# Does a Black Hole Ever Collapse?

#### CJames

Been a while since I posted here. I gave up on pursuing a career in engineering or physics so forgive me for any errors, but something occurred to me recently.

If you are familiar with general relativity then you know this:

$${t_o=t_f}$$$$\sqrt{1 - r_o/r}$$

where
$${t_o}$$ is time passed close to the gravitational field
$${t_f}$$ is time passed far from the gravitational field
$${r_o}$$ is the radius of the event horizon
$${r}$$ is the distance from the center of gravity

What this means is that the clock is ticking slower for an observer near a black hole (or any gravitational field) than for a distant observer. The weird part is the case where $${r_0 = r}$$. This means that zero time passes for an observer at the event horizon no matter how much time passes far from the black hole. But if that is true, how could anything ever get inside the black hole to begin with?

It seems to me that the black hole never really collapses in the first place, or rather that it takes infinite time for a black hole to collapse. In that case, if Hawking is right, black holes always evaporate before they form. So why is everybody so concerned about singularities? It doesn't sound like they form in the first place. A black hole is starting to sound more like a time-delayed implosion/explosion. The matter that forms the black hole and anything that fell in afterward is just crushed around the edge of the event horizon until it evaporates.

Am I missing something?

Related Special and General Relativity News on Phys.org

#### DaveC426913

Gold Member
This means that zero time passes for an observer at the event horizon no matter how much time passes far from the black hole. But if that is true, how could anything ever get inside the black hole to begin with?
As the infalling observer heads toward the centre, if he looked out the back of shis ship he would see the universe age rapidly. But this does not prevent him from falling to the centre of the BH, which he will of course experience over a normal length of time. The infalling obserever does not experience any sort of event horizon.

It is only from the distant observing point that the infalling ship is observed to slow down.

Though it may at first glance appear so, there isn't really a paradox here.

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#### pervect

Staff Emeritus
As the infalling observer heads toward the centre, if he looked out the back of shis ship he would see the universe age rapidly.
This is a common misconception. See for instance

Quiz question 5: As you fall freely into a black hole, you see the entire future of the Universe played out before your eyes. True or false?

Answer to the quiz question 5: False. You do NOT see all the future history of the world played out. Once inside the horizon, you are doomed to hit the singularity in a finite time, and you witness only a finite (in practice rather short) time pass in the outside Universe.

In order to watch the history of the Universe unfold, you would have to remain outside the horizon, the Schwarzschild surface. One way to watch all the history of the Universe would be to stay just above the horizon, firing your rockets like crazy just to stay put. The Universe would then appear not only speeded up, but also highly blueshifted (probably roasting you in gamma rays), and concentrated in a tiny piece of the sky just above you.

Personally if I had all that rocket power available, I wouldn't use it to hover just above the horizon of a black hole. I'd rather take a trip around the Universe. Thanks to special relativistic time dilation, I could travel vast distances in a modest time, at superluminal speeds - apparently faster than the speed of light.
The actuality is a bit more complicated. Note that the metric expression given by Cjames

$${t_o=t_f}$$$$\sqrt{1 - r_o/r}$$

only applies to stationary observers, not infalling observers. So, if you hover at the right distance away from a black hole, you can observe the universe age rapidly. But the same does not happen when you fall into a black hole, because you have to include the effects of your velocity (the relativistic doppler shift) as well as the gravitational time dilation.

Exactly what an infalling observer sees depends on their energy. The simplest case is when the infalling observer is in a free-fall from infniity, which is E=1. Such an infalling observer will actually see radially infalling light as redshifted (by a factor of 2:1) not blueshifted. This means that he'd see the universe directly behind him going by in slow-motion. I haven't calculated exactly what happens at other angles other than directly straight backwards.

Other online references:

The sci.physics.faq http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html

Won't it take forever for you to fall in? Won't it take forever for the black hole to even form?

Not in any useful sense. The time I experience before I hit the event horizon, and even until I hit the singularity-- the "proper time" calculated by using Schwarzschild's metric on my worldline-- is finite. The same goes for the collapsing star; if I somehow stood on the surface of the star as it became a black hole, I would experience the star's demise in a finite time.

On my worldline as I fall into the black hole, it turns out that the Schwarzschild coordinate called t goes to infinity when I go through the event horizon. That doesn't correspond to anyone's proper time, though; it's just a coordinate called t. In fact, inside the event horizon, t is actually a spatial direction, and the future corresponds instead to decreasing r. It's only outside the black hole that t even points in a direction of increasing time. In any case, this doesn't indicate that I take forever to fall in, since the proper time involved is actually finite.

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#### DaveC426913

Gold Member
As the infalling observer heads toward the centre, if he looked out the back of shis ship he would see the universe age rapidly.
This is a common misconception. See for instance

Quiz question 5: As you fall freely into a black hole, you see the entire future of the Universe played out before your eyes. True or false?

Answer to the quiz question 5: False. You do NOT see all the future history of the world played out. Once inside the horizon, you are doomed to hit the singularity in a finite time,
I don't see how these are contradictory. I did not suggest the entire universe would play out, I merely suggested that the outside universe would appear to speed up.

I then went on to point out that the infalling observer would indeed fall to the centre as normal.

#### pervect

Staff Emeritus
The universe may not appear to speed up depending on the details of your trajectory. A detailed calculation of the doppler shift, for instance shows redshift (which would be interpreted as the universe slowing down) for sources directly behind you for a free-fall from infinity. So at least in this particlar case, the universe does not appear to "speed up" at all. One might well quibble over what (if anything) the term "the universe speeds up" actually means in terms of physical observations, I'm not aware of any more compelling interpretation of what this might mean than an observation of the doppler shift, which determines whether things appear to happen in "slow motion" (associated with a redshift), or "faster than normal" (associated with a blueshift). But the doppler shift is not necessarily uniform for the entire universe.

#### yuiop

I would like to add a couple of observations to the conversation.

First, Hawking's radiation is a very small effect. The amount of mass lost due to Hawking's radiation, by a typical small black hole formed from a collapsed star, is much less than the mass absorbed by a black hole due to absorbing microwave background radiation. The universe would have to cool much more before Hawking's radiation resulting in black holes evaporating. In theory, very small black holes could evaporate but none have been observed yet.

Second, there is way that a "frozen star" could absorb mass. Imagine a large solid body that is almost a black hole. If it came into contact with another similar body, the gravitational field at the contact interface would cancel out and there would be no gravitational time dilation at the interface. The two bodies would absorb each other pretty much in real time (from the point of view of a distant observer) forming a body with twice the original mass and no significant increase in radius. In other words it would now be a full blown black hole. This is an extreme example, but the principle would work with any massive falling body that is not an exact perfect hollow spherical shell around the frozen star/ black hole. Even infalling gas tends to fall as a accretion disk rather than as a sphere.

#### CJames

Some interesting responses.

DaveC426913,
I understand that for an observer falling into the black hole the trip would be finite. My point is that from an external reference frame the collapse would take infinite time, while hawking radiation would cause the black hole to evaporate in finite time. In that case, why would the black hole ever form?

Pervect,
I hadn't thought of that. I don't know much about the math of general relativity, but I know enough to realize that a body falling uninhibited toward a gravitational body is actually in an inertial reference frame, NOT an accelerating reference frame. In that case, I suppose gravitational time dilation has no effect. Does this mean an external observer would only see a time delay if a body were held static outside the event horizon? Is there any visible time dilation for a body falling uninhibited into a black hole (besides time dilation caused by difference in velocity, of course)?

#### George Jones

Staff Emeritus
Gold Member
I understand that for an observer falling into the black hole the trip would be finite. My point is that from an external reference frame the collapse would take infinite time, while hawking radiation would cause the black hole to evaporate in finite time. In that case, why would the black hole ever form?
But, as you say, an observer can fall into a black hole in finite time. This means that an observer can find himself inside the event horizon before the black hole evaporates. This is the standard view.

A different opinion appears in a paper by Tanmay Vachaspati, Dejan Stojkovic, Lawrence M. Krauss,

http://www.arxiv.org/abs/gr-qc/0609024 (Accepted for publication in Phys. Rev. D.)

which generated a bit of controversy a few months ago.

Their FIG. 8 is the standard spacetime for an evaporating black hole. Note there is an event horizon, and once across this horizon, an observer cannot avoid hitting the spacelike singularity.

FIG. 9 shows their highly speculative opinion. Note that there is no event horizon and no singularity. This is the main (and controversial) point of their paper: *no* observer (hovering, freely falling, blasting away with a rocket *towards* the surface of the collapsing object, etc.) experiences an event horizon because there is no event horizon to experience.

From the paper: “Instead it may happen that the true event horizon never forms in a gravitational collapse ... The infalling observer never crosses an event horizon, not because it takes an infinite time, but because there is no event horizon to cross. As the infalling observer gets closer to the collapsing wall, the wall shrinks due to radiation back-reaction, evaporating before an event horizon can form. The evaporation appears mysterious to the infalling observer since his detectors don’t register any emission from the collapsing wall Yet he reconciles the absence of the evaporation as being due to a limitation of the frequency range of his detectors. Both he and the asymptotic observer would then agree that the spacetime diagram for an evaporating black hole is as shown in Fig. 9. In this picture a global event horizon and singularity never form. A trapped surface (from within which light cannot es cape) may exist temporarily, but after all of the mass is radiated, the trapped surface disappears and light gets released to infinity.”

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#### CJames

George,
Thank you very much for your response. This speculative paper sounds exactly like what I was talking about, although I hadn't spelled out the scenario for the infalling observer explicitly. According to them, the black hole does indeed evaporate before a singularity forms. It's exciting to know that intuition could lead me to the same result as somebody with the caliber of Lawrence M. Krauss and his colleagues (and published so recently too). I'm going to check out that paper right away (and see if I can comprehend it).

Pervect,
Despite this response, I'm still curious about time dilation for a free falling body. I can't find any info about it. The more I think about it, the less it makes sense there should be any gravitational time dilation for a falling body. The equivalence principle doesn't seem to apply at all! I find this very confusing because now it seems like matter falling into a black hole would do so in essentially the same amount of time regardless of the position of the observer, with some minor difference due to differing velocities (and possibly some gravitational time dilation, where friction between the infalling matter would cause it to accelerate "away" from the black hole). But that contradicts everything I've ever read on the subject!

#### Chris Hillman

Hi, CJames,

The suggestion by Krauss et al. (I would say this is extremely controversial, BTW) really has nothing to do with the elementary misconceptions mentioned by pervect. The best thing is to study a good gtr textbook such as D'Inverno and then to try to debunk your own post. If you have trouble, I have debunked similar misconceptions many times; see the list of posts in my sig for some older examples. The discussion in my posts of the physical experience of various families of observers (including Lemaitre observers who fall in freely and radially "from rest at $r=\infinity$" and Novikov observers who fall in freely and radially from rest at $r=r_0, \; r_0 > 2 m$) will be much easier to follow if you are already familiar with pictures of e.g. ingoing Eddington chart!

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#### pervect

Staff Emeritus
CJames:

The amount of proper time it's going to take for a body to fall into a black hole will depend on its trajectory.

The simplest case to compute is when an observer free-falls into a black hole from at rest at infinity. One can express the equation of this trajectory as

$$r = (9 m \tau^2 / 2) ^\frac{1}{3}$$

(This is from some worksheets I have on my computer, not directly from a textbook, so a typo could have snuck in somewhere).

Here $\tau$ is proper time, and r is the Schwarzschild radial coordinate. This isn't the full expression, you'd also need t, the Schwarzschild coordinate time, as a function of $\tau$ to get something complete enough to solve problems with.

You'll typically find a treatment detailed enough of such "falling in" problems to be useful in a GR textbook. (I use MTW, but I wouldn't recommend it for someone new to GR. "Exploring Black Holes" by the same author might be a good choice (at the link above some sample chapters are available online).

I also hear good things about Schutz and D'Inverno, and online there is Sean Caroll's lecture notes but Caroll is aimed at the graduate level reader because it requires tensors.

As far as working out what a falling observer sees, first you have to specify what you're interested in. The easiest thing to specify, compute, measure, and communicate is IMO the redshift factor.

Redshift is also equivalent to time dilation. "Gravitational redshift" and "Gravitational time dilation" are essentially the same.

For instance we can imagine sending a probe into a black hole, emitting a signal with a constant frequency. We know that as the probe falls into the black hole that we will receive a redshifted signal. We can plot the amount of redshift as a function of time, and if the probe also transmits its proper time, we can plot the redshift as a function of proper time.

We can do the reverse, as well, where we detect the incoming red or blueshift that an infalling probe recieves from some outside, stationary, observer.

One often sees people interested in the question "where is the infalling probe "now"". For instance, they want to know if the probe has crossed the horizon "now" yet, or not. The best answer to this question is the same as it was in special relativity - there is no universal notion of "now" - the question is ambiguous. It may be slightly annoying to attempt to think of everything in terms of the raw data that one will actually receive (such as curves of redshift vs time), but this is really the safest course. Thinking of things in terms of "where the probe is now" will inevitably lead to confusion, because there is no universal definition of what "now" means, different observers will regard different points as being simultaneous even in SR, and this does not change in GR.

The solution I suggest is to simply ignore the notion of now - to formulate the particular problem one is interested in in purely geometrical terms and stud9iously avoid even introducing the notion of "now".

The question may arise - how does one handle causality without a "now"? The answer is the same as in SR - via light cones. The question of whether A can cause B reduces to the question of whether the light cone of signals emitted from A includes B.

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#### DaveC426913

Gold Member
Right so the upshot is that the external observer will see in falling objects slow and stop, but that does not mean they do not reach the centre - they do.

The key here, I think, is that gravity is not "trapped" (like photons are) by the indefatigable maw of the BH. Gravity, in SR (which is the theory that allows BHs) is not made of particle-like gravitons; it is a bend in space-time, and thus is free to affect the space around the BH.

#### yuiop

http://www.mathpages.com/rr/s6-05/6-05.htm

It describes the proper time experienced by a falling object in terms of the time, distance and horizontal velocity as measured by a distance observer. If the ignore the horizontal components (orbital motion) the first equation simplifies to:

$$(d \tau)^2 = (1-\frac{2 G M}{r c^2}) (dT)^2 - \frac{1}{(1-\frac{2 G M}{r c^2})}(dr)^2$$

where dT and dr are the infinitesimal time and radial (falling) distance respectively, as measured by a distant observer and $$d\tau$$ is the proper time of the falling object. The distant observer is defined as an inertial observer sufficiently far away from the gravitational body that gravitational effects on that observer may be considered insignificant.

The proper time experienced by light can be considered to be zero. For a falling photon we can set $$d\tau$$ to zero. Simple algebra then gives the result:

$$\frac{dr}{dT}=(1-\frac{2 G M}{R c^2})$$

This implies that the vertical speed of light near a gravitational body, as measured by a distant observer, is slower by the square of the gravitational redshift factor. If we assume that local stationary clocks are slower by $$\sqrt{1-\frac{2 G M}{R c^2}}$$ it can immediately be deduced that vertical rulers length contract by the same factor $$\sqrt{1-\frac{2 G M}{R c^2}}$$ if the speed of light is measured as c by a local observer.

A similar manipulation of the first equation in the linked text also reveals that the horizontal speed of light as measured by the distant observer is

$$\frac{dr}{dT}=\sqrt{1-\frac{2 G M}{R c^2}}$$

hmmmmm….

#### CBLeffert

A 4-D black hole does not collapse and it does not evaporate. Spatial condensation continues on the mass inside to continue its growth. As mass slips over the edge to the inside, all motion inside that mass ceases and so does its time, but then quickly recovers as gravity approaches zero at the center.

#### Drakkith

Staff Emeritus
2018 Award
A 4-D black hole does not collapse and it does not evaporate. Spatial condensation continues on the mass inside to continue its growth. As mass slips over the edge to the inside, all motion inside that mass ceases and so does its time, but then quickly recovers as gravity approaches zero at the center.
Why would gravity approach zero at the center?

#### CBLeffert

Newton's iron-sphere theorem which is true for any spherically symmetric mass.

#### CBLeffert

Newton's iron sphere theorem which is true for any spherically symmetric Mass.

#### Passionflower

I don't see how these are contradictory. I did not suggest the entire universe would play out, I merely suggested that the outside universe would appear to speed up.
And you are correct, for a radially free falling observer the doppler factor at the EH is:

$$\left( 1+\sqrt {{\it rs}} \right) ^{-1}$$

Where rs is the Schwarzschild radius.

#### Drakkith

Staff Emeritus
2018 Award
Newton's iron-sphere theorem which is true for any spherically symmetric mass.
The net force (directional pull) is zero inside the sphere, but you ARE still within a gravitational field and would still experience normal time dilation.

#### agmateo

Does times stop in the black hole? Is time infinity equal to time stops the same.

#### Drakkith

Staff Emeritus
2018 Award
Does times stop in the black hole? Is time infinity equal to time stops the same.
No idea.

#### agmateo

does time stops at the black hole? from the observer outside the horizon. He see the object falling in the event horizon as being converted to velocity of light and time become infinity right. Is time equal infinity equal to time stop in the common interpretation.

#### Drakkith

Staff Emeritus
2018 Award
does time stops at the black hole? from the observer outside the horizon. He see the object falling in the event horizon as being converted to velocity of light and time become infinity right. Is time equal infinity equal to time stop in the common interpretation.
Time equal infinity equal to time stop? I don't know what that is supposed to mean.

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