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

  1. Mar 3, 2009 #1
    v=0 in E&F/Kerr coordinates

    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.
    _______________________________________________

    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 rs, 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 R0 in this paper- http://www.damtp.cam.ac.uk/user/sg452/black.pdf [Broken] pages 8 and 9.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 5, 2009 #2
    One of the reasons I ask this is that the equation to reproduce the fallout from a radiative tail is-

    [tex]m(v)=m_0-\delta m[/tex]

    where

    [tex]\delta m=av^{-(p-1)}[/tex]

    In the case of a black hole, v would tend to zero outside the EH, meaning [itex]\delta m[/itex] would blow up and m would become unbound.

    This may have been discussed in http://arxiv.org/PS_cache/gr-qc/pdf/9403/9403019v1.pdf" [Broken] where it's possibly referred to as the external potential barrier (page 4).

    The same applies to the outflux equation at the Cauchy horizon-

    [tex]m(v,r)\sim v^{-p}e^{\kappa_0\,v}\ \ \ \ \ \ (r<r_+)[/tex]

    simply being referred to as the potential barrier lying between the Cauchy horizon and event horizon (shallow region) where again, v (and u) tends to zero

    In http://arxiv.org/PS_cache/gr-qc/pdf/0209/0209052v1.pdf" [Broken] (page 14), though the equations are different, they seem to refer to something similar in the shallow region of a BH and even state it's related to [itex]u \rightarrow 0[/itex]. While this appears to address the issue in the shallow region, m becoming unbound just outside the EH still seems unresolved when calculating the radiative tail.
     
    Last edited by a moderator: May 4, 2017
  4. Mar 7, 2009 #3
    According to http://www.faqs.org/faqs/astronomy/faq/part4/section-10.html", v=0 appears to be the 'surface of last influence'-
    and http://www.sron.nl/~jheise/lectures/kruskal.pdf" [Broken] (page 66)-
    while this seems to justify the existence of v=0, it doesn't shed any light on the nature of the coordinate singularity that occurs in Price's power law [itex]\delta m=av^{-(p-1)}[/itex] just outside the event horizon.

    _______________________________________________

    UPDATE-
    A number of links refer to v=0 as r0, the peak of the potential barrier.

    '..wave scattering on the peak of the potential barrier..'-
    http://relativity.livingreviews.org/open?pubNo=lrr-1999-2&key=Chandra83 [Broken] eq 30

    A similar description applying to v=0 inside the event horizon, '..The radiation which crosses the event horizon gets scattered once again by the inner gravitational potential barrier..'-
    http://arxiv.org/PS_cache/gr-qc/pdf/9805/9805008v1.pdf page 6 fig. 2

    so it appears that v=0 isn't a coordinate singularity and has significance.
     
    Last edited by a moderator: May 4, 2017
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