While it's widely accepted that the photon sphere is equal to(adsbygoogle = window.adsbygoogle || []).push({});

[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

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# Position of photon sphere

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