Crossing the Event Horizon of a Black Hole

[DiamondGeezer]

Tending to infinity in the limit is not the same thing as reaching infinity.

A very, very large energy can be observed only by someone traveling at very, very high velocity relative to the apple being measured. The source of the high velocity is the rocket motor of the observer who has accelerated to that high velocity. No-one measures the apple's energy as infinite because no-one's rocket motors can accelerate the observer to high enough velocity. This applies to both my rocket example and a black hole. When you are close to an event horizon (or an apparent horizon) the natural tendency is to fall into it, and to resist that you need to expend a huge amount of energy.

Clearly that is incorrect. Unlike a rocket motor, the acceleration of any infalling particle to the event horizon involves arbitrarily high energies.

Or to put it another way, there's no single coordinate system in which energy conservation applies but in which the apple's energy changes from finite to infinite as it falls. The infinite value arising in this thread was a limit of energies in lots of different coordinate systems.

Except in the case of black holes, where any coordinate system regardless of whether it is inertial or not, measures the speed of an infalling particle to reach the speed of light and does so in the proper time of the particle (it would take an infinite amount of time from the perspective of an outside observer).

The energy comes from the observer's rocket, resisting the fall into the hole, and never reaches infinity in practice because the rocket eventually runs out of fuel. Shortly afterwards, the rocket falls through the horizon at a non-zero speed, and as it does so it measures the apple's energy as finite (i.e. a "physical speed" less than c).

The problem is that the energy of an infalling particle into a black hole appears to rise without limit.

So either the black hole possesses infinite energy or there's something fundamentally wrong with the theory of black holes because they possesses such unphysical properties.

 

[DrGreg]

Except in the case of black holes, where any coordinate system regardless of whether it is inertial or not, measures the speed of an infalling particle to reach the speed of light.

I don't think anyone ever said that in this thread, and it isn't true. In fact no locally inertial system measures the particle's speed as c. As I've said several times before, that value arose by considering the unrealisable limit of a family of different observers.

Unlike a rocket motor, the acceleration of any infalling particle to the event horizon involves arbitrarily high energies.

Er, a freely falling particle's acceleration (i.e. proper acceleration) is zero. The observers are accelerating, the falling particles are not. Maybe this is the cause of your confusion.

The problem is that the energy of an infalling particle into a black hole appears to rise without limit.

So either the black hole possesses infinite energy or there's something fundamentally wrong with the theory of black holes because they possesses such unphysical properties.

All of this almost seems to suggest you think energy is an absolute concept. I'm sure you know it isn't. You have to specify energy relative to something. And, in essence, that "something" has to be a particle with non-zero mass. You can't sensibly measure energy relative to an event horizon (if you try, you get an infinite answer). There is no coordinate system with 3 space axes and 1 time axis in which the event horizon is at rest. (NB. The event horizon lies outside the two Schwarzschild coordinate systems.)

 

[JesseM]

Except in the case of black holes, where any coordinate system regardless of whether it is inertial or not, measures the speed of an infalling particle to reach the speed of light and does so in the proper time of the particle (it would take an infinite amount of time from the perspective of an outside observer).

This obviously can't be true in Kruskal-Szekeres coordinates, since only null worldlines are depicted as diagonal in a Kruskal-Szekeres diagram, all timelike worldlines have a slope closer to the vertical time axis.

 

[DiamondGeezer]

This obviously can't be true in Kruskal-Szekeres coordinates, since only null worldlines are depicted as diagonal in a Kruskal-Szekeres diagram, all timelike worldlines have a slope closer to the vertical time axis.

The diagonals in the K-S diagram represent the "event horizon" aka "zero".

 

[JesseM]

The diagonals in the K-S diagram represent the "event horizon" aka "zero".

The event horizon is represented as a diagonal, but any other null geodesic would be represented as a diagonal too (and the event horizon is indeed a null geodesic itself, since a photon sent outward at the moment an object was crossing the horizon would remain on the horizon rather than falling in or escaping). That's just a property of how Kruskal-Szekeres coordinates work.

 

[DiamondGeezer]

I don't think anyone ever said that in this thread, and it isn't true. In fact no locally inertial system measures the particle's speed as c. As I've said several times before, that value arose by considering the unrealisable limit of a family of different observers.

I think I've repeated myself several times on this thread as to the difference between physical (observed) and coordinate speed.

Er, a freely falling particle's acceleration (i.e. proper acceleration) is zero. The observers are accelerating, the falling particles are not. Maybe this is the cause of your confusion.

Er, even a freely falling particle can ascertain that it is accelerating, even if it feels no net forces, by measuring tidal acceleration - a trick which won't work "inside" a black hole. It might also be able to ascertain that the Universe is accelerating relative to it.

All of this almost seems to suggest you think energy is an absolute concept. I'm sure you know it isn't. You have to specify energy relative to something. And, in essence, that "something" has to be a particle with non-zero mass. You can't sensibly measure energy relative to an event horizon (if you try, you get an infinite answer). There is no coordinate system with 3 space axes and 1 time axis in which the event horizon is at rest. (NB. The event horizon lies outside the two Schwarzschild coordinate systems.)

I don't think energy is absolute. I do think that 4-energies are invariant.

Unless you think that a black hole is massless, it too possesses an energy which is invariant.

By the way, have you spotted why the fudge of using two different metrics inside and outside the event horizon is incorrect? Because the square root of a negative number $\neq$ minus the square root of a positive number. They might meet at zero, but they are strangers after that.

 

[DrGreg]

I have no more time to comment today. But I've just time to ask this:

I do think that 4-energies are invariant.

What do you mean by that? If you'd said "4-momentum is covariant" I would have understood.

 

[DiamondGeezer]

I have no more time to comment today. But I've just time to ask this:

What do you mean by that? If you'd said "4-momentum is covariant" I would have understood.

http://en.wikipedia.org/wiki/Action_(physics)

 

[DrGreg]

Er, even a freely falling particle can ascertain that it is accelerating, even if it feels no net forces, by measuring tidal acceleration - a trick which won't work "inside" a black hole.

Tidal effects are irrelevant to what is being discussed here. The same things happen near the apparent horizon of the Rindler rocket, where there are no tidal effects. Tidal effects are relevant only to extended bodies, and here we are talking about a single particle. In relativity when we refer to acceleration without further qualification we usually mean "proper acceleration" i.e. acceleration measured by a comoving freefalling locally-inertial observer (the rate of change of "physical speed" with respect to proper time) or as measured by an acclerometer. Therefore by definition any freefalling particle's proper acceleration is zero. Any other type of acceleration is coordinate acceleration, and depends on your choice of coordinates and is therefore not "physical" in the sense we have been using that word.

Also your assertion that you can't measure tidal forces inside the event horizon is wrong; on the contrary, the tidal forces get ever stronger the further you fall in.

I do think that 4-energies are invariant.

What do you mean by that? If you'd said "4-momentum is covariant" I would have understood.

http://en.wikipedia.org/wiki/Action_(physics)

That response doesn't help. The article never uses the term "4-energy" or "invariant" (except for the phrase "adiabatic invariants" which doesn't seem to be relevant).

I have to guess that what you really meant was "4-momentum is conserved". Conservation means something quite different from invariance. Conserved quantities don't change over time (to put it simply; or more sophisticatedly they satisfy certain differential equations). Invariant quantities take the same value when measured by different coordinate systems.

If a particle has a 4-momentum P and an observer has a 4-velocity U, then the particle's energy relative to the observer is $g(\textbf{P},\textbf{U}) = P_\alpha U^\alpha$. The same argument used earlier in this thread to show the physical speed approaches c (as the height above the horizon of a hovering observer drops to zero) also shows that the energy, as given by this formula, diverges to infinity. That isn't because the 4-momentum becomes infinite, it's because of the change in U. In 4D-geometrical terms it's because of the "change in angle" between P and U.

The 4-momentum of a free-falling particle is constant (free falling means no external force, and force is rate of change of momentum). ("Constant" means the covariant derivative along the worldline is zero.) If you choose a family of Us approaching the speed of light (relative to some single inertial observer), then $P_\alpha U^\alpha$ will diverge to infinity even though P remains constant. The 4-momentum of the falling particle never changes but the 4-velocities of each of the local hovering observers are enormously different as you get very close to the horizon.

By the way, have you spotted why the fudge of using two different metrics inside and outside the event horizon is incorrect? Because the square root of a negative number $\neq$ minus the square root of a positive number. They might meet at zero, but they are strangers after that.

The metric equation ds² = ... is, in effect, a differential equation to be solved. There are two solutions, one valid only strictly outside the horizon, the other valid only strictly inside the horizon. The metric equation doesn't even make sense exactly on the horizon.

This is best understood through the analogous Rindler example; the same method applies in both cases. The Rindler example (see my earlier posts) gives explicit formulas for converting between each of the several local (T,R) coordinates and the single global Minkowski (t,x) coordinates. The formulas are different inside and outside the horizon, but the same metric equation applies to all of the different (T,R) coordinate systems.

I'm not sure why you think someone is implying that $\sqrt{-x} = -\sqrt{x}$, which obviously isn't true; you'll have to provide an explicit quote to where you think that's happening.

 

[DiamondGeezer]

Tidal effects are irrelevant to what is being discussed here. The same things happen near the apparent horizon of the Rindler rocket, where there are no tidal effects. Tidal effects are relevant only to extended bodies, and here we are talking about a single particle. In relativity when we refer to acceleration without further qualification we usually mean "proper acceleration" i.e. acceleration measured by a comoving freefalling locally-inertial observer (the rate of change of "physical speed" with respect to proper time) or as measured by an acclerometer. Therefore by definition any freefalling particle's proper acceleration is zero. Any other type of acceleration is coordinate acceleration, and depends on your choice of coordinates and is therefore not "physical" in the sense we have been using that word.

Also your assertion that you can't measure tidal forces inside the event horizon is wrong; on the contrary, the tidal forces get ever stronger the further you fall in.

The Devil is in the details Greg. How would you measure the spatial separation of two objects inside a black hole when

a) there is no space to be measured
b) all light cones point in the time direction only

In other words, any two spacially separated objects would be beyond the light horizon of the other.

How to measure the spatial separation of two objects like that? Impossible.

That response doesn't help. The article never uses the term "4-energy" or "invariant" (except for the phrase "adiabatic invariants" which doesn't seem to be relevant).

I have to guess that what you really meant was "4-momentum is conserved". Conservation means something quite different from invariance. Conserved quantities don't change over time (to put it simply; or more sophisticatedly they satisfy certain differential equations). Invariant quantities take the same value when measured by different coordinate systems.

If a particle has a 4-momentum P and an observer has a 4-velocity U, then the particle's energy relative to the observer is $g(\textbf{P},\textbf{U}) = P_\alpha U^\alpha$. The same argument used earlier in this thread to show the physical speed approaches c (as the height above the horizon of a hovering observer drops to zero) also shows that the energy, as given by this formula, diverges to infinity. That isn't because the 4-momentum becomes infinite, it's because of the change in U. In 4D-geometrical terms it's because of the "change in angle" between P and U.

The 4-momentum of a free-falling particle is constant (free falling means no external force, and force is rate of change of momentum). ("Constant" means the covariant derivative along the worldline is zero.) If you choose a family of Us approaching the speed of light (relative to some single inertial observer), then $P_\alpha U^\alpha$ will diverge to infinity even though P remains constant. The 4-momentum of the falling particle never changes but the 4-velocities of each of the local hovering observers are enormously different as you get very close to the horizon.

I'll get back to you on that one. This is a quick reply before I go to work.

The metric equation ds² = ... is, in effect, a differential equation to be solved. There are two solutions, one valid only strictly outside the horizon, the other valid only strictly inside the horizon. The metric equation doesn't even make sense exactly on the horizon.

Wrong. If the metric equation doesn't make sense on the horizon then the world line of an infalling particle is discontinuous at the EH.

This is best understood through the analogous Rindler example; the same method applies in both cases. The Rindler example (see my earlier posts) gives explicit formulas for converting between each of the several local (T,R) coordinates and the single global Minkowski (t,x) coordinates. The formulas are different inside and outside the horizon, but the same metric equation applies to all of the different (T,R) coordinate systems.

I'm not sure why you think someone is implying that $\sqrt{-x} = -\sqrt{x}$, which obviously isn't true; you'll have to provide an explicit quote to where you think that's happening.

That's exactly what you're implying by asserting that there "are two solutions, one valid only strictly outside the horizon, the other valid only strictly inside the horizon". It's mathematically invalid if there is meant to be a continuous function that leads to the singularity at r=0

According to you, the singularity occurs at s=-2M (using the Schwarzschild coordinate system where zero is the EH), whereas the continuous worldline of the Schwarzschild Metric shows the singularity at s=2iM

They are not the same place, and your "inner" curve may meet at zero, but is discontinuous with the Schwarzschild solution.

 

[DrGreg]

DiamondGeezer,

By your logic, it's impossible for anyone to travel through the Earth's North Pole because the latitude/longitude coordinate system has a discontinuity there.

All the issues you raise can be resolved if you make the effort the understand the analogous set of disjoint (T,R) coordinate systems for the Rindler rocket (follow the links in post #50), which also exhibit discontinuities at the horizon, where R is time and T is space behind the horizon, where the metric equation expressed in these coordinates makes no sense exactly at the horizon. And yet this is just a weird set of coordinate systems that, between them, describe the flat, continuous, gravity-free Minkowski spacetime of special relativity.

I've explained this the best I can. If you're not getting it, I'm not sure there's much more I can say. If anyone else is still reading this thread and would like to contribute, maybe they can find some different way of explaining it that will get the point across.

 

[DiamondGeezer]

DiamondGeezer,

By your logic, it's impossible for anyone to travel through the Earth's North Pole because the latitude/longitude coordinate system has a discontinuity there.

Nope. The lat/long coordinate system begins and ends at zero and not 200 miles above or below it.

All the issues you raise can be resolved if you make the effort the understand the analogous set of disjoint (T,R) coordinate systems for the Rindler rocket (follow the links in post #50), which also exhibit discontinuities at the horizon, where R is time and T is space behind the horizon, where the metric equation expressed in these coordinates makes no sense exactly at the horizon. And yet this is just a weird set of coordinate systems that, between them, describe the flat, continuous, gravity-free Minkowski spacetime of special relativity.

You keep repeating the Rindler metric without ever dealing with the Schwarzschild Metric. Am I meant to be impressed? One is a coordinate based singularity which is unreachable, the other is not only reachable but inevitable for all infalling particles leading to impossible speeds and impossible energies.

I've explained this the best I can. If you're not getting it, I'm not sure there's much more I can say. If anyone else is still reading this thread and would like to contribute, maybe they can find some different way of explaining it that will get the point across.

Actually you haven't. You've run out of excuses for the bizarre mathematical behaviour of the Schwarzschild Metric. You haven't bothered to explain how tidal acceleration can be measured inside the EH - you simply asserted this as an unassailable fact. You haven't explained how reversing the sign of the Schwarzschild Metric inside the EH is mathematically allowable, nor explained why Schwarzschilds coordinates and any other derived coordinate system such as Kruskal's become imaginary when r<2M.

I'm tired.

 

[Ich]

If anyone else is still reading this thread and would like to contribute, maybe they can find some different way of explaining it that will get the point across.

Some dozen posts ago, I considered jumping in, but thought better of it.

I'm tired.

As this thread comes to an end, maybe it's good to know for you, DiamondGeezer, that I'm backing DrGreg: you obviously suffer from some basic misunderstandings concerning the meaning of coordinates and the metric. Try to do some calculations, like proper time during infall or such, that show you that the singularity at the EH is removable.

 

[Dmitry67]

Some dozen posts ago, I considered jumping in, but thought better of it.

My first thought was "considered jumping in the Black Hole" to prove that we are right :)

 

[Ich]

My first thought was "considered jumping in the Black Hole" to prove that we are right :)

Next time there is one around, I'll give it a thought. Maybe we can persuade the colleagues from the LHC to save one for us. ;)

 

[atyy]

Schwarzschild coordinates have a coordinate singularity. If you start with it, you will always have the singularity. So do not look for an explanation of why Kruskal-Szekeres coordinates are good through the event horizon by starting from Schwarzschild coordinates. Start from the Kruskal-Szekeres metric, verify that it is a vacuum solution of Einstein's equations, and by calculating geometric invariants show that there is an event horizon, and that spacetime is normal all the way until the curvature singularity. Schwarzschild coordinates only cover bits of the Kruskal-Szekeres coordinates.

 

[JesseM]

Schwarzschild coordinates have a coordinate singularity. If you start with it, you will always have the singularity. So do not look for an explanation of why Kruskal-Szekeres coordinates are good through the event horizon by starting from Schwarzschild coordinates. Start from the Kruskal-Szekeres metric, verify that it is a vacuum solution of Einstein's equations, and by calculating geometric invariants show that there is an event horizon, and that spacetime is normal all the way until the curvature singularity. Schwarzschild coordinates only cover bits of the Kruskal-Szekeres coordinates.

Yeah, I think this may get to the heart of the problem--DiamondGeezer is probably starting from Schwarzschild coordinates (as a lot of textbooks do, I think) and seeing other systems as just weird and vaguely suspicious transformations of Schwarzschild coordinates. But you can just as easily start from some other system which has no coordinate singularity at the horizon, like Kruskal-Szekeres coordinates or free-fall coordinates (the latter has a nice physical definition of coordinate time in terms of proper time read by a set of infalling clocks, see the middle of this page, and the discussion on this thread), and derive Schwarzschild coordinates as a transformation of those.

 

[DiamondGeezer]

Schwarzschild coordinates have a coordinate singularity. If you start with it, you will always have the singularity. So do not look for an explanation of why Kruskal-Szekeres coordinates are good through the event horizon by starting from Schwarzschild coordinates. Start from the Kruskal-Szekeres metric, verify that it is a vacuum solution of Einstein's equations, and by calculating geometric invariants show that there is an event horizon, and that spacetime is normal all the way until the curvature singularity. Schwarzschild coordinates only cover bits of the Kruskal-Szekeres coordinates.

I can confirm that the Kruskal-Szekeres metric is a vacuum solution of Einstein's equations. Just like the Schwarzschild Metric. Unfortunately spacetime isn't normal all the way to the curvature singularity - it becomes imaginary.

Yeah, I think this may get to the heart of the problem--DiamondGeezer is probably starting from Schwarzschild coordinates (as a lot of textbooks do, I think) and seeing other systems as just weird and vaguely suspicious transformations of Schwarzschild coordinates. But you can just as easily start from some other system which has no coordinate singularity at the horizon, like Kruskal-Szekeres coordinates or free-fall coordinates (the latter has a nice physical definition of coordinate time in terms of proper time read by a set of infalling clocks, see the middle of this page, and the discussion on this thread), and derive Schwarzschild coordinates as a transformation of those.

The Kruskal-Szekeres coordinates and the free-fall coordinates display the same behavior as the Schwarzschild Metric coordinates - the line integral is continuous to $r=0$ but the spatial or time separations become complex when $r$<$2M$.

The line integral is continuous because it goes through zero at the event horizon regardless of the transformation of coordinate systems used.

All of this should be telling you something important, but unfortunately I've had intelligent people tell me that the square root of a negative number is the same as the negative root of a positive number and I shouldn't worry about this mathematical non sequitur.

 

[DrGreg]

Nope. The lat/long coordinate system begins and ends at zero and not 200 miles above or below it.

The point is when you pass through the North Pole, your longitude "magically" jumps through 180° and your coordinate velocity reverses. Apparently bizarre behaviour if you just look at the coordinates without understanding the bigger picture.

You keep repeating the Rindler metric without ever dealing with the Schwarzschild Metric. Am I meant to be impressed?

I keep mentioning the Rindler case because it's easier to understand than black hole. If you make the effort to understand that example and see just how similar the two cases are, you ought to find the black hole easier to understand too. I don't expect you to just take my word for it, but to investigate the maths further and take some time to make sense of it. But I can't get you to understand black holes until you first understand the Rindler example

One is a coordinate based singularity which is unreachable, the other is not only reachable but inevitable for all infalling particles leading to impossible speeds and impossible energies.

No, in both cases there's a coordinate based singularity in one coordinate system which isn't a singularity in another system. In both cases, the location is physically reachable, as can be seen in one system, but lies outside the other coordinate system and can't be reached in a finite coordinate time in the wrong system. You only get impossible speeds and impossible energies, in both cases, by taking a physically impossible mathematical limit.

I'm tired.

Me too. Merry Christmas!

 

[Dmitry67]

Diamond, Could you explain why time is imaginary inside the BH?
Light cone (for the falling observer) looks normal and oriented vertically.
Check the attachment

 

[atyy]

KS coordinates are something like

dss ~ (1/r)exp(-r/2GM)dtt + ...

For r>0, r/2GM changes from greater than one to less than one as r goes from r>2GM to r<2GM, but there is no change in sign of dss.

As long as a curve remains timelike, a single definition of proper time above and below the event horizon is fine. For a spacelike curve, then we do run into trouble, and have to put in a minus sign, but this is true even in flat spacetime, and is one of the peculiarities of pseudo-Riemannian geometry.

 

[DiamondGeezer]

The point is when you pass through the North Pole, your longitude "magically" jumps through 180° and your coordinate velocity reverses. Apparently bizarre behaviour if you just look at the coordinates without understanding the bigger picture.

But the spacetime separations between two points on the polar coordinate surface remain resolutely real and positive. Quite unlike the Schwarzschild solution below the EH.

I keep mentioning the Rindler case because it's easier to understand than black hole. If you make the effort to understand that example and see just how similar the two cases are, you ought to find the black hole easier to understand too. I don't expect you to just take my word for it, but to investigate the maths further and take some time to make sense of it. But I can't get you to understand black holes until you first understand the Rindler example

I keep mentioning the Schwarzschild Metric because its tougher and because its a static solution.

No, in both cases there's a coordinate based singularity in one coordinate system which isn't a singularity in another system. In both cases, the location is physically reachable, as can be seen in one system, but lies outside the other coordinate system and can't be reached in a finite coordinate time in the wrong system. You only get impossible speeds and impossible energies, in both cases, by taking a physically impossible mathematical limit.

Ah yes, when the Universe divides by zero. But the physically impossible mathematical limit happens regardless of the coordinate system used, and the "inner Schwarzschild metric" isn't continuous with the outside. Nor is the "inner Kruskal-Szekeres" continuous with the outer.

You'd think somebody might be concerned about this sort of problem but apparently its Christmas when the laws of mathematics get suspended for the holidays.

Merry Christmas!

Merry Christmas to you. And a Christmas present: http://www.springerlink.com/content/dl8mu550u7736567/" - its a hint.

 

[Dmitry67]

Here is my present:

http://en.wikipedia.org/wiki/Black_hole
Oppenheimer and his co-authors used Schwarzschild's system of coordinates (the only coordinates available in 1939), which produced mathematical singularities at the Schwarzschild radius, in other words some of the terms in the equations became infinite at the Schwarzschild radius. This was interpreted as indicating that the Schwarzschild radius was the boundary of a bubble in which time stopped. This is a valid point of view for external observers, but not for infalling observers.

In 1958, David Finkelstein introduced the concept of the event horizon by presenting Eddington-Finkelstein coordinates, which enabled him to show that "The Schwarzschild surface r = 2 m is not a singularity, but that it acts as a perfect unidirectional membrane: causal influences can cross it in only one direction

 

[Dmitry67]

... also

The apparent singularity at r = rs is an illusion; it is an example of what is called a coordinate singularity. As the name implies, the singularity arises from a bad choice of coordinates or coordinate conditions. By choosing another set of suitable coordinates one can show that the metric is well-defined at the Schwarzschild radius. See, for example, Lemaitre coordinates, Eddington-Finkelstein coordinates, Kruskal-Szekeres coordinates or Novikov coordinates.

 

[DiamondGeezer]

... also

The apparent singularity at r = rs is an illusion; it is an example of what is called a coordinate singularity. As the name implies, the singularity arises from a bad choice of coordinates or coordinate conditions. By choosing another set of suitable coordinates one can show that the metric is well-defined at the Schwarzschild radius. See, for example, Lemaitre coordinates, Eddington-Finkelstein coordinates, Kruskal-Szekeres coordinates or Novikov coordinates.

Have you actually read this thread or just trying to bore me into submission?

1. All of the above coordinate transformations show that space or time separations below $r=2M$ are imaginary.
2. I have never once claimed that the apparent surface at $r=2M$ is anything other than a coordinate singularity.

What astonishes me is the sheer inability to get to grips with a very simple set of arguments. Instead I have got repeated "proofs by assertion" combined with statements impugning my intelligence. The arguments given so far have been mathematically fallacious.

And please try not to quote Wikipedia on this forum unless you want to be serially ignored.

 

[Dmitry67]

Have you actually read this thread or just trying to bore me into submission? All of the above coordinate transformations show that space or time separations below $r=2M$ are imaginary.

In schwarsheild metrics? of course.
This is why you need to use OTHER metrics for the interior of the BH where these purely mathematical oddities do not exist.

In the metrics from the list (Eddington-Finkelstein are my favourite) it is clear that for the falling observer space and time look normal (even they might looc 'imaginary' in the coordinate system of free far observer)

 

[DiamondGeezer]

In schwarsheild metrics? of course.
This is why you need to use OTHER metrics for the interior of the BH where these purely mathematical oddities do not exist.

In the metrics from the list (Eddington-Finkelstein are my favourite) it is clear that for the falling observer space and time look normal (even they might looc 'imaginary' in the coordinate system of free far observer)

It's clear that changing coordinate systems like that are mathematically invalid.

 

[Dmitry67]

BTW I suggest using the Einstein approach
Forget about the metrics.
Lets talk about what observers will observer.

Say, we are falling inside the spaceship (without the winsows) into the big enough BH so tidal forces are low.
Do you agree that under the horizon austranauts will continue see each other normally?

 

[DiamondGeezer]

BTW I suggest using the Einstein approach
Forget about the metrics.
Lets talk about what observers will observer.

Say, we are falling inside the spaceship (without the winsows) into the big enough BH so tidal forces are low.
Do you agree that under the horizon austranauts will contanue see each other normally?

No. They won't see anything at all. In order to "see" something light has to travel from somewhere outside to the retina and then the electrical impulses travel to your brain.

Inside the EH, all light cones are timelike and nothing (not even light) will travel backwards to hit the retinas.

 

[Dmitry67]

I had drawn a diargam for you (sorry, drawing in notepad using a mouse is terrible). Also, angles are incorrect. So,

Black vertical line is singularity
Vertical red line is horizon
Blue and Green lines are Bob and Alice, they are sitting at the opposite sides of the spaceship.
I had drawn Alice's lightcones and where they intersect Bob's worldline.
You can do the the same for Bob and check that Alice can see Bob as well.
They see each other up to the moment when lightcones become too narrow in these coordinates (they interpret it is the increase of the tidal forces). Spaceship breaks apart. Soon Bob and Alice lose each other behind their apparent horizons (close to the singularity)

Note that lightcones are in fact timelike inside :)