculate the divergence in spherical coordinates from your covariant derivative to get the proof you require.