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First, I know this is related to the previous question Time inside a Black hole, but my focus here is more hypothetical.

The equation of time dilation stands for:

$$t_{0}=t_{f}{\sqrt {1-{\frac {2GM}{rc^{2}}}}}=t_{f}{\sqrt {1-{\frac {r_{s}}{r}}}}$$

According to this website, if an object has a mass above $10^{42}$ (the Schwarzschild radius), the equation will have a negative square root, or $i$.

So what does this mean for time?

Does the time pass in another dimension if the time is imaginary? Should I not use this equation for time dilation like this? (Time dilation only means difference of perceived time and not time itself.)

The question is, if the object is massive, more massive than light, does this mean the object is forced above the speed of light?

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Esdras Caleb
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1 Answers1

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From the perspective of an outside observer there is neither space nor time inside the black hole, since the hole is pinched into the fabric of spacetime (compare Flamm's paraboloid to Newtonian space). Nothing ever falls into the black hole since it takes an infinite amount of coordinate time to reach the event horizon, so it can never fall through (see Susskind 1 and 2).

Because of this real black holes do not form in finite coordinate time, so in the system of an outside observer they will always stay collapsars which converge to a black hole, but never really become one because of the time dilation slowing down the process.

For the infalling observer who crosses the horizon in a finite proper time it means that he would see the whole future of the universe pass in one moment if hovering at the horizon, which would mean infinite blueshift.

So the formula you quoted in your post holds only for locally stationary observers and breaks down at the event horizon because nothing can stay locally stationary when the escape velocity is c. If the infalling observer is behind the horizon, he is no longer causally related to the outside and can't transform his own proper time to outer coordinate time any more, while every coordinate time the outside observer transforms into proper time will give the result that the falling observer has not yet reached the horizon.

That was the relativistic viewpoint; what really happens to the infalling observer when he crosses the horizon if you also take quantum mechanics into account is still subject of research and not yet solved.

Yukterez
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  • well. If I understand rigth its like the speed of ligth paradigma, you cant reach it and you stand near it. So, the time pass really slow to you (bacause you are near speed of ligth). And if you could get the speed of ligth(what you can't) the time will pass so slow to you that all universy will over in a second. Is this? – Esdras Caleb Aug 02 '16 at 19:12
  • If you want your distance to the central singularity to stay constant for your formula to stay valid you have to move against the inflowing space which is c at the horizon in raindrop coordinates, see https://arxiv.org/pdf/gr-qc/0411060v2.pdf – Yukterez Aug 02 '16 at 19:16
  • well this aswers is okay to me bacause i cant acelerate a particle in the speed of ligth.

    But in case of a black hole there is already something inside it(mass and particles of stars) so to this particles(that already are inside the event horizon) how the time is passing if it is passing at all. Because with equation we got this particles are in a imaginary time.

     – Esdras Caleb Aug 02 '16 at 19:16
  • Who told you that there would be something inside a black hole? The consens is that everything that builds up the black hole is layered outside the horizon, haven't you seen the links to Susskind's lecture? If you have something in the center it is not yet a black hole, just a collapsar which converges to a black hole but never becomes one because of the gravitational time dilation slowing down the process. – Yukterez Aug 02 '16 at 19:18
  • ok, I got the model now, I gess. – Esdras Caleb Aug 02 '16 at 20:13
  • A mass colapses to itsels making a event horizon inside it (bot the mass is around it). util its particles evapore with hawking radiation. Correct? – Esdras Caleb Aug 02 '16 at 20:23
  • No, the event horizon only forms in the system of the infalling observer (in finite proper time). In the system of the outside observer there is no true event horizon, since all the mass stays outside the Schwarzschild- or Kerr-radius for all eternity (in infinite coordinate time). The hawking radiation is produced completely outside the event horizon. – Yukterez Aug 02 '16 at 22:42
  • Well this is what I tryed to say – Esdras Caleb Aug 03 '16 at 17:43