What happens to an astronaut who enters the event horizon of a black hole?

[Energystrom]

Sorry for what may be very basic and unscientific questions, I'm a brand new poster!

Regarding black holes:

It is my understanding (please correct me if I am wrong) that if an astronaut were to approach and enter the event horizon of the black hole, those watching from the outside would see the astronaut as taking infinite time to enter the black hole, while the astronaut himself would fall rapidly through and be torn apart by tidal forces. My questions are as follows:

1) What accounts for the fact that those watching outside never see the astronaut entering? My guess (rather, understanding, from what little I've read) is that the light from within the event horizon cannot escape to convey the descent of the astronaut; instead, only the light that barely escape from the black hole's clutches is transmitted, constantly showing the astronaut's entrance.

2) If those watching see the astronaut's entrance as taking an infinitely long time, could they leave and come back fifty years later and still see an image of an astronaut at the edge of the black hole? Could someone wandering along in a spaceship see the projection of the astronaut still at the event horizon?

 

[Nabeshin]

Hello Energystrom and welcome to PF!

1) Your intuition is correct, but here's a more detailed explanation. Refer to the spacetime diagram here:
http://www.askamathematician.com/wp-content/uploads/2009/11/U-V-297x300.jpg
So the "timelike worldline" represents our astronaut falling into the event horizon. The r=2.75M line represents an observer at constant radius, so we take this line to represent a stationary observer who's watching our poor astronaut. Note that time increases from the bottom of the graph to the top of the graph. One other thing to know about this diagram is that light rays always travel on 45º lines.

Okay, so now let's look at the worldline of our infalling astronaut. Imagine he is sending out light pulses (or equivalently, he is emitting light at some frequency which corresponds to what we would see as the outside observer). In his frame, he does this at a constant rate, so you can, along his world line, draw a set of equally spaced 45º lines. These liens represent the trajectories of the photons. Only when they intersect our r=2.75M observer is the astronaut seen.

As he approaches the horizon, we see that the light rays have to travel farther and farther away before they intersect with our distant observer, until at the horizon the light ray simply stays at the horizon and never is observed. This is why we see the man slow down. The time between reception of signals as measured by the outside observer increases to infinity. So, sure you could leave for 50 years and then come back and happen to "see" a photon that the guy shot off riiiiight before he crossed the horizon.

Hopefully my explanation of what the diagram is is clear. If my subsequent argument was not, open up MS paint or draw the picture yourself and draw the lines I mentioned and it should become clear to you.

Cheers!

 

[ThomasEdison]

Hello Energystrom and welcome to PF!

1) Your intuition is correct, but here's a more detailed explanation. Refer to the spacetime diagram here:
http://www.askamathematician.com/wp-content/uploads/2009/11/U-V-297x300.jpg
So the "timelike worldline" represents our astronaut falling into the event horizon. The r=2.75M line represents an observer at constant radius, so we take this line to represent a stationary observer who's watching our poor astronaut. Note that time increases from the bottom of the graph to the top of the graph. One other thing to know about this diagram is that light rays always travel on 45º lines.

Okay, so now let's look at the worldline of our infalling astronaut. Imagine he is sending out light pulses (or equivalently, he is emitting light at some frequency which corresponds to what we would see as the outside observer). In his frame, he does this at a constant rate, so you can, along his world line, draw a set of equally spaced 45º lines. These liens represent the trajectories of the photons. Only when they intersect our r=2.75M observer is the astronaut seen.

As he approaches the horizon, we see that the light rays have to travel farther and farther away before they intersect with our distant observer, until at the horizon the light ray simply stays at the horizon and never is observed. This is why we see the man slow down. The time between reception of signals as measured by the outside observer increases to infinity. So, sure you could leave for 50 years and then come back and happen to "see" a photon that the guy shot off riiiiight before he crossed the horizon.

Hopefully my explanation of what the diagram is is clear. If my subsequent argument was not, open up MS paint or draw the picture yourself and draw the lines I mentioned and it should become clear to you.

Cheers!

I have a related question that has puzzled me about the time dilation effects within a black hole.

Is this only in the event horizon area or is the "slowing down" heading towards a complete stop at the center of the Black Hole?

If so:



Is the singularity within a black hole a "snapshot" (frozen in time ) of the instant that the Black hole first formed... frozen in time?

 

[sylas]

I have a related question that has puzzled me about the time dilation effects within a black hole.

Is this only in the event horizon area or is the "slowing down" heading towards a complete stop at the center of the Black Hole?

It is at the event horizon. Furthermore, it depends on what co-ordinates you use. Time is not stopping in an absolute sense. The particle that falls into the black hole does actually cross the event horizon; it is just that it cannot send a signal back; and so co-ordinates reflecting what is observed from outside only describe photons which were emitted before the event horizon was crossed.

There's another point as well... you don't actually "see" a frozen image of the infalling particle, because the light coming back is redshifted without bound. The closer to the particle to the event horizon, the more the signal is redshifted -- and also the weaker the signal in terms of number of photons. So the particle becomes invisible to any possible detector before crossing the event horizon.

Consider this. Suppose you drop a large thermonuclear bomb (or even a whole star) into a black hole, so that the bomb detonates before crossing the horizon and is still emitting floods of photons as it crosses the horizon.

From outside, the signal of light from this thermonuclear explosion will be very weak, and it will become weaker and weaker and redder and redder without bound, until it is no longer detectable... and what you see from that vanishing signal is the progress of the explosion only from before crossing the event horizon.

Cheers -- sylas

 

[Ich]

So, sure you could leave for 50 years and then come back and happen to "see" a photon that the guy shot off riiiiight before he crossed the horizon.

Not really. The light gets exponentially redshifted in a matter of milliseconds (or seconds for a SMBH). After 50 years, there will be definitely nothing to see.

I have a related question that has puzzled me about the time dilation effects within a black hole.

Is this only in the event horizon area or is the "slowing down" heading towards a complete stop at the center of the Black Hole?

It's misleading to think of time dilation as something physically happening to the astronaut.
Time dilation exists only as a relation to an outside observer. It becomes infinite at the EH, and that means that everything inside the EH is completely disconnected with any outside observer (at least what concerns the observer). Disconnected means that there is no meaningful way to compare times inside the EH with times outside, so you can't even define time dilation.
The astronaut's proper time is independent of such comparisons anyway. It keeps ticking until it hits the singularity.

 

[Energystrom]

First off, thanks to all for the extremely concise and helpful responses; everything made sense and helped to answers my questions.

There's another point as well... you don't actually "see" a frozen image of the infalling particle, because the light coming back is redshifted without bound. The closer to the particle to the event horizon, the more the signal is redshifted -- and also the weaker the signal in terms of number of photons. So the particle becomes invisible to any possible detector before crossing the event horizon.


Cheers -- sylas

This bit especially cheers me up, as it had been my own hypothesis when thinking about this phenomenon that the light would become red-shifted as it more and more slowly left the area around the event horizon. This confirmation and the knowledge that there is no bound on the redshifting that occurs is awesome!

I appreciate all the responses once again, and look forward to any related responses or information anybody would like to post!

 

[ThomasEdison]

Not really. The light gets exponentially redshifted in a matter of milliseconds (or seconds for a SMBH). After 50 years, there will be definitely nothing to see.


It's misleading to think of time dilation as something physically happening to the astronaut.
Time dilation exists only as a relation to an outside observer. It becomes infinite at the EH, and that means that everything inside the EH is completely disconnected with any outside observer (at least what concerns the observer). Disconnected means that there is no meaningful way to compare times inside the EH with times outside, so you can't even define time dilation.
The astronaut's proper time is independent of such comparisons anyway. It keeps ticking until it hits the singularity.

I was only asking about the black hole, not about objects falling into it or even the event horizon. The black hole itself.. is it still in the instant moment in time when it was made?

 

[Nabeshin]

Not really. The light gets exponentially redshifted in a matter of milliseconds (or seconds for a SMBH). After 50 years, there will be definitely nothing to see.

I suppose it is indeed a question of scale. I simply meant that, at least theoretically, it is possible, if practically impossible. I suppose at some limit the energy of the photon gets drowned out by other sources including the CMB and eventually just the quantum noise in the measuring apparatus itself, which I suppose provides a "theoretical" cutoff.

 

[Ich]

The black hole itself.. is it still in the instant moment in time when it was made?

What exactly is "the black hole itself"? All the matter that formed it? That matter is broken, all in a singularity. Further, it is out of reach. What more do you want to know?

I suppose it is indeed a question of scale. I simply meant that, at least theoretically, it is possible, if practically impossible.

Just to be sure, the scale I'm talking about is ~10^-100000000. That's nothing to do with CMB or quantum noise. The wavelength outranks the size of the observable universe by a ridiculous amount.