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If I had a directional photon-emitting source and placed it inside a black hole pointing upward and out towards the visible universe, I assume the photons traveling at the speed of light would slow and reverse direction back into the center.

So if I took the same source and placed it outside of the black hole pointing inward towards the black hole center, can I assume that an emitted photon would travel towards the center faster than the speed of light it is already traveling at?

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    $\begingroup$ I voted you up because I love the imagery, but light doesn't behave like that. Light travels in a straight line, only curving when space is curved which it is, significantly, inside the event horizon of the black hole. All the possible paths of light from your source curve towards the singularity, there isn't any up, slowing or reverse, there's only all directions point at the singularity. That said, I'd rather not make this an answer as I still find space-time diagrams confusing. $\endgroup$ – userLTK Jan 22 '18 at 8:51
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    $\begingroup$ Small warning: All answers here rely on the assumption that our understanding of black holes (i.e. general relativity, along with our models of a black hole, most notably Schwarzschild and Kerr), keeps describing the inside of the event horizon with the same accuracy that it describes the outside. We have no reason to believe anything else, but assuming these models are indeed (close to) the truth, we can never know the actual answer. $\endgroup$ – Arthur Jan 22 '18 at 13:51
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    $\begingroup$ I was under the impression that the concepts "up" and "out" do not make sense in the context of inside the event horizon of a black hole. $\endgroup$ – emory Jan 22 '18 at 17:53
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    $\begingroup$ From my limited understanding, concepts like "speed" and "direction" don't really make sense inside black holes. $\endgroup$ – user253751 Jan 22 '18 at 22:46
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    $\begingroup$ @Fandango68 We don't. When you see "round non-rotating black hole", imagine "a black hole in which the event horizon is a sphere's surface". It doesn't say much about the internal structure of the black hole, which is usually imaged as a sort of a "funnel" (taking on crazy shapes with rotation and electromagnetism involved) of heavily distorted space-time that may or may not "end" in a singularity (which cannot quite be described by general relativity). And why round event horizon? It's easier on the math. We also have models for rotating and charged black holes, which aren't spherical. $\endgroup$ – Luaan Jan 23 '18 at 9:47
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It doesn't work like that. An observer at the light source (and indeed any observers anywhere else) will always see light travelling (in vacuum) at the speed of light locally.

There is also a major problem with your thought experiment. It is not possible for you to have a stationary light source within the event horizon of a black hole. It, and everything else in its vicinity, must be moving inwards. This is as inexorable and unavoidable as is the passage of time for an observer outside the event horizon.

In my opinion, the best "visual" way of thinking about the situation inside the event horizon is to imagine your photons of light like salmon trying to swimming upstream, whilst you are on a boat flowing with the stream and releasing the salmon into the water. You will always see the salmon swimming at some speed with respect to your boat. Unfortunately if the stream flows fast enough then the salmon will make no progress and you will both be swept over a waterfall (the singularity) a little further downstream.

Likewise, your common-sense fails with the situation of firing light towards a black hole. Light is always measured to have a speed of $c$ locally. It is following through with the consequences of this principle that leads to all the weird behaviour that black holes exhibit.

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    $\begingroup$ @tnt-rox There is no such thing as a static observer (sometimes called a shell observer) inside the event horizon. $\endgroup$ – Rob Jeffries Jan 22 '18 at 12:16
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    $\begingroup$ Isn't the word "vacuum" missing? Light doesn't always travel at $c$. $\endgroup$ – Eric Duminil Jan 22 '18 at 21:35
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    $\begingroup$ Do not try this at home. $\endgroup$ – Strawberry Jan 22 '18 at 23:36
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    $\begingroup$ I remember Leonard Susskind using that same "fish swimming upstream" in his lectures. It does capture the essence well. $\endgroup$ – TT. Jan 23 '18 at 5:46
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    $\begingroup$ @EricDuminil I was under the impression that light DOES always travel at c, but in a non-vacuum it ricochets off particles, causing it to take a longer path and making it SEEM slower. $\endgroup$ – Feathercrown Jan 23 '18 at 16:18
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You cannot exceed the speed of light "locally". But you can -see- imagine* distances increase quicker than the speed of light.

If by traveling you mean "moving compared to local space time", then light cannot travel quicker than the speed of light. In your example, the distance increases faster than the speed of light, because spacetime is dragged along inside the black hole by its gravity.

*You in fact cannot "see" it, because you would need some information to be transferred somehow to do the same! This is not possible, because of this damn c limitation.

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    $\begingroup$ @j-chomel your answer is brilliant, i especially like the "distances increase quicker than speed of light" now that i am starting to grasp this concept. $\endgroup$ – tnt-rox Jan 22 '18 at 22:57
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    $\begingroup$ @tnt-rox, I had similar question some time ago, and learned a lot from the answers I got: astronomy.stackexchange.com/questions/19909/… . $\endgroup$ – J. Chomel Jan 23 '18 at 14:07
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    $\begingroup$ This is exactly why the diameter of the visible universe (93 billion light years) can be larger than the number of years since its creation (13.8 billion years). $\endgroup$ – vsz Jan 23 '18 at 21:15
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The speed of light will remain constant. Though the way it is perceived near a black hole changes with where and how it is perceived, it will remain constant. The speed of light does not increase or decrease just because it is near a black hole.

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