Crossing the Event Horizon of a Black Hole

[DiamondGeezer]

To better understand a black hole's event horizon, and what happens at the horizon and the difference between above and below, it is well worth the effort to consider a different example of horizon.

If you are in an accelerating rocket in deep space, far from any gravitation, nevertheless an "apparent horizon" forms behind you. There are some minor technical differences between this "apparent horizon" and the "absolute horizon" of a black hole, but for the purposes of this discussion, the differences are irrelevant.

But that IS the difference. The absolute horizon of a black hole is quite unlike a coordinate based "apparent" horizon in your example.

If the rocket has a constant proper acceleration of a, then the apparent horizon forms at a distance of c²/a behind you (as measured by yourself). Nothing, not even light, can cross the horizon towards you. If you drop an apple out of your rocket, you'll see the (red-shifted) apple slow down as it approaches the horizon and it never crosses it. That's what you see with your eyes and it's also what you calculate in your own coordinate system.

What makes this scenario easier to understand than a black hole is that you can examine what is happening in inertial Minkowski coordinates, or in your own accelerated "Rindler" coordinates. In inertial coordinates the apple reaches the horizon in a finite time and crosses it without incident.

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.

We can ignore two space dimensions and consider inertial (t,x) coordinates. These are related to the rocket's Rindler coordinates (T,X) by

x = X \cosh \frac{aT}{c}
ct = X \sinh \frac{aT}{c}​

(for X > 0). The rocket is located at a constant X = c²/a in these coordinates, and the horizon is the limit as X \rightarrow 0. In inertial coordinates, the horizon is located at x = ct. There is a space-time diagram to illustrate tbis in post #9 of the "about the Rindler metric" thread[/color].

Post #15[/color] of that same thread shows an extra change of space-coordinate you can introduce

R = \frac{1}{2} \left[1 + \left(\frac{aX}{c^2}\right)^2 \right]​

which (in the simplified case where c = 1 = a) gives you a metric equation

ds^2 = (2R-1) dT^2 - \frac{dR^2}{2R-1}​

somewhat similar to the Schwarzschild metric. The post also shows how you can define a different coordinate transformation behind the horizon which gives rise to the same metric equation as above. And behind the horizon (which is at R=½), T measures distorted distance (not imaginary time) and R measures distorted time (not imaginary distance).

The point of all this is to show that almost all the strange things about horizons are due to the use of accelerating coordinate systems, and in particular using a coordinate system that fails at the horizon itself. In the accelerating rocket example, we can use ordinary inertial coordinates to show there's nothing weird "really" happening at all, only the accelerated coordinates making it seem weird.

Spend some time working through the maths of the accelerating Rindler rocket and its "Rindler horizon", and you should find a black hole's horizon easier to understand after that.

But we're not using "accelerated coordinate systems" so I'm sorry but your analogy, fascinating as it may be, does not explain 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.

The coordinate system you construct does not allow the rocket to cross the imagined "horizon" at r=1/2 at any time.

But the Schwarzschild Metric (in its classic form) does predict precisely such an occurrence. If black holes exist and Hawking radiation happens, then particles arising out of quantum fluctuations appear to be gaining infinite energy regardless of how the coordinate system is created, when they intersect the event horizon.

 

[JesseM]

But we're not using "accelerated coordinate systems" so I'm sorry but your analogy, fascinating as it may be, does not explain 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.

There is no physical "change from real to imaginary space and time" at the horizon, it is simply the fact that the Schwarzschild "time" coordinate becomes spacelike past the horizon, and the "radial" coordinate becomes timelike. But this is just a weakness of how Schwarzschild coordinates are defined! If you use Kruskal-Szekeres coordinates, for example, in these coordinates the time coordinate is always timelike both inside and outside the horizon, and the radial coordinate is always spacelike. Also, in many ways Kruskal-Szekeres coordinates are to Schwarzschild coordinates as inertial coordinates are to Rindler coordinates in flat spacetime; just as the Rindler horizon is seen as the diagonal surface of a future light cone when viewed in inertial coordinates, and objects at constant position coordinate in Rindler coordinates are seen as having worldlines that look like hyperbolas in inertial coordinates (see the second diagram on this page), so it is similarly true that the black hole event horizon looks like a diagonal line in Kruskal-Szekeres coordinates (these coordinates have the special property that all lightlike surfaces appear diagonal on a graph), and objects at constant Schwarzschild radius have worldlines that look like hyperbolas as well. See my post #4 on this thread for a quick rundown on the basics of Kruskal-Szekeres coordinates and a few illustrative Kruksal-Szekeres diagrams scanned from the textbook Gravitation.

 

[DiamondGeezer]

There is no physical "change from real to imaginary space and time" at the horizon, it is simply the fact that the Schwarzschild "time" coordinate becomes spacelike past the horizon, and the "radial" coordinate becomes timelike.

No they both become imaginary.

But this is just a weakness of how Schwarzschild coordinates are defined! If you use Kruskal-Szekeres coordinates, for example, in these coordinates the time coordinate is always timelike both inside and outside the horizon, and the radial coordinate is always spacelike.

No, they both become imaginary as well.

The Kruskal-Szekeres tranformed coordinates are:

u=\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}\exp\left(\frac{r}{4m^{*}}\right)\cosh\left(\frac{T}{4m^{*}}\right)

and

v=\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}\exp\left(\frac{r}{4m^{*}}\right)sinh\left(\frac{T}{4m^{*}}\right)

which also become imaginary because of

\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}

when r<2M

Also, in many ways Kruskal-Szekeres coordinates are to Schwarzschild coordinates as inertial coordinates are to Rindler coordinates in flat spacetime; just as the Rindler horizon is seen as the diagonal surface of a future light cone when viewed in inertial coordinates, and objects at constant position coordinate in Rindler coordinates are seen as having worldlines that look like hyperbolas in inertial coordinates (see the second diagram on this page), so it is similarly true that the black hole event horizon looks like a diagonal line in Kruskal-Szekeres coordinates (these coordinates have the special property that all lightlike surfaces appear diagonal on a graph), and objects at constant Schwarzschild radius have worldlines that look like hyperbolas as well. See my post #4 on this thread for a quick rundown on the basics of Kruskal-Szekeres coordinates and a few illustrative Kruksal-Szekeres diagrams scanned from the textbook Gravitation.

Neither Kruskal in his original paper, nor Misner/Wheeler/Thorne in Gravitation ever deal with the fact that however the Schwarzschild Metric is changed, the worldlines become imaginary when r<2M. Instead they take great comfort from the fact that the function is continuous, as if that meant anything. If you want to see how waffley some brilliant scientists become when faced with the obvious, then re-read Gravitation and how it deals with the event horizons.

This was something of a surprise when I read Kruskal's paper that he appeared to be oblivious that after so much effort, the result was the same.

In the K-S diagram, the west and east are outside the event horizon and real, north and south are inside the EH and imaginary.

As an example of this phenomenon, consider the simplest possible function:

y= \sqrt{x}

Conventionally, people will say that the function is only real (in both senses of the word) when x > 0 and that's where most people stop. But not so.

When you allow y to be imaginary then the function y= \sqrt{x} is continuous through zero from - \infty &lt;x&lt; \infty

What the K-S diagram does is flatten the real and imaginary parts of the graph onto a 2-d diagram.

 

[George Jones]

The Kruskal-Szekeres tranformed coordinates are:

u=\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}\exp\left(\frac{r}{4m^{*}}\right)\cosh\left(\frac{T}{4m^{*}}\right)

and

v=\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}\exp\left(\frac{r}{4m^{*}}\right)sinh\left(\frac{T}{4m^{*}}\right)

which also become imaginary because of

\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}

when r<2M.

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 &gt; 2m. For r &lt; 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\left(\frac{T}{4m}\right)

and

v = \left( 1 - \frac{r}{2m}\right)^{\frac{1}{2}}\exp\left(\frac{r}{4m}\right)\cosh\left(\frac{T}{4m}\right).

Neither Kruskal in his original paper, nor Misner/Wheeler/Thorne in Gravitation ever deal with the fact that however the Schwarzschild Metric is changed, the worldlines become imaginary when r<2M. Instead they take great comfort from the fact that the function is continuous, as if that meant anything. If you want to see how waffley some brilliant scientists become when faced with the obvious, then re-read Gravitation and how it deals with the event horizons.

So, you can see the "obvious," but thousands of professional relativists can't?

 

[JesseM]

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)

 

[DrGreg]

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.)

But we're not using "accelerated coordinate systems"

Oh yes we are! Do you not realize 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.

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 traveling 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.

...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.

.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.

 

[DiamondGeezer]

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 &gt; 2m. For r &lt; 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\left(\frac{T}{4m}\right)

and

v = \left( 1 - \frac{r}{2m}\right)^{\frac{1}{2}}\exp\left(\frac{r}{4m}\right)\cosh\left(\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 &gt; 2m" and there's another one for r &lt; 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?

 

[DiamondGeezer]

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 traveling 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.

 

[atyy]

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.

?

 

[DrGreg]

But their measurement of 4-energy will be the same because that is invariant and goes to infinity.

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 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.

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.

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).

 

[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 :)

 

[Dmitry67]

And yet another example.
Look at our world from the SUPERLUMINAL rest frame.
You see the same - imaginary time etc.
But it does not affect how we observe things, right?

 

[Ich]

I found a weird metric with imaginary time and space:
ds² = -dx² + dt² +dy² + dz²
Obvoiusly such spacetimes can't exist, as you can't see your nose there.

 

[stevebd1]

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.

I recommend you take a look at http://en.wikipedia.org/wiki/Gullstrand-Painlev%C3%A9_coordinates" which can be expressed in two forms, free-fall rain frame and global rain frame (r_s=2M)-

Free-fall rain frame-

c^2d\tau^2=c^2dt_r^2 - dr_r^2 - r^2\left(d\theta^2 + \sin^2\theta \, d\phi^2\right)

Global rain frame-

c^2d\tau^2=\left(1-\frac{r_s}{r}\right)c^2dt_r^2 - 2\sqrt{\frac{r_s}{r}}\ cdt_rdr - dr^2 - r^2\left(d\theta^2 + \sin^2\theta \, d\phi^2\right)

where

<br /> \begin{flalign}<br /> &amp;dt_r=dt-\beta\gamma^2dr\\[6mm]<br /> &amp;dr_r=\gamma^2dr-\beta dt<br /> \end{flalign}<br />

and

<br /> \begin{flalign} <br /> &amp;\gamma=\frac{1}{\sqrt{1-\beta^2}}\\[6mm]<br /> &amp;\beta=-\sqrt{\frac{r_s}{r}} <br /> \end{flalign}<br />

where \gamma is the Lorentz factor and \beta is the velocity of the rain frame relative to the shell frame.

The principle behind the form is-

\gamma=\frac{1}{\sqrt{1-(v/c)^2}}\equiv \frac{1}{\sqrt{1-(r_s/r)}}

Basically length contraction induced by velocity (which in turn is induced by curvature) balances out the length expansion induced by gravity. This is identical to 'Metric for the Rain Frame' shown on page B-13 of Exploring Black Holes by Taylor & Wheeler.

The free-fall rain frame proper time remains time-like all the way to the singularity, representing the frame of the in-falling observer while the global rain-frame proper time becomes negative at r<2M but with no geometric singularity at the event horizon (dr remains 1).

 

[JesseM]

1. All of the above coordinate transformations show that space or time separations below r=2M are imaginary.

Do you claim that the time separation of events along a timelike worldline is imaginary in Kruskal-Szkeres coordinates? If so I'm pretty certain your wrong, perhaps you'd care to address atyy's post #81 above. As atyy noted, it's true you'll get an imaginary value if you try to calculate the proper time along a spacelike curve, but this is always true in relativity, even in flat SR spacetime (and in KS coordinates it should be just as true outside the horizon as inside it).

 

[DiamondGeezer]

Do you claim that the time separation of events along a timelike worldline is imaginary in Kruskal-Szkeres coordinates? If so I'm pretty certain your wrong, perhaps you'd care to address atyy's post #81 above. As atyy noted, it's true you'll get an imaginary value if you try to calculate the proper time along a spacelike curve, but this is always true in relativity, even in flat SR spacetime (and in KS coordinates it should be just as true outside the horizon as inside it).

No I don't. If you read carefully you'll find I talk about the traditional spacetime separations between events. I know what the difference is between timelike and spacelike and its irrelevant to the arguments.

What I point out (tediously repeating myself I know) is that "inside" a black hole event horizon, spacelike and timelike separations are always imaginary. The various cited coordinate transformations end up with the same result, and curiously nobody seems to be perturbed by the mathematical gyrations which appear to explain what happens when a particle reaches the event horizon in order to allow it to reach the center of curvature.

All of this should tell you lots about the Schwarzschild Metric, but unfortunately nobody's paying any attention.

Nothing I have heard on this thread has shown any insight into the question of black holes or event horizons. Instead I read repeatedly the same arguments which are ipso facto mathematically and physically absurd when examined.

It's not arrogance on my part to say that the mathematical treatment which gives rise to the theory of black holes is badly flawed.

 

[Dmitry67]

DiamondGreezer, did you check my diagram?
Do you agree with it or not?

 

[DiamondGeezer]

DiamondGreezer, did you check my diagram?
Do you agree with it or not?

Yes.

No.

 

[Dmitry67]

Yes.

No.

Then put your version of events on
http://www.valdostamuseum.org/hamsmith/DFblackIn.gif

(Bob and Alice are freely falling into BH looking at each other)

 

[DrGreg]

I know what the difference is between timelike and spacelike and its irrelevant to the arguments.

What I point out (tediously repeating myself I know) is that "inside" a black hole event horizon, spacelike and timelike separations are always imaginary.

If you really did know the difference between timelike and spacelike then you would understand that "imaginary spacelike" means "real timelike" and "imaginary timelike" means "real spacelike".

It's not arrogance on my part to say that the mathematical treatment which gives rise to the theory of black holes is badly flawed.

Just because you don't understand something means you are right and the many, many thousands of relativists who have studied this, some of them eminent physicists and mathematicians, are all wrong?

 

[DiamondGeezer]

If you really did know the difference between timelike and spacelike then you would understand that "imaginary spacelike" means "real timelike" and "imaginary timelike" means "real spacelike".

No. It's irritating that I have to keep repeating myself to people who really should know better.

viz.,

(the square root of a negative quantity)\neq(the negative square root of a positive quantity)

Just because you don't understand something means you are right and the many, many thousands of relativists who have studied this, some of them eminent physicists and mathematicians, are all wrong?

If science was really decided by majority voting then relativity would have been rejected more than 100 years ago, when only Einstein understood it. You choose a really poor subject to make such a majoritarian fallacy.

It wasn't so long ago that the majority of experts knew that there were WMDs in Iraq - completely wrong of course.

I don't compare myself to Einstein - he was a genius. Nor do I criticize General Relativity in general. I don't claim to understanding of all of General Relativity.

But on this one small point, the majority on this particular issue are making a mathematical case which is false - and I believe I do have an answer but first there has to be an acknowledgment of a problem.

It does take a certain amount of arrogance to say that "I am right and everyone else is wrong" but there are so many examples of an outsider pointing out something that has somehow escaped the experts in the field that its no longer seen as weird.

I don't know of a single scientist who doesn't have a least 10 hypotheses that goes against what "the majority of experts in the field" believe to be true.

I am perfectly prepared to be wrong (and I may well be). But the argumentation on the issue of black holes and event horizons is mathematically weak and physically contrived - in my humble opinion.