Deriving the Metric Connection Equation

[PhyPsy]

Homework Statement

Show that $\frac{\partial\overline{\mathcal{L}}_G}{{\partial}\mathfrak{g}^{ab}_{,c}}=\Gamma^c_{ab}-\frac{1}{2}\delta^c_a\Gamma^d_{bd}-\frac{1}{2}\delta^c_b\Gamma^d_{ad}$.

Homework Equations

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

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.

 

[mark726]

Hi there,

I went through your solution and found a few mistakes that may have led to the incorrect result. Here are some suggestions for how you can improve your solution:

1. In the beginning, you write that \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}}. This is not correct. The correct expression should be \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}} + \frac{\partial g_{ef,d}}{\partial g^{ab}_{,c}} \frac{\partial g^{ab}_{,c}}{\partial \mathfrak{g}^{ab}_{,c}}.

2. In the same step, you write that g^{ab}_{,c} = (-g)^{-\frac{1}{2}} \mathfrak{g}^{ab}_{,c}. However, this is not correct either. The correct expression should be g^{ab}_{,c} = (-g)^{-\frac{1}{2}} \mathfrak{g}^{ab}_{,c} + (-g)^{-\frac{1}{2}} \frac{\partial g^{ab}}{\partial \mathfrak{g}^{ab}_{,c}}.

3. In the next step, you write that \frac{\partial g^{hi}}{\partial g^{ab}} = \frac{1}{2} (\delta^h_a \delta^i_b + \delta^h_b \delta^i_a). This is not correct either. The correct expression should be \frac{\partial g^{hi}}{\partial g^{ab}} = \frac{1}{2} (\delta^h_a \delta^i_b + \delta^h_b \delta^i_a) + \frac{1}{2} (\delta^h_a \delta^i_b + \delta^h_b \delta^i_a