5
$\begingroup$

I had a lively discussion with a person about black holes recently, and was making the point about gravitational acceleration in GR being paralleled by speed in SR. One thing that I know people talk about with special relativity, is that if you could reach a speed faster than the speed of light then time would reverse (I don't get this as the Lorentz factor gives an imaginary number, and who knows what that means). This is impractical for things with mass due to mass increasing asymptotically with velocity. However, since the event horizon for black hole behaves exactly like asymptotic barrier to reaching the speed of light (we'll say for a stationary non-charged black hole to keep things simple), shouldn't the effects translate to a black hole? Ie once you're in a black hole (ie the mass that was there when the black hole formed), wouldn't the mass be essentially traveling backwards in time, fastest the further it was from the event horizon?

For clarity sake, for matter, approaching the speed of light and approaching the event horizon of black hole are essentially the same with respect to redshift/blueshift, space and time dilation. There for approaching an event horizon from the inside of a black hole would be like de-accelerating toward the speed of light ($v>c$). So matter in a black hole should be falling outward toward the event horizon.

All of this I am of course imagining from the reference frame of a distant observer.

Just curious if I'm way off base on this.

$\endgroup$
  • $\begingroup$ All this is pretty off base. Time and space according to a distant observer are measured with Schwarzschild coordinates, which don't correspond to any timelike observer's reference frame in the vicinity of the black hole. They only correspond locally to the distant observer's proper time and distance measurements. They have no physical significance inside the event horizon. As far as the distant observer is concerned, inside the event horizon just isn't part of the universe. $\endgroup$ – user27578 Mar 4 '14 at 22:24
  • $\begingroup$ I guess the inside of the black hole is in an "imaginary" universe... I don't suppose there is ever a time when another imaginary number might be in the equation with the lorentz factor to give rise to a real number again. $\endgroup$ – jeffpkamp Mar 4 '14 at 22:58
  • $\begingroup$ Imaginary numbers don't have anything do with it, nor does the Lorentz factor. The coordinates adapted to a distant observer are still real inside an event horizon, they just aren't physical. $\endgroup$ – user27578 Mar 4 '14 at 23:01
  • $\begingroup$ related physics.stackexchange.com/questions/142193/… $\endgroup$ – Paul May 5 '15 at 14:06
4
$\begingroup$

The event horizon of a black hole is a very weird and complicated place. According to my undergrad GR course, at the event horizon of a black hole, time becomes a space-like coordinates, and space becomes a time-like coordinate. From what I remember, a time-like coordinate is one in which you absolutely have to move forward.

Therefore, beyond the black hole horizon, it is impossible to move backwards in space (away from the black hole) in the same manner as it is impossible to move back in time outside of the horizon.

Does it also mean that it is possible to move through time in both direction (time being a space-like coordinate inside the horizon)? Probably, but then, it's still impossible to get out of the black hole, so no one outside would know!

As a side note, according to the holographic principle, information on all events occuring beyond an event horizon is encoded into the event horizon itself, such that it is irrelevant to try to understand what goes on beyond the event horizon. In fact, the very physical existence of something beyond an event horizon is questionable, since it cannot affect us any much more than the event horizon itself can.

$\endgroup$
  • 1
    $\begingroup$ I have heard it said many times that the radial coordinate and time coordinate switch signs (and you can see from the metric that they clearly do in Schwarzschild coordinates) but what is the interpretation? If timelike radius means that radius ticks down to zero inexorably, what does a spacelike time coordinate mean? $\endgroup$ – JohnnyMo1 May 5 '15 at 14:00
-1
$\begingroup$

So as we from the outside of the black hole observe matter approach the event horizon, we will see time slow for the falling matter. Conversely, an observer falling into a black hole would see external events speed up (possibly, not sure about light travelling from external objects).

At the event horizon objects will appear stationary for us, frozen in time. Observers at the event horizon must logically see an infinite time for external events.

One can then use a reciprocal to show that after passing the event horizon an observer would then observe a time reversal from the infinite past. So if it was possible to observe events inside a back hole an external observer would maybe see a universe and big bang in time reverse. Conversely the observer inside the black hole would see our universe in time reverse disappearing back to the singularity.

All having symmetry around the event horizon.

If we "observe" objects in time reverse then they must be v>c.

But this can only be theoretical, as basically I'm saying that anything beyond the event horizon is outside our universe, so we cannot in reality observe it.

$\endgroup$
-1
$\begingroup$

According to one coordinate system for a Schwarzschild black hole, space can be treated as though it is flowing towards the singularity in an approximately Newtonain way and is flowing faster than light beyond the event horizon. Since we can derive anything from a contradiction, we can deduce from the assumption that an object is travelling faster than light though the space it's in that time is going backwards for it. We can only derive that property for an object hovering inside the event horizon but can't derive it for an object in a black hole moving through space slower than light and getting closer to the singularity. One way to prove that things go out of the black hole might be to assume that something is coming out of the black hole and from that derive that time is going backwards for it and from that in turn, derive that it's coming out of the black hole. The problem with that argument is that it uses circular reasoning.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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