Question about black holes

[OmegaGuy]

I was thinking about black holes lately and I'm confused about something. I did an internet search but couldn't find anything on it.

I'll start with what I do know (feel free to correct any of this information if it's wrong).

The faster an object is travelling, the more time dilation it will experience, which means time will pass slower for it. If an object is travelling at the speed of light, it will be completely frozen in time. However, since it takes infinite energy to accelerate an object with mass to the speed of light, an object with mass can never be completely frozen in time.

Near the event horizon of a black hole, (or near any other gravity well), objects also experience time dilation. When they get to the event horizon, the escape velocity is the speed of light, and thus they can never escape since nothing can go faster than light. Due to the lightspeed time dilation, everything past the event horizon of a black hole is completely frozen in time.

My question is, if the above is true, then what happens when an object with mass gets sucked into a black hole? Does it get frozen, even though that shouldn't be possible, or does time still move for it, albeit very slowly, even though that shouldn't be possible either?

Can someone please help me?

[Surlethe]

You're confusing reference frames. Think in terms of special relativity: if we're travelling relative at something like 0.99c, then you're going to see my clock moving more slowly, and I'm going to see your clock moving more slowly. To me, my clock stays the same; I don't see myself as having slowed-down time.

Similarly, to an object falling into a black hole, its clock stays the same all through the descent, because the clock is at rest relative to the falling object. However, to an outside observer, the gravitational time dilation will make the object seem to slow down as it approaches the event horizon. In fact, the object will grow dimmer and redder because of gravitational redshift, and it will appear to get closer and closer to the event horizon, but will never enter it. It will, instead, fade out of the visible spectrum.

So the passage time of an object falling into a black hole is relative to the observer, just as we would expect from a theory called relativity.

[OmegaGuy]

But what happens to the object once it enters the event horizon?

[Kuroneko]

Surlethe's right. With the confusion of reference frames out of the way, some corrections can be added.

The faster an object is travelling, the more time dilation it will experience, which means time will pass slower for it. If an object is travelling at the speed of light, it will be completely frozen in time. However, since it takes infinite energy to accelerate an object with mass to the speed of light, an object with mass can never be completely frozen in time.

Close enough, although that is an oversimplification. Let's take motion in one dimension for simplicity. Think of a (t,x)-plane, with an object's position x plotted for each time t, such that at t = 0 it is at x = 0 (just to simplify matters). If the object has a constant velocity v, its trajectory will be the line x'(t) = vt, which has (Euclidean) angle θ = atan(v) to the x-axis. The object naturally measures space along x'. It will measure time along t', which is at the same angle θ to to the t-axis. If v→1, then atan(v)→45°, and the lines x' and t' become the same line. In other words, if v = 1, there is no longer any difference between space and time. That's why massive objects cannot be luminal.

Code: Select all

^t   t'(t)           
|   /     _x'(t)
|  /   _--
| / _--
|_-- θ = atan(v)
+--------->x

Now, as at v=1, x' and t' collapse into one line, but as v>1, we can formally continue the process. The t',x' axes rotate past one other--for superluminal speeds, space and time switch places. As v→∞, atan(v)→90°, so that for "infinite" speed, the object's time axis t' is the observer's space axis x, and vice versa.

Near the event horizon of a black hole, (or near any other gravity well), objects also experience time dilation. When they get to the event horizon, the escape velocity is the speed of light, and thus they can never escape since nothing can go faster than light. Due to the lightspeed time dilation, everything past the event horizon of a black hole is completely frozen in time.

Not so. Let's say the object is infalling radially from rest and there is a sequence stationary fiducial observers along the object's path. For the observers some distance ε from the event horizon, the object's speed v→1 as ε→0. That's quite natural--since the escape velocity as the horizon is lightspeed, an infalling object's speed must approach lightspeed as it nears the horizon, relative to a stationary obsever. As the goes past the fiducial observer's location, the fido will see the it decelerate and come to a crawl before the horizon. We can, however, formally continue the process--the object's speed becomes superluminal [1], its light cone tips over in the same manner as above. Just as one can't go back in time, past the horizon, one can't go back in space.

[1] What this actually means is that spacetime is nonstationary past the horizon--space itself "flows into" the singularity.

But what happens to the object once it enters the event horizon?

Neglecting tidal forces, nothing at all. If the black hole is large enough relative to the you, the tidal force at the horizon will be quite low, so you may not even realize you've crossed the event horizon until later, when the tidal forces start ripping you apart.

[Surlethe]

If the black hole is large enough relative to the you, the tidal force at the horizon will be quite low, so you may not even realize you've crossed the event horizon until later, when the tidal forces start ripping you apart.

If an observer falls into a very large black hole (so tidal forces are negligible close to the event horizon), what will he see when he looks out (i.e., in the direction away from the singularity) after he's past the event horizon?

[Kuroneko]

If an observer falls into a very large black hole (so tidal forces are negligible close to the event horizon), what will he see when he looks out (i.e., in the direction away from the singularity) after he's past the event horizon?

The outward view will be same as ever. Infalling singals are affected approximately the same as the observer, so this is to be expected. Looking inward, any matter that previously fell in could be seen, redshifted and apparently not yet reaching the horizon. Nearer the singularity, however, the view in all directions will be quite distorted--contracted almost to a line, in fact.

[Surlethe]

The outward view will be same as ever. Infalling singals are affected approximately the same as the observer, so this is to be expected.

Okay. This comes from having an inertial frame, as opposed to an accelerating frame?

Looking inward, any matter that previously fell in could be seen, redshifted and apparently not yet reaching the horizon. Nearer the singularity, however, the view in all directions will be quite distorted--contracted almost to a line, in fact.

Wait -- if he's already inside the event horizon, how can the objects beneath him appear not to have yet reached the horizon? Or do I misunderstand your use of the word horizon?

[Kuroneko]

Okay. This comes from having an inertial frame, as opposed to an accelerating frame?

Right. Both the observer's trajectory and the infalling signal are geodesics. Tidal force is represented by geodesic deviation, which will be low under the conditions of the scenario.

Wait -- if he's already inside the event horizon, how can the objects beneath him appear not to have yet reached the horizon? Or do I misunderstand your use of the word horizon?

Well, why not? The event horizon is a null hypersurface representing escape velocity of the speed of light. This means that a light signal emitted outward at the event horizon will be stationary with respect to distant stars. A light signal emitted outward some distance within the horizon will eventually reach the singularity, as it must, but do so after any inward-bound signal from the same point; more strongly, looking at it in reverse, an outward lightlike geodesic is asymptotic to the horizon in t→-∞. Suppose another object that fell in the past emits an outward luminal signal at some distance ε within the horizon. For small enough ε, this signal will be detectable by the observer after crossing the horizon no matter how long ago the object fell in, just more red-shifted. And since it is directed outward, it will appear in front of the observer. Therefore, all past objects that fell in will still appear in front the observer even after crossing the horizon. The horizon itself, ε→0, appears in front of those objects. The result is that no observer at all, whether external or infalling, sees another object cross the event horizon.

[Surlethe]

Well, why not? The event horizon is a null hypersurface representing escape velocity of the speed of light. This means that a light signal emitted outward at the event horizon will be stationary with respect to distant stars. A light signal emitted outward some distance within the horizon will eventually reach the singularity, as it must, but do so after any inward-bound signal from the same point; more strongly, looking at it in reverse, an outward lightlike geodesic is asymptotic to the horizon in t→-∞.

This means that if you run time backwards, the light beam's path will be asymptotic to the event horizon, correct?

Suppose another object that fell in the past emits an outward luminal signal at some distance ε within the horizon. For small enough ε, this signal will be detectable by the observer after crossing the horizon no matter how long ago the object fell in, just more red-shifted.

As an aside, could the observer determine how long ago the object fell in from the red-shift of the signal?

And since it is directed outward, it will appear in front of the observer. Therefore, all past objects that fell in will still appear in front the observer even after crossing the horizon. The horizon itself, ε→0, appears in front of those objects. The result is that no observer at all, whether external or infalling, sees another object cross the event horizon.

This will, effectively, cause the event horizon to always remain in the observer's future, from his own point of view, even after he's crossed it in an external frame?

[Kuroneko]

This means that if you run time backwards, the light beam's path will be asymptotic to the event horizon, correct?

Right--for outward-directed signals.

As an aside, could the observer determine how long ago the object fell in from the red-shift of the signal?

If the original signal wavelength is known, of course.

This will, effectively, cause the event horizon to always remain in the observer's future, from his own point of view, ...

Right, although this is only illusory.

... even after he's crossed it in an external frame?

That never happens.

[Surlethe]

This means that if you run time backwards, the light beam's path will be asymptotic to the event horizon, correct?

Right--for outward-directed signals.

What is the behavior of an inward-directed signal as t→-∞? I'm having trouble visualizing it.

This will, effectively, cause the event horizon to always remain in the observer's future, from his own point of view, ...

Right, although this is only illusory.

Why is it only illusory? Is it because if he tries to turn around and escape, he will find he is unable?

... even after he's crossed it in an external frame?

That never happens.

Of course; my mistake. So in both frames, he appears never to cross the event horizon?

[Kuroneko]

What is the behavior of an inward-directed signal as t→-∞? I'm having trouble visualizing it.

Well, physically, they should have r→∞, since the event horizon presents no barrier for inward-directed signals. However, in Schwarzschild coordinates, they're a reflection of the outbound signals, so that they do not exist as t→-∞, but they are asymptotic to the horizon as t→+∞. Explicitly, the null congruences are t = ±(r+2m log|r-2m|+c). This is actually an artifact of the coordinates (Scwarzschild coordinates are singular at the horizon)--they're tailored for an external stationary observer at infinity, which never sees any signal cross the horizon. The Eddington-Finkelstein coordinates make the transformation t' = t+2m log(|r-2m|). These coordinates are nonsingular everywhere on the manifold; inward-directed null geodesics become straight diagonal lines (thus r→∞ as t→-∞) and outward-directed are still asymptotic to the horizon.

Why is it only illusory? Is it because if he tries to turn around and escape, he will find he is unable?

Right. The horizon appears in front, but it is also behind, even if it does not distort external images in the large-scale limit.

Of course; my mistake. So in both frames, he appears never to cross the event horizon?

Right.

[Surlethe]

What is the behavior of an inward-directed signal as t→-∞? I'm having trouble visualizing it.

Well, physically, they should have r→∞, since the event horizon presents no barrier for inward-directed signals. However, in Schwarzschild coordinates, they're a reflection of the outbound signals, so that they do not exist as t→-∞, but they are asymptotic to the horizon as t→+∞.

Wouldn't they be asymptotic to the singularity as t→+∞? Or is this because the horizon is always ahead of an infalling observer?

Explicitly, the null congruences are t = ±(r+2m log|r-2m|+c). This is actually an artifact of the coordinates (Scwarzschild coordinates are singular at the horizon)--they're tailored for an external stationary observer at infinity, which never sees any signal cross the horizon.

So outward-directed signals run into a singularity in both time directions because of the coordinate system?

The Eddington-Finkelstein coordinates make the transformation t' = t+2m log(|r-2m|). These coordinates are nonsingular everywhere on the manifold; inward-directed null geodesics become straight diagonal lines (thus r→∞ as t→-∞) and outward-directed are still asymptotic to the horizon.

In this case, if we run time backwards, then "infalling" objects and light appear to come out of the event horizon, but signals aimed outwards treat the event horizon as a singularity?

Why is it only illusory? Is it because if he tries to turn around and escape, he will find he is unable?

Right. The horizon appears in front, but it is also behind, even if it does not distort external images in the large-scale limit.

Is it possible for there to be an effective event horizon -- for example, if an observer is only capable of achieving speeds of 0.9c, then he will be unable to pull out of the black hole even if he hasn't yet actually crossed the event horizon, so it will seem to be in his past as well as his future?

[Ariphaos]

Well, why not? The event horizon is a null hypersurface representing escape velocity of the speed of light. This means that a light signal emitted outward at the event horizon will be stationary with respect to distant stars. A light signal emitted outward some distance within the horizon will eventually reach the singularity, as it must, but do so after any inward-bound signal from the same point; more strongly, looking at it in reverse, an outward lightlike geodesic is asymptotic to the horizon in t→-∞. Suppose another object that fell in the past emits an outward luminal signal at some distance ε within the horizon. For small enough ε, this signal will be detectable by the observer after crossing the horizon no matter how long ago the object fell in, just more red-shifted. And since it is directed outward, it will appear in front of the observer. Therefore, all past objects that fell in will still appear in front the observer even after crossing the horizon. The horizon itself, ε→0, appears in front of those objects. The result is that no observer at all, whether external or infalling, sees another object cross the event horizon.

Would this apply even as the black hole grows significantly?

[Kuroneko]

Would this apply even as the black hole grows significantly?

Yes. The images of the hovering, redshifted objects will be seen as moving outward with the horizon as the black hole grows larger.