The correct Einstein-Hilbert action for the metric signature (+,-,-) is,

  • $S=1/2k\int R\sqrt{g}d^3x$


  • $S=1/2k\int R\sqrt{-g}d^3x$

just like in the (-,+,+)?

As you can see I'm interested in the $\sqrt{-g}$ part, does it change signs or does it stay the same for both instances?

Thank you very much!!

  • $\begingroup$ Both need the minus. You want a positive quantity inside the square root and in 1+3 dimensions you'll need a minus to fix this. $\endgroup$
    – secavara
    Commented Feb 15, 2018 at 23:56
  • $\begingroup$ Thank you I meant it at 1+2 dimensions, sorry I need to fix it, what about then? @secavara $\endgroup$ Commented Feb 16, 2018 at 0:02
  • 3
    $\begingroup$ The thing under the square root is always $\sqrt{ |\det g|}$. You can fix your sign accordingly. $\endgroup$
    – Prahar
    Commented Feb 16, 2018 at 0:04
  • 1
    $\begingroup$ Great so $\sqrt{g}$ in this case to make it positive, right? @Prahar $\endgroup$ Commented Feb 16, 2018 at 0:06
  • 2
    $\begingroup$ In $(+,-,-)$ signature, $\det g > 0$ so there's no sign and in the $(-,+,+)$ signature, $\det g < 0$ so there needs to be a sign. $\endgroup$
    – Prahar
    Commented Feb 16, 2018 at 0:07


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