Black Hole Event Horizon and the Observable Universe

[skeptic2]

Thank you xanox and Chalnoth. I saw the Penrose diagram but I still fail to understand why the path of the infalling object leads to the young black hole instead of asymptotically approaching the event horizon.

Nevertheless I recognize there is a problem with claiming that a black hole must evaporate before an object can penetrate the event horizon. Evaporation itself depends upon one of the two virtual particles created at the event horizon falling past the event horizon.

 

[stevebd1]

Regarding Hawking radiation and crossing the event horizon, you might also find this of interest from Usenet Physics FAQ-

'What about Hawking radiation? Won't the black hole evaporate before you get there?'
http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html
(scroll down to the fouth FAQ)

 

[Chalnoth]

Thank you xanox and Chalnoth. I saw the Penrose diagram but I still fail to understand why the path of the infalling object leads to the young black hole instead of asymptotically approaching the event horizon.

Nevertheless I recognize there is a problem with claiming that a black hole must evaporate before an object can penetrate the event horizon. Evaporation itself depends upon one of the two virtual particles created at the event horizon falling past the event horizon.

Well, if you look at the Penrose diagram in fig. 3 of the paper I linked above, you should see a wavy, light-blue dashed line on the bottom of the graph. This is the event horizon of the evaporating black hole. The second point of interest is the singularity, which is the nearly-horizontal red line on the left: anything that hits that is squashed.

The thing to recognize here is that the event horizon is a horizon for outgoing light rays: if you start a light beam somewhere along that horizon traveling outward, it will travel straight along the horizon. If, however, I start a light beam traveling inward, it will just go and smack right into the singularity. So, what if I'm just a plain object? Well, if I was traveling inward I would just go right on through the event horizon and smack the singularity. The only way to prevent this occurring would be if the event horizon were also a horizon for inward-traveling light rays as well as outward-traveling ones. This clearly does not appear to be the case.

 

[Phrak]

Thank you xanox and Chalnoth. I saw the Penrose diagram but I still fail to understand why the path of the infalling object leads to the young black hole instead of asymptotically approaching the event horizon.

Nevertheless I recognize there is a problem with claiming that a black hole must evaporate before an object can penetrate the event horizon. Evaporation itself depends upon one of the two virtual particles created at the event horizon falling past the event horizon.

Virtual particles are not constrained to time-like world lines.

 

[Chalnoth]

Update on the falling into a realistic black hole with Hawking radiation:

My old GR professor got back to me, and pointed out something that I had forgot to consider: it depends upon whether or not a singularity forms before the black hole evaporates. If the singularity forms, then yes, some of the matter will travel in and smack the singularity before evaporation. If not, then indeed, infalling matter will see the black hole evaporate before it. Apparently the question as to whether or not the singularity will form in an evaporating black hole is still open.

 

[Phrak]

Update on the falling into a realistic black hole with Hawking radiation:

My old GR professor got back to me, and pointed out something that I had forgot to consider: it depends upon whether or not a singularity forms before the black hole evaporates. If the singularity forms, then yes, some of the matter will travel in and smack the singularity before evaporation. If not, then indeed, infalling matter will see the black hole evaporate before it. Apparently the question as to whether or not the singularity will form in an evaporating black hole is still open.

By 'infalling', I hope you mean 1) matter that is present in the interior when the event horizon forms, or 2) exterior matter that will find itself within the horizon if the Swarzchild radius increases, for instance.

 

[Chalnoth]

By 'infalling', I hope you mean 1) matter that is present in the interior when the event horizon forms, or 2) exterior matter that will find itself within the horizon if the Swarzchild radius increases, for instance.

Naturally, yes. At least it would in the idealized black hole that was already fully-formed. If there is no singularity, apparently, there is no true event horizon.

 

[MuggsMcGinnis]

If you accept in general that observers in different reference frames don't always agree on observed time,distance,mass,etc, elsewhere seems like this one should also also be accepted.

Yes, this is the crux of my problem with this. I can't get my head around the idea that reality is different for different observers. By this I mean that, the laws of physics are different, which is what this requires.

If anything can fall through an event horizon, then a single universe can have multiple realities that are logically, causally inconsistent.

For one observer, all causal contact is lost with the outside universe while, from another perspective, the in-falling object can be interacted with forever (until the black hole evaporates).

 

[MuggsMcGinnis]

By 'infalling', I hope you mean 1) matter that is present in the interior when the event horizon forms, or 2) exterior matter that will find itself within the horizon if the Swarzchild radius increases, for instance.

Presuming that it's possible to have "matter that is present in the interior <of an> event horizon" sort of bypasses the whole core of any discussion of whether it's possible to fall through an event horizon.

 

[MuggsMcGinnis]

Naturally, yes. At least it would in the idealized black hole that was already fully-formed. If there is no singularity, apparently, there is no true event horizon.

Chalnoth

How are these things incompatible?

Why can you not have an event horizon (which isn't really a "thing" so much as a geometrical definition) if you don't have a singularity?

Thanks

 

[MuggsMcGinnis]

Update on the falling into a realistic black hole with Hawking radiation:

My old GR professor got back to me, and pointed out something that I had forgot to consider: it depends upon whether or not a singularity forms before the black hole evaporates. If the singularity forms, then yes, some of the matter will travel in and smack the singularity before evaporation. If not, then indeed, infalling matter will see the black hole evaporate before it. Apparently the question as to whether or not the singularity will form in an evaporating black hole is still open.

Yes. I think that your old GR professor has described the problem nicely.

If an event horizon is something that anything can fall through, then it's inevitable that a singularity will form.

 

[Chalnoth]

Yes. I think that your old GR professor has described the problem nicely.

If an event horizon is something that anything can fall through, then it's inevitable that a singularity will form.

Well, the issue is that there's no such thing as a black hole that hasn't formed from infalling matter. So if infalling matter can't form an event horizon in the first place, then clearly it's not going to make it to the singularity.

But if you already have a black hole sitting there, singularity and event horizon and all, then matter can most certainly fall through the event horizon.

 

[MuggsMcGinnis]

The photon emitted at the limit of the horizon will be received in infinite time measured by the distant observer, ie. never (supposing the purely classical case without evaporation). You're apparently considering this problem within special relativity, but that only allow to talk about flat spacetime with no gravity, where black holes do not exist indeed. I friendly recommend a reading of the beautiful and accessible introductory text by James Hartle, "Gravity: An Introduction to Einstein’s General Relativity", Addison Wesley (2002) (available on Amazon).


I was supposed to mean that in terms of the distant observer clock, the photons which he would expect to receive much sooner if the spacetime was flat, will be received much later. This is again a manifestation of spacetime curvature, not a modification of properties of light which always move locally at the speed of light c. That the travel time of light increases within a strong gravitational field is one of the classic tests of general relativity which has been confirmed experimentally with a precision of about 0.1%.

Got it! Thanks.

I don't know why I was being so dense on this... it would take longer than the 5.8 days for the light-pulse round-trip.

So, at what point in time will the distant observer be unable to receive a reflected return-pulse? If the reflective ball ("object") is tossed at the black hole at .5 c, from a distance of 2.9 light-days, then you're saying the object will have crossed the event horizon in less than 5.8 days (using an unaccelerated clock, stationary WRT the black hole). (correct?)

You're saying that, if the distant observer sends a light pulse 6 days after tossing the object, he will never receive a return pulse? The light pulse will pass through the event horizon, because there will be nothing for it to reflect from?

 

[MuggsMcGinnis]

Naty1

I raised the question of how an object is able to cross the event horizon because I don't understand how it happens and what seems logical to me is at odds with the widely accepted interpretation of black hole geometries. Certainly I accept that observers in different reference frames don't always agree on observed time,distance,mass,etc. but I also believe that events in one reference frame can be mathematically transformed into any other reference frame to explain what those observers see. To suggest that simply because observers in different reference frames disagree about time, distance or mass, is sufficient reason to accept an ad hoc instance of differing observations without providing some sort of transformation between the reference frames is less than scientific.

I also raised the question hoping that someone here could point out the errors in my logic. The references I've seen, like xantox's posts, don't address the issue of the evaporation of the black hole during the extremely dilated time an object experiences as it falls towards the event horizon. Briefly put, is time infinitely dilated at the event horizon and if so, how does an object cross that event horizon in finite time? If it crosses in finite time in one frame of reference but not in another, what is the transformation between those reference frames that permits that?

skeptic2:

You and I are in the same place, on this.

I cannot (yet) see how two logically incompatible solutions can both be correct within any logically consistent system.

This requires that one problem has two correct solutions: BOTH (Yes = Falls through event horizion) AND (No = Does not fall through EH).

Doesn't this require that there is some statement, within the physical laws of our universe, for which "BOTH A AND NOT A" is true?

I'm not at all comfortable with that.

 

[MuggsMcGinnis]

Your suspicions are valid for a test particle. A massive object that perturbs the event horizon may be different.

In addition to this, a central singularity is often invoked, but not proven nor motivated.

Using the contemporary view of what happens when things fall to black holes, the apparent velocity (with respect to the black hole) of an infalling particle approaches the speed of light.

Does this not yield a Relativistic mass increase?

Since the net mass of the system will be unchanged, wouldn't a considerable amount of energy (mass) be transferred from the black hole to every infalling object?

In fact, it seems (to me) entirely possible that all of the mass of the black hole would be transferred to the infalling matter surrounding the black hole.

I expect that xantox might have something to say about this.

 

[MuggsMcGinnis]

Imagine you are halfway between two black holes that are approaching each other. To a observer the horizons merge, with you inside. Apparently the size of the black hole must increase for something to cross an event horizon--which, come to think about it, is nearly a tautalogy, anyway.


Why should it be dense? At the time of collapse, the mass is not all stuck at one point in space. Put enough air together at standard density and it's a black hole.

The Schwarzschild metric relates radius to mass: R = M (2G/c²)

Since radius and mass are directly proportional, the mass density of a Schwarzschild black hole drops with radius. Density is proportional to R^-2.

Interestingly, (according to the WMAP 5-year results) the mass density of a Schwarzschild black hole with a radius equal to that of the observable universe would equal the density of the observable universe.

In fact, our observable universe satisfies all requirements for a Schwarzschild black hole:

• Spherically symmetrical
• No electrical charge.
• No spin.
• R = M (2G/c²)

Wilkinson Microwave Anisotropy Probe:
http://map.gsfc.nasa.gov/universe/WMAP_Universe.pdf

 

[Chalnoth]

Using the contemporary view of what happens when things fall to black holes, the apparent velocity (with respect to the black hole) of an infalling particle approaches the speed of light.

Does this not yield a Relativistic mass increase?

Basically there's no such thing in General Relativity of an object's velocity that is far away from the observer. Relative velocities are only well-defined if computed at the same point in space-time. You just can't subtract velocities at different points, so this sort of question is meaningless.

Furthermore, the idea of the relativistic mass is no longer used, as it leads to too many mistakes. This would be one of them, because the energy of the infalling object certainly does not diverge at it crosses the event horizon.

 

[Chalnoth]

The Schwarzschild metric relates radius to mass: R = M (2G/c²)

Since radius and mass are directly proportional, the mass density of a Schwarzschild black hole drops with radius. Density is proportional to R^-2.

Er, in a Schwarzschild black hole, all of the mass is located at the singularity in the center.

Interestingly, (according to the WMAP 5-year results) the mass density of a Schwarzschild black hole with a radius equal to that of the observable universe would equal the density of the observable universe.

In fact, our observable universe satisfies all requirements for a Schwarzschild black hole:

• Spherically symmetrical
• No electrical charge.
• No spin.
• R = M (2G/c²)

Wilkinson Microwave Anisotropy Probe:
http://map.gsfc.nasa.gov/universe/WMAP_Universe.pdf

Actually the Schwarzschild radius of a black hole with the mass of everything within our current Hubble volume would only have a "radius" of about 1/4th of our Hubble volume (the R parameter of a black hole isn't actually a radius, but is directly related to the area of the event horizon...there is no well-defined radius, it turns out). So a black hole the size of our Hubble volume would be quite a bit more "dense", on average, than our universe.

 

[Phrak]

Just in case anyone missed it, The Schwazschild metric applies to space without any matter in it. The Schwarzschild metric is equally applicable to the space around the Earth, for instance. It's just not applicable within the Earth itself. All that is required of it, is that the mass be mass be spherically symmetric and unchanging in qauantity over time.

There is nothing that requires the central mass to occupy a single point.

 

[Phrak]

Presuming that it's possible to have "matter that is present in the interior <of an> event horizon" sort of bypasses the whole core of any discussion of whether it's possible to fall through an event horizon.

Not at all. The matter could be within the horizon at the time the horizon was created. Or it could be the result of a changing event horizon. Two blachholes merging in finite time would be an example.

A test particle, where by convention the mass of the test particle has no perturbative effect, is very small. It won't cross a static horizon.

 

[Phrak]

Using the contemporary view of what happens when things fall to black holes, the apparent velocity (with respect to the black hole) of an infalling particle approaches the speed of light.

I don't know where you are getting your facts. The velocities at assymtotic infinity are infinite in the reference frame of the horizon.

 

[Chalnoth]

Just in case anyone missed it, The Schwazschild metric applies to space without any matter in it. The Schwarzschild metric is equally applicable to the space around the Earth, for instance. It's just not applicable within the Earth itself. All that is required of it, is that the mass be mass be spherically symmetric and unchanging in qauantity over time.

There is nothing that requires the central mass to occupy a single point.

Very true. The demand that everything at the center rests upon the observation of the propagation of light within the black hole: if even outgoing light rays emitted within the event horizon impact the center in a finite (and very short) time, then it is necessarily true that all matter will as well.

But as I said, this may not be the case for a realistic black hole if the horizon fails to form before it evaporates.

 

[MuggsMcGinnis]

Basically there's no such thing in General Relativity of an object's velocity that is far away from the observer. Relative velocities are only well-defined if computed at the same point in space-time. You just can't subtract velocities at different points, so this sort of question is meaningless.

Furthermore, the idea of the relativistic mass is no longer used, as it leads to too many mistakes. This would be one of them, because the energy of the infalling object certainly does not diverge at it crosses the event horizon.

I'm not suggesting that the velocities of infalling objects would subtract, somehow, from each other. I'm suggesting that relativistic mass increase is inevitable for an infalling object.

Regarding "the idea of the relativistic mass is no longer used", perhaps I should refer to "momentum" or "energy"?

 

[MuggsMcGinnis]

I don't know where you are getting your facts. The velocities at assymtotic infinity are infinite in the reference frame of the horizon.

I apologize for any confusion, but, I don't refer to any reference frame at the horizon or within the horizon; only near the horizon or far from it.

The model I use is no less valid than the one Stephen Hawking used in lowering a box filled with thermal energy, via a rope, to the event horizon.

 

[Chalnoth]

I'm not suggesting that the velocities of infalling objects would subtract, somehow, from each other. I'm suggesting that relativistic mass increase is inevitable for an infalling object.

Except you have to subtract the velocities for the relativistic mass to turn any different from its rest mass. If you're in the infalling object's frame of reference, after all, you consider your own velocity to be precisely zero.

Regarding "the idea of the relativistic mass is no longer used", perhaps I should refer to "momentum" or "energy"?

Well, right. The proper way to compare it would be to look at the following:
1. What is the total energy of the particle very far away from the black hole?
2. What is the potential energy of a particle that falls from very far away to the event horizon?

Add the two and you get the mass-energy that is added to the black hole (assuming it's not spinning...things get a bit more interesting for spinning black holes).

 

[MuggsMcGinnis]

Well, the issue is that there's no such thing as a black hole that hasn't formed from infalling matter. So if infalling matter can't form an event horizon in the first place, then clearly it's not going to make it to the singularity.

But if you already have a black hole sitting there, singularity and event horizon and all, then matter can most certainly fall through the event horizon.

There is no reason you can't get an event horizon without matter falling through the event horizon.

Consider a massive, hollow spherically symmetrical shell of matter. Suppose that the shell initially formed as a point, but mass has been added and the shell has grown.

The external perspective is the first one to consider because it is, initially, the only valid perspective. All valid models will be consistent with models from external perspectives.

The gravitational characteristics of all spherically symmetrical objects of equal mass are identical:

• A spherical volume with all of the mass at the center point
• A spherical, homogeneous mass, with the mass uniformly distributed throughout
• A hollow, spherical shell, with all mass uniformly distributed about the shell

The astronomical characteristics would be the same.

 

[Chalnoth]

There is no reason you can't get an event horizon without matter falling through the event horizon.

If I'm understanding this correctly, the very reason why the horizon doesn't form in the first place is because matter doesn't pass it. After all, if the matter did pass the pre-horizon, then it would have enough density to produce an actual horizon.

Consider a massive, hollow spherically symmetrical shell of matter. Suppose that the shell initially formed as a point, but mass has been added and the shell has grown.

The thing to remember here is that if you have spherically-symmetric infalling matter, the matter inside any given shell cannot feel the gravitational effects of any of the matter outside of it. The only effect that does make a difference is the pressure, but in the picture of the black hole that evaporates before the horizon forms, the time dilation is so strong as it approaches the pre-horizon that the matter doesn't feel the pressure for long enough to actually be pushed anywhere.

 

[skeptic2]

When I picture the formation of a BH, I see a star collapsing through the white dwarf stage, through the neutron star stage and continuing to collapse until the matter, most likely near the center, reaches a density sufficient to form an EH. Matter at that radius cannot collapse further because of the contraction of space and dilation of time. The matter above that radius may continue to collapse until it reaches its own radius for an event horizon. The star essentially freezes from the inside out. This is a little different from MuggsMcGinnis' example because it doesn't involve a shell but still is similar. This permits the formation of an EH without any matter falling through it. Perhaps there is no singularity and no inside to the black hole.

There is another issue that MuggsMcGinnis just touched on. If the escape velocity is the velocity that would propel a projectile to come to rest at infinity, then matter infalling from infinity to a BH should just reach c at the EH. Another reason to consider why matter cannot fall into a BH is the same as why matter cannot escape a BH. To do so means exceeding c.

 

[xantox]

You're saying that, if the distant observer sends a light pulse 6 days after tossing the object, he will never receive a return pulse? The light pulse will pass through the event horizon, because there will be nothing for it to reflect from?

Yes.

This requires that one problem has two correct solutions: BOTH (Yes = Falls through event horizion) AND (No = Does not fall through EH).

A master of relativity that I know, when reading what you say would say what follows. This is the most common of all misconceptions concerning black holes as treated in GR. (Now even in galilean relativity, you could as well find strange that if you're in a boat and throw a ball vertically you will see it falling straight while someone on the shore will see it falling parabolically : which is the "correct solution" then? Of course, both are equivalent when taking into account the different system of coordinates of the two observers.)

The so-called "gravitational redshift" of signals sent from an observer hovering near the horizon to an observer hovering farther away is due simply to spacetime curvature. One of the fundamental interpretations of spacetime curvature is that initially parallel geodesics lying in a negatively (positively) curved two-surface will diverge (converge). So if we consider the ideal Schwarzschild vacuum outside some isolated massive object and suppress the angular coordinates, the "t, r plane" has negative curvature, so initially parallel null geodesics diverge as they head radially outward. That means that if we draw two radially outgoing null geodesics (ie light rays, in the geometric optics approximation), intersecting the world line of a static observer at r1, where horizon < r1, and see how they intersect the world line of a static observer at r2, where r1 < r2, we see that the "proper time interval" between two wave crests of a radio transmission (as measured by the ideal clock of the r1 observer) will be separated by a larger proper time interval when received by the r2 observer, because the two initially parallel null geodesics diverge. That's why the r2 observer measures a lower frequency at reception than the r1 observer measures at emission.

Notice that in GR, discussions of a "frequency shift" always refer to specific signals sent by an observer with one world line and received by an observer with another world line. That's two distinct timelike world lines plus at least two null geodesics intersecting both of these. Frequency shifts are never well-defined in GR unless you specify all this information.

As we imagine asking the lower observer to hover closer and closer to the horizon, the redshift becomes more and more extreme, and we see that distant observers cannot watch anything fall through the horizon because as an object crosses the horizon signals from it to the outside are infinitely redshifted.

In fact, our observable universe satisfies all requirements for a Schwarzschild black hole: Spherically symmetrical. No electrical charge. No spin. R = M (2G/c2)

No, our observable universe, modeled by a FRW solution, is a completely different beast than a Schwarzschild black hole. If you drop the word "Schwarzschild", still the statement is technically incorrect re the definition of a black hole. By dropping even more constraints it would start to get similar but you will not be quite speaking about black holes anymore. See http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/universe.html for some details.

Using the contemporary view of what happens when things fall to black holes, the apparent velocity (with respect to the black hole) of an infalling particle approaches the speed of light.

There are many distinct operational notions of "distance in the large" and thus "speed in the large", mostly not even symmetric. That you can obtain a single nice notion in flat spacetime is a very very special property, not shared by any curved spacetime.

When I picture the formation of a BH...

You're not using the correct picture. Take a look to the excellent book by R. Geroch, "General relativity from A to B" (available on Amazon).

The gravitational characteristics of all spherically symmetrical objects of equal mass are identical

The potentials in the interior of the spherical shell are not the same for each example you give, even in Newtonian gravity.

 

[Phrak]

No, our observable universe, modeled by a FRW solution, is a completely different beast than a Schwarzschild black hole. If you drop the word "Schwarzschild", still the statement is technically incorrect re the definition of a black hole. By dropping even more constraints it would start to get similar but you will not be quite speaking about black holes anymore. See http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/universe.html for some details.

I went. I read. That as the most dedctively incohesive piece I've read on the Baez FAQ. In multiple places, it doen't hang together. The final argument may be correct, but the path is just wrong. I think something less dated may be better.