The Shapiro delay and falling into a black hole

[Zan24C]

I realize that this question has been asked many times on this forum, however, I have yet to come across a satisfying/understandable answer that takes into account gravitational time dilation.

Premise:
The speed of light inside a gravitational field is slowed down relative to a distant observer by gravitational time dilation as proven by the Shapiro delay experiment. Therefore the speed of any object is slowed down in a gravitational field as seen by a distant observer. The speed of an object/light is related to the strength of the gravitational field.

Question:
How can anything 'fall' through the event horizon of a black hole within the time span of the entire future of the universe if its speed asymptotically approaches zero as it approaches the event horizon.

In the proper time of an object falling into a black hole, the object will cross the event horizon in a finite amount of time. For a distant observer, however, the time it takes is infinite. How does one reconcile this paradox?
Let us also ignore the effect of redshift by presuposing that the distant observer has access to infinitely sensitive observational equipment.

Logically I know I must be misunderstanding something fundamental as general relativity has been studied by very intelligent physicists for decades and this has not flagged up as an unresolvable paradox.

Having said all of that, black holes clearly do grow as they come in different sizes. Somehow matter must pass through the event horizon. However, I don't think this necessarily disproves my reasoning.
If the event horizon of a black hole is determined by the mass / energy within. Is it not the event horizon which grows to swallow/reestablish itself above the approaching object rather than the object traveling through the event horizon?

 

[mfb]

How does one reconcile this paradox?

Give up the idea of a global reference frame that would give useful results.

 

[Zan24C]

Give up the idea of a global reference frame that would give useful results.

Imagine that the Alcubierre drive were possible and we could travel through space faster than light. We could theoretically continuously travel between both the reference frames of the in falling object and the distant observer almost instantaneously. Assume that this third person inside the Alcubierre drive spaceship is also in a distant reference frame - the global reference frame.
Because both frames of reference can now be accessed virtually simultaneously, they must reconcile. I don't understand how giving up a global reference frame helps in any way.

 

[mfb]

Imagine that the Alcubierre drive were possible and we could travel through space faster than light.

Then we can also travel backwards in time and break causality, and not even the question "what caused what" works any more. It will depend on how you travel.

 

[Zan24C]

Then we can also travel backwards in time and break causality, and not even the question "what caused what" works any more. It will depend on how you travel.

I'm not an expert on the Alcubierre drive but I believe all the problems you mentioned have been previously debunked on this very forum. Regardless, I only introduced the Alcubierre drive as a thought experiment. Theoretically the distant observer could also quantum tunnel to and from the in falling object extremely rapidly, again requiring both reference frames to reconcile.

 

[pervect]

I realize that this question has been asked many times on this forum, however, I have yet to come across a satisfying/understandable answer that takes into account gravitational time dilation.

Premise:
The speed of light inside a gravitational field is slowed down relative to a distant observer by gravitational time dilation as proven by the Shapiro delay experiment. Therefore the speed of any object is slowed down in a gravitational field as seen by a distant observer. The speed of an object/light is related to the strength of the gravitational field.

The premise itself is a bit iffy. Unfortunately, even perfectly sound logic will give you misleading results if your premises are wrong.

There are some foundational issues about the speed of a distant object even being definable. See for instance Baez's remarks in "The Meaning of Einstein's equation" <<link>>.

Before stating Einstein’s equation, we need a little preparation. We assume the reader is somewhat familiar with special relativity — otherwise general relativity will be too hard. But there are some big differences between special and general relativity, which can cause immense confusion if neglected.

I have no idea of your background, but I will mention that I have a certain learned caution for assuming that posters on PF are really familiar with special relativity (SR) and not just jumping into general relativity (GR) without a good understanding of SR. But that's not the main point, so I'll skip on to what is.

In special relativity, we cannot talk about absolute velocities, but only relative velocities. For example, we cannot sensibly ask if a particle is at rest, only whether it is at rest relative to another. The reason is that in this theory, velocities are described as vectors in 4-dimensional spacetime. Switching to a different inertial coordinate system can change which way these vectors point relative to our coordinate axes, but not whether two of them point the same way.

In general relativity, we cannot even talk about relative velocities, except for two particles at the same point of spacetime — that is, at the same place at the
same instant. The reason is that in general relativity, we take very seriously the notion that a vector is a little arrow sitting at a particular point in spacetime.

To compare vectors at different points of spacetime, we must carry one over to the other. The process of carrying a vector along a path without turning or stretching it is called ‘parallel transport’. When spacetime is curved, the result of parallel transport from one point to another depends on the path taken! In fact, this is the very definition of what it means for spacetime to be curved.

Thus it is ambiguous to ask whether two particles have the same velocty vector unless they are at the same point of spacetime.

Unfortunately, if you are not already familiar with parallel transport, this explanation may be difficult to follow. And I'm not sure if I have any good non-technical references on the topic. Perhaps another poster can suggest some.

If I feel ambitious, I may post a bit on parallel transport, but in a separate thread - it would be too derailing to post it to this thread, I think, it's really a separate topic. The point I wanted to mention was that it is dangerous to assume that distant objects have a well-defined velocity as a premise for an argument. So my response would be that even if your logic is sound, there are some issues with your starting premise.

I'll add something that I hope may be more helpful. It is not necessary to assume that the speed of light is slowed down at all by gravity. In fact, if you look at the purely local speeds, the local speed of light is not affected at all, it is in fact a premise of both SR and GR that the local speed of light is constant.

So it is not necessary (and in my opinion not desirable) to formulate GR in the manner you have in your premise. It's better (and also more standard) to formulate GR as a geometric theory in which the local speed of light is always constant and equal to "c". By not introducing the concept of global speeds, but restricting ourself to local speeds, we can avoid the pitfalls Baez mentions.

The standard approach to GR and the complexities introduced by curvature is similar to how we handle the problems of making a map of the curved surface of the Earth. No map drawn on a flat sheet of paper of the globe can be perfectly to scale, but if we consider a small enough area, we can draw an approximate map that's good enough to cover some local region. So we can utilize a street atlas with a lot of flat maps, to plan a route in a city, without learning about spherical geometry. If we wanted to plan an ocean voyage, though, we would probably have to make the effort to learn a bit about spherical geometry first.

Similar remarks can be made about GR. If we look at a small enough region, we can avoid having to learn about Riemannian geometry.

It's really, really helpful if one appreciates the geometric formulation of SR first, before one tackled GR. In the geometric formulation of SR, we view space-time as having a unified geometry, a non-Euclidean geometry, called a Lorentian geometry. So studying how to go from Euclidean geometry to Lorentzian geometry is a way to learn SR that is a good step for learning GR. The next larger step goes from Lorentzian geometries (which are flat) to Riemannian and pseudo-Riemannian geometries, which are in general not flat.

 

[mfb]

Theoretically the distant observer could also quantum tunnel to and from the in falling object extremely rapidly, again requiring both reference frames to reconcile.

It cannot. Quantum tunneling cannot lead to superluminal information transfer. Quantum field theory is local, and while some interpretations of it are not local, none of them can allow superluminal information transfer (because that is an interpretation-independent process).

pervect explained the problem with the global reference frame nicely.

 

[Zan24C]

Quantum tunneling cannot lead to superluminal information transfer.

That may be so. I'm not very familiar with quantum tunnelling and was probably a little too hasty in my reply. Even so, as I understand it, quantum tunnelling is instantaneous and there is no upper limit on the distance that can be traversed. What fundamentally stops an entire person from tunnelling immense distances fully intact?

I have no idea of your background, but I will mention that I have a certain learned caution for assuming that posters on PF are really familiar with special relativity (SR) and not just jumping into general relativity (GR) without a good understanding of SR.

I'm a second year physics undergrad. We have only lightly covered special relativity but I'll try my best.

Unfortunately, if you are not already familiar with parallel transport, this explanation may be difficult to follow

Thank you for the link, it's an interesting read. But your right about it being confusing for someone who hasn't heard about parallel transport before. I am confused.
As far as I understand, each vector makes sense in describing the velocity of a particle in the context of its own local region of space. To compare vectors you first need to convert one of the vectors from one language to another(one context to another). To do this you parallel transport which can be ambigous because the conversion you get depends on the method you used(path you took).

What I don't understand is why there exists a need for a conversion in the first place. Isn't the whole point of a vector to convey all the necessary information about an object simply, by using a standardised coordinate system?
Using your analogy of the map of the earth: You wouldn't need to parallel transport across the surface of the map to map out an intercontinental journey if your coordinate system was in a higher dimension.
Equivalently, using the analogy of the ball in the link you posted: Parallel transport wouldn't be needed if you had a 3 dimensional coordinate system to describe the curvature of the 2 dimensional surface of the ball.

So I guess what I am trying to say is, could general relativity be made less ambiguous by introducing a higher reference frame? A reference frame of some sort beyond 4D space time.

I'll add something that I hope may be more helpful. It is not necessary to assume that the speed of light is slowed down at all by gravity. In fact, if you look at the purely local speeds, the local speed of light is not affected at all, it is in fact a premise of both SR and GR that the local speed of light is constant.

So it is not necessary (and in my opinion not desirable) to formulate GR in the manner you have in your premise. It's better (and also more standard) to formulate GR as a geometric theory in which the local speed of light is always constant and equal to "c". By not introducing the concept of global speeds, but restricting ourself to local speeds, we can avoid the pitfalls Baez mentions.

I'm not sure I understand what you mean by 'it is not necessary to assume the speed of light slows down'. Do you mean necessary to show that the in falling object crosses the event horizon? I know it does, I said as much my self. But by sticking with local reference frames are we not just choosing to ignore the perspective of the outside observer and the resulting inconsistency in events?
Just because introducing a global reference frame complicates things doesn't make unworthy of investigating

Thank you both for your input thus far.

 

[PeterDonis]

I realize that this question has been asked many times on this forum, however, I have yet to come across a satisfying/understandable answer that takes into account gravitational time dilation.

That's because gravitational time dilation, as a concept, only applies to static observers (observers who are at a constant altitude above the horizon), and there are no static observers at or below the horizon. So you are trying to apply a concept that simply has no meaning in the domain in which you are trying to apply it.

as I understand it, quantum tunnelling is instantaneous and there is no upper limit on the distance that can be traversed

Your understanding is incorrect. Where did you get this from?

Isn't the whole point of a vector to convey all the necessary information about an object simply, by using a standardised coordinate system?

Yes, but with a key additional aspect that you did not state: a vector only describes something in a particular small neighborhood of spacetime. A vector is "attached" to a particular event in spacetime (a point in space at an instant of time), and only has meaning in the immediate neighborhood of that event (i.e., at points/times infinitesimally close to the event the vector is attached to). Having a "standardised coordinate system" does not change that--all a coordinate system does is attach an agreed upon set of 4 numbers to every event, so we can unambiguously refer to events and know which ones we are talking about. It doesn't let you compare vectors attached to events that are separated from each other.

Using your analogy of the map of the earth: You wouldn't need to parallel transport across the surface of the map to map out an intercontinental journey if your coordinate system was in a higher dimension.

It's not a matter of the coordinate system being in a higher dimension: it's a matter of the Earth being embedded in a higher dimensional space which happens to be flat (at least in the Earth's vicinity, to a good enough approximation). So in that higher dimensional space, parallel transport is unambiguous (since the space is flat), and you can compare vectors at widely separated points. But that's not because parallel transport isn't required; it's because you are working in a flat space.

There is no analogue of this for spacetime. The spacetime of our universe is not embedded in any higher dimensional flat space. (Even in theories like string theory that postulate extra dimensions, the higher dimensional space that includes those extra dimensions is not flat.) So there is no way to avoid the ambiguities in parallel transporting vectors between distant events.

 

[Zan24C]

That's because gravitational time dilation, as a concept, only applies to static observers (observers who are at a constant altitude above the horizon), and there are no static observers at or below the horizon. So you are trying to apply a concept that simply has no meaning in the domain in which you are trying to apply it.

Please look up the Shapiro delay. It's in the title of this thread.

Your understanding is incorrect. Where did you get this from?

It may well be. I deduced it from the limited understanding I have of quantum mechanics. A wavefunction represents a probability distribution of finding a particle in a particular place along that wavefunction. Wavefunctions span the entire universe. When you are not observing the particle, it is in a superposition of states. When you begin to observe it it can appear at any position depending on the probability distribution of the wavefunction. What stops it from materialising on the other side of the universe?
If my understanding is wrong can you explain why?

It's not a matter of the coordinate system being in a higher dimension: it's a matter of the Earth being embedded in a higher dimensional space which happens to be flat

That's what I alluding to when I suggested using a higher dimension. I should have made that more clear. If the higher dimension is not flat, why invoke it at all?

There is no analogue of this for spacetime. The spacetime of our universe is not embedded in any higher dimensional flat space.

Why? Is it impossible to invent a new theory that embeds general relativity in a flat higher dimensional reference frame?
I don't know. I'm asking.

 

[Nugatory]

Please look up the Shapiro delay. It's in the title of this thread.

You misunderstand Peter's point about the Shapiro delay. He's not saying it doesn't exist, he's saying that gravitational time dilation is only a meaningful concept for static observers so won't help visualize what's going on with someone falling towards and through the event horizon.

Wavefunctions span the entire universe. When you are not observing the particle, it is in a superposition of states. When you begin to observe it it can appear at any position depending on the probability distribution of the wavefunction. What stops it from materialising on the other side of the universe?
If my understanding is wrong can you explain why?

You are thinking of non-relativistic quantum mechanics, QM formulated without considering the effects of relativity. That's the only kind you'll encounter in the first few years of a college-level physics degree program so that's the only kind that most people ever hear about. The proper relativistic formulation of quantum mechanics, which you probably won't ever meet unless you're in a PhD program, doesn't have these universe-spanning relativity-defying global-frame wave functions. The apparent contradiction you're seeing is the result of trying to use a theory that ignores relativistic effects in a situation where they are to large to ignore.

 

[Zan24C]

You misunderstand Peter's point about the Shapiro delay. He's not saying it doesn't exist, he's saying that gravitational time dilation is only a meaningful concept for static observers so won't help visualize what's going on with someone falling towards and through the event horizon.

So far no one has even attempted to answer the question. Everyone has decided to concern themselves with explaining the conventions in examining general relativity.
I'm not interested in the conventional approach, the question is much simpler than this.
I understand that you don't need a global reference frame to "predict" what happens in any particular reference frame. I am bringing into the question the validity of this very assumption. The penrose diagram also describes what happens to a person falling into a black hole. But only from the perspective of the falling person. It does not work at all at describing what an observer would see because they are two very different reference frames. It completely neglects time dilation.

In one reality (the distant observer's), the in falling person never crosses the horizon. In another (the in falling person's), he does. Which is correct? There is a concept called the relativity of simultaneity which explains how events that are simultaneous in one reference frame may not be in another. The end result however must be agreed upon. And crossing the event horizon is a very clear event because it has very clear implications. If the person crosses, he will never get out. If he does not cross he will.
In fact you could design a test to prove whether or not there grows a virtually infinite time difference between the two reference frames. When the in falling person gets extremely close to the black hole, he fires rockets to get himself back out. If from the outside observer's perspective it takes him quintillions upon sextillions of years (just a very large number) to go down and come back up but only minutes from the infalling person's perspective, you have proven the existence paradox.

Can you or can you not run this test?

ou are thinking of non-relativistic quantum mechanics, QM formulated without considering the effects of relativity. That's the only kind you'll encounter in the first few years of a college-level physics degree program so that's the only kind that most people ever hear about. The proper relativistic formulation of quantum mechanics, which you probably won't ever meet unless you're in a PhD program, doesn't have these universe-spanning relativity-defying global-frame wave functions. The apparent contradiction you're seeing is the result of trying to use a theory that ignores relativistic effects in a situation where they are to large to ignore.

Noted, thanks.

 

[Ibix]

Can you or can you not run this test?

Of course. We've done it by taking clocks to the top of a mountain, waiting, and coming back down. The clocks disagree with ones left in the lab. That's less extreme than your experiment, but it's the scenario you are talking about.

The result is not paradoxical. One clock took a longer route through spacetime than the other, that's all. It's not really any different from two cars starting at the same factory and ending at the same scrapyard but having different odometer readings.

 

[Zan24C]

The result is not paradoxical. One clock took a longer route through spacetime than the other, that's all.

It is paradoxical because unlike the route up and down a mountain, the route to the event horizon is infinitely long.

 

[Ibix]

It is paradoxical because unlike the route up and down a mountain, the route to the event horizon is infinitely long.

There is no timelike route touching the event horizon and returning to the outside. That's pretty much the definition of an event horizon.

 

[Nugatory]

In one reality (the distant observer's), the in falling person never crosses the horizon. In another (the in falling person's), he does. Which is correct?

It's not correct to say that "in one reality (the distant observer's), the in falling person never crosses the horizon". What's going on here is that the outside observer is using a particular (and locally sensible to him) convention for assigning time values to distant events; and this convention doesn't assign any time value to the event "infaller crosses horizon" so there's no way to describe that event in those coordinates. That doesn't mean the event didn't happen, it means that we've made a poor choice of coordinates for analyzing this problem.

Let's consider one way that the outside observer could be assigning time values to events on the infaller's worldline. The method I'm going to use ("radar coordinates") is different from the Schwarzschild coordinates that you've been assuming throughout this thread, but it is equivalent to them in the local patch of spacetime around the observer, has the same problem with assigning a time to the horizon-crossing event, aand is easier to visualize. You are free to choose some other method (but be sure that you can explain how it's implemented - it doesn't work to, for example, assert that you and the infaller can both just look at the same big clock in the sky!) but no matter which one you choose you'll conclude that the infaller does cross the horizon and light from that event doesn't reach the outside observer.

Let's say that I am the outside observer and at time $T$ according to my wristwatch I send a radio message to the infaller and the infaller immediately replies (he might be receiving then transmitting a message back, or we might just be bouncing my signal back from a mirror). I receive the reply at time $T+\Delta{T}$. This tell me that he received the signal at the same time (using this particular convention for defining "at the same time" - your point about relativity of simultaneity is well-taken) that my wristwatch read $T+\frac{\Delta{T}}{2}$; the distance out is the same as the distance back so the time out has to be equal to the time back.

Now consider what happens as I send a continuous stream of such signals. The $\Delta{T}$ values become longer and longer, and I never stop receiving replies even if I wait forever (although the replies may be redshifted beyond the ability of my receiver to detect them, they never stop). However, there is a moment when I send the signal that reaches the infaller just as he passes through the horizon - his reply never makes it back to me, so I don't have a $\Delta{T}$ value and the $T+\Delta{T}$ convention stops working as a means of assigning times to events on the infaller's worldline. However, that's telling us something about the behavior of light signals between me and the infaller, not about what's happening to the infaller. If I want to describe events on the infaller's worldline, I have to choose some other convention for assigning times.

 

[Nugatory]

It is paradoxical because unlike the route up and down a mountain, the route to the event horizon is infinitely long.

It is not infinitely long. It has a finite length that is easily calculated.

The apparent infinity on the inbound path is the result of trying to use Schwarzschild coordinates, which have a coordinate singularity at the event horizon so do not apply in this situation (at a handwaving level, using them across the event horizon is a mathematical error akin to dividing both sides of an equation by zero). Kruskal, Eddington-Finkelstein, and Painleve coordinates will all work and will give you the right answer. (The radar coordinates I described in my previous post won't work, but I deliberately chose them because they have the same sort of pathology as Schwarzschild coordinates).

 

[Zan24C]

There is no timelike route touching the event horizon and returning to the outside.

I think you missed my point by about a million miles here.
I meant simply just to the horizon in the first place.

Now consider what happens as I send a continuous stream of such signals. The ΔTΔT\Delta{T} values become longer and longer, and I never stop receiving replies even if I wait forever (although the replies may be redshifted beyond the ability of my receiver to detect them, they never stop). However, there is a moment when I send the signal that reaches the infaller just as he passes through the horizon - his reply never makes it back to me, so I don't have a ΔTΔT\Delta{T} value and T+ΔTT+ΔTT+\Delta{T} convention stops working as a means of assigning times to events on the infaller's worldline. However, that's telling us something about the behavior of light signals between me and the infaller, not about what's happening to the infaller.

Is the theory that explains why the light signals are delayed not exactly the same theory that explains the motion of the falling person? Regardless of what analogy you want to use - either there is more space so the light has more distance to travel or the speed of the beam itself slows down - the same should apply to the falling observer. The closer to the event horizon you are, the slower you would travel out and the slower you would travel in.

The apparent infinity on the inbound path is the result of trying to use Schwarzschild coordinates, which have a coordinate singularity at the event horizon so do not apply in this situation (at a handwaving level, using them across the event horizon is a mathematical error akin to dividing both sides of an equation by zero). Kruskal, Eddington-Finkelstein, and Painleve coordinates will all work and will give you the right answer.

By changing the coordinate system all you're doing is changing reference frames and ignoring time dilation again.
Clearly the Schwarzschild coordinates singularity has real physical significance because it predicts the increase in signal delay that you mentioned.

It is not infinitely long. It has a finite length that is easily calculated.

But only using a coordinate system that ignores time dilation? I'm getting out of my depth here.

 

[Ibix]

I think you missed my point by about a million miles here.
I meant simply just to the horizon in the first place.

Then your point is not correct. From any point outside the event horizon, there exist timelike paths to the event horizon that have finite proper time along them.

 

[Zan24C]

Then your point is not correct. From any point outside the event horizon, there exist timelike paths to the event horizon that have finite proper time along them.

Could you please give an example of this? I'm not actually very familiar with timelike paths.

 

[Ibix]

Could you please give an example of this? I'm not actually very familiar with timelike paths.

Free-falling into the black hole is following a timelike path. You are currently following a timelike path. Anything with non-zero mass always follows a timelike path.

With all due respect, if you don't know what a timelike path is, do you really think you understand this well enough to be criticising? It's kind of like complaining that Euclidean geometry doesn't make sense and then saying you're not very familiar with Pythagoras' Theorem.

 

[PeterDonis]

Please look up the Shapiro delay.

I don't need to, I know what it is. I also know that it is only applicable outside the horizon.

A wavefunction represents a probability distribution of finding a particle in a particular place along that wavefunction. Wavefunctions span the entire universe. When you are not observing the particle, it is in a superposition of states. When you begin to observe it it can appear at any position depending on the probability distribution of the wavefunction. What stops it from materialising on the other side of the universe?

First of all, what you describe is not quantum tunneling, it's just measuring the position of an object.

Second, not all wave functions span the entire universe. The wave function of a particle in a potential well can be confined to a finite spatial extent. And the particles inside a macroscopic object like a person (or a rock, for that matter) are certainly confined in a potential well, since they are all interacting strongly with each other (which is why the object acts like a single object).

Is it impossible to invent a new theory that embeds general relativity in a flat higher dimensional reference frame?

It's not a question of inventing a theory, it's a question of direct observation. We know the Earth is embedded in a higher dimensional flat space because we can directly observe it. We don't directly observe any higher dimensional space that the universe is embedded in.

Mathematically, of course you can embed any curved space in a flat space of some higher dimensionality. But so what? That's not a theory, it's just math with no physical meaning.

In one reality (the distant observer's), the in falling person never crosses the horizon. In another (the in falling person's), he does.

This is wrong. There is only one reality, and in it the falling person does cross the horizon. The distant observer can't see it happen, but that doesn't mean it doesn't happen, just as if an object goes beyond your horizon on Earth, you can't see it, but it still exists. You should read my Insights article on this:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

 

[Nugatory]

But only using a coordinate system that ignores time dilation? I'm getting out of my depth here.

There is no time dilation unless you choose a coordinate system that has time dilation - it is a property of the coordinate system you've chosen. It's easiest to see this in the flat spacetime of special relativity: If you and I are moving relative to one another and we choose coordinates in which you are at rest my clock will be dilated; but if we choose coordinates in which I am at rest your clock will be dilated.

If you work through the Lorentz transforms, you will see that's what really going on is that in one coordinate system we assigned the time "1:00 PM" to the events "my clock reads 1:00 PM" and "your clock reads 12:30 PM" while in the other coordinate system we assigned the label "12:30 PM" to the events "my clock reads 12:15 PM" and "your clock reads 12:30 PM". Note that that is three events, and the two that are simultaneous using one coordinate system are not simultaneous in the other, and vice versa.
(To actually work this example out with the Lorentz transforms, you'll need a fourth event "both clocks were at the same place at the same time and we set them both to read 12:00 noon", which is labeled "12:00 noon" in both coordinate systems. The relative speed between us will be about .87c).

(This might be a good time to mention that the Twin Paradox, in which one twin embarks on a high-speed round trip journey and returns having aged less than his sibling who stays home on Earth is NOT a time dilation problem and cannot be properly understood in terms of time dilation).

 

[PeterDonis]

I'm not actually very familiar with timelike paths.

Then, as Ibix has pointed out, you are lacking a very critical piece of knowledge that is highly relevant to the topic at hand. Knowing what a timelike path is is necessary background for any "I" level discussion of this topic.

That said, the formula for the finite time $\tau$, by the falling observer's clock, that it takes for him to fall from some radius $R$ outside the horizon, to the horizon, is simple; it's this:

$$
\frac{\tau}{2M} = \frac{2}{3} \left[ \left( \frac{R}{2M} \right)^{\frac{3}{2}} - 1 \right]
$$

where I have used units in which $G = c = 1$. For a numerical example, a black hole with the mass of the Sun has $2M$ equal to approximately 3 kilometers, so an observer who falls into such a hole from a radius of, say, 3 million kilometers (several times the radius of the actual Sun) would reach the horizon in about 2 billion kilometers of light travel time, or about 6,667 seconds by the observer's clock. (And he would hit the singularity in about 3 kilometers of light travel time after that, or about 10 microseconds--the time from the horizon to the singularity is just $\tau = 2M$.)

Regardless of what analogy you want to use - either there is more space so the light has more distance to travel or the speed of the beam itself slows down - the same should apply to the falling observer.

No, this is not correct. You are mixing up outgoing objects with ingoing objects. Gravity is attractive, so, heuristically, things falling towards the hole are sped up and things rising away from the hole are slowed down. (I say "heuristically" because light changes frequency, not speed, as seen by local observers, and because there are coordinate-dependent effects here as well.) So an observer, or a light ray for that matter, that is going inwards, towards the hole, has no trouble reaching the horizon and falling on inward. But a light ray, or an observer for that matter, that is shot upwards away from the hole will have more and more trouble getting away the closer they are to the horizon--and at the horizon it is impossible to escape. There's no mystery to this: it's just simple attractive gravity.

Clearly the Schwarzschild coordinates singularity has real physical significance because it predicts the increase in signal delay that you mentioned.

This is wrong too. The signal delay is an invariant--you can compute it in any coordinates you like. What predicts the increase in signal delay is the curvature of the spacetime (and the presence of the event horizon). But the curvature and the presence of the event horizon are also invariants--they are there in all coordinates.

The problem with Schwarzschild coordinates is that they get more and more distorted as you get closer to the horizon, and at the horizon, they are infinitely distorted, so to speak (that is the coordinate singularity). That means that the Schwarzschild time coordinate, which is what you are basing your reasoning on, is simply misleading you; it is not giving you an accurate picture of what the spacetime geometry looks like near and at the horizon. I discuss this in the series of Insights articles that I linked to above.

When the in falling person gets extremely close to the black hole, he fires rockets to get himself back out. If from the outside observer's perspective it takes him quintillions upon sextillions of years (just a very large number) to go down and come back up but only minutes from the infalling person's perspective, you have proven the existence paradox.

Can you or can you not run this test?

Yes, you can run it, and GR predicts that the result will be exactly what you describe--it will take a very long time by the distant observer's perspective, but a much, much shorter time from the infalling person's perspective. There is no paradox at all; it's just spacetime geometry. It's no different from one person going from New York to Washington, DC by going south on Interstate 95, while another takes a boat to Europe, crosses the continent of Eurasia, takes another boat across the Pacific to San Francisco, and then takes a train cross country to Washington. They cover very different distances between the same two points, not because of any paradox, but because of the geometry of Earth's surface.

The counterintuitive part of the scenario is that the distant observer, who appears to just "stay in the same place", is the one who has the much longer elapsed time. That is a consequence of the geometry of spacetime being locally Minkowskian instead of locally Euclidean. There's no paradox at all; you just have to get used to the effects of the different metric signature.

 

[Nugatory]

Clearly the Schwarzschild coordinates singularity has real physical significance because it predicts the increase in signal delay that you mentioned.

The Schwarzschild coordinate singularity does not have anything to do with the change in signal delay, because the light beam is never traversing the region of spacetime where the Schwarzschild metric is singular; and when you calculate that delay (which is more rigorously defined as "the amount of proper time that elapses on my worldline between the events "I sent a signal" and "I received the reply") you will get the same answer no matter what coordinate system you use.

The coordinate singularity has no more physical significance than any other coordinate singularity, such as at the one at the origin when you use polar coordinates on an ordinary sheet of paper, or when you try to find the longitude of the north and south pole. There's nothing physically strange about the paper at the origin, there's nothing physically strange about the surface of the Earth at the poles and there's nothing physically strange about the spacetime at the event horizon.

 

[Zan24C]

PeterDonis, oh man.

First of all, what you describe is not quantum tunneling, it's just measuring the position of an object.

The chance of a particle appearing on the other side of a barrier is given by the wavefunction which spills over to the other side of the barrier.

the particles inside a macroscopic object like a person (or a rock, for that matter) are certainly confined in a potential well, since they are all interacting strongly with each other (which is why the object acts like a single object).

LOL no. They are not certainly combined because a probability they may escape no matter how small is still a probability. This is true even if it were < 1/10^100^100^100^100^100^100 chance. This is terribly abstract, I know, but hey, theoretical physics generally is.
Either way, it's not the subject of this thread and I would like to stop talking about it now.

It's not a question of inventing a theory, it's a question of direct observation.

IT ABSOLUTELY IS A QUESTION OF INVENTING A THEORY. That's what every mathematical theory is. No one has ever observed spacetime curvature. They have observed that a mathematical model has adequately described a physical phenomenon. Get the same result using a mathematical model with a different interpretation of reality and you will not know which one is more correct. Be careful, this is philosophy and goes into metaphysics - something I sense you are not comfortable with.

Mathematically, of course you can embed any curved space in a flat space of some higher dimensionality. But so what? That's not a theory, it's just math with no physical meaning.

Basically what I wrote immediately above.

There is only one reality, and in it the falling person does cross the horizon.

My experiment suggests that with respect to the greater universe, he never crosses the event horizon.

The distant observer can't see it happen, but that doesn't mean it doesn't happen,

My experiment suggests that with respect to the greater universe, he never crosses the event horizon.

read my Insights article on this:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

Quoted from your article:
"What is going on here? We have two seemingly contradictory predictions: the equation for dr/dtdr/dt is telling us that an infalling object slows down more and more and never quite reaches r=2Mr=2M; while the equation for dr/dτdr/dτ is telling us that the object continues to speed up as it falls, and goes right through r=2Mr=2M and beyond. Which one is right?"
"The idea is to try to rescale the time coordinate so that it reflects the proper time of the radially infalling object. The transformation that does that is:"

Ignore the reference frame of the observer. AGAIN.

"So the behavior of dr/dtdr/dt in Schwarzschild coordinates as r→2Mr→2M was in fact an artifact of the coordinates and was not telling us anything physical."

My experiment suggests that with respect to the greater universe, he never crosses the event horizon.

That said, the formula for the finite time ττ\tau, by the falling observer's clock, that it takes for him to fall from some radius RRR outside the horizon, to the horizon, is simple; it's this:

τ2M=23[(R2M)32−1]​

You've done it again. See my comments about your article immediately above.

No, this is not correct. You are mixing up outgoing objects with ingoing objects. Gravity is attractive, so, heuristically, things falling towards the hole are sped up and things rising away from the hole are slowed down. (I say "heuristically" because light changes frequency, not speed, as seen by local observers, and because there are coordinate-dependent effects here as well.)

'as seen by local observers'... And again.

This is wrong too. The signal delay is an invariant--you can compute it in any coordinates you like. What predicts the increase in signal delay is the curvature of the spacetime (and the presence of the event horizon). But the curvature and the presence of the event horizon are also invariants--they are there in all coordinates.

Because you are now factoring in a change in reference frame from local to global, where there is a change in spacetime curvature.

That means that the Schwarzschild time coordinate, which is what you are basing your reasoning on, is simply misleading you; it is not giving you an accurate picture of what the spacetime geometry looks like near and at the horizon.

I'm pretty sure it does because:
My experiment suggests that with respect to the greater universe, he never crosses the event horizon.

Yes, you can run it, and GR predicts that the result will be exactly what you describe--it will take a very long time by the distant observer's perspective, but a much, much shorter time from the infalling person's perspective. There is no paradox at all

Do you know what a paradox is?
My experiment suggests that with respect to the greater universe, he never crosses the event horizon.

It's no different from one person going from New York to Washington, DC by going south on Interstate 95, while another takes a boat to Europe, crosses the continent of Eurasia, takes another boat across the Pacific to San Francisco, and then takes a train cross country to Washington. They cover very different distances between the same two points, not because of any paradox, but because of the geometry of Earth's surface.

τ2M=23[(R2M)32−1]

If the person traveling around the world took an infinite amount of time from the perspective of the person going south on the interstate - which is the more appropriate analogy by the way - then yes it would be.

Please don't misunderstand me. It really is simple. I'm not saying what GR predicts happens from the perspective of the infalling person is wrong. I'm saying:
My (thought)experiment suggests that with respect to the greater universe, he never crosses the event horizon.

 

[mfb]

You can find coordinate frames where the object never crosses the event horizon.
You can also find coordinate frames where a falling object never hits the surface of Earth. That is a silly example, but it shows the point: It is just a matter of coordinates. Coordinates are arbitrary, they don't have a physical meaning.

PeterDonis, oh man.

You could re-consider your attitude here. If an expert like PeterDonis tells you you are wrong, what is more likely? (a) the expert who worked on that topic for years is wrong, or (b) the student who just started learning about relativity is wrong? Telling the expert he should look up what the Shapiro delay is has some irony.

 

[Zan24C]

To be clear, the global frame of reference frame of the greater universe IS important because it is the ONLY perspective that matters when "WE ON EARTH" want to determine whether:

• A black hole has grown,
• How its effect on its surroundings will change,
• That infalling person can still come back,
• etc

 

[Ibix]

My experiment suggests that with respect to the greater universe, he never crosses the event horizon.

There's a difference between "never sees it happen" and "never happens". Your experiment concerns the former - you're just sending somebody down close to the horizon to do the observing and waiting for him to report back.

Can I suggest dialling back the attitude? You've admitted not being familiar with basic concepts in relativity, but are being patronising and bordering on outright rude to people who do understand it and are trying to help you improve your understanding.

 

[Zan24C]

You could re-consider your attitude here.

You are right. Peter I sincerely apologise. I truly appreciate that you took time to reply to this thread and are trying to help me.
I get too worked up about physics too often.

 

[Zan24C]

There's a difference between "never sees it happen" and "never happens". Your experiment concerns the former - you're just sending somebody down close to the horizon to do the observing and waiting for him to report back.

But according to my thought experiment, there is always a time for the distant observer which corresponds with the infalling person being able to reverse course and come back.
So, by logic, as far as the distant observer is concerned it never happens because there is never a time when the infalling observer cannot come back. Except at infinity. Yet in his own perspective the infalling person could activate his rocket a fraction of a second too late and cross the event horizon. That fraction of a second has somehow bridged the gap between all of the observer's future time and infinity.

Just to illustrate the change in speed of light near a black hole, I have found this graphic posted by John Rennie as a reply in http://physics.stackexchange.com/questions/77227/speed-of-light-in-a-gravitational-field

The variation of the velocity of light with distance from the black hole looks like:

calculated for a light ray heading directly towards a black hole in the reference frame of a distant observer, where v is the speed of light as a fraction of c and r is the distance from the centre of the black hole as a fraction of the radius of the event horizon.

 

[PeterDonis]

it's not the subject of this thread and I would like to stop talking about it now.

Fair enough, but you should be aware that we have no experimental evidence that extremely small probabilities like 1 / googol have any physical meaning. It is one thing to say that non-relativistic quantum mechanics mathematically includes such probabilities (and even there the "non-relativistic" is key--as has already been pointed out, when you include relativity you are using quantum field theory, which works differently); it is quite another to insist that such probabilities are real. You can open a new thread in the QM forum if you want to ask about the limitations of this viewpoint.

I truly appreciate that you took time to reply to this thread and are trying to help me.

Thanks for the appreciation. I understand that this is a difficult topic, and what GR says in this case is very counterintuitive. I'll limit my response to your previous posts to pointing out a few specific statements that seem to me to pinpoint where the counterintuitiveness is concentrated, so to speak.

Ignore the reference frame of the observer. AGAIN.

There is a very important point to be made here, which I think is made in the Insights series but perhaps I need to go back and stress it more. Every observer's "reference frame", strictly speaking, is local. In other words, every observer can construct a "reference frame" that describes events in his local vicinity. And in his local vicinity, he has a valid argument for treating the description of his observations in terms of his local reference frame as "real".

But any attempt to extend that beyond the local vicinity of that observer no longer can be justified by such an argument in the general case. In flat spacetime, there happens to be a set of reference frames--the inertial frames--that can be so justified; but that is a particular feature of flat spacetime and inertial frames and cannot be extended beyond that. For example, even in flat spacetime, non-inertial frames will not, in general, cover the entire spacetime. (Rindler coordinates, for example, which are the natural "reference frame" for an observer with uniform proper acceleration, have a horizon that has a number of similarities with the event horizon in Schwarzschild spacetime.) And in curved spacetime, even inertial frames (i.e., frames in which an object at rest is in free fall--note that Schwarzschild coordinates are not such a frame, btw) are only local and cannot cover an extended region of spacetime, let alone the entire spacetime.

So when you say that "the reference frame of the distant observer says the falling observer never crosses the horizon", you are trying to extrapolate the distant observer's frame beyond the region where it is valid. You simply can't treat that reference frame, or any reference frame, in curved spacetime the way you treat an inertial frame in flat spacetime. This goes for the falling observer's frame too, btw: strictly speaking, the falling observer's frame, in which it is simple to show that he reaches the horizon in finite time, only covers his local region and cannot be extrapolated to cover the distant observer. So he can't assume, for example, that the coordinate time in his reference frame (which is basically Painleve coordinates) will be meaningful in the distant observer's vicinity.

The upshot of all this is that there is no "reference frame" that you can use to describe the entire geometry of a curved spacetime like Schwarzschild spacetime. The only way to do it is to use coordinate charts, which in general will not correspond everywhere to any observer's "reference frame". And there is no single chart that can represent the entire geometry without distortion, so coordinate charts have limitations too if you're trying to visualize what's going on. We don't have to confront this in everyday life because we are used to limited domains, like the vicinity of the Earth, in which we can find single "reference frames" or coordinate charts that cover everything we are interested in. But that's because we live our everyday lives in a limited domain, not because it's always possible.

Ultimately, the physics comes down to invariants--quantities that do not depend on your choice of reference frames or coordinate charts. When I say that the falling observer reaches the horizon in finite time, that is an invariant: the invariant length of a particular timelike curve in the spacetime geometry. It's no different from saying that the great circle distance from New York to London is an invariant on the Earth's surface, and doesn't depend on whether you use spherical coordiates, Mercator coordinates, or stereographic coordinates, or even if you choose a reference frame or coordinates that include New York but don't include London. London is there in the geometry whether your coordinates include it or not; you can compute geometric invariants that show that. Similarly, the horizon and the region inside it are there in the Schwarzschild geometry, whether your coordinates include it or not; you can compute geometric invariants that show that.

Because you are now factoring in a change in reference frame from local to global, where there is a change in spacetime curvature.

No, I'm not. The signal delay is an invariant, as I said. Invariants are independent of any choice of reference frame or coordinates. See above.

Also, the spacetime curvature is itself an invariant: it also doesn't depend on your choice of reference frame or coordinates. So saying that changing reference frames changes the spacetime curvature is wrong.

If the person traveling around the world took an infinite amount of time from the perspective of the person
going south on the interstate - which is the more appropriate analogy by the way

No, it isn't. In your thought experiment about an observer going down close to the horizon, staying there for a while, then coming back up to the distant observer, both of their elapsed times are finite.

If, OTOH, the falling observer actually reaches the horizon, then he can never get back out to the distant observer again, so there is no way for them to compare clock readings. And since both of their reference frames are local (see above), neither one can extrapolate his own clock readings to the other. In other words, there is no way to make the comparison you are trying to make unless the two observers can come back together, and if the falling observer reaches the horizon, they can't.

 

[PAllen]

I'm not an expert on the Alcubierre drive but I believe all the problems you mentioned have been previously debunked on this very forum. Regardless, I only introduced the Alcubierre drive as a thought experiment. Theoretically the distant observer could also quantum tunnel to and from the in falling object extremely rapidly, again requiring both reference frames to reconcile.

It is rigorously proven that two alcublerre drives can be used to construct closed time like curves, allowing general time travel. Similarly, it is established that a traversible wormhole, if it exists, can be converted to wormhole back in time.

 

[PeterDonis]

according to my thought experiment, there is always a time for the distant observer which corresponds with the infalling person being able to reverse course and come back.

Yes, but that's because all of the distant observer's "times" correspond to the falling observer being somewhere above the event horizon. And as long as the falling observer is above the horizon, then yes, he can reverse course and come back.

But once the falling observer reaches the horizon, he can no longer reverse course and come back, and there is also no longer any "time" according to the distant observer that can be extrapolated to his location. See my previous post.

The variation of the velocity of light with distance from the black hole

The variation of the coordinate velocity of light, in Schwarzschild coordinates. But no observer ever measures light moving at this velocity locally. Any observer measuring the velocity of a light beam passing him will measure it to be $c$.

 

[PeterDonis]

@Zan24C I am closing this thread because if it continues as it has been you are going to get a misinformation warning. Your posts have been addressed, and I don't see the point of continuing to repeat the same responses to the same statements by you. If you have something new to suggest that has not been already discussed in this thread, PM me and I will consider it.