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Escaping the Schwarzchild Radius

  1. Jul 28, 2011 #1

    DeG

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    There's something bothering me about the event horizons of black holes. The Schwarzchild radius (as I see it) is basically the distance from a center of mass at which the escape velocity is the speed of light. The way escape velocity is defined though is the speed a body must have to "reach infinity," or leave the gravitational field, from a given radius in the proximity of a gravitating body. Couldn't a photon at or inside the event horizon leave the vicinity (say to a nearby galaxy) with corresponding red shift, then "re-gravitate" to the black hole? The red shift equation says a photon emitted from the Schwarzchild radius will be infinitely red-shifted at infinite distance (as r->inf), but it doesn't have to travel to infinity to be observed or absorbed. Am I thinking of this to classically? Do photons not make elliptical orbits in gravitational fields like planets?
    Thanks
     
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  3. Jul 29, 2011 #2
    DeG: Would it be correct in assuming this all comes from mulling over what I pointed you to in your other recent entry https://www.physicsforums.com/showthread.php?t=517172, ie at http://en.wikipedia.org/wiki/Gravitational_redshift? The redshift formula there is formulated only in terms of 'as seen at infinity', and it's not hard to think this may suggest that if the photon 'dies out' only at infinity then 'nearer in' there ought to be a finite frequency to measure. According to Schwarzschild metric space and time measure, this is not the case. A more general expression is shown here: http://en.wikipedia.org/wiki/Redshift , see the bottom formula inside the boxed region under "Redshift formulae". From that expression it is evident that at any distance beyond the EH (ie Schwarzschild radius rs), measured redshift is infinite. Note that owing to the infinite compression of radial distance at rs, 'further out' carries a different sense than one might initially think.
    Best to think in terms of differing clock rates (as pointed out in that other thread). As the clock has stopped completely at rs, any light source (ie oscillator) there is simply not putting anything out - referenced to coordinate time. Whether the observer's clock rate further out is at infinity or closer in is mute in that respect - provided it is non-zero (all referenced to coordinate time 'at infinity'), then zero divided by any non-zero factor is still zero!
    One thing to keep in mind here is that it is not possible to remain at a static distance arbitrarily close to the EH, and 'realistically' the situation there is one of infalling observers trying to maintain communication. Consequently another viewpoint is that space itself is infalling at light velocity at the EH (see eg. http://arxiv.org/abs/gr-qc/0411060), and hence any light ray emitted there cannot propagate outward at all (light cones tilted inward).
    Having recently suffered deletion of an entry and censure for questioning the validity of Schwarzschild geometry, -> BH, let me just say the above is I believe consistent with consensus opinion.
     
    Last edited by a moderator: Apr 26, 2017
  4. Jul 29, 2011 #3

    DeG

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    Alright, let me see if I understand this. Within the EH all light cones bend inward making it impossible for and light within the EH to reach even the Schwarzschild radius. Whereas light outside of the EH does have the ability to escape the gravitational field. Does this mean the light cones of em-waves at the EH have "flattened" (i.e. not bend inward or outward)?
     
  5. Jul 29, 2011 #4
    A photon on the EH in a Schwarzschild solution directed outward 'stands still'.
     
  6. Jul 29, 2011 #5
    First part is agreeing with the standard viewpoint, but not the bit about light cone 'flattening'. Once again, try a trip to Wikipedia land: http://en.wikipedia.org/wiki/Event_horizon. Replete with nice diagrams that more or less show what tilting means. There is another link with perhaps better pics but I can't recall where just now. EDIT: Just found a better link re light cone & BH - http://www.phy.syr.edu/courses/modules/LIGHTCONE/schwarzschild.html [Broken]
     
    Last edited by a moderator: May 5, 2017
  7. Jul 29, 2011 #6

    pervect

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    I'm not sure what you mean by "flattening". As Passionflower mentioned, the outward bound light cone for a photon is r=r_s, where r_s is the Schwarzschild radius.

    Finding the equations for the inward light cone isn't very hard, but potentially confusing in Schwarzschild coordinates. The interval between (t,r) and (t,r+dr) for small dr is time-like. Or more formally [itex]\frac{\partial}{\partial r}[/itex] is a timelike vector. Similarly, [itex]\frac{\partial}{\partial t}[/itex] is space-like. Informally, one says that r and t switch places, r becoming timelike and t becoming spacelike.

    The only really important feature of the inbound light cones is that [itex]\frac{\partial}{\partial r}[/itex] is negative when the light cone is directed towards the future and one is in the black hole region.

    Kruskal coordinates are particularly well behaved for drawing light cones- light rays always have a 45 degree slope on a Kruskal diagram, and finding and studying a Kruskal diagram (Wiki has said diagrams) is a good way to understand the causal structure. The Penrose diagram of a black hole is another way to understand it, though the Penrose diagram is basically an abstract and simplified version of the Kruskal diagram.
     
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