# Crossing the Event Horizon of a Black Hole

by DiamondGeezer
Tags: black hole, event horizon
 Sci Advisor P: 8,430 To be clear, DiamondGeezer, are you claiming that any coordinate-invariant quantities determined by the metric, like the proper time between two events on a given worldline, become imaginary once you cross the horizon? (of course if you try to calculate the proper time along a spacelike path you get an imaginary number, but this is equally true outside the horizon)
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P: 1,772
 Quote by DiamondGeezer The absolute horizon of a black hole is quite unlike a coordinate based "apparent" horizon in your example.
I thoroughly disagree. In the context being discussed here, the two scenarios are very similar indeed.

(Specifically, we are talking here about near the horizon, above, at and below, and not what happens near the central singularity, which, of course, exists only in the case of the black hole.)

 Quote by DiamondGeezer But we're not using "accelerated coordinate systems"
Oh yes we are! Do you not realise that an observer who is hovering at constant height above a black hole is undergoing outward proper acceleration? That is a rather fundamental application of the equivalence principle. The "acceleration due to gravity" that you feel is because you are properly-accelerating upwards. The Schwarzschild coordinate system is a non-inertial accelerated coordinate system.

 Quote by DiamondGeezer If by "without incident" you mean "gains infinite energy by accelerating to the speed of light", then I'd like to ask you what you consider an "incident" to look like.
OK, you've mentioned the "infinite energy" issue before, and I don't think anyone gave an adequate reply.

In the case of a black hole, we have already seen, earlier in this thread, that different observers hovering at different constant heights above a horizon will measure the energy of an infalling particle with a number that gets ever greater the closer the observer is to the horizon. Remember, energy is an observer-dependent quantity. Even in non-relativistic theory, ½mv2 depends on what you measure v relative to. So there's no need to ask "where the extra energy comes from" when you switch from one hovering observer's measurement to another, as it's an irrelevant question: energy isn't conserved when you are comparing one observer's measurement against another's.

(N.B. the observers at different heights really are in different frames because although they are at fixed distances from each other, each has a different notion of radial distance and time -- gravitational time-dilation. Therefore their definitions of energy disagree.)

If you consider a sequence of hovering observers getting closer to the horizon, yes, in the limit as the distance tends to zero then the "physical speed" (as defined earlier in the thread) of the infalling particle tends to c and its energy tends to infinity. But these are mathematical limits and can never actually be observed by a hovering observer. No observer can actually hover on the horizon itself.

And if you want, instead, to imagine an oberver who slowly descends to the horizon, that observer isn't hovering, so the equations derived earlier are no longer valid. The difference may be tiny at a significant height, but very close to the horizon the difference will be substantial (in fact, diverging to infinity). Any observer will only ever measure a finite energy (and a "physical speed" less than c). George Jones chose his words carefully in post #16 to avoid saying that the particle's physical speed was equal to c.

Everything I've said above also applies to my example of an accelerating rocket. The only difference is that whenever I speak of a "hovering observer" above, that should be interpreted as meaning "stationary relative to the rocket, as measured by the rocket". In the diagram I referred to in my previous post, an apple dropped from the rocket when T=t=0 follows a vertical worldline on the spacetime diagram. As the apple gets closer to the horizon at x=ct, the local Rindler observers are travelling faster and faster relative to the apple, approaching c in the limit. These observers, each at rest in the rocket's frame, measure a kinetic energy of the apple that diverges to infinity in the limit. So ask yourself, where does this energy come from? Does the question make sense?

In both cases, the rocket and the black hole, the reason the falling object "gains infinite energy" (as you put it) is because the observer is accelerating towards the speed of light, relative to the non-accelerating object, not because of something happening to the falling object.

 Quote by DiamondGeezer ...the change from real to imaginary space and time which happens as the worldline of an infalling particle reaches a geometric surface called the "event horizon" of a black hole.
This has already been explained. You need to check the definitions of "spacelike" and "timelike". If you have an equation that seems to be giving you imaginary time, you have made a mistake and what you have actually found is a real distance. If you have an equation that seems to be giving you imaginary distance, you have made a mistake and what you have actually found is a real time. The same issues apply to both the accelerating rocket and the black hole.

 Quote by DiamondGeezer .The coordinate system you construct does not allow the rocket to cross the imagined "horizon" at r=1/2 at any time.
I don't really understand what you are getting at here. Everything is measured relative to the rocket, so of course the rocket cannot be a non-zero distance away from itself (which is at R=1).

But this is the point. The construction of the (T,R) coordinate system is such that an object cannot cross the Rindler horizon at finite coordinate values, but it certainly does cross the horizon at finite (t,x) coordinates. This is an artefact of the (T,R) coordinate system. Similarly, an object cannot cross a black hole's event horizon at finite Schwarzschild coordinates, but it certainly does cross the horizon at finite coordinates in other coordinate systems. Again, this is an artefact of the Schwarzschild coordinate system.

Incidentally the "Rindler horizon" has something equivalent to a black hole's Hawking radiation. It is called the Unruh effect.

A final comparison between my rocket example and a black hole.

My inertial coords (t,x) are equivalent to Kruskal-Szekeres coords.

My (T,R) coords are equivalent to Schwarzschild coords.

All of this is discussed in Rindler's book which I referenced in my old post that I previously linked to.
P: 126
 Quote by George Jones No, MTW does not state this. Look at the bottom of page 833. Have you deliberately misrepresented MTW? The coordinate transformation which you have given is valid only when $r > 2m$. For $r < 2m$, the transformation (given in MTW and dozens, if not hundreds, or relativity books) is $$u = \left( 1 - \frac{r}{2m}\right)^{\frac{1}{2}}\exp\left(\frac{r}{4m}\right)\sinh\lef t(\frac{T}{4m}\right)$$ and $$v = \left( 1 - \frac{r}{2m}\right)^{\frac{1}{2}}\exp\left(\frac{r}{4m}\right)\cosh\lef t(\frac{T}{4m}\right).$$ So, you can see the "obvious," but thousands of professional relativists can't?
No. What I point out is that there is no mathematical justification for doing so if the line integral is continuous.

It's a fudge, George.

MTW fudges it by rolling out two different equations which avoid the fact that the line integral becomes complex when r<2M. Taylor and Wheeler do the same. Kruskal tried it with a different coordinate transformation but got the same result.

Are you seriously arguing that thousands of professional scientists cannot be wrong and therefore the laws of mathematics can be suspended by popular vote?

If the line integral is continuous then "The coordinate transformation which you have given is valid only when $r > 2m$" and there's another one for $r < 2M$ then the line integral is discontinuous and you are talking about two different universes.

Or that I cannot argue that this is mathematically invalid lest I get a from George Jones?
P: 126
 Quote by DrGreg In the case of a black hole, we have already seen, earlier in this thread, that different observers hovering at different constant heights above a horizon will measure the energy of an infalling particle with a number that gets ever greater the closer the observer is to the horizon. Remember, energy is an observer-dependent quantity. Even in non-relativistic theory, ½mv2 depends on what you measure v relative to. So there's no need to ask "where the extra energy comes from" when you switch from one hovering observer's measurement to another, as it's an irrelevant question: energy isn't conserved when you are comparing one observer's measurement against another's. (N.B. the observers at different heights really are in different frames because although they are at fixed distances from each other, each has a different notion of radial distance and time -- gravitational time-dilation. Therefore their definitions of energy disagree.)
But their measurement of 4-energy will be the same because that is invariant and goes to infinity.

 If you consider a sequence of hovering observers getting closer to the horizon, yes, in the limit as the distance tends to zero then the "physical speed" (as defined earlier in the thread) of the infalling particle tends to c and its energy tends to infinity. But these are mathematical limits and can never actually be observed by a hovering observer. No observer can actually hover on the horizon itself.
This is exactly what I have said, except that I didn't limit myself to "observers on shells near the event horizon". Every calculation of speed and energy from any inertial or non-inertial POV comes to the same answer about the speed of an infalling particle and the energy of same at the event horizon.

The relativistic energy as you know is: $E = \gamma mc^2$ so as $r \rightarrow 2M$ then $\gamma$ reaches $\infty$

 And if you want, instead, to imagine an oberver who slowly descends to the horizon, that observer isn't hovering, so the equations derived earlier are no longer valid. The difference may be tiny at a significant height, but very close to the horizon the difference will be substantial (in fact, diverging to infinity). Any observer will only ever measure a finite energy (and a "physical speed" less than c). George Jones chose his words carefully in post #16 to avoid saying that the particle's physical speed was equal to c.
I have never claimed that the infalling particle will be observed to be moving at c. George danced around this point because he knows that the implication of a speed of c implies quite a few unphysically realistic results if the black hole event horizon exists.

I do know the difference. I'm not stupid.

 Everything I've said above also applies to my example of an accelerating rocket. The only difference is that whenever I speak of a "hovering observer" above, that should be interpreted as meaning "stationary relative to the rocket, as measured by the rocket". In the diagram I referred to in my previous post, an apple dropped from the rocket when T=t=0 follows a vertical worldline on the spacetime diagram. As the apple gets closer to the horizon at x=ct, the local Rindler observers are travelling faster and faster relative to the apple, approaching c in the limit. These observers, each at rest in the rocket's frame, measure a kinetic energy of the apple that diverges to infinity in the limit. So ask yourself, where does this energy come from? Does the question make sense?
Yes it does. The rocket produces energy. The rocketeers will not "see" the apple reaching c.

Neither will they or anything else reach c (which requires infinite energy that the Universe does not have). They will not reach r=1/2, which isn't on their worldline.

On the other hand, any infalling particle (or apple) apparently gets infinite energy from falling into a black hole. Every free falling particle reaches r=2M.

Or if you like, as the apple approaches the event horizon, the kinetic energy of the apple rises without limit (even past the total energy of the Universe itself).

I argue that that points to a fundamental flaw in black hole theory, one that cannot be transformed away.

 In both cases, the rocket and the black hole, the reason the falling object "gains infinite energy" (as you put it) is because the observer is accelerating towards the speed of light, relative to the non-accelerating object, not because of something happening to the falling object.
If the observer is accelerating towards the speed of light, then that energy must come from somewhere. Unless the black hole has infinite energy then something is wrong.

 You need to check the definitions of "spacelike" and "timelike". If you have an equation that seems to be giving you imaginary time, you have made a mistake and what you have actually found is a real distance. If you have an equation that seems to be giving you imaginary distance, you have made a mistake and what you have actually found is a real time. The same issues apply to both the accelerating rocket and the black hole.
No, what I have found is that the classical Schwarzschild Metric has been mathematically fudged to permit an infalling particle to gain infinite energy, achieve light speed and access a Universe at right angles to our own.

I question those fudges that avoid those questions.

I think there IS a transformation of the Schwarzschild Metric which leads to a mathematically consistent non-contradictory solution for the region about a mass which undergoes infinite collapse.

But if so, then black holes do not exist. Something else does.
 Sci Advisor P: 7,403 While we're discussing energy conservation, can someone tell me please tell me if this is true: 1) Outside the horizon there is a time-like Killing vector so energy is conserved for freely falling particles. 2) Inside the horizon there is no time-like Killing vector, so energy is not conserved for freely falling particles. ?
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 Quote by DiamondGeezer But their measurement of 4-energy will be the same because that is invariant and goes to infinity.
 Quote by DiamondGeezer The relativistic energy as you know is: $E = \gamma mc^2$ so as $r \rightarrow 2M$ then $\gamma$ reaches $\infty$
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 travelling 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.

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.

 Quote by DiamondGeezer If the observer is accelerating towards the speed of light, then that energy must come from somewhere. Unless the black hole has infinite energy then something is wrong.
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).
P: 126
 Quote by DrGreg 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 travelling 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 possess such unphysical properties.
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 Quote by DiamondGeezer 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.
 Quote by DiamondGeezer 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.
 Quote by DiamondGeezer 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 possess 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.)
P: 8,430
 Quote by DiamondGeezer 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.
P: 126
 Quote by JesseM 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".
P: 8,430
 Quote by DiamondGeezer 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.
P: 126
 Quote by DrGreg 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.
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I have no more time to comment today. But I've just time to ask this:
 Quote by DiamondGeezer 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.
P: 126
 Quote by DrGreg 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_%28physics%29
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 Quote by DiamondGeezer 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.

Quote by DiamondGeezer
Quote by DrGreg
 Quote by DiamondGeezer 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_%28physics%29
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.

 Quote by DiamondGeezer 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.
P: 126
 Quote by DrGreg 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 spacial 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 spacial 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.
 PF Patron Sci Advisor P: 1,772 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.
P: 126
 Quote by 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.
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.

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