Crossing the Event Horizon of a Black Hole

[Dmitry67]

What's interesting is that no-one has pointed out that if a particle travels at the speed of light at the EH in one frame of reference then it travels at the speed of light in all of the others.

This is true only in SR, not in GR.

 

[Dmitry67]

i.e., the universe outside the EH comes to an end

No, infalling observer will NOT see all future history of the Universe.

 

[stevebd1]

No, infalling observer will NOT see all future history of the Universe.

I didn't say that Dmitry, I simply said that the universe outside the EH ceases to exist for the in-falling observer.

 

[Dmitry67]

I didn't say that Dmitry, I simply said that the universe outside the EH ceases to exist for the in-falling observer.

In what sense?
Some of the light from the outside still catches him, so he can see a history of what happened outside few seconds after he crossed the horizon. Even more, he continues to receive that light until he desitegrates in the singularity.

 

[George Jones]

Might it be fair to say that as an object crosses the event horizon, the time dilation is zero at exactly 2M and according to Gullstrand-Painleve coordinates, time outside the EH increases to infinity (i.e., the universe outside the EH comes to an end) so the only frames of reference the in-falling object relate to once at/past 2M are those inside the EH.

Let event A be the event for which an in-falling observer is on the event horizon, i.e., A is part of both the observer's worldline and the event horizon. The observer can use the radar method to establish a local inertial coordinate system that has A as the origin. Part of this local system will label events outside the horizon, and part of the system will label events inside the horizon. The radar method requires the observer to make wristwatch recordings both before and after A, but this is the same as for special relativity.

 

[DrGreg]

Here is another (slick) way to calculate relative physical (not coordinate) speed between two observers who are coincident at an event. Suppose the two 4-velocities are u and u'. Then, \gamma = \left( 1 - v^2 \right)^{-1/2} = g \left( u , u' \right) = g_{\alpha \beta}u^\alpha u'^\beta. This is an invariant quantity, and, consequently, can be calculated using any cordinate system/basis.

Is the terminology "physical speed" fairly standard amongst authors? Are there any other names for \sqrt{1 - |u_\alpha u'^\alpha|^{-2}}?

 

[George Jones]

Is the terminology "physical speed" fairly standard amongst authors?

I don't know; probably not.

Are there any other names for \sqrt{1 - |u_\alpha u'^\alpha|^{-2}}?

Relative speed.

For a short (maybe too short to be understandable) derivation of the result, see

https://www.physicsforums.com/showthread.php?p=1111069#post1111069.

Again, for coincident observers, this derivation works in both special and general relativity. In this thread, I do not agree with Oxymoron's interpretation of the relative velocity between two non-coincident observers in general relativity.

 

[DiamondGeezer]

I don't know; probably not.


Relative speed.

Is there any other kind?

For a short (maybe too short to be understandable) derivation of the result, see

https://www.physicsforums.com/showthread.php?p=1111069#post1111069.

Again, for coincident observers, this derivation works in both special and general relativity. In this thread, I do not agree with Oxymoron's interpretation of the relative velocity between two non-coincident observers in general relativity.

Its striking to me that this thread is short on explanation and really long on proof by assertion.

At the event horizon, there is no space separation between two events (or come to think of it, thereafter) so how can there be an inertial frame of reference?

Also, and you've yet to explain it, any free-falling particle getting asymptotically close to an event horizon will gain kinetic energy which eventually will be larger than the gravitational energy of the black hole. What happens then?

 

[DiamondGeezer]

It exceeds speed of light only in the coordinate system of a distant observer. GR has no problems with superluminal speed of far objects in curved spacetime.

So the superluminal speed has no physical consequences: you can not observe objects passing you faster then light, and all relative speeds, even inside the BH, are <c

An assertion without any proof. I haven't spoken about superluminal speed at all.

But just for the record, observing anything at all inside a black hole is going to be difficult, since light cones inside a BH all point away from you, so how will you see anything?

Secondly since there is no space-like separations, only time-like, how will you breathe when every atom in your lungs has no spatial separation?

Thirdly, how will you think when the electric signals in your brain can only move forward along the time axis?

 

[Dmitry67]

Also, and you've yet to explain it, any free-falling particle getting asymptotically close to an event horizon will gain kinetic energy which eventually will be larger than the gravitational energy of the black hole. What happens then?

No
As explained, you are mixing coordinate speed and physical speed.
free falling particle gains 'c' only the the refernce frame of external observer.
It is coordinate speed, not physical speed.

 

[Dmitry67]

Thirdly, how will you think when the electric signals in your brain can only move forward along the time axis?

For a freely falling observer, nothing interesting happens when he crosses the event horizon. Even more, if you fall into supermassive black hole, you can even miss the 'point of no return'. Ooops, I am already inside the horizon! Huston, we have a pro... aaaa!

Inside the BH time is flowing towards the center of the BH, but the observer's mframe rotates this way too, so for the observer time and space do not chnage their places - it happens only in the coordinate system of external observer.

http://www.valdostamuseum.org/hamsmith/DFblackIn.gif

 

[stevebd1]

Also, and you've yet to explain it, any free-falling particle getting asymptotically close to an event horizon will gain kinetic energy which eventually will be larger than the gravitational energy of the black hole. What happens then?

Unless I'm mistaken, gravitational potential is synonymous with the the time dilation, so as an object increases in speed in free fall towards an object of mass, the kinetic energy is only relative to the inertial frame and observers outside the gravitational field (i.e. at infinity) would not perceive/be aware of the the total increase in energy. It's a bit like an object that fell to Earth would have kinetic energy as it struck the ground but observers outside the gravitational field will observe that the activity of the object has slowed down (due to the time dilation). In order for the object to share the same frame as the observers outside the gravity field, work would have to be done to remove it from the gravity field, meaning the kinetic energy gained would be paid back, complying with the first law of thermodynamics. Hence a predicted free fall velocity of c results in a time dilation of zero.

 

[DiamondGeezer]

No
As explained, you are mixing coordinate speed and physical speed.
free falling particle gains 'c' only the the refernce frame of external observer.
It is coordinate speed, not physical speed.

No I'm not. Physical speed is measured relative to a coordinate system using light as a measuring stick. If there is no spatial dimension to create a coordinate system, then "physical speed" means nothing. Quite how George manages to create an inertial coordinate system without a space dimension at the EH is one of those things I'm hoping to discover.

Since we're talking about invariants here, the 4-energy of a free-falling particle increases without limit as the particle reaches the event horizon, regardless of who is doing the measurement.

 

[DiamondGeezer]

For a freely falling observer, nothing interesting happens when he crosses the event horizon. Even more, if you fall into supermassive black hole, you can even miss the 'point of no return'. Ooops, I am already inside the horizon! Huston, we have a pro... aaaa!

Inside the BH time is flowing towards the center of the BH, but the observer's mframe rotates this way too, so for the observer time and space do not chnage their places - it happens only in the coordinate system of external observer.

This is what annoys me in this thread. "Proof by assertion" run riot.

"Inside the BH" whatever that means, the Schwarzschild metric for both space and time shows that spacelike and timelike separations are IMAGINARY, and no changing of coordinate systems changes that salient (and for some reason completely unexplained) fact.

Change the coordinate system after Kruskal and Szekeres and the same thing happens. Change the coordinate system to Taylor and Wheeler's "rain frame" and the same thing happens.

I note that Exploring Black Holes makes the same statement given without proof. But what does imaginary time and imaginary space (at right angles to our spacetime) actually mean?

http://www.valdostamuseum.org/hamsmith/DFblackIn.gif [/QUOTE][/URL]

Ooh look a picture!

Did you show how you can see anything inside a black hole yet?

 

[Dmitry67]

Look at the trajectory of infalling particle. Look at the point where it crosses the horizon. look at the lighcone. Now rotate the picture clockwise so the lightcone becomes vertical. This is falling observer perspective.

"the Schwarzschild metric for both space and time shows that spacelike and timelike separations are IMAGINARY"

This is wrong! It is just a ROTATION! So time becomes spacelike (in the coordinates of the distant observer) and one of the space coordinates becomes timelike (this part you are probably missing) but for the observer itself everything looks normal, again, just rotate the picture.

 

[Dmitry67]

Look at the infalling particle. I added red line - a path of another particle, say, another side of a spaceship. Initially they are spacially separated. Both particles are mutually inside the lightcones of each other, so they can 'see' each other (green lines).

Always - except the very final moments close to the singularity. Note that lightcones becomes narrower and narrower, it is a result of increasing tidal forces inside. At some point they look causial contact.

Timelike is NOT vertical, and spatial IS NOT horizontal. Time (for the freefalling observer) is in the middle of the lightcone.

 

[George Jones]

This is wrong! It is just a ROTATION! So time becomes spacelike (in the coordinates of the distant observer) and one of the space coordinates becomes timelike (this part you are probably missing) but for the observer itself everything looks normal, again, just rotate the picture.

I wouldn't say this. I would say this that Schwarzschild coordinates really make up two different coordinates systems. One system is valid outside the black hole, but not on the event horizon or inside the black hole. The other system is valid inside the black hole, but not on the event horizon or outside the black hole. The metric unambiguously determines the nature of each coordinate in each coordinate system.

So, inside the event horizon, space doesn't get rotated into a time direction, it's just that human-chosen labels (coordinates) have names (chosen by humans, not by nature) that are very misleading. See

https://www.physicsforums.com/showthread.php?p=1146536#post1146536.

 

[Dmitry67]

Yes, I meant - rotation when you look at the diagram. SO we agree that when 2 sides of a spaceship are spacially separated, it is still valid inside the BH, so I still see everything inside normally (until it is ripped apart by tidal forces)

 

[DiamondGeezer]

I wouldn't say this. I would say this that Schwarzschild coordinates really make up two different coordinates systems. One system is valid outside the black hole, but not on the event horizon or inside the black hole. The other system is valid inside the black hole, but not on the event horizon or outside the black hole. The metric unambiguously determines the nature of each coordinate in each coordinate system.

It's this "one is valid for outside the black hole but not on the event horizon or inside the black hole" that interests me. Either the metric describes a continuous line integral that constitutes the worldline of an infalling particle or it doesn't.

In whatever case, the fact that the spacelike and timelike separations become imaginary beyond the event horizon means something, and it ain't rotation.

The classical form of the Schwarzschild Metric shows a singularity in the future of every spherically symmetric mass, not simply black holes.

So, inside the event horizon, space doesn't get rotated into a time direction, it's just that human-chosen labels (coordinates) have names (chosen by humans, not by nature) that are very misleading. See

https://www.physicsforums.com/showthread.php?p=1146536#post1146536.

Its abundantly clear that the Schwarzschild Metric unambiguously claims that inside of r=2M, whatever is there is at right angles to space and time in this Universe. However the coordinate system is transformed, that's what happens.

Another possibility is that the classical form of the Schwarzschild Metric which gives rise to the unphysical event horizon (that is, infalling particles gain infinite energy and reach the speed of light at the EH) is wrong.

I think I might know what that is.

 

[DrGreg]

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.

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.

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.

 

[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).