Re-Examining Black Holes and the Standard Model

[TheScienceOrca]

I was thinking about this and either I have a misunderstanding of black holes or they are simply not how the standard model proposes them to be.

Lets start out by setting a few a statements from the standard model that you agree with.

If you disagree about any of these points please comment so we can fix that misunderstanding early.

As soon as an object hits the event horizon according to the standard model it will appear as if it is locked in time right there.

Well if that's true why do we just see a black hole? Aren't we supposed to see all the matter that's inside the black hole right at the event horizon?

 

[PeterDonis]

As soon as an object hits the event horizon according to the standard model it will appear as if it is locked in time right there.

No, it won't. To someone outside the horizon, it won't appear at all; light emitted by the object when it is just crossing the horizon stays at the horizon forever, it never gets any further out, so it never reaches someone who is further out.

To someone who falls through the horizon, they see the object cross the horizon just as they cross the horizon themselves; that is, the light emitted by the object when it crossed the horizon stays at the horizon, so that's where the observer will see it.

Well if that's true why do we just see a black hole?

Meaning, why do we see nothing from inside the black hole? I'll assume that's what you mean; see below.

Aren't we supposed to see all the matter that's inside the black hole right at the event horizon?

No. No one outside the horizon ever sees anything that's at or inside the horizon. See above.

 

[Matterwave]

When an object moves closer to the horizon, the light from this object gets more and more red-shifted until you can no longer detect this light. Even if you want to take the time-dilation approach, an object which is "frozen in time" won't be something that you can see. Imagine an object that emits flashes of light every 1s (according to its own clock), you, far from the horizon, begin by seeing 1 flash per second. As this object gets closer and closer to the horizon, you begin seeing fewer and fewer flashes per second. You start off maybe seeing 1 flash every 2 seconds, and then every 3 seconds, and then every hour...and when the object gets "to" the horizon, you just stop seeing flashes. That last flash will never reach you.

 

[TheScienceOrca]

Just beyond the event horizon by 1m time will move extremely slow so couldn't you see it then?

 

[Matterwave]

Yep, if the object is outside the horizon, you could still technically "see" it, but the light would be very very red-shifted (so perhaps you would see only radio waves or some such).

 

[TheScienceOrca]

Yep, if the object is outside the horizon, you could still technically "see" it, but the light would be very very red-shifted (so perhaps you would see only radio waves or some such).

Ok so why can't we see all the objects even with very very red shifted light.

I mean we can see galaxies extremely far away even though the light is very red shifted as well

 

[Matterwave]

Ok so why can't we see all the objects even with very very red shifted light.

I mean we can see galaxies extremely far away even though the light is very red shifted as well

As long as the object is outside the event horizon, we can technically "see" it (depending on limits to our detection apparatus), but it gets more and more red-shifted until we can't see it as it passes through the event horizon.

 

[TheScienceOrca]

As long as the object is outside the event horizon, we can technically "see" it (depending on limits to our detection apparatus), but it gets more and more red-shifted until we can't see it as it passes through the event horizon.

Correct so why can't we see mass outside the event horizon? The gravitational 1m outside of the event horizon is still VERY great

 

[Matterwave]

Correct so why can't we see mass outside the event horizon? The gravitational 1m outside of the event horizon is still VERY great

How much time-dilation there is is directly proportional to how much red-shift there is. The more time-dilated, the more redshift. By the time you reach "freezing time" kind of dilation, you're at redshifts where all the visible light has been shifted to well outside the visible range.

 

[TheScienceOrca]

How much time-dilation there is is directly proportional to how much red-shift there is. The more time-dilated, the more redshift. By the time you reach "freezing time" kind of dilation, you're at redshifts where all the visible light has been shifted to well outside the visible range.

It may not be in the very tiny visible range, but can't we still measure it and with software convert it into visible images.

 

[Nugatory]

It may not be in the very tiny visible range, but can't we still measure it and with software convert it into visible images.

Not only does it get redder, it gets weaker, because a red shift reduces the amount of energy carried by the light per unit time. No matter how good our detection equipment, the light from the infalling object will eventually (actually this happens quite quickly) be dimmer than the equipment can detect.

If the light is going to become infinitely dim, then you would need infinitely sensitive equipment to detect it - and there's no such thing.

 

[Matterwave]

It may not be in the very tiny visible range, but can't we still measure it and with software convert it into visible images.

Realize that it also gets dimmer since its light is losing energy being red-shifted. But sure, you can use a radio telescope to perhaps see a very bright object as it gets close to a black hole.

EDIT: sniped by nug

 

[PAllen]

Well if that's true why do we just see a black hole? Aren't we supposed to see all the matter that's inside the black hole right at the event horizon?

All other parts of you post have been well addressed, but I didn't notice this one being addressed as I would.

Imagine you are watching a ball of luminescent, transparent, dust collapsing into a BH. The point of this conceit is to allow seeing throughout the body as it collapses. First, ask carefully, what you see before it is close to horizon. You see matter throughout the body, but the light you image from the center was emitted earlier than the light from the surface closest to you. This is unsurprising and you still image as seeing throughout the body. As it gets close to the horizon, the whole body (without any dramatic change in appearance - especially if it is isolated so you don't see extreme lensing of bodies behind the collapsing body) grows dimmer, redder. At all times, until it is blacker than anything else in the universe except another BH, you image (at longer and longer wavelengths) the whole body, matter throughout. As was the case well away from the horizon, the image of matter inside represents earlier emission such that this this light escaped the surface closest to you just before horizon formation. Thus, in this highly idealized collapse, you simply see the whole body vanishing throughout, fairly quickly.

To get an idea how black an isolated BH would be (if there is nearby matter, infall, even at slow rates, makes a BH bright), you need to define some 'blackest possible' state. For example, a box at 10^-15 degrees above absolute zero that is perfect opaque. Then, in quite a short time, for any given definition you pick, a BH will become a trillion times blacker than that within a remarkably short period of time.

 

[jartsa]

As soon as an object hits the event horizon according to the standard model it will appear as if it is locked in time right there.

Well if that's true why do we just see a black hole? Aren't we supposed to see all the matter that's inside the black hole right at the event horizon?

I'll try to answer the question without almost any physics:

Yes, objects appear to freeze at the event horizon.

When an object appears to freeze, it appears to stop doing anything. If an object is emitting light, it appears to stop emitting light, when it appears to freeze.

It seems possible that a frozen object would still reflect light. But there appears to be no reflection because:

A system consisting of an object and light approaching that object seems to become frozen at the event horizon. Light-object distance was changing, that change seems to stop happening, because light-object system seems to freeze at the even horizon.

 

[PeterDonis]

It seems possible that a frozen object would still reflect light.

No, it isn't, because the object isn't "frozen". The light emitted by the object before it reaches the horizon is being redshifted by the hole's gravity. The object itself falls through the horizon; that's why it can't reflect light.

 

[TheScienceOrca]

Thanks for the answers guys!

 

[lukesfn]

No, it isn't, because the object isn't "frozen". The light emitted by the object before it reaches the horizon is being redshifted by the hole's gravity. The object itself falls through the horizon; that's why it can't reflect light.

I'm not sure that is quite right. In the frame of a falling object as it cross the horizon, it can be reflecting light. In the frame of the distant observer, you could imagine the object frozen on the horizon reflecting infinitely redshifted light. However, the math at the horizon from the frame of a distant observer can be become a little confusing more maybe even ill defined, so maybe it is better to talk about light reflecting just before crossing the horizon, which would be too redshifted and faint to see.

I have a thought experiment that helped me a little exploring this kind of question from a different angle.. Imagine we have a powerful telescope observing from a distance a clock as it approaches the horizon of a symmetrical Black Hole. Now imagine what would happen if the observed light was not actually redshifted and was intense enough to see. As the clock approaches the horizon imagine seeing the clocks time slow down to an apparent stand still. You might expect to also observe the clock to stop falling and to stay stuck at the horizon, however, counter intuitively, I don't think this is the case. Because of the geometry of space is so curved, the closer an object gets to the horizon, the more light reflecting from it will diverge. I would image his may create the appearance that the clock is continuing to fall into the hole, even while it's time will remain frozen. If the light diverges to the point it is evenly spread around the horizon, then it will appear that the clock has fallen into a singular point at the center. The clock could be imagined as frozen across the entire surface of the BH, but giving the optical illusion that it is a a singular point a the center.

Of course, once you take into account light intensity and redshfts, all the above because invisible or perhaps even nonsense.

 

[PAllen]

I'm not sure that is quite right. In the frame of a falling object as it cross the horizon, it can be reflecting light. In the frame of the distant observer, you could imagine the object frozen on the horizon reflecting infinitely redshifted light. However, the math at the horizon from the frame of a distant observer can be become a little confusing more maybe even ill defined, so maybe it is better to talk about light reflecting just before crossing the horizon, which would be too redshifted and faint to see.

I have a thought experiment that helped me a little exploring this kind of question from a different angle.. Imagine we have a powerful telescope observing from a distance a clock as it approaches the horizon of a symmetrical Black Hole. Now imagine what would happen if the observed light was not actually redshifted and was intense enough to see. As the clock approaches the horizon imagine seeing the clocks time slow down to an apparent stand still. You might expect to also observe the clock to stop falling and to stay stuck at the horizon, however, counter intuitively, I don't think this is the case. Because of the geometry of space is so curved, the closer an object gets to the horizon, the more light reflecting from it will diverge. I would image his may create the appearance that the clock is continuing to fall into the hole, even while it's time will remain frozen. If the light diverges to the point it is evenly spread around the horizon, then it will appear that the clock has fallen into a singular point at the center. The clock could be imagined as frozen across the entire surface of the BH, but giving the optical illusion that it is a a singular point a the center.

Of course, once you take into account light intensity and redshfts, all the above because invisible or perhaps even nonsense.

Peter is right and you are not. Imagine an external observer sending a signal. It reaches the surface at some event. The reception of the signal by the surface is one event, and the interaction is frame invariant. It occurs either before the surface crosses the horizon, or at the horizon, or inside, but whichever it is invariant. In particular, I have explained, in many threads here, that for a given external observer world line, sending a sequence of signals to the surface of catastrophically collapsing body, there is a precise event on the external world line whose signal reaches the body at horizon crossing. Signals sent after this event by the external observer reach the surface inside the horizon. These features are frame invariant. Signals sent after this critical event on the external observer world line cannot be reflected, in principle.

Relating this to an image of a frozen clock, imagine the last visible time on a collapsing surface clock is 3:00 pm - this is the time the clock is seen to freeze at for the external observer. A signal sent after the critical external event will arrive, say, at 3:01 PM at the surface clock, but the external observer can never see this arrival, nor can the signal be returned or reflected because this clock time is inside the horizon.

 

[PeterDonis]

In the frame of a falling object as it cross the horizon, it can be reflecting light.

Yes, I should have been clearer: once an object reaches the horizon, it can't reflect light that is visible outside the horizon. It can still reflect light, yes, but that reflected light will only be visible to other observers inside the horizon (and since all such observers must be falling towards the singularity, there are limits to which observers can see reflected light from which others).

There are other issues with your thought experiment as well, but I'll address those in a separate post if I get a chance.

Signals sent after this critical event on the external observer world line cannot be reflected, in principle.

I think you mean, such signals can't be reflected back to the observer outside the horizon, correct? A light signal could still reach the falling object inside the horizon, be reflected by it, and be seen by some other observer inside the horizon. It just can't get back outside the horizon once it's inside.

(Of course, if the signal is sent after another critical time on the external observer's worldline, it will not reach the falling object before the latter hits the singularity; in that case, yes, there is no reflection, period. The ingoing light signal hits the singularity, not the falling object.)

 

[PAllen]

I think you mean, such signals can't be reflected back to the observer outside the horizon, correct? A light signal could still reach the falling object inside the horizon, be reflected by it, and be seen by some other observer inside the horizon. It just can't get back outside the horizon once it's inside.

(Of course, if the signal is sent after another critical time on the external observer's worldline, it will not reach the falling object before the latter hits the singularity; in that case, yes, there is no reflection, period. The ingoing light signal hits the singularity, not the falling object.)

Yes, of course, that is what I meant.

 

[PeterDonis]

Because of the geometry of space is so curved, the closer an object gets to the horizon, the more light reflecting from it will diverge.

Why do you think this?

 

[.Scott]

I think there is a missed point here. From the point of view of an outside observer, material accumulates at an event horizon that is growing as material is added. Certainly an actual attempt at observing material close to the event horizon is fraught with technical difficulties - extreme red-shifts, extreme light tunneling, extreme effects from Heisenberg Uncertainty, etc, but that doesn't stop the arithmetic of how an object is transformed as "seen" from an outside frame of reference.

But even with this view, does material ever make it passed the event horizon? Say I drop an apple onto the black hole - allowing it to freeze at the event horizon. Then I drop a planet over it. The planet would cause the radius of the event horizon to expand and at first thought, this would fully bury the apple.

Say we deposited material on the far side of the black hole - say we drop the apple onto the South Pole and then start piling material up at the North Pole? This would cause the event horizon at the South Pole to move in response to the overall momentum of the system and to reduce it curvature due to the increase in mass - but neither of those actions should change our "perception" of what is happening to the apple at the event horizon.

Part of the picture is that the apple becomes spread across the entire event horizon as it freezes in time and Heisenberg Uncertainty forces it to be less spatially specific.

 

[PAllen]

I think there is a missed point here. From the point of view of an outside observer, material accumulates at an event horizon that is growing as material is added. Certainly an actual attempt at observing material close to the event horizon is fraught with technical difficulties - extreme red-shifts, extreme light tunneling, extreme effects from Heisenberg Uncertainty, etc, but that doesn't stop the arithmetic of how an object is transformed as "seen" from an outside frame of reference.

This is not correct. A particle of matter has one world line, that may be represented in multiple coordinates. If this world line crosses a horizon and end in a singularity, these are invariant facts about the world line. Its history doesn't change according what observer is looking at it, or what coordinates are used. What can happen is that part of its history cannot be seen by some observer, or that some coordinate system does not cover all of its history. The phrasing you use is just wrong.

In the case of collapse, this wrong view is absurd on its face because there is matter at the center throughout the collapse, until the singularity forms. Do you honestly believe the matter in the center jumps to outside the horizon for some external observer? See my post #13 for a sense in which the matter at the center is seen to be frozen at the center, not at the horizon.

 

[.Scott]

Would you also argue that when a mass is accelerated to a significant portion of the speed of light, it does not increase in mass? Lot's of things vary based on the reference frame of the observer. Those effects are very real.

There is no doubt that, from the reference frame of the falling object, there is no event horizon, there is no red-shift, there is no time dilation - only an inescapable fall. That in no way invalidates what a black hole is from an inertial frame of reference at a healthy distance outside the black hole.

 

[phinds]

Would you also argue that when a mass is accelerated to a significant portion of the speed of light, it does not increase in mass?

I can't speak for Pallen, but I would certainly argue that, yes. I am, of course, speaking about the object's rest mass, that is, the mass that is seen by a co-moving observer in the object's frame of reference. If the object's actual mass varied by observer (and speed is observer dependent) then the mass would take on an infinite range of values all at the same time. Do you think that would be a real effect, applicable to the object itself?

 

[PAllen]

Would you also argue that when a mass is accelerated to a significant portion of the speed of light, it does not increase in mass? Lot's of things vary based on the reference frame of the observer. Those effects are very real.

You need to understand the difference between coordinate dependent facts and invariant facts. Energy of an object is coordinate dependent. Rest mass or invariant mass is not. Two objects colliding is an invariant fact, even if some Rindler observer cannot see the collision.

There is no doubt that, from the reference frame of the falling object, there is no event horizon, there is no red-shift, there is no time dilation - only an inescapable fall. That in no way invalidates what a black hole is from an inertial frame of reference at a healthy distance outside the black hole.

You are using 'is' for things that are coordinate dependent choices. For an external observer, there is a point on its world line where the horizon changes from being (all) in the causal future to being NOT (all) in the causal future - part of it has spacelike separation from later points in the external observer's world line. Once this is true, it is purely a matter of choice of simultaneity convention as to whether the horizon is past, present, or future.

[edit: There is no such thing as a global inertial frame in GR. There is no preferred set of coordinates that correspond to 'global reality' for any observer in GR. There are only locally inertial frames, and a good case can be made that Fermi-Normal coordinates best represent a local inertial frame.]

 

[.Scott]

I can't speak for Pallen, but I would certainly argue that, yes. I am, of course, speaking about the object's rest mass, that is, the mass that is seen by a co-moving observer in the object's frame of reference. If the object's actual mass varied by observer (and speed is observer dependent) then the mass would take on an infinite range of values all at the same time. Do you think that would be a real effect, applicable to the object itself?

If an object is traveling at near the speed of light relative to your inertial reference frame and you measure the amount of gravitational pull from it, it will correspond to a mass much larger than its rest mass. Certainly, everyone from every reference frame can deduce its rest mass. But this doesn't make the relativistic mass any less real or valid.

Similarly, a view of the black hole from a frame of reference drifting a light-year away is a valid view of the black hole. From that reference frame, objects do not drop below the event horizon - even though people in that outer reference frame can deduce that someone in the falling reference frame would experience a continued fall.

 

[.Scott]

There is no such thing as a global inertial frame in GR. There is no preferred set of coordinates that correspond to 'global reality' for any observer in GR. There are only locally inertial frames, and a good case can be made that Fermi-Normal coordinates best represent a local inertial frame.

I'm certainly aware of that. My point is that every inertial reference frame provides a valid interpretation of what is occurring. That would be the whole point behind the notion of "no preferred" inertial reference frame.

 

[PAllen]

I'm certainly aware of that. My point is that every inertial reference frame provides a valid interpretation of what is occurring. That would be the whole point behind the notion of "no preferred" inertial reference frame.

But you're missing that there is no global inertial frame at all. When describing global features of the universe, there are just coordinate choices. An external observer can choose perfectly reasonable simultaneity conventions such that questions like "when did the horizon form?" or "when did the baseball cross the horizon?" have specific answers e.g. 3:15PM today. However, there is a specific, earliest point on an external world line that can possibly considered simultaneous to e.g. baseball crossing the horizon. That earliest point when the horizon crossing event ceased being in the causal future.

Similarly, a view of the black hole from a frame of reference drifting a light-year away is a valid view of the black hole. From that reference frame, objects do not drop below the event horizon - even though people in that outer reference frame can deduce that someone in the falling reference frame would experience a continued fall.

And this is false. You continue to believe there is such a thing as "a (global) reference frame of the external observer". There is no such thing in GR. Instead, for global questions, there are only coordinate choices, and the lack of coverage of one such choice is not saying anything about "reality for external observers".See the difference between true versus false:

1) External observer never can see a horizon crossing event. (true)

2) Object never crosses a horizon from the point of view or external observer. This false, because point of view depends simultaneity choice, and this is purely arbitrary.

 

[PeterDonis]

If an object is traveling at near the speed of light relative to your inertial reference frame and you measure the amount of gravitational pull from it, it will correspond to a mass much larger than its rest mass.

No, it won't. I realize it seems intuitively like it should, but this is a case where intuition can lead you astray.

To see the issue, consider the scenario as it is seen in the rest frame of the gravitating object. (Note that, as PAllen pointed out, this frame is not actually an inertial frame. But we can still construct a valid coordinate chart centered on the object, as long as we keep in mind that it won't have all the properties we expect in a global inertial frame.) In that frame, you are traveling near the speed of light (and we assume your mass is negligible so you don't produce any significant gravity yourself). You fly by the object; how much does its gravity bend your trajectory?

If your intuition were correct, the object should always capture you (i.e., the bending of your trajectory should increase without bound, so that it is bent right into the object), since you, traveling at near the speed of light, should measure the object to have near-infinite gravity. But that's not what will actually happen. The GR prediction is that your trajectory will only be bent twice as much (in the limit as your velocity goes to $c$) as a Newtonian calculation using the gravitating object's rest mass would suggest. This prediction has been verified by measuring the bending of light passing close to the Sun.