# Derivative with respect to partial derivative of contravariant metric tensor density

1. Dec 1, 2011

### PhyPsy

1. The problem statement, all variables and given/known data
Show that $\frac{\partial\overline{\mathcal{L}}_G}{{\partial}\mathfrak{g}^{ab}_{,c}}=\Gamma^c_{ab}-\frac{1}{2}\delta^c_a\Gamma^d_{bd}-\frac{1}{2}\delta^c_b\Gamma^d_{ad}$.

2. Relevant equations
$\overline{\mathcal{L}}_G=\mathfrak{g}^{ab}(\Gamma^d_{ac}\Gamma^c_{bd}-\Gamma^c_{ab}\Gamma^d_{cd})$
$\mathfrak{g}^{ab}=\sqrt{-g}g^{ab}$
$\Gamma^a_{bc}=\frac{1}{2}g^{ad}(\partial_bg_{dc}+{\partial}_cg_{db}+\partial_dg_{bc})$

3. The attempt at a solution
This is one uuuuuuuuuuugly problem. In its simplest terms, $\overline{\mathcal{L}}_G$ is an equation of the metric tensor and its derivatives. So the first thing I do is figure out $\frac{{\partial}g_{ef,d}}{\partial\mathfrak{g}^{ab}_{,c}}$.
$\frac{{\partial}g_{ef,d}}{\partial\mathfrak{g}^{ab}_{,c}} = \frac{{\partial}g_{ef,d}}{{\partial}g^{ab}_{,c}} \frac{{\partial}g^{ab}_{,c}}{\partial\mathfrak{g}^{ab}_{,c}}$
$g^{ab}_{,c}=(-g)^{-\frac{1}{2}}\mathfrak{g}^{ab}_{,c}$, so $\frac{{\partial}g^{ab}_{,c}}{\partial\mathfrak{g}^{ab}_{,c}}=(-g)^{-\frac{1}{2}}$
$g_{ef,d}=\delta^c_d\partial_c(g_{eh}g_{fi}g^{hi})$
$\frac{{\partial}g^{hi}}{{\partial}g^{ab}}=\frac{1}{2}(\delta^h_a\delta^i_b+\delta^h_b\delta^i_a)$
Using the product rule and substituting in the above, I find $\frac{{\partial}g_{ef,d}}{{\partial}g^{ab}_{,c}} = \delta^c_d\frac{1}{2}(\delta^h_a\delta^i_b +\delta^h_b{\delta}^i_a) g_{eh}g_{fi} + g_{fi}g^{hi} \frac{{\partial}g_{eh,d}}{{\partial}g^{ab}_{,c}} +g_{eh}g^{hi} \frac{{\partial}g_{fi,d}}{{\partial}g^{ab}_{,c}}$
$= \frac{1}{2}\delta^c_d(g_{ea}g_{fb} + g_{eb}g_{fa}) + \delta^h_f \frac{{\partial}g_{eh,d}}{{\partial}g^{ab}_{,c}} + \delta^i_e\frac{{\partial}g_{fi,d}}{{\partial} g^{ab}_{,c}}$
$\implies - \frac{{\partial}g_{ef,d}}{\partial g^{ab}_{,c}} = \frac{1}{2}\delta^c_d(g_{ea}g_{fb} + g_{eb}g_{fa})$
$\frac{{\partial}g_{ef,d}}{\partial\mathfrak{g}^{ab}_{,c}}=-\frac{1}{2}\delta^c_d(-g)^{-\frac{1}{2}}(g_{ea}g_{fb}+g_{eb}g_{fa})$

Now comes the fun part. First, I write out $\overline{\mathcal{L}}_G$ in terms of the metric and its derivatives: \begin{align} & \mathfrak{g}^{ab}[\frac{1}{2}g^{db}(\partial_ag_{bc}+\partial_cg_{ba}-\partial_bg_{ac})][\frac{1}{2}g^{ca}(\partial_bg_{ad}+\partial_dg_{ab}-\partial_ag_{bd})]\\ & \ \ \ \ - \mathfrak{g}^{ab}[\frac{1}{2}g^{cd}(\partial_ag_{db}+\partial_bg_{da}-\partial_dg_{ab})][\frac{1}{2}g^{da}(\partial_cg_{ad}+\partial_dg_{ac}-\partial_ag_{cd})]\end{align}
If I multiply this out, the first term would be $\frac{1}{4}\mathfrak{g}^{ab}g^{db}g^{ca}\partial_ag_{bc}\partial_bg_{ad}$, and all the other terms would be similar, just with different indices. Since there are two partial derivatives of the metric in this term, the derivative of the term with respect to $\mathfrak{g}^{ab}_{,c}$ is $\frac{1}{4}\mathfrak{g}^{ab}g^{db}g^{ca} (g_{bc,a}\frac{{\partial}g_{ad,b}}{\partial \mathfrak{g}^{ab}_{,c}} + g_{ad,b}\frac{{\partial}g_{bc,a}}{{\partial} \mathfrak{g}^{ab}_{,c}})$ according to the product rule. However, instead of going through term by term like this, I am going to work it more like $$\frac{1}{4} \mathfrak{g}^{ab} g^{db} g^{ca} (g_{bc,a} + g_{ba,c} + g_{ac,b}) (\frac{{\partial}g_{ad,b}} {{\partial}\mathfrak{g}^{ab}_{,c}} +\frac{{\partial}g_{ab,d}}{{\partial}\mathfrak{g}^{ab}_{,c}} + \frac{{\partial}g_{bd,a}}{{\partial} \mathfrak{g}^{ab}_{,c}}) + \frac{1}{4}\mathfrak{g}^{ab}g^{db}g^{ca} (g_{ad,b}+g_{ab,d}+g_{bd,a}) (\frac{{\partial}g_{bc,a}}{{\partial}\mathfrak{g}^{ab}_{,c}} +\frac{{\partial}g_{ba,c}}{{\partial}\mathfrak{g}^{ab}_{,c}} +\frac{{\partial}g_{ac,b}}{{\partial}\mathfrak{g}^{ab}_{,c}})+...$$ so that the terms will come out in a way that makes it easier to simplify to metric connections $\Gamma$.

I come up with
\begin{align} & -\frac{1}{8}(-g)^{-\frac{1}{2}}\mathfrak{g}^{ab}g^{db}g^{ca} [(\delta^c_bg_{aa}g_{db} +\delta^c_bg_{ab}g_{da} +\delta^c_dg_{aa}g_{bb} +\delta^c_dg_{ba}g_{ab} -\delta^c_ag_{ba}g_{db} -\delta^c_ag_{bb}g_{da}) (\partial_ag_{bc} +\partial_cg_{ba} -\partial_bg_{ac})\\ & \ \ \ \ + (\delta^c_ag_{ba}g_{cb} +\delta^c_ag_{bb}g_{ca} +\delta^c_cg_{ba}g_{ab} +\delta^c_cg_{bb}g_{aa} -\delta^c_bg_{aa}g_{cb}-\delta^c_bg_{ab}g_{ca}) (\partial_bg_{ad} +\partial_dg_{ab}-\partial_ag_{bd})]\\ & \ \ \ \ +\frac{1}{8}(-g)^{-\frac{1}{2}}\mathfrak{g}^{ab}g^{cd}g^{da} [(\delta^c_cg_{aa}g_{db} +\delta^c_cg_{ab}g_{da}+\delta^c_dg_{aa}g_{cb} +\delta^c_dg_{ab}g_{ca}+{\delta}^c_ag_{ca}g_{db} +\delta^c_ag_{cb}g_{da})(\partial_ag_{db} +\partial_bg_{da}-\partial_dg_{ab})\\ & \ \ \ \ +(\delta^c_ag_{da}g_{bb}+\delta^c_ag_{db}g_{ba} +\delta^c_bg_{da}g_{ab}+\delta^c_bg_{db}g_{aa} -\delta^c_dg_{aa}g_{bb}-\delta^c_dg_{ab}g_{ba})(\partial_cg_{ad}+\partial_dg_{ac} -\partial_ag_{bd})]\end{align}
This simplifies to
\begin{align} & -\frac{1}{4}g^{cc}(\partial_ag_{bc}+\partial_cg_{ba} -\partial_bg_{ac})-\frac{1}{4}g^{cd}(\partial_bg_{ad} +\partial_dg_{ab}-\partial_ag_{bd})\\ & \ \ \ \ + \frac{1}{4}g^{cd}(\partial_ag_{db} +\partial_bg_{da}-\partial_dg_{ab}) +\frac{1}{4}g^{cd}(\partial_bg_{ad}+\partial_dg_{ac} -\partial_ag_{bd})\\ & =\frac{1}{2}\Gamma^c_{ab}-\frac{1}{2}\delta^c_b\Gamma^c_{ac} \end{align}

It looks like two of the four terms in the penultimate step above are correct. I'd like someone to just go through the math and try to find where I messed up. I've checked it several times and can't find anything.

Last edited by a moderator: Dec 1, 2011