I Final part of proof of timelike Killing w/ Frobenius --> static

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If ##V## is timelike Killing with Frobenius condition ##V_{[\alpha} \nabla_{\mu} V_{\nu]} = 0## then you can derive the equation:$$\nabla_{\mu} (|V|^2 V_{\nu}) - \nabla_{\nu} (|V|^2 V_{\mu}) = 0$$which has the solution$$V_{\alpha} = \partial_{\alpha} \phi \quad \mathrm{where} \quad \phi = x^0 + f(x^i)$$The final part of the proof is to show that you can use this function to transform the coordinates ##\bar{x} = \bar{x}(x)## such that ##\bar{g}_{0i} = 0## and ##\partial_0 \bar{g}_{\mu \nu} = 0##. I don't see it?
 
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ergospherical said:
The final part of the proof is to show that you can use this function to transform the coordinates ##\bar{x} = \bar{x}(x)## such that ##\bar{g}_{0i} = 0## and ##\partial_0 \bar{g}_{\mu \nu} = 0##. I don't see it?
I don't know the details of such proof, however the result is that the timelike congruence at rest in such a chart with coordinates ##\bar x## (i.e. the set of timelike worldlines described by fixed spacelike coordinates in this chart) turns out to be hypersurface orthogonal.
 
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From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
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