0

I'm trying to determine the equation of divergence for the given metric

$$ g_{ij} = \begin{bmatrix} u^2+v^2 & 0 & 0 \\ 0 & u^2+v^2 & 0 \\ 0 & 0 & u^2v^2 \end{bmatrix} $$

Which is the metric for a paraboloidal space.

The divergence of some vector, given in my textbook, is

$$ \nabla_{i}V^{i} = \partial_i V^{i} + \Gamma^{i}_{i j}V^{j} = \frac{1}{\sqrt{|g|}} \partial_i(\sqrt{|g|}\ V^{i} ) $$

Where $g = \det{g_{ij}}$.

If I where to work it out, how would some parts in the derivative cancel out with the inverse determinant?

I tried this formula to determine the divergence in spherical coordinates, but I also run into the same problem to how I cancel coefficients. What am I missing?

iron2man
  • 485

1 Answers1

2

Well, the relation you're looking for is easy enough. If you have:

$$\Gamma_{ab}{}^{b}$$, you can show:

$$\begin{align} \Gamma_{ab}{}^{b}&=\frac{1}{2}g^{bc}\left(g_{ac,b} + g_{bc,a} - g_{ab,c}\right)\\ &=\frac{1}{2}g^{bc}g_{bc,a} \end{align}$$

thanks to the fact that the other two terms amount to multiplying a symmetric tensor in bc by an antisymmetric tensor.

Then, we have the determinant $g$ (where the ellipsis means to continue until you get to the number of dimensions of your space:

$$\begin{align} \partial_{\alpha}g &=\partial_{\alpha}\left( \frac{1}{d!}\epsilon^{abc...}\epsilon^{ABC...}g_{aA}g_{bB}g_{cC}...\right)\\ &=\partial_{\alpha}g_{aA}\left(\frac{1}{(d-1)!}\epsilon^{abc...}\epsilon^{ABC...}g_{bB}g_{cC}...\right)\\ &= \partial_{\alpha}g_{aA}\left(g\,g^{aA}\right) \end{align}\\ $$

Therefore, we have:

$$\begin{align} \partial_{a}\sqrt{|g|} &=\frac{1}{2\sqrt{|g|}}\partial_{a}g\\ &=\frac{1}{2\sqrt{|g|}}|g|g^{bc}\partial_{a}g_{bc}\\ \frac{\partial_{a}\sqrt{|g|}}{\sqrt{|g|}}&= \frac{1}{2}g^{bc}g_{bc,a}\\ &= \Gamma_{ab}{}^{b} \end{align}$$

Zo the Relativist
  • 42,237