Photons on a perpendicular bisector path from a black hole's core

In summary, the trajectory suggested by the author does not exist and would require the presence of repulsive gravity instead of attractive gravity.
  • #1
jaketodd
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Not directly from the core, but a trajectory that goes to the event horizon, and gets corrected to a perfect, perpendicular bisector path by the gravity of the core, when it reaches the event horizon.

Would they escape the event horizon, since they have to always move at c? On this trajectory, they would have to either stop and reverse direction, or escape. And for the photon to stop, I think, would violate relativity, since they can't stop; they always move at c.

All other particles, other than photons, would never escape the event horizon, even on the trajectory I suggest, because they can stop, and reverse direction, without violating relativity or any other approach. We know this.

I don't think a paper needs to be cited here, because most of us know that traditional theory says photons cannot escape the event horizon.

But as I present here, either they do, or they violate relativity.

Has a photon ever been observed escaping a black hole? Maybe it happens so rarely that we just haven't noticed yet?

1687568932727.png


Thanks,

Jake
 
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  • #2
jaketodd said:
a trajectory that goes to the event horizon, and gets corrected to a perfect, perpendicular bisector path by the gravity of the core, when it reaches the event horizon
Such a path doesn’t exist. It would require repulsive gravity instead of attractive gravity.

jaketodd said:
But as I present here, either they do, or they violate relativity.
Or you just made a mistake. Which option do you think is most likely?
 
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  • #3
jaketodd said:
Not directly from the core, but a trajectory that goes to the event horizon, and gets corrected to a perfect, perpendicular bisector path by the gravity of the core, when it reaches the event horizon.
You don't mean photons because they don't have paths, but we can substitute "flash of light" for "photons" and try the question that way. And then....

The diagram you draw makes no sense because it does not properly represent the geometry inside the horizon - it's not a sphere with a singularity in the center, and there is no such thing as moving outwards towards the horizon from inside. To get the geometry right, you will want to start with a Kruskal diagram; draw the paths of light signals in that diagram (they will be moving up on the page at a 45-degree angle, just as they do in an ordinary Minkowski diagram) and you will see why light cannot escape.

My advice: Kruskal diagrams are one of the easiest ways to visualize the spacetime geometry around a black hole; until you understand them you are wasting your time thinking about black holes.
 
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  • #4
Neither the singularity nor the event horizon are places in the naive sense. The singularity is more like a point in time than a place in space, and it lies in the future of everything that crosses the horizon. The horizon is a null surface, meaning that it lies exactly on the dividing line between being a place in space and a point in time, and has some of the characteristics of both but not all of them. The result of this is that talking about a path from the singularity towards the event horizon makes no sense; your diagram assumes that the center and the horizon are places and neither one is. Can you draw a line away from tomorrow morning?

It is, in principle, possible for a light ray to "hover" at the event horizon. For example, consider light from a spaceship's taillight as it falls into a black hole. Light emitted before it reaches the horizon escapes. Light emitted after it reaches the horizon strikes the singularity. Light emitted as it reaches the horizon can "hover" because, from the perspective of anyone in a positiom to find out, the event horizon and any "hovering" light will pass through them at the speed of light. Such outgoing paths are unstable, though, very much like trying to balance a pencil on its point it's possible only in principle.

The interiors of black holes are strange places. They do not behave remotely like the interior of a normal sphere.
 
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  • #5
jaketodd said:
a trajectory that goes to the event horizon, and gets corrected to a perfect, perpendicular bisector path
Perpendicular bisector of what? If your answer is "the hole", there is no such thing. Space in the interior of the hole is not like ordinary space, and ##r = 0## is not a place in space at all. It's a moment of time.

The rest of your question is nonsense because you do not grasp the basic fact about the spacetime geometry of black holes that I have just stated, and which others have also stated here.

jaketodd said:
I don't think a paper needs to be cited here,
You are way off base here. You have posted multiple threads now in which you show no understanding whatsoever of the topic you are asking about, with the result that (a) your questions are not even wrong (to use Pauli's famous phrase), and (b) the time of other PF members is taken up in responding to questions that are not even wrong, which is not a good use of their time. The fact that you are also claiming that relativity is wrong makes it even worse.

You would be much better served by taking the time to learn more about the topics you are asking about, before posting further questions. That is what textbooks and papers are for. Your blithe assumption that you have no need of such things in this thread illustrates how very, very far you have to go, and how much you do need such things.

This thread is closed.
 
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