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How does one see that

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

is equal to

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

where $\eta$ is the Minkowski metric $(+,-,-,-)$ and $g_{ab,cd} = \frac{\partial ^2 g_{ab}}{\partial x^d \partial x^c}$, I keep getting lost :(

bobby
  • 689

1 Answers1

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First throw away the $c$ since it multiplies both expressions, then distribute the derivatives (i.e. $\partial_a g_{be,c} = g_{be,ca}$). This gives you six scalar terms of the form $\eta^{..} \eta^{..} g_{..,..}$, and due to symmetries there are only two unique such terms: $\eta^{ab} \eta^{cd} g_{ab,cd}$ and $\eta^{ac}\eta^{bd} g_{ab,cd}$. You just need to relabel dummy indices to get each term into one of these forms and then count the signs.

  • How do I start from $c g_{ab,cd}\left(\eta^{ac}\eta^{bd} - \eta^{ab}\eta^{cd}\right)$ and end up with the other thing, I don't know what dummy indice switches are valid and get lost, sorry :( – bobby Oct 19 '14 at 15:21