Sciencemadness Discussion Board

Does anything actually enter a black hole?

Fulmen - 5-1-2016 at 05:35

Trying to wrap my head around general relativity and black holes, one thing has me confused. How can anything actually enter a BH when time must stop at the event horizon? There was a program on Discovery where all the brainiacs talked about what would happen as an object enters the BH, but does that even make sense?
I know the an object falling in would experience time differently, but from our perspective it can't enter. And unless that information can escape a BH, anything happening on the inside must be pure conjecture, right?

If time stops at the event horizon, wouldn't matter "simply" be deposited at the horizon, frozen in time? And doesn't Hawkins Radiation also support this? According to Hawkins BH's radiate by capturing one part of a virtual particle pair allowing the other to escape as real matter. But this is also an effect tied to the horizon, so from the outside shouldn't one see matter smeared over it's surface and slowly escape due to quantum fluctuations?

In reality matter will never reach the event horizon, it will continue to fall inwards more and more slowly without ever reaching it. As the thermal radiation is redshifted towards infinity the radiated energy will also diminish, but it can never be zero. Is this in any way related to Hawkins radiation?

There is however one problem with my arguments: The BH (or at least the event horizon) grows as matter falls in. Does this mean that matter can enter, not by falling in but by being "absorbed" by the growing BH?

hissingnoise - 5-1-2016 at 07:44

I posted this one earlier ─ thereafter the complex nature of the topic kinda rendered me effectively speechless . . .

http://www.sciencemadness.org/talk/viewthread.php?tid=62554

Fulmen - 5-1-2016 at 10:02

How embarrassing, I'll make sure to read that thread before continuing here.

Fulmen - 5-1-2016 at 10:32

My question is the opposite of that one, rather than viewing it from the perspective of the viewer I want to understand what happens viewed from the outside.

As far as I can understand the physics as viewed from the outside is within the realm of general relativity, so won't the event horizon effectively be a two-dimensional bubble?

phlogiston - 5-1-2016 at 12:20

I can't say that I fully understand relativity myself, but your description of what an outside observer sees matches with descriptions that I have read in various other sources. The outside observer never sees the object crossing the event horizon from the outside. The image of the object 'frozen' very close to the event horizon fades over time.

Mesa - 5-1-2016 at 12:22

The problem is you are assuming exists independently of space, It doesn't(Simple proof, measure a length of time without using space.)

As the object is pulled further toward the center of the hole, it will appear to an outside observer to be slowing down. If the observer were to plot their subjective data on a graph, they'd find the object "stops" at the point of the event horizon.

Fulmen - 5-1-2016 at 15:50

Quote: Originally posted by Mesa  
The problem is you are assuming [time] exists independently of space, It doesn't

I agree, both space and time should be "compressed" into a two-dimensional bubble. From our perspective the notion of an "inside" doesn't make any sense, space-time as we know it must end at the horizon. I think...

But I also realized that as matter never reaches the horizon it must also radiate out indefinitely. The red-shift will reduce the energy towards zero, but energy will always escape. But I'm unsure if this is in any way related to the Hawkins-radiation.

The more I think about it the less sense it makes to discuss what an observer would see falling in. It's more sci-fi than science. How would we ever be able to verify it?

annaandherdad - 5-1-2016 at 18:49

Quote: Originally posted by Fulmen  
My question is the opposite of that one, rather than viewing it from the perspective of the viewer I want to understand what happens viewed from the outside.

As far as I can understand the physics as viewed from the outside is within the realm of general relativity, so won't the event horizon effectively be a two-dimensional bubble?


The event horizon is three-dimensional. One of the dimensions is light-like, which means that the EH is swept out by a 2-dimensional family of light rays.

If the outside observer watches something falling into the black hole, he/she will see the thing seem to hover near the event horizon, growing dimmer exponentially. The "something" needs to be illuminated to be seen; or it could be a light source, a flashlight, for example, shining back out at the outside observer. Not only does the light get dimmer, but it also red shifts, at the same exponential rate.

Fulmen - 6-1-2016 at 02:26

Quote: Originally posted by annaandherdad  

The event horizon is three-dimensional. One of the dimensions is light-like, which means that the EH is swept out by a 2-dimensional family of light rays.


I'm sorry, you've lost me there.

As for the need for illumination, remember that any object will give off thermal radiation, so they are in fact luminous and thus observable. The red-shift will of course make it increasingly harder to observe, but in theory a black hole should radiate ever so slightly.

Mesa - 7-1-2016 at 02:03

Black body radiation(a.k.a. heat) can only emit at visible frequencies at temperatures exceeding 480*C. An object below that threshhold is observably not able to be "luminous."(Ignoring photo-excitation and other non-thermal "visible" spectrum radation.)
This has nothing to do with doppler effect/red shift or black holes though.

Honestly, discussing theoretical phyics/black holes etc. requires at least some formal education or legitimate study to be worthwhile. Many concepts are simply too difficult to effectively explain on this platform of discussion.

Fulmen - 7-1-2016 at 03:36

Why the focus on visible light? I agree that the term "luminous" was somewhat inaccurate, but that's more semantics than physics. We can observe the CMB @ 2.7K, and besides it's not about what we can observe today but what could be observed in theory.

Eddygp - 7-1-2016 at 04:15

From the Earth we can detect muons coming in from space, albeit, because of their short half-life, technically they no longer exist from their perspective. Both points of view are correct. That last statement is what makes relativity so insanely amazing.

Fulmen - 7-1-2016 at 04:59

Quote: Originally posted by Eddygp  
technically they no longer exist from their perspective.

Do you have any references for that claim? Detection means interaction, you cannot have interaction if the particle doesn't exist. I know that GR means that 'distant simultaneity' is relative, but only when events are separated in space.

MrHomeScientist - 7-1-2016 at 06:41

Surprised this hasn't been corrected yet, but it's "Hawking" radiation, not Hawkins. As in Stephen Hawking, the black hole guy.

I think annaandherdad has had the best explanation so far. As the object falls towards the horizon, it will appear to an outside observer to slow down (as time slows) and grow dimmer (as there are fewer light rays able to bounce off the object and escape to your eyes). Even if it were emitting light, like a flashlight, at the horizon the gravity is so immensely strong that any light rays emitted by the object simply can't escape. They make a U-turn and fall back into the black hole. So eventually the object disappears from view as no light from it's surroundings can escape the pull of gravity. If things simply appeared to get stuck at the horizon, I would think black holes wouldn't really be black at all. You would see a pileup of objects (asteroids, gas, dust, etc.). Especially if a black hole has an accretion disk, I would think, all of that bright hot gas would pile up at the horizon and make the hole look like a regular star.

Coincidentally, right now I'm reading The Science of Interstellar, where Kip Thorne goes into detail on how the black hole images from that movie were created. They are, as far as we can tell and with reasonable confidence, what actual black holes would really look like. Really, really interesting read. Especially in the difference between a rotating and non-rotating black hole.


Edit: Fixed misspelling of Hawking's first name. Oh the irony!

[Edited on 1-7-2016 by MrHomeScientist]

hissingnoise - 7-1-2016 at 07:43

Quote:
Surprised this hasn't been corrected yet, but it's "Hawking" radiation, not Hawkins. As in Stephen Hawking, the black hole guy.

I did post a correction, but a few remarks, uncomplimentary to the OP inveigled their way in, so I deleted it so as not to appear overly sadistic!

Fulmen - 7-1-2016 at 08:28

Quote: Originally posted by MrHomeScientist  
it's "Hawking" radiation

My bad. Thank you for correcting me.

Quote:
Even if it were emitting light,... at the horizon the gravity is so immensely strong that any light rays emitted by the object simply can't escape.


That assumes that the object actually falls in, which I'm having a hard time understanding. The object would dim to the point where it no longer could be detected, but if it never reaches the EH it must also radiate some (no matter how little) energy. Right?

A BH consuming large amounts of matter will indeed produce a hot accretion disk, it will even expel some of it as immense jets from it's poles (quasars). But these are easily distinguishable from regular stars.

Quote:

I'm reading The Science of Interstellar

I remember finding the BH (and the physics in general) in Interstellar both impressive and believable. I'll see if I can't find that book one of these days.

wg48 - 7-1-2016 at 09:24

Quote: Originally posted by MrHomeScientist  
snip

I think annaandherdad has had the best explanation so far. As the object falls towards the horizon, it will appear to an outside observer to slow down (as time slows) and grow dimmer (as there are fewer light rays able to bounce off the object and escape to your eyes). Even if it were emitting light, like a flashlight, at the horizon the gravity is so immensely strong that any light rays emitted by the object simply can't escape. They make a U-turn and fall back into the black hole. So eventually the object disappears from view as no light from it's surroundings can escape the pull of gravity. If things simply appeared to get stuck at the horizon, I would think black holes wouldn't really be black at all. You would see a pileup of objects (asteroids, gas, dust, etc.). Especially if a black hole has an accretion disk, I would think, all of that bright hot gas would pile up at the horizon and make the hole look like a regular star.

snip

[Edited on 1-7-2016 by MrHomeScientist]


Apparently things do get stuck at the horizon but as you have said they are not visible because any light would be infinitely red shifted to infinite wave lengthens . So the horizon would still be black no matter how much stuff has accumulated.

Its probably inaccurate to say that light does a U-turn if you mean it stops and falls back because that would imply that there was some location that a local observer could observe the light stopping and reversing. Of cause relativity assumes that local the velocity of light is always c.

I often wondered how a black hole grows if all the in falling matter is trapped at the event horizon and it occurred to me that the horizon much enlarge and engulf the matter. But how can it when even time is frozen at its location? I eventually realised that the horizon is a not a thing its a mathematical surface. |When in falling matter passes the surface of an event horizon defined by the mass of the in falling matter plus the mass of the black the new horizon will be created which is large than the original horizon of the black hole (the in falling mass/energy). This is confirmed by various computer simulations that can be found on the web.




MrHomeScientist - 7-1-2016 at 10:27

Quote: Originally posted by wg48  
Its probably inaccurate to say that light does a U-turn if you mean it stops and falls back because that would imply that there was some location that a local observer could observe the light stopping and reversing. Of cause relativity assumes that local the velocity of light is always c.

Well, yes I agree with that. What I meant to say is that ultimately, light heading away from the black hole ends up bending back around and falling back towards it. Space is so severely warped near the horizon that to an outside observer a light ray appears to bend rather sharply, and possibly orbit the horizon several times before hitting the horizon. It never stops at "U-turn" implies, of course. Thanks.

I also like your interpretation of how a hole grows. You're definitely right that the horizon is not a physical thing, but a mathematical surface that simply defines the point at which light cannot escape any more. This thread is making me want to speed read that book! I'll be sure to come back with any insights I glean.

Fulmen - 7-1-2016 at 10:56

Quote: Originally posted by wg48  

I often wondered how a black hole grows if all the in falling matter is trapped at the event horizon and it occurred to me that the horizon much enlarge and engulf the matter. But how can it when even time is frozen at its location?

This has troubled me as well, but I think it can be explained by the way space curves.
Imagine a rubber sheet, stretched out. Now pierce a hole and watch what happens. The hole will expand, but relative to the rubber sheet there is no movement as it's the sheet itself that moves.

With a growing EH the space around will "compress outwards", giving the appearance of growth. But in reality the EH is the edge of a singularity,and while it might seem to have a volume it does not occupy any space. All of the space remains on the outside, compressed towards infinity.

I find that my vocabulary is woefully inadequate for topics such as this. I have a mental image of what I believe happens, but putting it into words is challenging. really

MrHomeScientist - 7-1-2016 at 13:04

Well, the singularity really does have no volume (by definition). The event horizon is a mathematical surface enclosing the singularity; the surface of a sphere (slightly deformed for a rotating black hole, interestingly) within which nothing can escape. The wiki article on Event Horizon (not the movie) is pretty good. An interesting excerpt:
Quote:
Often, this is described as the boundary within which the black hole's escape velocity is greater than the speed of light. However, a more accurate description is that within this horizon, all lightlike paths (paths that light could take) and hence all paths in the forward light cones of particles within the horizon, are warped so as to fall farther into the hole. Once a particle is inside the horizon, moving into the hole is as inevitable as moving forward in time, and can actually be thought of as equivalent to doing so, depending on the spacetime coordinate system used.

There's a short discussion in the Science of Interstellar that mentions this time notion too. Neat!

wg48 - 7-1-2016 at 14:03

Quote: Originally posted by Fulmen  
snip but I think it can be explained by the way space curves.
Imagine a rubber sheet, stretched out. Now pierce a hole and watch what happens. The hole will expand, but relative to the rubber sheet there is no movement as it's the sheet itself that moves.

With a growing EH the space around will "compress outwards", giving the appearance of growth. But in reality the EH is the edge of a singularity,and while it might seem to have a volume it does not occupy any space. All of the space remains on the outside, compressed towards infinity.
snip


The event horizon (EH) is not the edge of the singularity. Its believed that space continues through the event horizon to its centre where the singularity is or at least where our theory breaks down.

Locally and either side of the EH an observer would find nothing special about the space. That's one of the tenants of relativity: space is homogeneous (smooth, except the singularity). Your punctured rubber sheet analogy implies an edge at the EH which is misleading.
.

aga - 7-1-2016 at 14:09

Jeez Fulmen, you don't know already ?

Don't feel bad : Nobody actually knows, and cannot.

The Time thing wrecks the maths, so the mathematical model falls apart at at that point, and there is no other 'probe' to send in.

Edit:

... apart from an agaspace probe, however it's merely an unproven prototype, and may give unpredictable results.

[Edited on 7-1-2016 by aga]

Fulmen - 7-1-2016 at 15:54

As I understand it everything outside the EH falls under the domain of general relativity, so while we haven't observed it directly we have a pretty good idea of what happens. And as far as I can understand nothing will ever cross the EH (perhaps light, I haven't figured that part out yet). And if BH's do indeed "evaporate" this also occurs on the outside of the EH. So from our POV, what else is there? Can we even talk about BH's having an interior beyond the EH? Does that even make sense?

This is the reason why I focus solely on what can be observed from the outside. Even if objects can enter (from their perspective) and even if information somehow can exit through Hawking radiation I have little faith in our ability to ever make useful observations of it. And without observations, can it be called science? I have no problem accepting that some parts of physics will be forever out of bounds. The future might prove me wrong, but we'll be dust long before that happens...