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The metric of a Vaidya black hole in outgoing/retarted null coordinates are $$ds^2=-\left(1-\frac{2m(u)}{r^2}\right)du^2-2dudr+r^2\Big(d\theta^2+\sin^2\theta d\phi^2 \Big)$$ The eveolving horizon locates at $r=2m(u)$ and thus $dr=2\dot{m}du$ with dot denoting $\frac{d}{du}$. Hence, I conclude that the degenerate metric of the Vaidya black-hole horizon should be $$ds^2=-4\dot{m}du^2+4m(u)^2\Big(d\theta^2+\sin^2\theta d\phi^2 \Big)$$

Am I right? Thank you very much!

I am unsure because the null time coordinate $u$ is NOT orthogonal to $r$.

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  • $\begingroup$ Do you have a reason to believe you are wrong? $\endgroup$ – ACuriousMind Aug 25 '14 at 21:14
  • $\begingroup$ @ACuriousMind I guess I am right, but not sure. $\endgroup$ – David Aug 25 '14 at 21:16
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    $\begingroup$ What do you mean by "the degenerate metric?" Are you talking about a metric whose signature has a zero in it? $\endgroup$ – Ben Crowell Aug 26 '14 at 1:45
  • $\begingroup$ ...or the induced metric on the surface $r=2m$? $\endgroup$ – user10851 Aug 26 '14 at 9:00
  • $\begingroup$ Hello @Ben Crowell and Kris White! I have re-edited the title to avoid ambiguity. By "the degenerate metric", I mean the horizon is only 2+1 dimensional, instead of 3+1 dimensional like the full spacetime. $\endgroup$ – David Aug 26 '14 at 13:11

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