Do Black Holes Really Make Objects Disappear or Just Stretch Them Out?

[derek.basler]

Well, I'm currently reading The Black Hole War by Leonard Susskind, and one thing to me seriously doesn't add up. It says that when someone falls into a black hole, and observer from the outside would not see them fall into the singularity, but they would seem to stretch out and move slower and slower until they stop entirely. However, if they did cross the event horizon, wouldn't their light not escape, therefore they would completely disappear? And if the light is just being stretched out, it seems that black holes would in fact be very luminous objects, since the light from everything that fell in would be stretched out as well. Maybe its just a mistake in taking the visualization to seriously. Could someone perhaps clear this up? Thank you in advance.

 

[Naty1]

In a local free falling frame a particle approaching a black hole crosses the event horizon, disappears, and is torn apart as it approaches what is believed to be a singularity of immense gravitational strength. From the perspective of an outside stationary observer at great distance ,the particle appears to never reach the event horizon.

Locally everything seems approximately flat like special relativity; over greater distances curvature makes things appear different as time dilates and space contracts (General relativity).

However, if they did cross the event horizon, wouldn't their light not escape, therefore they would completely disappear?

yes, but you can't see that from a great distance. It is what you see in free falling frame..before you disappear in a hail of radiation.

And if the light is just being stretched out, it seems that black holes would in fact be very luminous objects, since the light from everything that fell in would be stretched out as well.

It never escapes gravity..it's curved back in along the event horizon.

Try reading http://math.ucr.edu/home/baez/physics/

under the section BLACK HOLES

 

[derek.basler]

But that doesn't make sense. If someone at a great distance never see's the particle cross at the event horizon, then it never disappears completely does it? so wouldn't the black hole illuminate, or at least be visible from all the particles that has fallen into it?

 

[DiamondGeezer]

No observer outside the black hole would see a person cross the event horizon because it would take an infinite amount of time from the point of view of any observer.

 

[michelcolman]

But that doesn't make sense. If someone at a great distance never see's the particle cross at the event horizon, then it never disappears completely does it? so wouldn't the black hole illuminate, or at least be visible from all the particles that has fallen into it?

Here's how I understood it:

Suppose a black hole exists somewhere, and you see a spaceship falling into it. Let's assume someone on board the spaceship has calculated that they will cross the event horizon at noon, measured on the spaceship clock.

From a distance, we see the spaceship approaching the event horizon but never quite reaching it, and we see the spaceship clock slowing down so it never quite reaches noon. On board the spaceship, though, the clock just ticks normally and the horizon is crossed at noon. A bit of a non-event for them, really.

Now, as more and more objects approach the black hole and its event horizon, they increase the amount of mass in that area. So the total distortion of space-time caused by all this mass, old and new together, becomes greater. Therefore, the event horizon moves outward and eventually includes the spaceship!

In other words, the spaceship could never cross the original event horizon but, as more objects later on contributed to the total mass in the vicinity of the black hole, the event horizon moved outward and passed the spaceship instead of the other way around.

 

[skeptic2]

Now, as more and more objects approach the black hole and its event horizon, they increase the amount of mass in that area. So the total distortion of space-time caused by all this mass, old and new together, becomes greater. Therefore, the event horizon moves outward and eventually includes the spaceship!

In other words, the spaceship could never cross the original event horizon but, as more objects later on contributed to the total mass in the vicinity of the black hole, the event horizon moved outward and passed the spaceship instead of the other way around.

I think there's a problem with this interpretation. If you drop one object and then another 1 second later, the two objects may be only 4.9 meters apart when the second object is dropped but that distance will continue to increase as the objects fall. Objects falling in behind the spaceship continue get farther and farther behind the spaceship and their influence on the radius of the event horizon will become less and less from the perspective of the crew of the spaceship. Nevertheless, the objects behind the spaceship will see an event horizon with a slightly larger radius than that seen by the spaceship.

Note that not only is time increasingly dilated closer to the event horizon but space is increasingly contracted as well. There always is an infinite distance between the spaceship and the EH. To the crew of the spaceship everything seems normal because time is dilated by the same amount that space is contracted. The dilation of time and contraction of space are expressed by the Lorentz Transformation where escape velocity is substituted for velocity and are not merely an optical illusion due to the increasing time it takes their photons to reach a distant observer

 

[JesseM]

But that doesn't make sense. If someone at a great distance never see's the particle cross at the event horizon, then it never disappears completely does it? so wouldn't the black hole illuminate, or at least be visible from all the particles that has fallen into it?

The closer objects get to the horizon, the more and more redshifted the radiation from them will be when it reaches an outside observer at a fixed distance, meaning the energy of the light escaping gets weaker and weaker, and the wavelength gets larger and larger, making the radiation practically impossible to detect after a certain time. There's a good discussion in this section of the Usenet Physics FAQ:

So if you, watching from a safe distance, attempt to witness my fall into the hole, you'll see me fall more and more slowly as the light delay increases. You'll never see me actually get to the event horizon. My watch, to you, will tick more and more slowly, but will never reach the time that I see as I fall into the black hole. Notice that this is really an optical effect caused by the paths of the light rays.

This is also true for the dying star itself. If you attempt to witness the black hole's formation, you'll see the star collapse more and more slowly, never precisely reaching the Schwarzschild radius.

Now, this led early on to an image of a black hole as a strange sort of suspended-animation object, a "frozen star" with immobilized falling debris and gedankenexperiment astronauts hanging above it in eternally slowing precipitation. This is, however, not what you'd see. The reason is that as things get closer to the event horizon, they also get dimmer. Light from them is redshifted and dimmed, and if one considers that light is actually made up of discrete photons, the time of escape of the last photon is actually finite, and not very large. So things would wink out as they got close, including the dying star, and the name "black hole" is justified.

As an example, take the eight-solar-mass black hole I mentioned before. If you start timing from the moment the you see the object half a Schwarzschild radius away from the event horizon, the light will dim exponentially from that point on with a characteristic time of about 0.2 milliseconds, and the time of the last photon is about a hundredth of a second later. The times scale proportionally to the mass of the black hole. If I jump into a black hole, I don't remain visible for long.

 

[Naty1]

But that doesn't make sense.

Bravo! you are correct: in terms of everyday experience , classical knowledge and it doesn't make sense...neither does time dilation and length contraction in gravitational potential/accelerating frame. THAT was the genius of Einstein! He decided that only the speed of light is constant...NOT space and time as had been assumed for thousands of years. His genius was not especially mathematics but in conceptual thinking.

It's so crazy it takes physicists and mathematicians to figure out what this stuff "looks" like.

 

[derek.basler]

Actually JesseM, that makes perfect sense to me now. I somewhat thought that may be it, but I understand now that the ever increasing redshift causes it to disappear to plain sight. Its really interesting how black holes work and its odd to think how many contradictions to everyday life occur with them. One last question, would the wavelength be so stretched out that it has an wavelength of 0? what would that be? would it still be able to be seen with some sort of radiowave vision?

 

[JesseM]

Actually JesseM, that makes perfect sense to me now. I somewhat thought that may be it, but I understand now that the ever increasing redshift causes it to disappear to plain sight. Its really interesting how black holes work and its odd to think how many contradictions to everyday life occur with them. One last question, would the wavelength be so stretched out that it has an wavelength of 0? what would that be? would it still be able to be seen with some sort of radiowave vision?

The wavelength is increasing, not decreasing, so for a given object falling into the black hole, the wavelength seen by a distant observer would be going to infinity in the limit as time goes to infinity, not to zero. At any finite time there'd be some very large finite wavelength though. Of course this is assuming an idealization where radiation is emitted continuously, since radiation really is quantized into photons, there well be some last photon ever to escape the object before it crosses the horizon, after that no further photons will reach the distant observer.

 

[derek.basler]

Ok, that makes a lot of sense. Thank you for clearing that up!