On the nature of the infinite fall toward the EH

[PeterDonis]

"we are attributing physical meaning to the directly observable proper time on the infaller's clock." But this is exactly what I said. And this is exactly the meaning i attributed to the word subjective. I.e. Pertaining to and relevant only to that frame.

And I am saying you are wrong when you say it is "subjective" in that sense. It is a geometric invariant, the length of a curve; it is not "pertaining to and relevant only to that frame", any more than the distance from New York to London is "pertaining to and relevant only to" a particular set of coordinates for charting the Earth's surface. The fact that the curve happens to be the worldline of a particular observer does not make its length "subjective"; it just means that particular observer is the one who can read the curve length directly off his clock, while other observers have to calculate it from other observations.

I have question regarding the mathematical application of the limit in this case , which by the way I mentioned in another post to you, to which you did not respond.

Can you quickly point me at the post? There are so many threads running on this topic that I can't keep track, so I must have missed it.

You, et al. are not asserting that the infaller's clock will read some relatively short elapsed
proper time at some impossibly distant future time in the real world ( the static world outside the EH) to which I would have no logical problem.

No, because such an assertion would not have an invariant meaning, since it requires adopting a simultaneity convention, and those are not invariant.

On the contrary you all are asserting that the proper time of the infaller, per se, has physical meaning in the world at large. I.e. That it reaches the horizon in some short finite time.

Yes, because this proper time is the invariant length of a curve, as I said above.

But of course the physics along the infaller's worldline does suddenly start working differently at the horizon. Light cannot escape outward from inside this point.

That doesn't mean physics starts working differently. The Einstein Field Equation is as valid at the horizon as it is outside it. Light can't escape outward from the horizon because the light cone at the horizon is tilted inward just enough that its outgoing side is vertical, i.e., outgoing light stays at the same radius. But that behavior of the light cone is part of the solution of the EFE; it's not a sign that physics is working any differently.

The radial speed of light is zero at this point.

Correction: the radial speed of outgoing light is zero at this point. The radial speed of ingoing light is not.

On what do you base an assumption that this geometry has no effect on the motion of the infaller?

Who said it had no effect on the infaller's motion? All I have said is that it doesn't prevent the infaller from falling in, because the infaller is moving inward, not outward.

On what do you base an assumption that even if the falling clock reaches this point that it would in fact continue ticking at all? Is this something that is explicitly derived directly from the EFE?

Yes, as I've said a number of times.

Regarding the fundamental black hole formed from supercondenced mass , is it not somewhat controverisial whether or not a final singularity would form at r=0 as infered from the EFE?

Not if you are talking about the classical solution to the EFE for a spherically symmetric spacetime in which a massive object surrounded by an exterior vacuum region collapses, no. There are proven theorems that guarantee that a singularity will form in this case; Penrose, Hawking, and others proved them in the 1960's and early 1970's. There is no controversy whatever on this point.

What is still an open question is what difference quantum effects make. But from the point of view of the EFE, any difference made by quantum effects that is enough to either prevent the singularity from forming after a horizon has formed, or to prevent even a horizon from forming, will show up as a change to the stress-energy tensor, so that it is no longer vacuum. That means the classical solution that I referred to above would no longer describe the actual collapse of a massive object with quantum effects taken into account.

My understanding was that we were only talking about the standard classical solution in this thread, which is why I haven't said anything about the quantum versions. It's hard to say much about them anyway since they're all still speculative, and will remain so until we have an accepted theory of quantum gravity.

So you agree that the metric does apply and have meaning above the horizon.

As it stands, this statement is too vague for me to either agree or disagree. See further comments below.

So virtually any time, however distant in the future, we may choose, you would agree that according to the metric the infaller has not reached the horizon

No, because "according to the metric" as you are using it here does not say anything about invariants, only about coordinates.

even though there is a vanishingly small difference between the elapsed time at this point and the hypothetical delta time at the horizon? Both frames would agree on this.. yes?:??

I don't understand what you mean by this.

if in fact it is already determined and agreed that at exceedingly distant future times the faller has NOT YET reached the horizon?

This is a coordinate-dependent statement, not an invariant one, so it doesn't tell us anything about the physics.

 

[Austin0]

Quote by Austin0

Isn't it actually only AT the horizon where there are no possible hypothetical static observers to use as a basis for an evaluation of relative dilation?

No. There are no timelike world lines that can maintain a fixed radial position anywhere inside, or at, the horizon. This is trivially verifiable from the metric.

Yes I was talking within the context of this thread. I.e., Approaching the horizon from the outside. The point being that although you said "The ability to define gravitational time dilation disappears on approach to (and past) the horizon because there are no static (hovering) observers " this was in fact not correct. There are static observers right up to actually reaching the horizon YES?





Quote by Austin0

You in other threads have stated that the infaller's clock in the vicinity of the horizon would be ticking at roughly the same rate as the infinity observer's and that signals received from the distant observer would not be blue shifted , in fact would be roughly equivalent or slightly red shifted .
Could you explain the basis for this evaluation?

The redshift observed looking out as you cross the horizon basically depends on where you started free fall from (assuming you start with no initial radial velocity). If you free fall from 'infinity', then you see redshift from the distance as you cross the horizon. If free fall starting from a static position near the horizon, you see blueshift as you cross. One thing you never see is infinite blueshift.

Well this does not really explain the basis. It is just more statements without the reasoning behind them..

 

[PAllen]

Quote by Austin0




Yes I was talking within the context of this thread. I.e., Approaching the horizon from the outside. The point being that although you said "The ability to define gravitational time dilation disappears on approach to (and past) the horizon because there are no static (hovering) observers " this was in fact not correct. There are static observers right up to actually reaching the horizon YES?

Yes, in principle there are static observers up to the horizon. A slightly loose formulation on my part.

Quote by Austin0




Well this does not really explain the basis. It is just more statements without the reasoning behind them..

The real reasoning is math. To justify it formally, I need to know if you are familiar with parallel transport equations, which are the fundamental basis of all redshift calculations in GR. If not, there is an intuitive justification:

Consider a static, near horizon observer, measuring high blue shift from a distant source. Imagine a free faller going past them, away from said source (relative to adjacent static observer) at near c. The blue shift can be redshifted any amount depending on speed of infaller, which depends on how 'high' they started their fall from. If you take the limit of this computation on approach to the horizon (increasing speed of free faller relative to adjacent static observers; increasing blue shift seen by static observers), you get a finite value on approach to the horizon. The limit can be any result from extreme (but finite) blue shift, to high redshift (for high inward speed at start of inertial fall). Such a limiting process can get you accurately to the horizon figure. However, for inside the horizon, there is no way to use such a procedure, because there are no static observers. Instead you have no choice but to use the only fundamental basis red/blue shift calculation in GR: parallel transport of source 4-velocity over null path to receiver; then express the transported source 4-velocity in local frame of receiver world line at reception event, and use SR Doppler formula with this local frame 3-velocity and light propagation vector.

One alternative to parallel transport is to consider a pair of signals emitted a small proper time interval apart on the distant source static world line; track these to the interior free fall world line; compute proper time between reception events on the free fall world line; take the limit of ratio emitter and receiver proper time intervals, as interval size goes to zero. This will give you the same result.

Whatever method you use, you find that the interior free faller sees increasing redshift (not blue shift) for the distant source as they approach the singularity.

 

[pervect]

\tau= t(1-2M/r)^1/2(1-v²/c²)^1/2

Do you think that this is not a valid equation relating distant static time t to infalling proper time?

It's not a valid equation, at least not for a fall from infinity with zero velocity. The correct expression for this can be worked out by solving the geodesic equations for t(\tau) and r(\tau). The result was given in post 12 of this thread (for a black hole of mass 2). This gives t as a function of \tau, according to the simultaneity convetions of a static observer (the opposite order of what you gave).

https://www.physicsforums.com/showpost.php?p=4185014&postcount=12

To rewrite it in slightly clearer notation
<br /> -\infty &lt; \tau &lt; 0<br />
<br /> u = \left( -3 \tau\right)^ {\frac{1}{3}}<br />
<br /> t = \tau -4 u +4 \ln \left(u+2\right)- 4 \ln \left( u - 2 \right)<br />
<br /> r = u^2<br />

The way one knows this is correct is that it satisfies the geodesic equations, which for a free fall from infinity with zero velocity in the Schwarzschild metric are:

\frac{dr}{d\tau} = \sqrt \frac {2m}{r}
\frac{dt}{d\tau} =\frac{1}{1-2m/r}

I've assumed m=2 to simplify the calculations.

Post #13 gives the observed frequency the infalling observer sees from a monochromatic beam from infinity that's also falling into the black hole as a function of proper time:

<br /> z = \frac{\sqrt{r}}{\sqrt{r}+2}<br />

The event horizon here is at r=2m, and since m=2 this means it's at r=4. one can see that the doppler shift there is exactly 1/2.

 

[Mike Holland]

Sure.
So, how, in terms of these coordinates, does one explain the fact that signals sent once per millisecond from an observer at sea level arrive on top of a mountain at a rate lower than that? Well, in the free-falling coordinate system, the two observers are accelerating upward. Each signal sent by the observer at sea level must travel farther than the last to reach the observer on the mountain. So the free-falling coordinate system attributes the difference in send rates and receive rates purely to Doppler shift, not to time dilation. (At least initially.)

Thanks, Stevendaryl. That has helped me understand how the observations can be explained in different ways by different coordinate systems. So by changing coordinate systems the relative dilation doesn't go away, but the interpretation changes.

 

[Dale]

No, they're not. The curves of constant r, theta, phi in SC coordinates are also curves of constant r, theta, phi in GP coordinates. This is also true of Eddington-Finkelstein coordinates. Shell observers would be "moving" in Kruskal coordinates.

D'oh, of course you are right. GP coordinates use the rain observer's time and the shell observer's spatial coordinates. I always forget that.

So, the difference in gravitational time dilation between GP and SC would be accounted for due to the cross terms, not the spatial terms.

 

[Dale]

Near the surface of the Earth, the metric can be described approximately using ...

So the free-falling coordinate system attributes the difference in send rates and receive rates purely to Doppler shift, not to time dilation. (At least initially.)

This is exactly what I was thinking.

 

[Dale]

Wouldn't observers in other frames also see the 1% difference in signal frequencies, and come to the same conclusion about relative dilation between the clocks, even though both clocks might be running fast or slow for them?

Yes, all observers in all frames will see the same 1% dilation between the clocks. That is an invariant, so all coordinate systems will agree.

What is not invariant is whether that 1% dilation is attributable to gravitation, velocity, cross-terms, or some combination.

 

[Dale]

Thanks, Stevendaryl. That has helped me understand how the observations can be explained in different ways by different coordinate systems. So by changing coordinate systems the relative dilation doesn't go away, but the interpretation changes.

Yes, that was my point, which stevendaryl made much more effectively than I did. Thanks, Stevendaryl!

 

[PeterDonis]

So, the difference in gravitational time dilation between GP and SC would be accounted for due to the cross terms, not the spatial terms.

If you're talking about shell observers, who are on worldlines of constant r, theta, phi, only the dt^2 term in the line element is relevant. g_tt is the same for GP and SC coordinates, so the equation for gravitational time dilation of shell observers is identical.

If you're talking about different "time dilation" for observers who are moving radially, compared to shell observers, you can compute that in either GP or SC coordinates, because dr is nonzero as well as dt.

 

[PeterDonis]

If you're talking about shell observers, who are on worldlines of constant r, theta, phi, only the dt^2 term in the line element is relevant. g_tt is the same for GP and SC coordinates, so the equation for gravitational time dilation of shell observers is identical.

Perhaps I should expand on this a little more, since it does seem a bit fishy that GP and SC coordinates give the same time dilation equation for shell observers, even though their time coordinates are obviously not the same.

The key thing to note is that gravitational time dilation for shell observers only depends on the *ratio* of the "rates of time flow" at the two different altitudes; that ratio is the actual observable. GP and SC coordinates parameterize the curves of constant r, theta, phi differently, but when calculating a ratio, the parameterization drops out and all you're left with is the g_tt values. So since g_tt is the same (as a function of r) for GP and SC coordinates, they both give the same equation for gravitational time dilation for shell observers.

 

[Dale]

OK, so we have to go to a rain frame like Stevendaryl explained. Or any coordinate system that uses a different Spatial coordinate.

 

[Mike Holland]

Is the Rain frame the frame of a free-falling observer? In which case he sees himself as hanging motionless in his frame while a great big condensing mass is accelerating towards him!

That makes sense, but I am still trying to visualize what he sees regarding an object falling in behind him. Will it be blue-shifted or red-shifted? He will be accelerating away from it, so he should see it as accelerating away from himself in his frame, and it should be red-shifted. Is that correct? In fact, there is no gravity in his frame, and he sees all remote objects accelerating in the same direction as the approaching mass. So he should see everything else as red-shifted. Of course, some distant objects would have velocities/accelerations relative to the approaching BH, which could override this red shift.

So I gather that only a suspended observer sees a blue shift in the outside world, and a falling observer sees everything red-shifted. (Again ignoring motion of outside objects relative to the BH).

Edit: No, I got that wrong. You see a red shift either way, whether an object is approaching or receding. So the falling guy only sees red shifts.

 

[PAllen]

So I gather that only a suspended observer sees a blue shift in the outside world, and a falling observer sees everything red-shifted. (Again ignoring motion of outside objects relative to the BH).

No, I discussed this in my last post. It depend on the 'initial conditions' of free fall. You can have free fall from infinity, or starting at zero radial speed from any point outside the horizon. You can also have, at some starting radius, an inward radial speed greater than free fall from infinity. The result is that at any radius you choose, their is some inertial radial path that achieves any blue or redshift you care to name (it may be one that is inertial but moving radially outwards at that point).

 

[zonde]

I would call the things one can measure along a single worldline "measurements". An example of what I am calling a measurement would be something similar to this. "At a proper time of xx.xxx by my clock, a signal of frequency yyy was recorded , identified as being from object zzz. The signal was decoded as having a timestamp (from object zzz) of uu.uuu.

Without going completely into the definition of an observer, I'll relate one quantity of interest that's relevant to the discussion that is not in the form of such a measurement.

This is "Event P is simultaneous with event Q".

Making such a statement requires more than just a "measurement" as I have described it. One could say that one received a signal (as above) from P and a signal from Q at the same time, but it's easy to see that this does not imply that P and Q are simultaneous - for instance P might be further away from you than Q, in which case the simultaneous receipt of signals would show that Q occurred before P.

I'm saying that making such a statement requires more structure than a "measurement" does. I was going a bit into the detail of what sort of extra structure was required - I'll repeat myself on this point a bit later.

Yes. I hope the example I've given above explains the specific point in mind. I'll take the opportunity to describe in detail the set of measurements and the extra structure needed to say that "event P is simultaneous with event Q" beyond specifying the worldline of a single observer.

The particular suggestion I made (which is more or less the standard way of defining simultaneity) was that one had a chain of observers, all synchronizing their clocks by exchanging signals and using the Einstein Convention. This process of synchronzing also in general requires rate-adjusting in GR. When, according to this chain of observers , the adjusted reading for the observer in the chain co-located with P is the same as the adjusted reading for the observer in the chain colocated with Q is the same, the events are simultaneous.

The sub-point is that this statement is NOT in general independent of what chain of observers you use between P and Q. So one way of defining this extra structure, needed to talk about simultaneity, is to define this chain of observers. Which requires more than specifying the worldline of a single observer.

All the things you say are more or less clear. These are SR things so there are not much questions about that.

It certainly doesn't demonstrate that to me! I'm not quite sure what you are thinking here. I will try to resist the obvious interpretation of "I don't like it when you bring up things that are contrary with my position."

If "overwhelming majority of the relativists" can quickly change their viewpoint after noticing some theoretical argument then there is quite a possibility that this can happen again.

 

[Dale]

If "overwhelming majority of the relativists" can quickly change their viewpoint after noticing some theoretical argument then there is quite a possibility that this can happen again.

Definitely. Changing scientists minds is the whole point of doing physics, both theoretical and experimental. Does that surprise you in any way?

The point is that the opposition to the existence of the interior of a BH is not based on a sound understanding of the theory. It is based on an unsound elevation of a particular coordinate chart to some priveliged status.

 

[Dale]

Is the Rain frame the frame of a free-falling observer? In which case he sees himself as hanging motionless in his frame while a great big condensing mass is accelerating towards him!

Yes, that is correct.

That makes sense, but I am still trying to visualize what he sees regarding an object falling in behind him. Will it be blue-shifted or red-shifted? He will be accelerating away from it, so he should see it as accelerating away from himself in his frame, and it should be red-shifted. Is that correct?

In the rain frame both shell observers are accelerating away from the mass (up). So during the time that light goes from the lower shell to the upper shell, the upper shell observer has accelerated away from the light, leading to velocity redshift on the way up. Conversely, during the time that the light goes from the upper shell to the lower shell, the lower shell observer has accelerated towards the light, leading to velocity blueshift on the way down.

Both shifts are entirely attributed to velocity, but they match quantitatively the shift attributed to gravitation in the SC frame. This is what stevendaryl quantified mathematically up above. I hope this helps.

 

[pervect]

All the things you say are more or less clear. These are SR things so there are not much questions about that.If "overwhelming majority of the relativists" can quickly change their viewpoint after noticing some theoretical argument then there is quite a possibility that this can happen again.

It could happen. Note, though, that here at PF, we have an educational evnvironment, not a research one. Our goal is not to advance the state of science. While this is of course an important task, it's not our goal. According to the PF guidelines, our goal is:

Our mission is to provide a place for people (whether students, professional scientists, or others interested in science) to learn and discuss science as it is currently generally understood and practiced by the professional scientific community.

Thus people who want to advance the current understanding of science need to publish papers in the literature and journals, not present arguments here. (The first step would generally be getting the paper past peer review.)

The way I'd summarize the current situation is that in my mind there isn't any real doubt that about the position of the scientific community with regard to the "finite proper time to reach the event horizon" issue as long as the black hole is classical. This is standard textbook stuff nowadays.

It's not nearly as clear to me what the prevailing view is in the non-classical case, specifically with regards to the question as to whether or not the black hole evaporates before someone falls in. I was a bit surprised, but even the author who claims that black holes will NOT evaporate says that the calculation is "non-trivial".

[add].Note that if there was some dramatic self-inconsistency that arose when saying that black holes did not evaporate, the situation would be clear rather than unclear. But here isn't any such contradiction internal to GR. There may be an external conflict with some posters personal wordlviews and GR. There may (or may not) be issues with the famous "information loss" issues as well, this isn't an issue internal to GR, but an issue in quantum gravity of how to make GR and QM work together.

 

[grav-universe]

τ= t(1-2M/r)^1/2(1-v²/c²)^1/2

Do you think that this is not a valid equation relating distant static time t to infalling proper time?

It's not a valid equation, at least not for a fall from infinity with zero velocity. The correct expression for this can be worked out by solving the geodesic equations for t(τ) and r(τ).

...

\frac{dt}{d\tau} =\frac{1}{1-2m/r}

Just a quick mention that in Schwarzschild coordinates, sqrt(1 - 2 m / r) = K sqrt(1 - (v'/c)^2), where v' is the speed that is locally measured by a static observer at r and K is a constant of motion, with K = 1 for a freefall from rest at infinity, so those two statements would be equivalent in terms of dt and dτ in that case.

 

[A.T.]

We can always do the same thing with Zeno time by judicious choice of our reference clock and our simultaneity convention. For instance, we can use a Rindler-like simultaneity convention.

Thanks. I see know that we can make any process "take infinite time" at an actual physical clock.

 

[A.T.]

An interesting intuitive approach for this problem is presented by Epstein in his book "Relativity Visalized". He uses space-propertime embeddings of the Schwarzshild-Metric like this one:

http://www.adamtoons.de/physics/gravitation.swf

"space" : radial Schwarzshild coordinate r
"proper time" : the proper time elapsed along the cyan world line
length of the world line : Schwarzshild coordinate time t
inflation of the "pipe" at certain r : time dialtion of a static clock at that r relative to t

For a BH the funnel would inflate infinitely into a plateau at the event horizon:

The cyan world line here can have an infinite length (coordinate time t) but will still cover a finite angular displacement around the pipe (proper time)

 

[Austin0]

Just a quick mention that in Schwarzschild coordinates, sqrt(1 - 2 m / r) = K sqrt(1 - (v'/c)^2), where v' is the speed that is locally measured by a static observer at r and K is a constant of motion, with K = 1 for a freefall from rest at infinity, so those two statements would be equivalent in terms of dt and dτ in that case.

Thank you.
As free fall from rest at infinity was in fact the context of the thread I was correct in my understanding then , right??
Am I correct in assuming that K=1 in this case is related to free fall velocity at a particular r being equivalent to the escape velocity at that location, which is the case when starting from v=0 at r=∞?

 

[grav-universe]

Thank you.
As free fall from rest at infinity was in fact the context of the thread I was correct in my understanding then , right??
Am I correct in assuming that K=1 in this case is related to free fall velocity at a particular r being equivalent to the escape velocity at that location, which is the case when starting from v=0 at r=∞?

Actually, you were correct regardless of where the object falls from or the value of K. K is the constant of motion for freefall here, so with the initial conditions v' = 0 at r = ∞, we get

K = sqrt(1 - 2 m / ∞) / sqrt(1 - (0/c)^2) = 1

and K will then remain constant for any r during freefall, even non-radially, so one can find the speed locally measured by a static observer at some other r with

v' = c sqrt[1 - (1 - 2 m / r) / K^2]

Since SR is valid locally, we have

dτ = dt' sqrt(1 - (v'/c)^2)

= dt sqrt(1 - 2 m / r) sqrt(1 - (v'/c)^2)

= dt (1 - 2 m / r) / K

Only the last statement depends upon the value of K. Vice versely if an object were thrown upward from r with the corresponding speed v' where K = 1, that would be its escape velocity also, right.

 

[Austin0]

The underlying thought process here is that there is some physically meaningful way to define a "local rate of time". Relativity doesn't necessarily say this. (I think one can make even stronger claims, but it'd start to detract from my point, so I'll refrain from now).

One can certainly say that Alice appears to freeze according to the coordinate time "t". But is this physically significant?

It might be instructive to consider Zeno's paradox. I'll use the wiki definition of the paradox.
Let's define a "zeno time" as follows. At a zeno time of 0, Achillies is 100 meters behind the tortise.

At a zeno time of 1, Achilles is 50 meters behind the tortise.

At a zeno time of 2, Achillies is 25 meters behind the tortise

At a zeno time of n, Achillies is 100/(2^n) meters behind the tortise.

Then, as n goes to infinity, Achillies is always behind the tortise.

So, in "zeno time", Achilles never does catch up with the tortise, even as "zeno time" appoaches infinity.

Are we therefore justified in claiming that Zeno was right, and that Achilles never catches the tortise? I don't think so, and I'd be more than a bit surprised if anyone really believed it. (I could imagine someone who likes to debate claiming they believed it as a debating tactic, I suppose - and to my view this would be a good time to stop debating and do something constructive).So in my opinion, the confusion arises by taking "zeno time", which is analogous to the Schwarzschild coordinate time "t", too seriously. While it is correct to say that as t-> infinity Alice never reaches the event horizon, just as Achilles never reaches the tortise in zeno time, it still happens. It's just that that event hasn't been assigned a coordinate label.

The analogy is actually quite apt regarding light chasing an accelerating system with only the slight modification that the tortoise has some finite constant acceleration. In which case at and beyond some magnitude of head start Achilles (light) can never catch up.

But wrt approaching an EH I think both the paradox and the related theorem that resolves the same kind of unbounded series to a finite value
1/2 + 1/4 + 1/8 + ... = 1

are not validly applicable.

In the first case (Zeno) as the distance incrementally reduces, the velocity of Achilles remains constant. So for each reduction in distance, the time for the next reduction in distance becomes shorter.

So it is obvious, even without a formal mathematical proof, that the difference between an evaluation of some large but finite number of iterations and the self evident ultimate value after an infinite number of iterations effectively disappears.

In approaching the horizon this is not true. Each reduction in distance results in a reduction in speed so increases the time interval for the next distance. Etc etc
As the speed approaches zero nearing the horizon the time approaches infinite which is clearly a whole other ball game. Or at least seems so to me.
Neither the theorem nor the paradox apply.

 

[PeterDonis]

Each reduction in distance results in a reduction in speed

The "speed" that is reduced is just a coordinate "speed". It doesn't have any physical meaning. For example, there is no observer who observes the infalling object moving at this "speed".

 

[Dale]

In the first case (Zeno) as the distance incrementally reduces, the velocity of Achilles remains constant. So for each reduction in distance, the time for the next reduction in distance becomes shorter.

In Zeno coordinate time the time for the next reduction is constant, by definition. So the Zeno coordinate velocity in fact reduces.

It is the proper time which reduces. And the velocity in some unspecified inertial coordinate system which remains constant.

In approaching the horizon this is not true. Each reduction in distance results in a reduction in speed so increases the time interval for the next distance. Etc etc
As the speed approaches zero nearing the horizon the time approaches infinite which is clearly a whole other ball game. Or at least seems so to me.
Neither the theorem nor the paradox apply.

No, the two scenarios are very closely analogous on this point. Again in SC coordinate time the time for the next reduction is constant, by definition. So as you mention the SC coordinate velocity reduces.

Similarly, the proper time reduces in the SC case, and the velocity in a local inertial frame remains constant. Exactly analogously to Zeno.

 

[zonde]

It could happen. Note, though, that here at PF, we have an educational evnvironment, not a research one. Our goal is not to advance the state of science. While this is of course an important task, it's not our goal.

And ... ? I know that. Why do you think you have to remind me that?

 

[Austin0]

In Zeno coordinate time the time for the next reduction is constant, by definition. So the Zeno coordinate velocity in fact reduces.

It is the proper time which reduces. And the velocity in some unspecified inertial coordinate system which remains constant.

No, the two scenarios are very closely analogous on this point. Again in SC coordinate time the time for the next reduction is constant, by definition. So as you mention the SC coordinate velocity reduces.

Similarly, the proper time reduces in the SC case, and the velocity in a local inertial frame remains constant. Exactly analogously to Zeno.

Actually I was talking within the context of the original statement of the paradox where the distance was the basis parameter so I didn't look at the specifics of Pervect's Zeno time .

Having done so it appears that it was not explicitly stated that the intervals were equivalent. And in fact they would not correspond to time on any normal clock with a constant rate.

According to such a constant clock the interval between events Zeno t=1 and t=2 would be smaller than between events Zeno t=0 and t=1
Or do you disagree??

So are you talking about an arbitrary clock that speeds up over time ??

Could you explain where you get this "Again in SC coordinate time the time for the next reduction is constant, by definition." In the original, the next reduction was reducing the remaining distance to the horizon by half so how do you get a constant time interval for each of these increments?

 

[Dale]

Actually I was talking within the context of the original statement of the paradox where the distance was the basis parameter so I didn't look at the specifics of Pervect's Zeno time .

Yes, he adapted the original paradox deliberately in order to make the analogy with SC time more exact.

Having done so it appears that it was not explicitly stated that the intervals were equivalent. And in fact they would not correspond to time on any normal clock with a constant rate.

According to such a constant clock the interval between events Zeno t=1 and t=2 would be smaller than between events Zeno t=0 and t=1
Or do you disagree??

I agree completely. Again, the whole point of the analogy is that the exact same thing happens with SC time. For Zeno time, the proper time on Achilles' clock between Zeno t=1 and t=2 is indeed smaller than between Zeno t=0 and t=1. For SC time, the proper time on the free-falling clock between SC t=1 and t=2 is also smaller than between SC t=0 and t=1.

So are you talking about an arbitrary clock that speeds up over time ??

No, I just mean that coordinate time proceeds at a rate of one coordinate second per coordinate second, by definition. It is a simple tautology. If you are using coordinate time as your standard (as the SC proponents want to do) then coordinate time is uniform, by definition, i.e. tautologically.

In SC coordinate time each successive SC coordinate time interval for the free-faller is tautologically constant. In Zeno coordinate time each successive Zeno coordinate time interval for Achilles is tautologically constant. Neither correspond to the proper time on the falling/Achilles' clock.

Could you explain where you get this "Again in SC coordinate time the time for the next reduction is constant, by definition." In the original, the next reduction was reducing the remaining distance to the horizon by half so how do you get a constant time interval for each of these increments?

The analogies diverge quantitatively, but not qualitatively. In SC coordinates each successive SC coordinate time interval does not correspond to half the distance to the horizon, that is a feature of the construction of Zeno coordinates. But in both SC and Zeno coordinates the coordinate distance traveled by the free-faller/Achilles decreases for each successive coordinate time interval. This obviously means that their coordinate velocity is reducing, which you already recognized and pointed out above.

 

[pervect]

Just a quick mention that in Schwarzschild coordinates, sqrt(1 - 2 m / r) = K sqrt(1 - (v'/c)^2), where v' is the speed that is locally measured by a static observer at r and K is a constant of motion, with K = 1 for a freefall from rest at infinity, so those two statements would be equivalent in terms of dt and dτ in that case.

As long as v actually is the speed that's locally measured by a static observer, I believe that's correct. I usually use E for K, many sources use ~E.

It wasn't clear to me how v was being defined - I should have asked. I should still ask, because it's still not clear to me how the OP is defining v, and it's very common not to use the correct formula or defintion of v.

The relation between v and the derivatives of the coordinates is moderately messy, but in https://www.physicsforums.com/showpost.php?p=602558&postcount=29

I get (and another poster also gets) in geometric units.

v= \frac{\sqrt{E^2 - (1 - \frac{2M}{r})}}{E}(The natural way to do this is via frame fields, but I choose to introduce locally Lorentz coordinates instead. THe intent was to make it easier to follow, I'm not sure how successful it was. But the intent is to use local coordinates rr and tt that agree with the local clocks and rulers.)

This expression for v is also what I get when I solve your equation for v/c (which is just v in geometric units as c is assumed to be 1).