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

[tiny-tim]

nothing happens until it happens

We know that the infalling observer crosses the horizon in finite time simply because the theory predicts it, ie because we can calculate it.

Exactly … predicts!

We can only predict that there will be a horizon! ​

 

[xantox]

Exactly … predicts!

We can only predict that there will be a horizon! ​

We can also postdict horizon formations in the distant past (eg primordial black holes, etc).

 

[rjbeery]

The very speculative proposal by Vachaspati et al. suggests that an horizon never forms not because it takes infinite time, but because some quantum effect would supposedly radiate away the mass quickly before the horizon forms – this is different from your argumentation in this thread, which seems based on the relation between infinite coordinate time and finite proper time – this relation is uncontroversial within general relativity.

Vachaspati's proposal is equivalent to mine because, as you say, quantum effects radiate the mass quickly before the horizon forms, but "quickly" is relative to the free-falling body's frame. In fact, the further one is from the apparent horizon, the more slowly this evaporation seems to occur, yet it always seems to occur before anybody is able to reach it (according to their calculations). Quickly or not, I've never seen a proposal suggesting that the evaporative process is infinite from any frame, and if that's true then the two arguments are equivalent in my opinion.

Let's make it simpler...the distant frame calculates the event horizon to be formed at t=infinity, while complete evaporation of the black hole due to Hawking radiation is calculated to be t = 5120*pi*G^2*M^3/(hc^4).
Isn't that sufficient to be able to say that the free falling body will be destroyed before it reaches the event horizon? Surely the ordering of local events cannot change (regardless of what kind of axes motls uses in his graph )

 

[xantox]

Vachaspati's proposal is equivalent to mine because, as you say, quantum effects radiate the mass quickly before the horizon forms, but "quickly" is relative to the free-falling body's frame.

To say that black hole horizons are not forming, some explanation is required. Vachaspati's speculative and controversial explanation is that they do not form because of some physical mechanism able to radiate away the mass during the extremely short time the black hole would take to collapse. Your explanation since post #1 seems to be another one, ie you seem to consider that the reason they do not form is because distant observers always register them forming in their infinite time and that you consider this to be inconsistent with a formation in finite time in the infalling frame.

Let's make it simpler...the distant frame calculates the event horizon to be formed at t=infinity, while complete evaporation of the black hole due to Hawking radiation is calculated to be t = 5120*pi*G^2*M^3/(hc^4).
Isn't that sufficient to be able to say that the free falling body will be destroyed before it reaches the event horizon? Surely the ordering of local events cannot change (regardless of what kind of axes motls uses in his graph )

When Hawking radiation is included in the picture, the distant observer does no longer calculate the event horizon to be forming at infinite coordinate time, but instead at a finite time corresponding with the end of the evaporation process. Nonetheless, the infalling body still had entered the black hole quickly, at the beginning of the incredibly long time it would take for an astronomical black hole to Hawking-evaporate.

 

[pesto]

Sorry to interject so late, and please let me know if my questions are unwelcome, but did I read correctly that an outside observer would not be able to see the creation of a black hole, but an observer inside the gravitational field would? This would be essentially the same situation as watching an object enter a black hole, and being the object?

Thank you.

 

[rjbeery]

during the 0.3 milliseconds a 10-solar mass black hole would take to collapse.

But this calculated .3 milliseconds is local to the formation! Why do you seem to assign such an absolute value to this time calculation when from most frames the collapse takes "forever"?

When Hawking radiation is included in the picture, the distant observer does no longer calculate the event horizon to be formed at infinite coordinate time, but instead precisely when the evaporation is completed.

Fair enough, but we both agree that relativity does not allow for a reordering of local events due to a transition in frames, correct? If the collapse happens after the evaporation is complete (or precisely as it completes) as calculated from the distant frame, then how can the local frame claim that the horizon was ever available to be crossed?

Sorry to interject so late, and please let me know if my questions are unwelcome, but did I read correctly that an outside observer would not be able to see the creation of a black hole, but an observer inside the gravitational field would? This would be essentially the same situation as watching an object enter a black hole, and being the object?

The outside observer (sitting at "an infinite distance" because it is mathematically simpler if he cannot feel any gravitational pull from the black hole) sees, or equivalently calculates, that the black hole never forms. Any body (someone called it a beer can) approaching the area where the event horizon supposedly resides would slow to a stop and dim into nothingness. The distant observer would calculate that the beer can will cross the event horizon at a time = infinity. Yet the beer can, from its perspective according to relativity, would cross just over just fine and quite quickly. I claim that the beer can's perspective is irrelevant because before "infinity" occurs other things happen such as the beer can and whole ball of wax evaporating into space over many billions of years (from the distant observer's perspective). Who knows what the beer can would actually experience? I'm sure it wouldn't be pleasant regardless...

 

[xantox]

But this calculated .3 milliseconds is local to the formation! Why do you seem to assign such an absolute value to this time calculation when from most frames the collapse takes "forever"?

Because the physical mechanism proposed by Vachaspati et al. must also act local to the formation, in order to radiate the local mass before it's locally too late and the horizon locally forms.

Fair enough, but we both agree that relativity does not allow for a reordering of local events due to a transition in frames, correct? If the collapse happens after the evaporation is complete (or precisely as it completes) as calculated from the distant frame, then how can the local frame claim that the horizon was ever available to be crossed?

Note that things happening in some place only happen in that place and nowhere else, and they do not ask permission to distant observers in order to happen. If an horizon was crossed there, then it was crossed – distant observers just receive or not receive some signals such as light rays carrying an information about that event. If the spacetime is too curved those light rays will get slowed down, and eventually remain trapped, then too bad for the distant observer who will get no photograph of the events, but in no way this means the events did not happen. Events are not getting reordered, as the evaporation happens behind the event horizon so it is less redshifted than signals at the horizon which can only get released when the horizon is no more.

Yet the beer can, from its perspective according to relativity, would cross just over just fine and quite quickly. I claim that the beer can's perspective is irrelevant because before "infinity" occurs other things happen such as the beer can and whole ball of wax evaporating into space over many billions of years (from the distant observer's perspective). Who knows what the beer can would actually experience? I'm sure it wouldn't be pleasant regardless...

The infalling body perspective is not irrelevant, to the contrary it's the only relevant one if we want to talk about what happens to the infalling body. You seem to suggest that during the few milliseconds of the body infall, by some trick infinite time has passed and these few milliseconds are just an illusion, but this is certainly not the case. What a distant observer calls "time" is not what the infalling observer calls "time", even if they share the same name – it is not possible to compare them more than it is possible to compare apples with oranges. What matters is that the free-falling observer has a watch, which is as good as any other in measuring proper time on his worldine, and under that worldline the proper time of the horizon crossing is short.

 

[rjbeery]

xantox: To me, if a black hole completes its evaporation precisely as the event horizon forms then at its final evaporative moment it is but a point, and the supposed event horizon is also a mere singularity which never came to be and therefore offers nothing to cross...

Anyway I've enjoyed this discussion, xantox, thank-you, this is how I learn. It seems we disagree on nothing but the interpretation of the facts and I think it's clear that our differences have boiled down to subjective opinion.

 

[qraal]

Just a thought, but if the horizon nevers forms there can be no Hawking radiation either. One follows from the other.

 

[xantox]

Just a thought, but if the horizon nevers forms there can be no Hawking radiation either. One follows from the other.

Correct. The two entirely different scenario commented above are:
a) Framework theory: classical general relativity. Horizons form at any cosmological time. They evolve. They can collide and merge. There are primordial black holes, and (in the semiclassical formalism) they eventually disappear by Hawking evaporation so there can be also young and old black holes, and some may form at cosmological times which are subsequent to the disappearance of prior ones. In case they form by gravitational collapse of a dense body, they form in a short proper time.
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.

 

[xantox]

xantox: To me, if a black hole completes its evaporation precisely as the event horizon forms then at its final evaporative moment it is but a point, and the supposed event horizon is also a mere singularity which never came to be and therefore offers nothing to cross...

Anyway I've enjoyed this discussion, xantox, thank-you, this is how I learn. It seems we disagree on nothing but the interpretation of the facts and I think it's clear that our differences have boiled down to subjective opinion.

In fact I failed to precisely grasp your argumentation. It seemed to me to be at moments an epistemological statement (such as "if a black hole doesn't form in my own frame then I may consider it doesn't exist", similar to "if nobody looks at the moon then the moon doesn't exist"), at moments a discussion within the general relativistic theory of black holes where you seemed to consider by comparing times measured on different frames that its conclusions about the formation of horizons are inconsistent and that horizons cannot form (this is however wrong), and at moments a statement of support of Vachaspati theory (where horizons cannot form for other reasons than relativistic time dilation and spacetime curvature). It is always nice to finish a discussion upon scientific agreement, however unless the above is clarified I cannot comment further.

 

[tiny-tim]

nothing happens until it happens

… If the spacetime is too curved those light rays will get slowed down, and eventually remain trapped, then too bad for the distant observer who will get no photograph of the events, but in no way this means the events did not happen. …

Yes it does …

in no way does this mean the events will not happen …

(though perhaps the most accurate way of describing it is to say: "the events will happen … never!")

but they certainly did not happen, in any sense of the word "did"!

Who knows what the beer can would actually experience? I'm sure it wouldn't be pleasant regardless...

From The Hitch-hiker's Guide to the Galaxy …

It would be unpleasantly like being drunk! ​

Just a thought, but if the horizon nevers forms there can be no Hawking radiation either. One follows from the other.

But wouldn't we see the same amount of radiation from the strong gravitational field outside an almost-black-hole? …

the "inward" half of the radiation, which with a black hole falls through the event horizon to vanish, will instead fall onto the surface of the almost-black-hole and vanish!

There would be no way to tell the difference.

 

[rjbeery]

This is true, I was fudging the math when I assumed that "pre"-Hawking radiation would occur at the same rate as Hawking radiation because I simply don't know that, and then I began referring to both generically as radiation or evaporation. This isn't a foundation of my point though, and my intuition (as well as tiny-tim's) is that the evaporation rate would be the same between a certified black hole and an "almost" black hole.

Xantox and I will have to agree to disagree because my interpretation is that t=infinity from our perspective means that black holes do not exist from our perspective. Is this different from saying that they don't exist at all? I don't know, but this thread has evolved and my ORIGINAL question was...

I'm pretty sure the SR time dilation math shows that the "outside world" clocks move to infinity during a black hole's formation, yet we seem to readily postulate that black holes currently exist...

Can someone that understands black holes please explain?

This question has been answered to my satisfaction, and the answer is that they do NOT currently exist in our frame. The rest is semantics and opinion.

 

[sloughter]

How do we know if a neutron star under collapse (going from closest packing configuration, being crushed, going to even more dense state), isn't capable of producing an event horizon where no light escapes? In other words, there is no need for a singularity.

To say that black hole horizons are not forming, some explanation is required. Vachaspati's speculative and controversial explanation is that they do not form because of some physical mechanism able to radiate away the mass during the extremely short time the black hole would take to collapse. Your explanation since post #1 seems to be another one, ie you seem to consider that the reason they do not form is because distant observers always register them forming in their infinite time and that you consider this to be inconsistent with a formation in finite time in the infalling frame.


When Hawking radiation is included in the picture, the distant observer does no longer calculate the event horizon to be forming at infinite coordinate time, but instead at a finite time corresponding with the end of the evaporation process. Nonetheless, the infalling body still had entered the black hole quickly, at the beginning of the incredibly long time it would take for an astronomical black hole to Hawking-evaporate.

 

[DrGreg]

...and the answer is that they do NOT currently exist in our frame.

That pre-supposes we all share the same frame. We don't. For those of us who choose to hover at a fixed distance above a collapsing black hole, we won't see it form. For those of us who choose to drop into the hole, we will see it form.

 

[George Jones]

How do we know if a neutron star under collapse (going from closest packing configuration, being crushed, going to even more dense state), isn't capable of producing an event horizon where no light escapes? In other words, there is no need for a singularity.

The Penrose-Hawking singularity theorems show that, under reasonable hypotheses, if there is a trapped surface, as there is inside an event horizon, then spacetime is singular. If you think spacetime isn't singular in this case, you need to show which of the hypotheses of the Penrose Hawking singularity theorems are violated.

 

[skeptic2]

I think the argument in this thread is not that spacetime is not singular inside the event horizon but that there is no inside to the event horizon.

 

[George Jones]

I think the argument in this thread is not that spacetime is not singular inside the event horizon but that there is no inside to the event horizon.

Not sure if you're referring to me, but my previous post was a response to sloughter's scenario of event horizon and no singularity.

 

[rjbeery]

Originally Posted by rjbeery View Post

...and the answer is that they do NOT currently exist in our frame.

That pre-supposes we all share the same frame. We don't. For those of us who choose to hover at a fixed distance above a collapsing black hole, we won't see it form. For those of us who choose to drop into the hole, we will see it form.

"Just when I thought I was out...they pull me back in!"
Tell me DrGreg, which among us "currently" chooses a frame dropping into a black hole? Someone at the LHC perhaps?

 

[xantox]

Yes it does …
in no way does this mean the events will not happen …
(though perhaps the most accurate way of describing it is to say: "the events will happen … never!")
but they certainly did not happen, in any sense of the word "did"!

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.

Xantox and I will have to agree to disagree because my interpretation is that t=infinity from our perspective means that black holes do not exist from our perspective. Is this different from saying that they don't exist at all? I don't know, but this thread has evolved and my ORIGINAL question was...This question has been answered to my satisfaction, and the answer is that they do NOT currently exist in our frame. The rest is semantics and opinion.

I guess the question is ill-posed. What "black holes do not exist from our perspective" may mean here?

• Either it means that we do not receive light signals from them (in that case, the statement is purely epistemological, and yes, this is different from saying that they don't exist at all, since at least some observers in the full manifold can receive light signals emitted at the horizon).
• Either it means that we describe the dynamics of their formation in terms of some "time" variable valid for us and find they don't form. I realize these two alternatives are what you called respectively "seeing" and "calculating" at the beginning of the thread. Unfortunately, this second approach is not unique, since there are many possible definitions of time. I emphasize that if we just say "time" without stating which definition is used, then the discussion is entirely devoid of meaning. Also, Newtonian or special relativistic notions of time cannot be used when dealing with black holes. In general relativity we can for example call time one of these:
  
  1. Coordinate time, which is a projection onto a global system of coordinates, allowing to map spacetime events just like longitude and latitude allow to map Earth locations. Here it is possible to locate a distant event in terms of a vector in the coordinate system. However, some coordinate systems, such as Schwarzschild coordinates, are singular at the horizon, so they give t=infinity. This simply means we need to use a different coordinate system which is not broken at the horizon and which gives a finite value there.
  2. Proper time measured on a worldline in spacetime. Here it is critical to understand that generally, each proper time is specific to its own worldline, and that it is not possible to express global dynamics as evolution in proper time, which is not spatially global, thus you cannot use this definition of time for the above argument. That is, the sentence "the event horizon of a black hole will happen at the distant observer's infinite proper time", eg where you attempt to define the dynamical evolution in terms of a generic distant observer's proper time, has absolutely no meaning whatsoever. However it is possible to say "light rays emitted by a body while crossing the event horizon of a black hole will be received from a distant observer after an infinite amount of his own proper time", here it is correct to measure the wordline of the observer until it "sees" a photon coming from the event horizon.
  3. 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.

So to resume, and using your terms of "seeing vs calculating":

• If you use coordinate time you will "calculate" that the horizon forms at a finite "t" upon choice of suitable coordinates. This "t" is just a coordinate.
• If you use cosmological time you will also "calculate" that the horizon forms at finite "t". This is a metric time obtained by operating a preferred slicing of the whole manifold.
• But, you cannot "calculate" anything about the horizon by using the proper time of a distant observer, as proper time on a specific worldline cannot be used to describe the evolution of the universe. You can merely say the distant observer will "see" a photon emitted at the horizon after infinite proper time on his worldline – which is not inconsistent with a horizon "calculated" as forming in finite proper time for the infalling body.

 

[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.