Value of g near a black hole (re-visited)

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The discussion centers on the behavior of gravitational acceleration (g) near simple, non-rotating, uncharged black holes, specifically regarding its asymptotic nature as one approaches the event horizon (EH) and singularity. Participants agree that gravity approaches infinity at the singularity, while the EH is defined as the radius where escape velocity equals the speed of light (c). The conversation emphasizes that a purely Newtonian approach is inadequate for calculations near black holes, particularly within the Schwarzschild radius. The consensus is that gravity's behavior is coordinate-dependent and requires a relativistic framework for accurate understanding.

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  • #91
Spin-Analyser said:
It's old light from outside the horizon coming at you from inside the horizon of light that hasn't reached the horizon yet?

I give up. I write English, you twist into word soup.
 
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  • #92
Spin-Analyser said:
It's old light from outside the horizon coming at you from inside the horizon of light that hasn't reached the horizon yet?
No, it's old light from outside the horizon that has never been inside the horizon.

If you and your partner are both falling into the black hole, then from your point of view, you and your partner are stationary and the event horizon is rushing towards you at the speed of light. You see your partner 10 feet in front of you at all times. The image you see has been delayed 10 nanoseconds; you see where your partner was 10 ns ago.

At exactly the moment you reach the event horizon (i.e. at a distance of zero, not a distance of 1 Planck length) you see, 10 feet in front of you, what your partner was doing 10 ns earlier, which was crossing the event horizon. (10 ns ago the event horizon was 10 ft in front of you, as was your partner.)

This illustrated in the left-hand spacetime diagram below.

attachment.php?attachmentid=45019&stc=1&d=1331603712.png


If you decide at the last minute to brake and hover at a small fixed distance outside the event horizon, you see your partner's image slow down, red-shift and darken and never actually cross the horizon. This illustrated in the right-hand spacetime diagram below.
 

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  • #93
DrGreg said:
If you and your partner are both falling into the black hole, then from your point of view, you and your partner are stationary and the event horizon is rushing towards you at the speed of light. You see your partner 10 feet in front of you at all times.
Sorry but I cannot agree with this.
I think you are forgetting the tidal forces between them.
 
  • #94
Passionflower said:
Sorry but I cannot agree with this.
I think you are forgetting the tidal forces between them.

If you go back to the post I introduced this scenario, I specified a supermassive black hole. Tidal forces can be made as small as desired by making the mass large enough, as you have noted in other discussions.
 
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  • #95
DrGreg said:
No, it's old light from outside the horizon that has never been inside the horizon.

If you and your partner are both falling into the black hole, then from your point of view, you and your partner are stationary and the event horizon is rushing towards you at the speed of light. You see your partner 10 feet in front of you at all times. The image you see has been delayed 10 nanoseconds; you see where your partner was 10 ns ago.

At exactly the moment you reach the event horizon (i.e. at a distance of zero, not a distance of 1 Planck length) you see, 10 feet in front of you, what your partner was doing 10 ns earlier, which was crossing the event horizon. (10 ns ago the event horizon was 10 ft in front of you, as was your partner.)

If you decide at the last minute to brake and hover at a small fixed distance outside the event horizon, you see your partner's image slow down, red-shift and darken and never actually cross the horizon. This illustrated in the right-hand spacetime diagram below.
As you approach you see objects in front of you crossing the horizon, and you're seeing light from the other side of the event horizon. When you move away the light from previous observers moves back across the event horizon. So what if you stop one plank length away from the horizon and the one in front of you never reached the horizon either? The event horizon is no longer in one place. If you moved alongside and hovered next to the one in front of you then you would still be outside the horizon. It doesn't work. The event horizon must be moving inwards at c, not outwards.
 
  • #96
Spin-Analyser said:
As you approach you see objects in front of you crossing the horizon, and you're seeing light from the other side of the event horizon. When you move away the light from previous observers moves back across the event horizon.

Why do you do this? So far now, a dozen or more times, 3 of us have now told you this is false, in all different words, and clear beautiful pictures from Dr. Greg. You respond with the opposite of what everyone says. You never see light emitted from inside the horizon unless you are also at or inside the horizon. Every other part of your statement is false as well, and this has been explained multiple times.

I think this thread is dead.
 
  • #97
If there's distance between them and you are one plank length away from the horizon then 'As you approach you see objects in front of you crossing the horizon, and you're seeing light from the other side of the event horizon. When you move away the light from previous observers moves back across the event horizon'. I thought there was "plenty of room". The only other alternative is that they all pile up at the horizon.
 

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