0
$\begingroup$

I have always been intrigued by what seems to be very similar characteristic of the phenomena of traveling very, very fast, (close to the speed of light , and the Schwartzchild radius. AS a body gets close to the speed of light, (relative to the rest of the local universe), two phenomena occur. First, time dilates. Clocks traveling at a high rate of speed relative to an observer in the body, would appear to run slower and slower, until, at the limit, if a body, (like a photon) reached the speed of light, all other clocks would appear to almost freeze, and stop moving completely. Second, the length of all objects moving at a high rate of speed relative to any frame of reference, (the length in the dimension parallel to the relative velocity of the frame of reference), would appear to shorten, until, again, at the limit, if the relative velocity reached the speed of light, (as it is for a photon), all lengths parallel to the velocity would appear to be zero. This implies that, to a photon, the entire rest of the universe, appears as a two dimensional surface, of zero thickness. And that to a photon, the travel from any point where it is emitted, to the point in the universe where it is absorbed, takes zero time and covers zero distance.

Now consider the Schwartzchild radius. It is a closed, two-dimensional curved surface, which, from the outside has the characteristics, that as a body approaches it, time slows down. To an observer on such a body, clocks far away from the Schwartzchild radius would appear to speed up, to an observer far away, a clock on the body would appear to slow down, until at the limit, as the body touched or crossed the surface, a clock on the body would appear to have stopped completely.

What about objects inside of a Schwartzchild radius (inside a black hole)? If a clock falling through a Schwartzchild radius would appear to have stopped as it crosses this threshold, does this mean that from our position far away from the surface, the object would never actually cross the boundary? Does this mean that all mass that falls into a black hole after it has initially formed, at least from the perspective of an observer far away, exists, forever, within the two-dimensional surface of the Schwartzchild radius? If so, this is remarkably similar to the 2D image I have of what the entire universe looks like to a photon traveling at the speed of light.

Is there any significance to this analogy?

... and does it imply that everywhere within our universe, we are in a black hole of enormous size, and that photons within our universe are traveling on the surface of its event horizon, and that we can "approach" it's event horizon by accelerating to a velocity close to the speed of light?

Improve this question
$\endgroup$
7
  • $\begingroup$ Both Schwartzchild metric and Lorentz transformation feature a factor divergent to infinity for a finite value (resp. $r$ and $c$). I guess that's about it (the divergences are not of the same shape). $\endgroup$
    – Stéphane Rollandin
    Jan 25, 2018 at 16:21
  • $\begingroup$ My previous comment is a little terse. I meant, in other words, that although the situations you describe may seem intuitively similar to you, mathematically they are not equivalent. $\endgroup$
    – Stéphane Rollandin
    Jan 25, 2018 at 16:24
  • $\begingroup$ In the hyperbolic geometry of our spacetime, objects (e.g. lines or surfaces) can be either time-like, space-like, or light-like. Both the light trajectory in vacuum and the event horizon are light-like. So your analogy holds (except for the last paragraph perhaps). $\endgroup$
    – safesphere
    Jan 25, 2018 at 16:26
  • $\begingroup$ Yeah, the last paragraph is entirely speculative, based on the similarities I have been wondering about. $\endgroup$
    – Charles Bretana
    Jan 25, 2018 at 17:42
  • $\begingroup$ Stephanie, is the Lorentz transformation understood to be applicable to photons? And if so, how is this divergent factor dealt with for the clearly real situation of a photon? $\endgroup$
    – Charles Bretana
    Jan 25, 2018 at 17:44

1 Answer 1

Reset to default
0
$\begingroup$

Let us compare the travelling close to the speed of light in SR (special relativity) and the approaching to the event horizon of a Schwarzschild black hole.
Here the correspondence is:
SR travelling observer - Schwarzschild approaching event horizon observer
SR stationary observer - Schwarzschild far away observer
1. In SR the stationary observer would measure the clock of the travelling observer to tick slower. However also the travelling observer would measure the clock of the stationary observer to tick slower. SR is symmetrical as for reference frames in relative velocity.
2. In Schwarzschild the far away observer would measure the clock of the approaching event horizon observer to tick slower. Instead the approaching event horizon observer would measure the clock of the far away observer to tick faster. Schwarzschild is not symmetrical.
Therefore there is no analogy between the travelling close to the speed of light in SR and the approaching to the event horizon of a Schwarzschild black hole.
Note:
As for the Schwarzschild far away observer the approaching event horizon observer will never reach the horizon if not at an infinite time. As for the last paragraph it looks like more a philosophical statement than a scientific speculation.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Well, the more appropriate analogy would be an observer approaching the Schwarzschild event horizon from the inside, not from the outside. And this is problematic, as from our perspective, all activity inside an event horizon has yet to occur, it will occur only in the infinitely far future in our space-time, right? $\endgroup$
    – Charles Bretana
    Jan 26, 2018 at 21:22
  • $\begingroup$ I am afraid you stated an inconsistency. No object, matter or light, entering a black hole horizon can ever come back. When the horizon is crossed the causal connection with the outside universe is lost. $\endgroup$
    – Michele Grosso
    Jan 27, 2018 at 21:17
  • $\begingroup$ I understand that, you are of course correct, as far as we understand it today. But I said approaching, not escaping from, and the rule is that no fermion can ever come back, (because it cannot reach the necessary escape velocity, (c). A photon is a boson, which by definition is always traveling at the speed of light , which is the escape velocity necessary to escape. And this is of course speculative, as it is not at all well understood what physics is actually valid on the surface of or inside of the event horizon. $\endgroup$
    – Charles Bretana
    Jan 28, 2018 at 2:34
  • $\begingroup$ This is a wrong answer. Both the event horizon and light trajectory are light-like objects in the hyperbolic geometry of spacetime. Thus both belong to the same class of geometrical objects with a very deep analogy between the two. This is exactly why approaching the event horizon is so much similar to approaching the speed of light. And also why both can't be reached in the view of a remote observer. $\endgroup$
    – safesphere
    Jan 28, 2018 at 7:37
  • $\begingroup$ @safesphere. I agree that the event horizon and a light ray path are both light-like trajectories, however to state an analogy you require that the description of a physical occurrence is symmetric between the two observers, which is not as for the time dilation. $\endgroup$
    – Michele Grosso
    Jan 29, 2018 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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