0
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ Every barionic object is more massive than light... $\endgroup$ – hebetudinous Aug 2 '16 at 18:49
  • $\begingroup$ well thanks. My question is about the time for this objects. Because if time don't pass the state can not change. Or can? $\endgroup$ – Esdras Caleb Aug 2 '16 at 18:51
  • $\begingroup$ The Schwarzschild radius is a length, i.e. meter. The mass is in "kg". Now think a little bit, how can be a "mass" greater as a "length"... $\endgroup$ – peterh Aug 3 '16 at 5:12
2
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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? $\endgroup$ – Esdras Caleb Aug 2 '16 at 19:12
  • $\begingroup$ 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 arxiv.org/pdf/gr-qc/0411060v2.pdf $\endgroup$ – Gendergaga Aug 2 '16 at 19:16
  • $\begingroup$ 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. $\endgroup$ – Esdras Caleb Aug 2 '16 at 19:16
  • $\begingroup$ 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. $\endgroup$ – Gendergaga Aug 2 '16 at 19:18
  • $\begingroup$ ok, I got the model now, I gess. $\endgroup$ – Esdras Caleb Aug 2 '16 at 20:13

Not the answer you're looking for? Browse other questions tagged or ask your own question.