On the nature of the infinite fall toward the EH

[Austin0]

While you're asking Vachaspati to make clarifications, since it appears he might be the rare person you might actually listen to (I'm sorry, but I don't think you've actually listened to any of the 3-4 SA's on this thread), you might ask him if he agrees that the proper time it takes for a free-falling observer starting at rest at a large (but finite) distance away from a black hole to reach the event horizon is finite.

You might also ask him if said proper time can be observed, not directly, but as a limit, by an external observer.

True or false:
The integrated proper time of an unbounded accelerating system to the limit as velocity approaches c is a finite Δ$\tau$?

 

[Austin0]

This is an illogical inference. The same observed facts fit with other coordinate systems where the gravitational time dilation is not 1%.

Could you give an example of other systems with the same observations but different conclusions??

 

[Austin0]

Quote by Mike Holland

Yes, in their local proper time they all run at the same rate.

But consider the following thought experiment.

I will make two clocks that emit light pulses every second, and then place one at the bottom of a deep hole and one at the top. For convenience, I will assume that the Earth is not rotating, and that its mass is concentrated near the centre, so that gravity is a lot stronger at the botton of the hole.

Now I observe the light pulses coming from the two clocks, and find that they are not synchronised. I get 99 pulses on the bottom clock for every 100 on my local clock. So I infer that I have 1% gravitational time dilation present. In what way is this conclusion dependent on a coordinate system? If I position myself next to the bottom clock, I will see the same difference in rates. If I position myself 100 miles above the top clock, I will again get the same result. All I have done is count light pulses. If I am moving towards the two clocks or away from them, I may see them both pulsing faster or slower, but I will still see this 1% difference - every 100 flashes of the top clock the photons from the two clocks will arrive side by side, wherever I am along the line joining the two clocks.

So where have I assumed a simultaneity convention?

This is direct observation. However, now Let's ask what Einstein's 1915 equations predict about a clock in inertial free fall towards a collapsing body of sufficient mass (inertial clock), that is engaged in communication with a distant clock. As the inertial clock nears the surface:

- The signals it gets from the distant clock may show the same rate as its own clock, be moderately slower than its own clock, or be faster - all depending on where it's fall started from, and any initial radial speed it had. This remains true all the way to the singularity - there will never be infinite blue shift measured by the the free fall clock based on signals it gets from the distant clock - right up to the singularity.

- The distant clock sees the inertial clock slow down, effectively stop, and suffer infinite red shift.

There is no contradiction because the distant observer can calculate that GR says the latter is due to gravity's effect on light and all other possible signals; and GR makes a completely unambiguous prediction about what the inertial clock will measure (even if the distant clock can never access those measurements). The ability to define gravitational time dilation disappears on approach to (and past) the horizon because there are no static (hovering) observers in reference to which it can be defined and separated from Doppler.

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?

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?

 

[Dale]

Could you give an example of other systems with the same observations but different conclusions??

Sure. Gullstrand Painleve coordinates. The shell observers are moving in GP coordinates, so some of the 1% change in frequency will be attributed to velocity time dilation rather than gravitational.

 

[PeterDonis]

The shell observers are moving in GP coordinates

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.

 

[PAllen]

Quote by Mike Holland


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.

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.

 

[Austin0]

Austin0, I gather that I did not reply to this post of yours. I have poseed this reply under the original topic, and am repeating it here.

Quote by Austin0
you say the falling observers clock is never stopped in either frame because the distant observers clock never reaches infinity.
I agree. but you seem to ignore the fact that this is only true in the region where the faller has NOT reached the singularity.
you then want to magically have the faller PASS the horizon without ever having reached it.
It appears you interpret time dilation in a way that creates alternate contradictory realities.
If your premise that reaching the horizon requires infinite coordinate time for the distant observer is correct, that means that at all points in that interval the times at the two locations will be related by the SC metric. Both observers will agree on these relative elapsed times and both observers will agree that the faller has not reached the horizon.


The answer here is that all points on the two time scales ARE related by the SC metric, all the points from 0 to infinity on the distant observer's clock are related to the points from 0 to T on the faller's clock, where T is his local time when he gets to the horizon. Obviously it is not a linear relationship, more like a tangent graph where tangent goes to infinity as angle goes to 90 degrees, and so they don't agree on relative elapsed times. Each sees the other's clock ticking at a different rate to his own, an ever increasing difference.

This is NOT an answer of any kind.
It is simply an agreement and restatement of exactly what I said which is not in controversy.
But you stopped short of the actual point.
I.e. " Both observers will agree on these relative elapsed times and both observers will agree that the faller has not reached the horizon."

SO in principle there is a finite point, short of the horizon, where both observers will agree that the distant clock reads 10¹²years and the inertial clock reads some relatively short time (in related threads approx. 1 day has been mentioned for freefall proper time to EH) correct?
This is a rational application of the metric as it pertains to and in both frames, agreed?

And of course there is in principle yet another point even closer to the horizon etc., etc.

SO again what is the significance of the 1 day elapsed time on the falling clock??

How do you manage to turn this into an idea that the free faller reaches the horizon in some relatively short time in the real world. I.e. the majority of the universe which is outside the EH and relatively static.

 

[PAllen]

This is NOT an answer of any kind.
It is simply an agreement and restatement of exactly what I said which is not in controversy.
But you stopped short of the actual point.
I.e. " Both observers will agree on these relative elapsed times and both observers will agree that the faller has not reached the horizon."

SO in principle there is a finite point, short of the horizon, where both observers will agree that the distant clock reads 10¹²years and the inertial clock reads some relatively short time (in related threads approx. 1 day has been mentioned for freefall proper time to EH) correct?
This is a rational application of the metric as it pertains to and in both frames, agreed?

And of course there is in principle yet another point even closer to the horizon etc., etc.

SO again what is the significance of the 1 day elapsed time on the falling clock??

How do you manage to turn this into an idea that the free faller reaches the horizon in some relatively short time in the real world. I.e. the majority of the universe which is outside the EH and relatively static.

I'm not sure the context, but a free fall observer will never see something like 10^12 years on distant clock. As I've explained, if they start free fall from relatively far away, they will see the distant clock fall behind theirs (but not by a lot).

 

[stevendaryl]

Could you give an example of other systems with the same observations but different conclusions??

Sure.

Near the surface of the Earth, the metric can be described approximately using the line element $ds^2 = (1+gX)^2 dT^2 - dX^2$ where $X$ is the height above the surface, and $g$ is the acceleration due to gravity.

In these coordinates, we can compute the "rate" $\dfrac{d \tau}{dT}$ for a clock at rest at height $X$:

$\dfrac{d \tau}{dT} = (1+gX)$

So higher clocks (larger $X$) have a higher rate. In particular, if an observer at sea level sends a signal once per millisecond (according to his clock) toward an observer on top of a mountain, the arrival times for the signals will be slower than one per second, according to the clock at the top of the mountain.

Now, transform coordinates to free-fall coordinates $x,t$ defined by:

$x = (1/g + X) cosh(gT) - 1/g$
$t = (1/g + X) sinh(gT)$

In terms of these coordinates, the metric looks like:

$ds^2 = dt^2 - dx^2$

This is the metric of Special Relativity. In these coordinates, there is no "gravitational time dilation". The locations of clocks have no effect on their rates. In particular, a clock at sea level will have the same rate as a clock on top of a mountain. Initially.

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.)

 

[Mike Holland]

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.

Now I am confused. Moving relative to what? Are you saying they would be moving relative to each other in Kruskal coordinates? I know exactly how far my clocks are from each other, and that is not changing.

I can see how one could "make" them move. One could place an elastic ruler between them, and then stretch it. Is this what Kruskal coordinates do?

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?

 

[PeterDonis]

Now I am confused. Moving relative to what?

Relative to the coordinate chart. I put "moving" in quotes because in the sense in which I was using that word (which was the sense in which I believe DaleSpam was using it) "moving" just means "has spatial coordinates that change with time". "Shell observers" have the same spatial coordinates for all time in the SC, GP, and EF (Eddington-Finkelstein) charts, but not in the Kruskal chart. However, this is obviously a coordinate-dependent notion and doesn't say anything in itself about the physics.

I can see how one could "make" them move. One could place an elastic ruler between them, and then stretch it. Is this what Kruskal coordinates do?

No. Kruskal coordinates are designed to make the causal structure of radial motion in the spacetime clear, by making light rays look like 45 degree lines. They also make the worldlines of shell observers, who stay at the same Schwarzschild r coordinate for all time, look like hyperbolas, similar to the way the worldlines of Rindler observers look in Minkowski coordinates. But the Kruskal spatial coordinate itself doesn't really have an easy interpretation in terms of "rulers", at least not globally. Locally, any small patch of the Kruskal chart looks like a small patch of a Minkowski chart, but the scaling of the time and radial coordinates in a small patch varies in the Kruskal chart, where it does not in the Minkowski chart.

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?

Observers at rest relative to both clocks would see the same signal difference. Observers moving relative to the clocks would not, because there would be a Doppler shift due to the relative motion in addition to the gravitational redshift/blueshift (at least, that's the interpretation that shell observers would put on the observations). Here "moving" means moving in an invariant sense; the simplest way to test for relative motion in this sense is to see if the round-trip travel time of a light beam sent between the observer and either one of the clocks changes with time.

 

[PeterDonis]

SO again what is the significance of the 1 day elapsed time on the falling clock??

Um, that it's a direct observable? This has been answered before.

How do you manage to turn this into an idea that the free faller reaches the horizon in some relatively short time in the real world.

I.e. the majority of the universe which is outside the EH and relatively static.

How do you justify your claim that this is all that "the real world" consists of?

 

[Mike Holland]

OK, I wasn't serious about the elastic ruler.

Observers at rest relative to both clocks would see the same signal difference. Observers moving relative to the clocks would not, because there would be a Doppler shift due to the relative motion in addition to the gravitational redshift/blueshift (at least, that's the interpretation that shell observers would put on the observations). Here "moving" means moving in an invariant sense; the simplest way to test for relative motion in this sense is to see if the round-trip travel time of a light beam sent between the observer and either one of the clocks changes with time.

That's what I meant about the clocks being red or blue shifted for other observers. They might be affected by local gravitational fields or the lack thereof, or by relatrive velocity. But these effects would apply to observations of the light pulses from the two clocks equally, and they would still see the 1% difference in clock rates, and come to the same conclusion about the relative time dilation between the two clocks.

 

[Austin0]

Quote by Austin0

But are you not attributing equal physical meaning to the subjective time of the infaller??

No, we are attributing physical meaning to the directly observable proper time on the infaller's clock. That is not "subjective", except in the trivial sense that it's that particular observer who directly observes it. But that directly observable number is an invariant; anyone can calculate it using any coordinate chart they like that covers the appropriate portion of the infaller's worldline, and they will get the same answer..

"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.
I was not suggesting that it was not invariant although 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.

Furthermore, the proper time on the infaller's clock is only being used to make assertions about what happens along the infaller's worldline, i.e., along the worldline where that proper time is directly observable. The coordinate time is being used, by those who make assertions about what it "means", to make assertions about what happens *elsewhere* than on the worldline of an observer "at infinity", for whom coordinate time = proper time. It's the fact that something that can only be observed on one particular worldline (and on an idealized one at that, since it's the worldline of the observer "at infinity") is being used to make assertions about the entire spacetime, that creates the problem.

But this is clearly not the case. 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.
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.

Quote by Austin0 View Post

Time dilation is inherently a relative evaluation. What difference does it make what the elapsed time on the falling clock is.

The assertion that's being made is not about "time dilation". It's not relative. It's an assertion that the infaller's worldline continues all the way down to the singularity, because the infaller's proper time is finite and the spacetime curvature in the infaller's vicinity is finite all the way down to the singularity. Those are physical invariants--direct observations that the infaller can make. For the claim not to be true, physics along the infaller's worldline would suddenly have to start working differently at the horizon, for no apparent reason. That's why it makes a difference what the elapsed time on the falling clock is.

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. The radial speed of light is zero at this point. On what do you base an assumption that this geometry has no effect on the motion of the infaller?
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?

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?

Quote by Austin0 View Post

Would you disagree with this?

The relationship between the elapsed time on the infaller's clock and the coordinate time is fine for the portion of the infaller's trajectory that is above the horizon. And yes, both observers will agree that the infaller has not yet reached the horizon, *on that portion of his trajectory*.

But when the infaller reaches the horizon, he "disappears" from the distant observer's coordinates, and from his "line of sight", since light rays emitted at or inside the horizon can't get back out to the distant observer. The problem arises when people try to translate "the infaller disappears from the distant observer's sight at the horizon" into "the infaller never reaches the horizon, period". That's not a valid translation.

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

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 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?:??
SO how do you arrive at a conclusion that the infaller reaches the horizon in a relatively short time as far as we are concerned if in fact it is already determined and agreed that at exceedingly distant future times the faller has NOT YET reached the horizon?

 

[Austin0]

I'm not sure the context, but a free fall observer will never see something like 10^12 years on distant clock. As I've explained, if they start free fall from relatively far away, they will see the distant clock fall behind theirs (but not by a lot).

$\tau$= 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?

 

[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
$$-\infty < \tau < 0$$
$$u = \left( -3 \tau\right)^ {\frac{1}{3}}$$
$$t = \tau -4 u +4 \ln \left(u+2\right)- 4 \ln \left( u - 2 \right)$$
$$r = u^2$$

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:

$$z = \frac{\sqrt{r}}{\sqrt{r}+2}$$

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.