# Tensors - Computing the Divergence formula for a given metric tensor

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?

• I think this formula is made for non-normalized bases, so you need to put in the corresponding non-unit vector in cases where the unit vector is used if you understand. Like $\partial_\theta=\hat\theta\cdot h_\theta$ (hope im correct) – Emil Aug 29 '16 at 16:12
• @JerrySchirmer I'm taking this straight from Carroll's 'Intro to Relativity' textbook on Pg 101 from eq (3.32-3.34). But I see where I probably stir confusion. I'll edit it. – iron2man Aug 29 '16 at 16:12
• @Emil I think I understand. I already have given coordinate relations, so I would just apply a transformation like $\hat{e}_{(\beta)} = \frac{\partial x^{\alpha}}{\partial x^{\beta}} \hat{e}_{(\alpha)}$ right? – iron2man Aug 29 '16 at 16:16
• When I tried to use the formula it didn't work because I tried to plug in the components using an orthonormal basis, but since $\hat a \hat A = \hat a / h_a * h_a * \hat A = \hat a / h_a * \partial_a$ I had to divide the component. Perhaps thats what you wrote there. But I am no wizard in this, I might be wrong. – Emil Aug 29 '16 at 16:33
• Emil is write. If you use this formula in spherical coordinates you arrive at correct result, but $V^i$ are to be treated as coefficients of expansion in terms of holonomic basis $\partial_i$. Result for divergence in spherical coordinates is usually reported in elementary physics textbooks (and sometimes not only there) in terms of coefficients of expansion in orthonormal basis, which is in the case of sphere of unit radius with usual coordinates is $e_{\theta}= \partial_{\theta}$, $e_{\phi}=\frac{1}{\sin \theta} \partial_{\phi}$. – Blazej Aug 29 '16 at 17:05

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}

• I appreciate the way you derived the result of the determinate metrics, but that was not my question. It was my fault for not being clearer. I was just wandering how it was possible to derive the divergence for my coordinate system (metric) from the definition. I've posted. – iron2man Aug 29 '16 at 22:45
• @DarthLazar: why can't you just calculate the determinant and then take the derivatives? You have it in the form of $\partial_{a}\left(f(t,x,y,z)V{a}\right)$. At that point, it's just partial derivatives and sums. – Jerry Schirmer Aug 29 '16 at 23:24
• @JerryChirmer Okay, lets say that I want to calculate the divergence formula in spherical coordinates (to compare for example). Where the metric tensor is $$g_{ij} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^{2} \sin^2\theta \end{bmatrix}$$ Then applying the same formula for divergence yields $$\frac{2}{r} V^{r} + \partial_{r}V^{r} - \cot \theta V^{\theta} + \partial_{\theta} V^{\theta} + \partial_{\phi} V^{\phi}$$ What then is left? – iron2man Aug 29 '16 at 23:51
• @DarthLazar: sure. What's the problem? – Jerry Schirmer Aug 29 '16 at 23:56
• And then just a reverse product rule? – iron2man Aug 29 '16 at 23:58