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Edit: I've realised the thread title is a bit misleading. The title suggests that v divided by u equals zero when in fact it's about the point where the ingoing and outgoing coordinates, v and u, become zero independently.

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Unless I've got something wrong, for static black holes, the ingoing null coordinates, [itex]v=t+r^\star[/itex], are v=∞ at large radii, -∞ at r_{s}, increasing back to zero inside the EH and becoming positive again before reducing back to zero at r=0

For non-static black holes, v=∞ at large radii, -∞ at r_{+}, ∞ at r_{-}and is finite at r=0; the opposite applies for outgoing null coordinates, [itex]u=t-r^\star[/itex] (i.e. u=-∞ at large radii, ∞ at r_{+}, etc.).

As v moves from ∞ to -∞ and back again, there are points where v=0, for static black holes this is at ~2.218M and ~1.594M. This changes as spin is introduced, for a spin of a/M=0.95, v=0 occurs at ~1.934M and ~0.906M (in the case of u, for a spin of 0.95, u=0 occurs at ~3.859M and ~0.834M).

Is there a significance to these radii or is it just a coordinate issue? v=0 appears to be referred to as R_{0}in this paper- http://www.damtp.cam.ac.uk/user/sg452/black.pdf [Broken] pages 8 and 9.

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# V/u=0 in E&F/Kerr coordinates

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