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An object traveling directly toward the center of a black hole crosses the Schwarzschild Radius, and then emits a photon pointing "directly away" from the center of the Black Hole about a nanometer inside and exactly perpendicular to Schwarzschild Radius. Since in the Schwarzschild radius, Space Time is bent such that all paths toward the future lead to the center of the black hole (so directly away can't exist), one can imagine how a photon emitted at the slightest angle would bend toward the center.

One issue with the analogy of a boat moving against a waterfall is that the boat would be traveling at a negative speed (backwards) of that of the boat minus that of the waterfall. But a photon traveling in a vacuum travels at the speed of light, c. So from the reference frame of the BlackHole, would the photon appear to immediately travel toward the center of the BlackHole at the the speed of light? I.e. no difference than if the photon had initially been emitted toward the center of the black hole?

--Regarding fish goes over the waterfall

OR, in this analogy, is the blackhole's reference frame not static, but is the waterfall and moving toward the center the BH due to gravity? So the photon can be traveling "away" from the center of the BH at the speed of light in a reference frame moving toward the BH's center at a speed faster than the speed of light --> and will eventually end up in the center.

Just Read (Hamilton & Lisle 2008), it exactly answered the concept I was driving at with my question. Thank you very much, ProfRob, for the reference. All Set.

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  • $\begingroup$ It would just go "backwards", no bending involved. Also no observer can sit on a photon! $\endgroup$ – m4r35n357 Mar 29 at 17:38
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    $\begingroup$ Light can go at speeds other than c in curved spacetime. $\endgroup$ – m4r35n357 Mar 29 at 21:19
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    $\begingroup$ Only a local inertial observer must measure the speed of light to be $c$. $dr/dt$ is not $c$ in general for Schwarzschild coordinates. The fish swims away from the boat at $c$ according to the observer on the boat. There are no "absolute speeds". $\endgroup$ – ProfRob Mar 29 at 21:43
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As far as the observer emitting the photon is concerned, it travels away from them, at the speed of light. That observer though is falling rapidly towards the singularity (indeed, a stationary observer is impossible at $r<r_s$) and will get there before the light they emitted.

The light is not bent. It makes not one iota of outward progress in terms of the Schwarzschild $r$ coordinate. Its $dr/dt$ is immediately negative.

I understand the conceptual difficulty. Perhaps the best way to think about this is the waterfall analogy (Hamilton & Lisle 2008). If you are on a boat travelling towards a waterfall, and you have a tame fish that is capable of swimming very fast, then you can release it upstream just before you reach the waterfall. It swims away from the boat, but if the water is flowing faster with respect to the waterfall then you and the fish (in that order) are going over the waterfall. In this analogy, the waterfall is the singularity, the speed of the fish with respect to the water is the locally measured speed of light, and the coordinates along the river bank are the Schwarzschild radial coordinates.

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  • $\begingroup$ Thank you for your answer. One issue with the above analogy is that the boat would be traveling at a negative speed (backwards) of that of the boat minus that of the waterfall. But a photon traveling in a vacuum travels at the speed of light, c. So from the reference frame of the BlackHole, would the photon appear to immediately travel toward the center of the BlackHole at the the speed of light? I.e. no difference than if the photon had initially been emitted toward the center of the black hole? $\endgroup$ – Clay Holdsworth Mar 29 at 18:32
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    $\begingroup$ @GuyInchbald there are no massless observers. $\endgroup$ – ProfRob Mar 29 at 19:07
  • $\begingroup$ There is no issue with the analogy. The fish swims away from the boat at whatever speed the fish can swim relative to the boat. It still goes over the waterfall. $\endgroup$ – ProfRob Mar 29 at 19:10
  • $\begingroup$ @GuyInchbald that part of the question has been edited away, so the bone of contention has been removed. $\endgroup$ – ProfRob Mar 30 at 11:47
  • $\begingroup$ @ProfRob So it has. I have deleted my comments on it too, now. $\endgroup$ – Guy Inchbald Mar 30 at 13:46

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