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

[rjbeery]

No, as the sentence described the horizon crossing by an infalling body. For the body crossing the horizon, this did happen, and too bad for distant observers.

Now we're arguing the definition of "did"...I feel like Bill Clinton. The infalling beer can calculates that our "now" expired an infinite time ago, and the distant observer calculates that the beer can crosses the event horizon an infinite time from our "now". From both perspectives the frame from which we are discussing this topic happens an infinite amount of time before the beer can crosses - remember, this is the frame that I designated in my OP - therefore the beer can "did not" cross. "Will" it? Well, now we're back to opinion.

Cosmological time, where the isotropic coordinates of comoving observers are singled out. And in general, we can single out some dynamical parameter such as the radius of the universe, so that evolution can be expressed in terms of that parameter. Here it is possible to locate the black hole formation in terms of such parameter (even if the whole manifold cannot be covered in general). When drawing the horizon on a conformal diagram using as vertical time coordinate such a time, eg a primordial black hole horizon segment will appear to begin at the bottom of the diagram eg in the young universe region, and not on the top, where is the infinite future.

I confess I am not familiar with Cosmological Time and I am very curious about it. You wouldn't possibly want to expand on it, would you? Or give me a couple of references to research? It sounds to me like a simple preferred frame which would clearly not resolve this problem - the radius of the universe will be infinitely small (in a crunch) or large (heat death), presuming it does not reach an equilibrium, before the beer can crosses that damn line!

 

[rjbeery]

Now we're arguing the definition of "did"...I feel like Bill Clinton. The infalling beer can calculates that our "now" expired an infinite time ago, and the distant observer calculates that the beer can crosses the event horizon an infinite time from our "now". From both perspectives the frame from which we are discussing this topic happens an infinite amount of time before the beer can crosses - remember, this is the frame that I designated in my OP - therefore the beer can "did not" cross. "Will" it? Well, now we're back to opinion.

To hammer home my point, replace my OP with the following...

Q: Some people speculate that space elevators are technically feasible. Do space elevators currently exist?

A: You didn't specify a frame. From some perspectives (for example, from the perspective of the person living 500 years from now), YES, space elevators exist.

 

[xantox]

The infalling beer can calculates that our "now" expired an infinite time ago, and the distant observer calculates that the beer can crosses the event horizon an infinite time from our "now".

Which definition of time are you using here? It seems again proper time of the observer. As I tried to explain above, in general relativity the evolution of the observed system cannot be described in terms of it. Either coordinate time or cosmological time must be used.

Q: Some people speculate that space elevators are technically feasible. Do space elevators currently exist?
A: You didn't specify a frame. From some perspectives (for example, from the perspective of the person living 500 years from now), YES, space elevators exist

There are rumors of space elevators "currently" existing on Andromeda ("currently" in terms of cosmological time).

I confess I am not familiar with Cosmological Time and I am very curious about it. You wouldn't possibly want to expand on it, would you? Or give me a couple of references to research? It sounds to me like a simple preferred frame which would clearly not resolve this problem - the radius of the universe will be infinitely small (in a crunch) or large (heat death), presuming it does not reach an equilibrium, before the beer can crosses that damn line!

In short, there is a way to slice a FRW universe in spatially isotropic slices. The indexes of the slices represent cosmological time. This slicing starts at the big-bang. Approximating our universe to a FRW model at large scale, it is found that current slices correspond to the 13.7th billion year. We could then look at some slice T corresponding to eg 10 minutes after the big-bang. If the slice contains (black) holes, they will have formed earlier than T.

Cosmological time is covered in any cosmology textbook, usually in the chapter presenting the FRW metric. As a beautiful undergraduate introduction to general relativity also presenting some basic cosmology I would recommend the book by James Hartle, "Gravity" (Addison Wesley, 2003). For an article representative of research on black holes created in the early universe, try A. M. Green, A. R. Liddle, "Constraints on the density perturbation spectrum from primordial black holes", Phys. Rev. D 56, 6166-6174 (1997).

 

[tiny-tim]

nothing happens until it happens

Hi rjbeery!

… The infalling beer can calculates that our "now" expired an infinite time ago …

No, I'm not following that …

deosn't the beer can calculate that our "now" expired when our "now" equalled infinity, which was only a millisecond ago for the beer can?

From both perspectives the frame from which we are discussing this topic happens an infinite amount of time before the beer can crosses …

Why are you bothering with the perspective of the beer can?

(hmm … is that where you get your best ideas from? )​

 

[qraal]

b) Framework theory: canonical quantum gravity. Vachaspati et al. speculate that horizons may never form, since the collapsing mass could be radiated by a quantum mechanism (not Hawking radiation) before it gets too dense.

By the "pre-Hawking" radiation? Does that mean there is no end point to collapse, because it just radiates away when it gets too dense? Could it be a power-source at the far-end of the Main Sequence?

 

[rjbeery]

deosn't the beer can calculate that our "now" expired when our "now" equalled infinity, which was only a millisecond ago for the beer can?

Maybe I worded it poorly, but what I was saying is that the distant observer calculates that the beer can freezes in time AND that the beer can calculates that the distant observer's clock infinitely speeds up. In other words, both bodies agree that there is an infinite time differential (from our perspective, the one we are discussing) between the distant observer seeing the beer can on this side of the event horizon and the beer can actually crossing it. I was clarifying that there is no contradiction, no paradox, and imagining that we are the beer can does not make the crossing event "happen" any faster for us on Earth.

Why are you bothering with the perspective of the beer can?

(hmm … is that where you get your best ideas from? )

The perspective of the beer can seems to be the sole argument for those claiming that it ever makes it across, and I was pointing out that even from that perspective an infinite amount of time has gone by for the people discussing this issue on Earth today.

And I prefer Scotch, The Macallan. If I ever say something truly Cranky please note the time (US Central), for I may be under the influence...

 

[George Jones]

AND that the beer can calculates that the distant observer's clock infinitely speeds up.

This isn't true.

Suppose observer A hovers at a large distance from a Schwarzschild black hole, and that observer B falls from rest from the same position. If observer B uses a telescope to observe A's watch, B will see A's watch continually slow down relative to his own watch. At the event horizon, B will see A's watch running at the rate of his own watch. For the math, see

https://www.physicsforums.com/showthread.php?p=861282#post861282

and the correction in post #7 of the same thread.

 

[rjbeery]

Suppose observer A hovers at a large distance from a Schwarzschild black hole, and that observer B falls from rest from the same position. If observer B uses a telescope to observe A's watch, B will see A's watch continually slow down relative to his own watch. At the event horizon, B will see A's watch running at the rate of his own watch.

Wait a minute. I want to understand this but I'm currently working and I don't have time to analyze your reference. Do we consider free-falling into the black hole and standing on the collapsing neutron star's surface as two different things which have different experiences? To me they are the same thing but maybe I'm mistaken because the body on the neutron star's surface is not "weightless". As I type this I think I've resolved the problem in my head, and my post should've read...

Maybe I worded it poorly, but what I was saying is that the distant observer calculates that the beer can freezes in time AND that the beer can, sitting on the neutron star's surface as the black hole forms, calculates that the distant observer's clock infinitely speeds up.

Do you agree with this statement, George?

 

[George Jones]

Do you agree with this statement, George?

No, an observer on the surface of the collapsing star will see either a redshift or a blueshift even on and inside the event horizon, depending on the speed of the collapse, but the shift will always be finite.

 

[rjbeery]

I don't think that's right, George. How could the local observer ever experience redshift? The ground is preventing his free fall, it isn't being "pulled out from under him". The observer on the surface would be experiencing incredible acceleration as the star radius approached the Schwarzschild radius and, analogous to the distant observer seeing the local one being redshifted into nothingness, I believe the local one would see the outside world blueshifted towards infinity, wouldn't it?

 

[George Jones]

I don't think that's right, George. How could the local observer ever experience redshift? The ground is preventing his free fall, it isn't being "pulled out from under him". The observer on the surface would be experiencing incredible acceleration as the star radius approached the Schwarzschild radius and, analogous to the distant observer seeing the local one being redshifted into nothingness, I believe the local one would see the outside world blueshifted towards infinity, wouldn't it?

No.

Consider freely falling observer B (that is about to splat on the surface) coincident with an observer C that is on the surface of the collapsing star. Just before splat, B is moving towards C with some local speed that is strictly less than the speed of light, and, consequently, there is a finite (Doppler) time shift between B and C. In my previous post, I pointed that B sees a finite redshift of the light emitted by A, an observer who hovers far from the collapsing star. The composition of two finite shifts is always finite, i.e., is never infinite.

If the relative speed between B and C is small, then C will see the light emitted by A to be redshifted by a somewhat smaller amount than does B. If the relative velocity between B and C is large, then C will see the light emitted by A to be blushifted by a finite amount.

 

[rjbeery]

George: thanks for explaining. I have more questions though:

strictly less than the speed of light

Doesn't this presume that B and C (and the surface) are all above the horizon? I thought that from A's perspective the velocity of the free falling beer can is c at the horizon, and dealing with mass traveling at c is the main source of all of these strange behaviors and peculiar explanations.

If the relative speed between B and C is small, then C will see the light emitted by A to be redshifted by a somewhat smaller amount than does B.

Whether B exists as an intermediary or not I do not believe that C will ever see redshifting of A's light. I'm using intuition here so I am obviously liable to be proven wrong mathematically.

 

[rjbeery]

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}

Xantox: I have a degree in Comp Sci and took Calc and Diff EQ but that was many moons ago. Could you hold my hand a bit in understanding the transition from the Schwarzschild metric to calculating the coordinate times for the local vs the distant (Minkowski metric, right?) frames? I'm back in school for Physics but it would be nice to get a good grasp on this before classes start (on Monday!)

 

[xantox]

Could you hold my hand a bit in understanding the transition from the Schwarzschild metric to calculating the coordinate times for the local vs the distant (Minkowski metric, right?) frames?

The first part is to understand that finite proper time intervals for the infalling body are equivalent to infinite coordinate time intervals for the distant observer. We may proceed by expressing the Schwarzschild metric in timelike form:

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

At fixed r and \Omega, dr=d\Omega=0, and the relation between proper time and Schwarzschild time is:

{d\tau \over dt} = \sqrt{{1-{2M \over r}}}

This shows that at infinite r, proper time intervals are equal to coordinate time intervals. The Schwarzschild metric is minkowskian and flat at infinity, ie Schwarzschild time has a special relativistic meaning in that limit. But this also shows that when the infalling particle approaches the horizon at r=2M, its wristwatch runs infinitely fast when expressed in Schwarzschild time coordinate units. This is the consequence of the curvature of the time dimension – signals emitted near the horizon at regular intervals get delayed longer and longer when they are projected onto curved away regions. Thus we can equivalently say that in this case the particle is both crossing a finite proper time and an infinite coordinate time.

The second part is to understand that the proper time elapsed on the free-falling particle worldline to reach the horizon from a finite distance is indeed finite. For this it is needed to determine its equation of motion, solve it and integrate on the worldline. You may find the full derivation in any general relativity textbook, see for example in Frolov and Novikov here.