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I'm starting with the Newman-Penrose formalism and have a very basic question that I'm very confused about. The standard Minkoswki metric is $\eta_{ab}=\mathrm{diag}(-1,1,1,1)$. Is then the null tetrad metric

$$g_{ab}=\left(\begin{array}{cccc}0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 1 & 0\end{array}\right)$$

the equivalent of the Minkowski metric in the tetrad formalism?

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  • $\begingroup$ Are you using the full null metric with $k^ak_a=-1$,$l^al_a=-1$ and $m^a\bar{m}_a=1$ where $m^a=x^a+iy^a$? In any case the metric must always be symmetrical, even in null tetrad formalism. In case you are using the basis I mentioned you almost got it right, just put an extra "1" in there to make things symmetrical $\endgroup$ – cesaruliana Oct 9 '15 at 12:21
  • $\begingroup$ @cesaruliana ah of course, it was a typo. Thank you for the correction! $\endgroup$ – user46446 Oct 9 '15 at 13:18

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