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Position of photon sphere

  1. May 1, 2008 #1
    While it's widely accepted that the photon sphere is equal to

    [tex]r_{c}=\frac{3Gm}{c^2}[/tex] or [tex]3r_{g}[/tex]

    [tex]r_{g}[/tex]= gravitational radius ([tex]Gm/c^{2}[/tex])

    while looking at the double photon sphere for the rotating black hole I found myself wanting to understand the actual specifics of the photon sphere. According to most sources it's the place where light's momentum can no longer counteract the gravitation pull of the black and spirals into the black hole i.e. the gravitational acceleration of the black hole is greater than the centripetal acceleration of the photon. Based on a 3 sol static black hole, I put this to the test-

    radius of photon sphere

    [tex]r_{c}=\frac{3Gm}{c^2}[/tex] = 13291.648 m

    gravity at photon sphere

    [tex]a_{g}=\frac{Gm}{r^{2}}\left(1-\frac{2Gm}{rc^{2}}\right)^{-\frac{1}{2}}[/tex]

    = 3.9039e12 m/s^2

    Required velocity at photon sphere based on [tex]a_{c}=v^{2}/r[/tex] and that [tex]a_{c}[/tex] (centripetal acceleration) needs to equal [tex]a_{g}[/tex] (gravitational acceleration)

    [tex]v=\sqrt{a_{g}r}[/tex] = 2.2779e8 m/s (0.76c)

    From the above it seemed that light could orbit closer without being captured, this appears to be at ~10.696.3 m or 2.414[tex]r_{g}[/tex] where the centripetal acceleration of light equals gravity acceleration.

    Is there something I've missed or does the [tex]3r_{g}[/tex] represent the point where light begins to bend due to gravity and the ~[tex]2.5r_{g}[/tex] the point of no return? I was under the impression that the photon sphere was where photons form a stable orbit around the black hole but the info above suggests it closer to the black hole.

    regards
    Steve
     
    Last edited: May 1, 2008
  2. jcsd
  3. May 1, 2008 #2
    I'm not sure if your latter equations are rigorously legitimate when applied to light.
    I mean: i don't think its valid to apply centripetal force on a photon, and perhaps not gravity either.
     
  4. May 2, 2008 #3
    Thanks for the response. I found an equation in the book 'Exploring Black Holes' by Edwin F Taylor & John A Wheeler, pages 5-11 - 5-13, which provides information regarding [tex]r_{c}=3r_{g}[/tex]'-

    [tex]\frac{1-2\left(\frac{r_{g}}{r_{c}}\right)}{r_{c}^{2}}=\frac{1}{b^{2}}[/tex]

    [tex]r_{g}[/tex]= gravitational radius, [tex]r_{c}[/tex]= photon sphere, [tex]b[/tex]= [tex]b_{critical}[/tex]=[tex]r_{g}\sqrt{27}=r_{g}5.196[/tex]

    from the book-
    'The RHS of the equation is the measure of the radial velocity of the particle which depends, through [tex]b[/tex], on the choice of orbit but not on the Schwarzschild geometry. The LHS equation depends on the Schwarzschild geometry but not on choice of orbit'.

    [tex]b[/tex] is the impact parameter, the perpendicular distance between the velocity vector of the photon and the black hole it's approaching. (basically, any photon approaching within this zone will collide with the black hole, any photon approaching on the edge of this zone, [tex]b_{critical}[/tex], will end up in the photon sphere

    As this applies to a static black hole, I'd be interested in knowing how it applies to a rotating black hole. Also while there are two photon spheres when looking at the rotating black hole from the equator, it seems they might converge at the poles (as with the ergosphere and outer event horizon), as there is no frame-dragging to create corotation & counterrotation. it would be interesting to know what would dictate the radius at the poles.

    There's probably a way of extracting [tex]r_{c}[/tex] but it looks like it might end up as a cubic equation.

    Steve
     
    Last edited: May 2, 2008
  5. May 3, 2008 #4
    photon sphere radius for static and rotating black holes-

    [tex]r_{ph}=2r_{g}\left[1+cos\left(\frac{2}{3}arccos(-a)\right)\right][/tex]

    [tex]r_{ph}[/tex]- photon sphere, [tex]r_{g}[/tex]- gravitational radius, [tex]a[/tex]- spin parameter between 0 - 1

    source-
    http://www.lsw.uni-heidelberg.de/users/mcamenzi/CObjects_06.pdf page 259

    (this appears to be an extract from the book 'Compact Objects in Astrophysics' which is available online through Max Camenzind's homepage at The Centre for Astronomy of Heidelberg University (ZAH)).

    According to the equation, the photon sphere reduces from [tex]3r_{g}[/tex] at [tex]a=0[/tex] to [tex]1r_{g}[/tex] at [tex]a=1[/tex] (the event horizon reducing to [tex]1r_{g}[/tex] at [tex]a=1[/tex] also).

    Interestingly there is no mention of the double photon sphere.
     
  6. May 5, 2008 #5
    While looking at various sources, there seems to be differing opinion on whether the rotating black hole has a double or single photon sphere. While there is the equation in post #4 that shows a singular photon sphere reducing to the event horizon at a=1, is it possible there is another photon sphere that reduces to the profile of the ergosphere which relates to light going in the opposite direction of the rotation; or does the equation in post #4 simply imply that light isn't affected by rotation (maybe in terms of blueshift and redshift but not in terms of capture). I'd be interested to hear other peoples points of view regarding this.

    regards
    Steve
     
    Last edited: May 5, 2008
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