What is the Relationship Between Proper Time and Coordinate Time in Black Holes?

[Naty1]

"But I question whether or not an event horizon, or even a singularity, is ever created."

Well if you mean " Is there incontrovertible proof that black holes and singularities inside them exist", the answer is "no"...

but most physicts believe that to be true based on theory and indirect experimental observations.

As others have noted, the existence/observation of an event horizon is relative...analogous to "who is moving, you or me?"...

Kip Thorne's BLACK HOLES AND TIME WARPS book provides detailed explanations of event horizons...a few excerpts that might help the poster:

General relativity insists that, if one falls into a black hole, one will encounter nothing at the event horizon except (continued) spacetime curvature.

And the black holes birth...as seen from outside, the implosion slowed and became frozen (in time) at the critical circumference, but as seen by someone on the stars surface (riding the implosion inward) the implosion did not freeze at all...The stars surface shrank right through the critical circumference and on inward...

..That the implosion freezes forever as measured in the statric external frame but continues rapidly on past the freezing point as measured in the frame of the stars surface was extremely hard to comprehend...

 

[skeptic2]

George,

Not being a physicist I didn't recognize your diagram. Please accept my apologies.

 

[rjbeery]

Well I didn't mean to make this a contentious thread. I'm not defending some radical idea that black holes don't exist. However, most Relativity-related responses PRESUME that the black hole already exists in order for there to be an applicable frame in which crossing the event horizon is possible.

Again, I don't disagree with the math. For the most part I think I understand GR theory as it relates to black holes, and my problem is pretty simple:

If a Physicist believes that a singularity exists somewhere out in deep space today, then when did the singularity form? This is not an attack on science or something. I am looking at the math involved, and to me it appears that in our frame of reference the singularity forms after an infinite amount of time. In other words, in my mind there is a logical disconnect between accepting the Schwarzschild metric and also claiming that singularities exist today.

I was hoping someone would help me understand why this is not the case but this has not happened yet. (And this may be my fault in not comprehending the answer so please try again!)

 

[malawi_glenn]

If a Physicist believes that a singularity exists somewhere out in deep space today, then when did the singularity form? This is not an attack on science or something. I am looking at the math involved, and to me it appears that in our frame of reference the singularity forms after an infinite amount of time. In other words, in my mind there is a logical disconnect between accepting the Schwarzschild metric and also claiming that singularities exist today.

This "when" in observer dependent..

we can measure the effect of BH's..

 

[ZikZak]

If a Physicist believes that a singularity exists somewhere out in deep space today, then when did the singularity form?

Your question is ill posed. There is no absolute "when" for anything in Relativity. Observers disagree on which set of events constitutes which moment in time. That is why it is called "spacetime."

Physicists believe that singularities exist NOT "somewhere out in deep space," but rather "somewhere out in deep spacetime." The particular locations in spacetime are inaccessible unless you are willing to pass through the event horizon.

But to answer the ill-posed question: not at any time given on your watch here on Earth. But that fact is irrelevant to its existence. The coordinate system containing your watch is separated by an enormous amount of curvature from it. You are insisting on being told the longitude of the north pole, and claiming that since we cannot provide it, it cannot exist anywhere.

 

[skeptic2]

This "when" in observer dependent..

I believe the observer rjbeery is interested in is us.

we can measure the effect of BH's..

Being able to measure the effect of an object too massive to be a neutron star may not necessarily mean it contains a singularity. I understand the basis for the belief that singularities exist is that after gravity overpowers electron degeneracy pressure and neutron degeneracy pressure there is nothing left to prevent a collapsing star from collapsing all the way to a singularity. However if nothing can cross the event horizon in finite time, doesn't the event horizon itself prevent the collapse to singularity?

 

[malawi_glenn]

I believe the observer rjbeery is interested in is us.

well I don't care what you believe, we need to be rigour here.

However if nothing can cross the event horizon in finite time, doesn't the event horizon itself prevent the collapse to singularity?

Stop claiming that! An observer free falling towards the horizon will pass it and reach the singularity within finite time according to his clock.

It is clear that both you and the OP have not done proper classes in relativity, simultaneity etc are subjective entities.

 

[rjbeery]

This "when" in observer dependent..

we can measure the effect of BH's..

I provided the observer frame from which to devise an answer...

and to me it appears that in our frame of reference the singularity forms after an infinite amount of time.

Also, by you stating that we can measure the effect of BH's (from OUR frame of reference) you seem to be defending the idea they fully exist (from OUR frame of reference). Is this what you believe?

singularities exist NOT "somewhere out in deep space," but rather "somewhere out in deep spacetime."

Yes, I agree with this. Why do I feel like some people are arguing the same points from different sides? ZikZak, do you agree that a person falling through an event horizon will calculate distant clocks as moving quickly towards infinity?

 

[skeptic2]

Stop claiming that! An observer free falling towards the horizon will pass it and reach the singularity within finite time according to his clock.

The frame of reference in this discussion is not and never has been that of a free falling observer. It seems that every time the discussion brings up some problems associated with the time for passing through the event horizon being infinite (for a distant observer) the issue is dodged by switching the frame of reference to that of an infalling observer.

 

[malawi_glenn]

The frame of reference in this discussion is not and never has been that of a free falling observer. It seems that every time the discussion brings up some problems associated with the time for passing through the event horizon being infinite (for a distant observer) the issue is dodged by switching the frame of reference to that of an infalling observer.

The problem is the inaccurate language.

see ZikZak's post #35

 

[rjbeery]

Being able to measure the effect of an object too massive to be a neutron star may not necessarily mean it contains a singularity. I understand the basis for the belief that singularities exist is that after gravity overpowers electron degeneracy pressure and neutron degeneracy pressure there is nothing left to prevent a collapsing star from collapsing all the way to a singularity. However if nothing can cross the event horizon in finite time, doesn't the event horizon itself prevent the collapse to singularity?

Skeptic you understand fully. At the very "instant" that gravity overcomes the neutron degeneracy pressure, and the first particle begins its descent into a singularity, I am questioning whether or not it ever gets there. I believe it does not. This does not prevent other mass from falling into the "neutron-star-just-on-the-brink-of-collapse", and I believe that Newton's Shell Theorem ensures that we could not tell the difference!

 

[malawi_glenn]

Skeptic you understand fully. At the very "instant" that gravity overcomes the neutron degeneracy pressure, and the first particle begins its descent into a singularity, I am questioning whether or not it ever gets there. I believe it does not. This does not prevent other mass from falling into the "neutron-star-just-on-the-brink-of-collapse", and I believe that Newton's Shell Theorem ensures that we could not tell the difference!

according to a distant observer it will reach it in infinite time.

 

[xantox]

I conclude that I am naive, and I want someone to explain how we can so easily accept that black holes currently exist if their formation takes an infinite amount of time from our perspective.

SR and GR can give little to no meaning to the expression "currently exist" for distant bodies, since there is no absolute time, so the word "now" is only valid in the observer reference frame, and there is no "distant now". The black hole is formed in a finite time in the frame of the collapsing body, as it can be easily computed. This fact is enough for it to exist "now" for the collapsing body itself, and this is enough for it to exist at all.

But, as a mere trick to answer your question, we can use cosmological time (which in GR defines one arbitrary foliation of spacetime), for example within the FRW dust solution – and see whether, given a cosmological time T corresponding to some observer frame, a black hole can form elsewhere at the same cosmological time T. A black hole forms behind the finite null boundary of the horizon. It is certainly possible to identify a finite cosmological time T when such horizon forms. So, in this sense, we can say a black hole is forming eg "1 minute after the big-bang", or "right now" – and not necessarily in the infinite cosmological future. This is unrelated to the fact we cannot receive signals from it and thus observe its formation.

 

[rjbeery]

And it is certainly possible to identify a finite cosmological time T when such horizon appears in the frame of the collapsing body. So, in this sense, we can have a black hole forming, say, 1 minute after the big-bang, or "right now", and not in the infinite cosmological future.

Are you saying that, given any arbitrary frame (including ours) we can calculate some singularity formation at any arbitrary finite time? If this is true, are those arbitrary frames restricted to those that are local to the formation?

Here's a related question: does a body under acceleration such that it "hovers" over the event horizon calculate the same fate (i.e. becoming "frozen in time") for bodies falling through the event horizon as the inertial distant viewer does? I've only seen the analysis involving an observer at infinite distance so as to avoid the BH acceleration, and I'm wondering if the "infinitely distant" conclusions are equivalent to those of the "local but hovering" conclusions.

Thanks for helping me understand Xantox

 

[xantox]

Are you saying that, given any arbitrary frame (including ours) we can calculate some singularity formation at any arbitrary finite time? If this is true, are those arbitrary frames restricted to those that are local to the formation?

If for example we draw a conformal diagram of the entire spacetime, we can mark the beginning of the black hole horizon at any possible time and space location of the map – according to when and where the horizon forms. Since we also suppose that black holes evaporate, we would need to also draw a second, future mark, again at a finite time, to represent when and where the horizon disappears. In-between, the horizon would evolve dynamically as a result of absorption of radiation and matter and evaporation. Since this diagram represents the entire spacetime, any observer willing to describe the entire spacetime must agree on those same marks, under appropriate coordinate transformations.

Here's a related question: does a body under acceleration such that it "hovers" over the event horizon calculate the same fate (i.e. becoming "frozen in time") for bodies falling through the event horizon as the inertial distant viewer does? I've only seen the analysis involving an observer at infinite distance so as to avoid the BH acceleration, and I'm wondering if the "infinitely distant" conclusions are equivalent to those of the "local but hovering" conclusions.

The photons emitted by the hovering body must climb the black hole gravitational well in order to reach the distant observer in flat spacetime, so they will be gravitationally redshifted to very low frequencies (though still not zero frequency eg. no photons at all like if the body was already crossing the horizon). The distant observer would see clocks on the body run slower, but at a constant rate (while they would be seen running exponentially slower and slower if the body was in free fall, and completely stopping if it was crossing the horizon).

 

[tiny-tim]

singularity forms after an infinite amount of our-time

Hi rjbeery!

If a Physicist believes that a singularity exists somewhere out in deep space today, then when did the singularity form? … to me it appears that in our frame of reference the singularity forms after an infinite amount of time.

You mean, in our frame of reference, events "on" the event horizon are at our-time = infinity, sooo … how did they get there?

I suspect you're confused about the concept of "now" …

we don't say "there are events happening on the event horizon now" …

events on, or very near, the event horizon by definition happen after the collapse, so there's no logical problem with them being "in the wrong order".

And consider the event of a photon being emitted from the surface of the star just before the collapse (so there's no event horizon "yet") … it may take a million billion years to reach us, because of the intense gravity … that photon was emitted before the collapse, but at a million billion AD in our-time …

the collapsing star "drags" our-time to infinity as it collapses!

(and just as it "gets to infinity", the event horizon forms)

 

[rjbeery]

Since this diagram represents the entire spacetime, any observer willing to describe the entire spacetime must agree on those same marks, under appropriate coordinate transformations.

But isn't this saying that an event happening at t=infinity for one frame has a transform that reduces t to < infinity for some other frame? If this is what you're saying it contradicts my understanding of transforms.

The photons emitted by the hovering body must climb the black hole gravitational well in order to reach the distant observer in flat spacetime, so they will be gravitationally redshifted to very low frequencies (though still not zero frequency eg. no photons at all like if the body was already crossing the horizon). The distant observer would see clocks on the body run slower, but at a constant rate (while they would be seen running exponentially slower and slower if the body was in free fall, and completely stopping if it was crossing the horizon).

You misunderstood my question here. I was asking if the hovering body would calculate that the body free-falling through the event horizon would become frozen in time. I believe the answer is yes but I don't know because I've only seen the inertial analysis of an infinitely distant observer.

(and just as it "gets to infinity", the event horizon forms)

Hi tiny-tim! It's possible that some think I'm splitting hairs or playing semantic games, but my point is that t=infinity from our frame of reference is equivalent to never.

 

[tiny-tim]

singularity forms after an infinite amount of our-time

Hi tiny-tim! It's possible that some think I'm splitting hairs or playing semantic games, but my point is that t=infinity from our frame of reference is equivalent to never.

Hi rjbeery!

Well, we can never know that there's an event horizon …

however long we wait, we'll never have the evidence.

Imagine we're hovering at the photon sphere (r = 3M, compared with the non-rotating event horizon radius of 2M) …

things are pretty counter-intuitive there, but there's still no difference in the geometry there between a black hole and a neutron star with a surface just outside r = 2M.

Since the geometry is the same, the only way we can tell the difference (between a black hole and a neutron star) is by looking …

if we continue to see photons from the surface, with the same wavelength, then we know the surface is stationary … but if the photons get fewer, and have increasingly longer wavelength, then we can say that the surface is shrinking …

but we can never say that it has shrunk beyond r = 2M (and therefore disappeared), because we would have to wait infinitely long to confirm that!

So in that sense, yes, the event horizon never happens, and the black hole never forms …

nothing happens until it happens, and the surface passing through r = (2 + x)M doesn't happen until our t = f(x) (can't be bothered to check what f is ) (and we don't see it until about t = 2f(x)), so, for us, passing through r = 2M never happens.

 

[rjbeery]

tiny-tim: thanks for explaining. You've verified what I was postulating in https://www.physicsforums.com/showpost.php?p=2217145&postcount=41". It appears some people agree with this and some people reject it.

 

[malawi_glenn]

tiny-tim: thanks for explaining. You've verified what I was postulating in https://www.physicsforums.com/showpost.php?p=2217145&postcount=41". It appears some people agree with this and some people reject it.

I think the main point is the logic you draw up:

Premise: the BH horizon form at time infinity in our/earth frame of reference (in principle never)

Claim: BH-singularities exists

According to this, you say that the claim is contradictory to the premise, but one has not specified in the claim in what frame of reference BH-singularities exists. BH-singularities exists in one class of reference frames.

 

[rjbeery]

Premise: the BH horizon form at time infinity in our/earth frame of reference (in principle never)

Claim: BH-singularities exists

According to this, you say that the claim is contradictory to the premise, but one has not specified in the claim in what frame of reference BH-singularities exists. BH-singularities exists in one class of reference frames.

The claim would be contradictory to the premise if, for example, a process of finite time from the Earth's frame evaporated the potential black hole, such as "pre"-Hawking radiation. As tiny-tim pointed out, a photon might take "a million billion years" to make it out of the gravitational grip, but the time involved is irrelevant as long as it escapes eventually.

 

[Naty1]

Much of the confusion here stems from utilizing relative horizons rather than absolute horizons. Roger Penrose utilized relative horizons in his theorem that all black holes contain singularities; Stephen Hawking later realized there were severe limitations to such relative horizons (such as discontinuous jumps in size when matter is swallowed by the black hole, and their frame dependency) and embarked on work utilizing absolute horizons which formed the basis of his entropy work on black holes. Absolute horizons evolve continuously and are NOT frame dependent...The absolute horizon is created at a stars center well before the star shrinks through the critical circumference and expands to the critical circumference just as the star shrinks through that critical circumference.

The above capsulizes Kip Thorne's discussion of absolute and relative horizons in his book BLACK HOLES AND TIME WARPS, 1994, PAGES 412 TO 419.

 

[skeptic2]

Naty1,

Can you expand a little on the difference between relative and absolute horizons?

 

[xantox]

But isn't this saying that an event happening at t=infinity for one frame has a transform that reduces t to < infinity for some other frame?

Yes. In curved spacetime, different definitions of time may lead to dramatically different results. If we use the distant observer coordinate time t1, then the infalling body crosses the limit of the horizon at t1=infinity. If we use the infalling body wristwatch time t2, then it crosses the horizon at a finite (and extremely quick) t2. If we use a cosmological time parameter T then all observers will agree the infalling body crosses the horizon at some finite T, and the evolution of the black hole can be studied by mapping its mass to T. Hence, if an event horizon forms -now- in my room (which could be entirely possible), we shall consider it has formed at the 13.75th billion year of cosmological time.

You misunderstood my question here. I was asking if the hovering body would calculate that the body free-falling through the event horizon would become frozen in time. I believe the answer is yes but I don't know because I've only seen the inertial analysis of an infinitely distant observer.

Yes, a hovering observer ("shell observer") will also find that the infalling body is crossing the limit of the horizon at his infinite coordinate time. This just means that his worldline in spacetime never crosses signals from the horizon.

 

[skeptic2]

How is cosmological time different from local time in flat spacetime or from very slightly curved spacetime such as what we have on earth?

 

[rjbeery]

If we use a cosmological time parameter T then all observers will agree the infalling body crosses the horizon at some finite T, and the evolution of the black hole can be studied by mapping its mass to T. Hence, if an event horizon forms -now- in my room (which could be entirely possible), we shall consider it has formed at the 13.75th billion year of cosmological time.

I'm going to sound very ungrateful for your explanation but my intuition tells me that this is not true. I am obviously not infallible, and you speak with more authority than I do, but it doesn't "feel right" that a time tx = infinity can be transformed to a finite time ty. Nevertheless, this appears to be what you are saying because an observer residing an infinite distance from your room would continue to claim that the event horizon "never" forms while you are claiming that all observers would concur on an absolute time T for the singularity formation.

 

[atyy]

Just in case the event horizon formed in xantox's bedroom , you may try http://motls.blogspot.com/2008/11/why-can-anything-ever-fall-into-black.html.

 

[ZikZak]

it doesn't "feel right" that a time tx = infinity can be transformed to a finite time ty.

Consider a polar system of coordinates (u,\theta), where u=1/r. This system has an infinite value of u at the origin. Clearly there is a coordinate transformation that undoes the infinite coordinate at the origin and makes it finite. How is this any more difficult to understand?

 

[atyy]

Melia's section 1.4 has a discussion of when black holes form in terms of cosmic time http://arxiv.org/abs/0705.1537.

 

[xantox]

I'm going to sound very ungrateful for your explanation but my intuition tells me that this is not true. I am obviously not infallible, and you speak with more authority than I do, but it doesn't "feel right" that a time tx = infinity can be transformed to a finite time ty.

You may intuitively consider the case of your shadow. At noon if the sun was perfectly above you, the length of your shadow is zero, and at sunset it becomes longer and longer and ideally infinite in the limit. This does not mean that your body has zero or infinite length, but just that its projection onto perpendicular coordinates is not very meaningful at noon and at sunset. Similar and nastier things may happen when proper time is projected onto some coordinate system in curved spacetime and you describe the events in term of the "shadow time" projected on the coordinates. For the ultimate proof, this must be actually computed. Here is the standard way to compute it:

The Schwarzschild line element describing the geometry outside a static black hole is:

ds^2 = - \left( {1-{2M \over r}} \right) dt^2 + \left( {1 - {2M \over r}} \right) ^{-1} dr^2 + r^2 d \Omega ^2

where d\Omega^2 = d\theta^2 + sin^2\theta\phi^2 and (t, r, \theta, \phi) the Schwarzschild coordinates.

A body free-falling from the far distance takes a finite proper time \tau of about 0.3 milliseconds to go from r0=100 km to the horizon of a 10-solar masses black hole at r=29km:

\delta\tau = {2 \over 3} {1 \over \sqrt{2M} } \left[ r_0^{3/2} - r^{3/2} \right] = 0.000348 s

On the other side, if we express the infall in terms of coordinate time by means of the following differential equation, it can be seen that the same body takes infinite coordinate time t to reach the limit of the horizon r=2M.

{dt \over dr} = {dt \over d\tau} {d\tau \over dr} = - \sqrt {r \over 2M} \left( 1- 2M \over r \right) ^{-1}