One step in deriving the Einstein-Hilbert action

Question

In this amazing first principles derivation of the Einstein-Hilbert action there is one small manipulation needed to show

$$c g_{ab,cd}\left(\eta^{ac}\eta^{bd} - \eta^{ab}\eta^{cd}\right)$$

is equal to

$$(c/2)\eta^{bc}\eta^{ae}\partial_{a}\left(g_{be,c} + g_{ce,b} - g_{bc,e}\right) - (c/2)\eta^{ae}\eta^{bc}\partial_{c}\left(g_{be,a} + g_{ae,b} - g_{ba,e}\right).$$

I keep messing the indices up and doing potentially non-valid manipulations, I just can't start with the first thing and end up with the second thing! Some help doing this would massively be appreciated, I can even see the intuition on why second derivatives of the metric should give the Ricci curvature scalar, but I have no intuition for doing the math.

Answer 1

Start with the lower expression: $$ (c/2)\eta^{bc}\eta^{ae}\partial_{a}\left(g_{be,c} + g_{ce,b} - g_{bc,e}\right) - (c/2)\eta^{ae}\eta^{bc}\partial_{c}\left(g_{be,a} + g_{ae,b} - g_{ba,e}\right).\\ = \frac{c}{2}\eta^{bc}\eta^{ae}\left(g_{be,ca} + g_{ce,ba} - g_{bc,ea}-g_{be,ac} - g_{ae,bc} + g_{ba,ec}\right)\\ =\frac{c}{2}\eta^{bc}\eta^{ae}\left(g_{ce,ba} - g_{bc,ea}-g_{ae,bc} + g_{ba,ec}\right) $$ Now, take each term separately and rename dummy indices such that the g-metric carries the correct indices, e.g. $$ \eta^{bc}\eta^{ae}g_{ce,ba}=\eta^{ca}\eta^{db}g_{ab,cd} $$ If you do this for all four terms you will get the first expression you have given. If you still have mismatches recall that you have symmetry under exchange of the first pair and second pair of indices in $g_{ab,cd}$.