What Happens Inside a Black Hole's Event Horizon?

[skeptic2]

Perhaps the simple answer is that if they went faster than c,
they would(could) escape.

That's a circular argument.

Or perhaps it is closely tied to V_escape = root(GM/R)*root(2) that can result in V_escape > c, for R < R_eh, and the photons have
petered-out to c when they get to the event horizon. IE, a misconception that escape velocity > c implies anything more than a more difficult job for photons going c to get out.

Photons always travel at c so there's no real argument about photons exceeding c. Is there any reason why massive particles cannot travel faster than c inside the EV?

 

[ClamShell]

Photons always travel at c so there's no real argument about photons exceeding c. Is there any reason why massive particles cannot travel faster than c inside the EV?

OK, for particles of matter:

Or perhaps it is closely tied to V_escape = root(GM/R)*root(2) that can result in V_escape > c, for R < R_eh, a misconception that escape velocity > c implies anything more than a more difficult job for matter to get out.

My personal favorite is that the hypothetical structure on the other side
of the horizon is an "independent coordinate system" and obeys
the very same rules that we have here. Circular again?

Or would you prefer the concept that the Schwarzschild Metric
really represents two "independent coordinate systems" with
different rules?

 

[nismaratwork]

OK, for particles of matter:

Or perhaps it is closely tied to V_escape = root(GM/R)*root(2) that can result in V_escape > c, for R < R_eh, a misconception that escape velocity > c implies anything more than a more difficult job for matter to get out.

My personal favorite is that the hypothetical structure on the other side
of the horizon is an "independent coordinate system" and obeys
the very same rules that we have here. Circular again?

Or would you prefer the concept that the Schwarzschild Metric
really represents two "independent coordinate systems" with
different rules?

Nothing that crosses the event horizon IS matter anymore, at best you're talking about radiation. Remember, a neutron star, or even a hypothetical quark star isn't dense enough to have an event horizon; by the time you reach that you've already gone beyond the limits of degenerate matter. MASS yes, but what is that in the context of an unobservable region?

 

[skeptic2]

Nothing that crosses the event horizon IS matter anymore, at best you're talking about radiation. Remember, a neutron star, or even a hypothetical quark star isn't dense enough to have an event horizon; by the time you reach that you've already gone beyond the limits of degenerate matter. MASS yes, but what is that in the context of an unobservable region?

I understand that when matter reaches the singularity it may no longer be matter but it's no longer traveling either. Between the EV and the singularity, how fast can massive particles travel and why? Is there some prohibition against exceeding c in that region?

 

[ClamShell]

Nothing that crosses the event horizon IS matter anymore, at best you're talking about radiation. Remember, a neutron star, or even a hypothetical quark star isn't dense enough to have an event horizon; by the time you reach that you've already gone beyond the limits of degenerate matter. MASS yes, but what is that in the context of an unobservable region?

There are some pretty smart posters on this thread; if we let your post
cook for awhile, you should get a decent answer.

 

[ClamShell]

Nothing that crosses the event horizon IS matter anymore, at best you're talking about radiation. Remember, a neutron star, or even a hypothetical quark star isn't dense enough to have an event horizon; by the time you reach that you've already gone beyond the limits of degenerate matter. MASS yes, but what is that in the context of an unobservable region?

Dense mass without a horizon seems to be like a brick wall to falling matter.
Dense mass with a horizon seems to be like a curtained stage with the brick
wall there or not there or both or neither. IE, if we linger long enough just
above the horizon, by the time we finally cross the horizon, the brick wall
will have evaporated.

 

[nismaratwork]

Dense mass without a horizon seems to be like a brick wall to falling matter.
Dense mass with a horizon seems to be like a curtained stage with the brick
wall there or not there or both or neither. IE, if we linger long enough just
above the horizon, by the time we finally cross the horizon, the brick wall
will have evaporated.

There's no wall, especially since you have to remember that everything falling into a black hole is ripped apart by gravitational tidal forces, and blasted by radiation. The EH has no physical existence, it's just the ever-changing (as long as there is infalling matter or HR) demarcation of the point of no return. Degenerate matter is dense, yes, but even that would be "sphagettified" as it fell into a BH. In a very real sense, anything on OUR side of the EH can never be observed by us to cross the EH, so there is the theoretical notion of a wall. Keep in mind that the infalling mass will not experience any such barrier, and crosses the EH without any resistance. I believe that your understanding of Einstein's view of gravity and spacetime is incomplete, and to grasp just what a black hole is, you need to understand that first.

I understand that when matter reaches the singularity it may no longer be matter but it's no longer traveling either. Between the EV and the singularity, how fast can massive particles travel and why? Is there some prohibition against exceeding c in that region?

I truly have no idea... certainly as Chalnoth has said earlier until you hit the singularity you can work out numbers with GR equations, but there is in my view, plenty of reason not to trust them. I don't think velocity and the notion of individual particles applies beyond the EH, but who knows? Truly, we just can't know anything about what happens past the EH; there is only theory that ceases to be meaningful at the most critical point (the singularity). I think that a theory of quantum-gravity should eliminate the singularity, and then you might have some reasonable predictions, but we're just not there yet.

 

[ClamShell]

There's no wall, especially since you have to remember that everything falling into a black hole is ripped apart by gravitational tidal forces, and blasted by radiation. The EH has no physical existence, it's just the ever-changing (as long as there is infalling matter or HR) demarcation of the point of no return. Degenerate matter is dense, yes, but even that would be "sphagettified" as it fell into a BH. In a very real sense, anything on OUR side of the EH can never be observed by us to cross the EH, so there is the theoretical notion of a wall. Keep in mind that the infalling mass will not experience any such barrier, and crosses the EH without any resistance. I believe that your understanding of Einstein's view of gravity and spacetime is incomplete, and to grasp just what a black hole is, you need to understand that first.

I would be the first to admit that my understanding of everything is incomplete.
Only fools are so confident as to think they know it all, and there are no fools
here. Not even Einstein would claim to have a complete knowledge of gravity.
I suspect you mean that my knowledge of Einstein is incomplete...yours is?

It is accepted by previous posters, that distant observers will never see a
a test mass cross the horizon. I take this to mean that when it does finally
happen(relative to the test mass), the stage and its contents will have evaporated.
Supposedly by Hawking radiation. And that a distant observer does not have a
long enough duration to observe this. But the test mass(by its own clock)
would experience nothing in particular because (after infinity by distant
observers clocks), the BH will have evaporated. A no show.

 

[George Jones]

Nothing that crosses the event horizon IS matter anymore, at best you're talking about radiation.

This isn't true. If a star collapses and forms a black hole, then matter falling towards the star, but above the star, will remain matter far inside the event horizon. Matter that falls into a black hole at the centre of a galaxy won't spaghettified until far inside the event horizon.

I understand that when matter reaches the singularity it may no longer be matter but it's no longer traveling either. Between the EV and the singularity, how fast can massive particles travel and why? Is there some prohibition against exceeding c in that region?

The speed of light is the local speed limit everywhere, even inside black holes.

There's no wall, especially since you have to remember that everything falling into a black hole is ripped apart by gravitational tidal forces, and blasted by radiation.

According to the book Quantum Fields in Curved Space by Birrell and Davies, pages 268-269,

These consideration resolve an apparent paradox concerning the Hawking effect. The proper time for a freely-falling observer to reach the event horizon is finite, yet the free-fall time as measured at infinity is infinite. Ignoring back-reaction, the black hole will emit an infinite amount of radiation during the time that the falling observer is seen, from a distance to reach the event horizon. Hence it would appear that, in the falling frame, the observer should encounter an infinite amount of radiation in a finite time, and so be destroyed. On the other hand, the event horizon is a global construct, and has no local significance, so it is absurd to0 conclude that it acts as physical barrier to the falling observer.

The paradox is resolved when a careful distinction is made between particle number and energy density. When the observer approaches the horizon, the notion of a well-defined particle number loses its meaning at the wavelengths of interest in the Hawking radiation; the observer is 'inside' the particles. We need not, therefore, worry about the observer encountering an infinite number of particles. On the other hand, energy does have a local significance. In this case, however, although the Hawking flux does diverge as the horizon is approached, so does the static vacuum polarization, and the latter is negative. The falling observer cannot distinguish operationally between the energy flux due to oncoming Hawking radiation and that due to the fact that he is sweeping through the cloud of vacuum polarization. The net result is to cancel the divergence on the event horizon, and yield a finite result, ...

This finite amount of radiation is negligible for observers freely falling into a black hole.

It is accepted by previous posters, that distant observers will never see a
a test mass cross the horizon. I take this to mean that when it does finally
happen(relative to the test mass), the stage and its contents will have evaporated.
Supposedly by Hawking radiation. And that a distant observer does not have a
long enough duration to observe this. But the test mass(by its own clock)
would experience nothing in particular because (after infinity by distant
observers clocks), the BH will have evaporated. A no show.

Consider two observers, observer A that falls across the the event horizon and observer B that hovers at a finite "distance" above the event horizon, and two types of (uncharged) spherical black holes, a classical black hole that doesn't emit Hawking radiation and a semi-classical black hole that does.

For the classical black hole case, B "sees" A on the event horizon at infinite future time, and B never sees the singularity.

For the semi-classical black hole case, at some *finite* time B simultaneously "sees": A on the event horizon; the singularity. In other words, the singularity becomes naked, and A winks out of existence at some finite time in the future for B.

In both cases, A crosses the event horizon, remains inside the event horizon, and hits the singularity. In both cases, B, does not see (even at infinite future time) A inside the event horizon, as this view is blocked by the singularity.

These conclusions can be deduced from Penrose diagrams, FIGURE 5.17 and FIGURE 9.3 in Carroll's text, and Fig. 12.2 and Fig, 14.4 in Wald's text, or

http://www.google.ca/imgres?imgurl=...a=X&ei=3pmdTP63FcaAlAexkYntAg&ved=0CBwQ9QEwAA.

 

[ClamShell]

Consider two observers, observer A that falls across the the event horizon and observer B that hovers at a finite "distance" above the event horizon, and two types of (uncharged) spherical black holes, a classical black hole that doesn't emit Hawking radiation and a semi-classical black hole that does.

For the classical black hole case, B "sees" A on the event horizon at infinite future time, and B never sees the singularity.

For the semi-classical black hole case, at some *finite* time B simultaneously "sees": A on the event horizon; the singularity. In other words, the singularity becomes naked, and A winks out of existence at some finite time in the future for B.

In both cases, A crosses the event horizon, remains inside the event horizon, and hits the singularity. In both cases, B, does not see (even at infinite future time) A inside the event horizon, as this view is blocked by the singularity.

I guess this holds even when B is a distant(but finite) observer. Is it because
A, in the Hawking radiation case, "pairs-up" with A' (a wave), that A can wink
out when A' escapes the grip of the black hole(becomes Hawking radiation) and
heads for infinity as A drops through the event horizon? Does A' come from
additional infalling matter or does A' come from the black hole? IE, does the
black hole in both cases, last forever? Do modern black holes "evaporate"?

 

[skeptic2]

The speed of light is the local speed limit everywhere, even inside black holes.

Thank you George. I understand why c is the limit outside of black holes but not why those reasons must also apply inside them.

 

[nismaratwork]

This isn't true. If a star collapses and forms a black hole, then matter falling towards the star, but above the star, will remain matter far inside the event horizon. Matter that falls into a black hole at the centre of a galaxy won't spaghettified until far inside the event horizon.

Yes, I realize that, but for the sake of this thread I didn't think that getting into the distinction between stellar mass and AGNs was such a good idea.

According to the book Quantum Fields in Curved Space by Birrell and Davies, pages 268-269,


This finite amount of radiation is negligible for observers freely falling into a black hole.


Consider two observers, observer A that falls across the the event horizon and observer B that hovers at a finite "distance" above the event horizon, and two types of (uncharged) spherical black holes, a classical black hole that doesn't emit Hawking radiation and a semi-classical black hole that does.

Well there shouldn't be ANY HR emitted if there's anyone around to fall into a black hole (background temps and all). I was thinking again, of a stellar mass black hole with an active accretion disk, not radiation emitted from the BH itself. Specifically a Kerr BH with a rapid rotation and a fairly robust ergoregion, probably with a companion star and constant infalling matter. Again, I didn't see the need to enter into those complexities when the basics seemed to be at issue. Thanks for the clarification however.

 

[JaredJames]

http://en.wikipedia.org/wiki/Wormhole

So here's the wiki page on wormholes.

Now, it describes two wormholes, one which may possibly be present by a black holes:

"The first type of wormhole solution discovered was the Schwarzschild wormhole which would be present in the Schwarzschild metric describing an eternal black hole, but it was found that this type of wormhole would collapse too quickly for anything to cross from one end to the other."

But these cannot be traversed as it explains and as such, I don't understand how particles as the article puts it are able to 'cross between the two universes'.

Wormholes which could actually be crossed, known as traversable wormholes, would only be possible if exotic matter with negative energy density could be used to stabilize them (many physicists such as Stephen Hawking[1], Kip Thorne[2], and others[3][4][5] believe that the Casimir effect is evidence that negative energy densities are possible in nature). Physicists have also not found any natural process which would be predicted to form a wormhole naturally in the context of general relativity, although the quantum foam hypothesis is sometimes used to suggest that tiny wormholes might appear and disappear spontaneously at the Planck scale.

This uses negative energy, enough said.

Now, you keep pointing us to the wikis and to read them, and I have done. I have also done some digging and following links provided in the wikipedia article (the articles I believe you are reading) I found this (http://casa.colorado.edu/~ajsh/schww.html):

Do Schwarzschild wormholes really exist?
Schwarzschild wormholes certainly exist as exact solutions of Einstein's equations.
However:
# When a realistic star collapses to a black hole, it does not produce a wormhole;
# The complete Schwarzschild geometry includes a white hole, which violates the second law of thermodynamics;
# Even if a Schwarzschild wormhole were somehow formed, it would be unstable and fly apart.

As the type of wormhole you are referring to is the above Schwarzschild wormhole, everything I have read so far is very clear in what it is saying regarding their existence - they only exist under the 'perfect' conditions of the equations. "when a realistic star collapses to form a black hole, it does not produce a wormhole.". So unless you can cite sources which show these wormholes can exist when a star collapses, this is overly speculative and against PF guidelines. It is pointless us discussing this if it just isn't possible and so far, nothing has shown it is.

 

[ClamShell]

Free falling observer does not observe the same Hawking radiation as observer located far from BH because for the falling observer event horizon is in different place. So the amount of hawking radiation he receives is very small. both position of the apparent Horizon and Hawking radiation are observer-dependent.

Time dilation is infinite only for the hovering observer, so true, observer hovering near the horizon would see the Universe accelerated and blue-shifted. However, falling observer would see the Universe red-shifted (surprise!)

I misunderstood this post first time through. Now I'm thinking Dmitry
has a clever way of minimizing the HR by allowing the HR evaporation
to reduce the BH mass and thereby reduce the horizon so the infalling
test mass has an even harder time getting to the horizon and minimizing
the HR for the free-falling test mass. Wait, is this a bit too circular?
Nevermind, whatever makes the HR non-lethal is OK by me.

 

[ClamShell]

http://en.wikipedia.org/wiki/Wormhole

So here's the wiki page on wormholes.

Now, it describes two wormholes, one which may possibly be present by a black holes:


But these cannot be traversed as it explains and as such, I don't understand how particles as the article puts it are able to 'cross between the two universes'.


This uses negative energy, enough said.

Now, you keep pointing us to the wikis and to read them, and I have done. I have also done some digging and following links provided in the wikipedia article (the articles I believe you are reading) I found this (http://casa.colorado.edu/~ajsh/schww.html):


As the type of wormhole you are referring to is the above Schwarzschild wormhole, everything I have read so far is very clear in what it is saying regarding their existence - they only exist under the 'perfect' conditions of the equations. "when a realistic star collapses to form a black hole, it does not produce a wormhole.". So unless you can cite sources which show these wormholes can exist when a star collapses, this is overly speculative and against PF guidelines. It is pointless us discussing this if it just isn't possible and so far, nothing has shown it is.

Yes, good work...wormholes are in the peer literature, so we can discuss
them. The GR model (guess it's not quantum mechanical) has them doing
such-and-such in GR metrics. QM is bound to be a better framework, but
pretty speculative...lets never mention them again...and let's never mention
quantum gravity either...and the strong force, what's that all about? Don't
mention it. And fringe physics and all the nuts in the basement doing it.
And alpha...who cares if it's changing. etc., etc. Can you add to the list
any more forbidden topics? I don't even like wormholes...I am much more
interested in a BH evaporating before matter can ever fall into it. Are
you going to forbid this too? Seriously, if a concept is only on somebody's
personal webpage, I rather not have it jammed down my throat either.

 

[JaredJames]

What are you talking about ClamShell? Seriously, I don't want to sound nasty here, but I find your posts to be full of metaphors and riddles and make little sense.

I have nothing against the concept of a wormhole, but so far everything I have read says they cannot be created when a star collapses into a black hole. So discussing it, unless I'm otherwise informed, is pointless.

You don't even like wormholes? A few posts back you were explaining how they were the potential answer to conditions inside the event horizon (something regarding entropy I believe and you not liking the idea of Hawking Radiation).

Stick to the black hole evaporation from now and and let's forget wormholes were ever brought into this particular topic of "What's inside the event horizon?".

 

[ClamShell]

...let's forget wormholes were ever brought into this particular topic of "What's inside the event horizon?".

Agreed, but I might forget every now and then.
And let's not mention Schroedinger's Cat either;
half the time when I open the box it's stiff as a
board. What is it that you said about metaphores
and riddles?

Consider this...a way to transport yourself into an infinitely
distant future, is to hover over the event horizon until the
BH finishes evaporating. Imagine all the cool stuff that would
just be lying around, free for the taking. And it should only
take a couple of minutes.

Somebody just hit on my *Wormholes?* thread...and my
dyslexia started acting up...need to take a Tum; or is it
dyspepsia...doesn't matter.

 

[JaredJames]

Consider this...a way to transport yourself into an infinitely
distant future, is to hover over the event horizon until the
BH finishes evaporating. Imagine all the cool stuff that would
just be lying around, free for the taking. And it should only
take a couple of minutes.

I believe this is the plot of the Andromeda TV show. A ship gets stuck in the event horizon of a black hole and experiences 300 years of time dilation.

Again, in reality, the gravity that causes the time dilation would cause your immediate destruction.

 

[Chalnoth]

I believe this is the plot of the Andromeda TV show. A ship gets stuck in the event horizon of a black hole and experiences 300 years of time dilation.

Again, in reality, the gravity that causes the time dilation would cause your immediate destruction.

Well, the idea here is that the ship was able to hover just above the event horizon (it is powered, after all). So it's not completely nuts (except for the fact that the power requirements would be astronomical). The real problem is that that degree of time dilation doesn't occur until you're just outside the event horizon, which means the ship itself was too large for it to work that way.

For an astrophysical black hole, I don't think the tidal forces outside the event horizon would have been enough to destroy the ship.

 

[ClamShell]

I believe this is the plot of the Andromeda TV show. A ship gets stuck in the event horizon of a black hole and experiences 300 years of time dilation.

Again, in reality, the gravity that causes the time dilation would cause your immediate destruction.

Remember, weightless, free-falling into a non-rotating BH, with Hawking
radiation of very low intensity inside the rocket ship. Infinity is much bigger
than 300; they must have only been stuck for a picosecond. This destruction
you speak of is only wishful thinking; plenty of sources disagree with
this. Anyway, what's immediate mean when you are approaching the
horizon? The important thing is what Alice(A) sees, not what Bob(B)
sees. Bob sees Alice wink out, that doesn't mean that Alice has past.
IE, Alice's past does not include the evaporation of the BH. And it's
the Carl Sagan movie 'Contact', not the kids show 'Andromeda'.

 

[Calimero]

For an astrophysical black hole, I don't think the tidal forces outside the event horizon would have been enough to destroy the ship.

But acceleration against gravity would be tremendous, and pretty unpleasant for anybody on board.

 

[Chalnoth]

Remember, weightless, free-falling into a non-rotating BH, with Hawking
radiation of very low intensity inside the rocket ship. Infinity is much bigger
than 300; they must have only been stuck for a picosecond. This destruction
you speak of is only wishful thinking; plenty of sources disagree with
this. Anyway, what's immediate mean when you are approaching the
horizon? The important thing is what Alice(A) sees, not what Bob(B)
sees. Bob sees Alice wink out, that doesn't mean that Alice has past.
IE, Alice's past does not include the evaporation of the BH. And it's
the Carl Sagan movie 'Contact', not the kids show 'Andromeda'.

If the ship were actually in the event horizon, the electromagnetic force would no longer be able to keep the ship's atoms together, so it would actually have been pulled apart. The only thing that rescues this scenario is the idea that it would have been just above the event horizon, not within it. But even then, as I mentioned earlier, it doesn't work numerically because the ship was just too big.

 

[Chalnoth]

But acceleration against gravity would be tremendous, and pretty unpleasant for anybody on board.

That's another good point, but they did have artificial gravity on board!

 

[Calimero]

That's another good point, but they did have artificial gravity on board!

Ah yes, good old artificial gravity, interstellar traveler's best friend.

 

[JaredJames]

That's another good point, but they did have artificial gravity on board!

Which "amplified the effects of the time dilation from the black holes gravity".

From the point of view of those on board the ship, no time passed, but for those outside 300 years went by.

But yes, the gravity would pancake everyone on board and you would need some wicked engine power to get away from the event horizon.

In the TV show they were 'towed' out with what I can only describe as the worlds greatest tow rope!

Ironically, I just finished watching all five seasons of it. So pretty sharp on it's content at the moment.

 

[ClamShell]

But acceleration against gravity would be tremendous, and pretty unpleasant for anybody on board.

Wait a gosh darn second, time goes very slow near the horizon as measured
by Bob. If Alice tried to hover longer by turning on her
retrorockets it would be in vain; better to conserve fuel for finding an earth-
like planet after the BH has evaporated. And she wouldn't retro very long,
anyway. Wait, Bob would see Alice's rockets fire for a long time; he might
even think she is going to run out of fuel. No way, Alice would only use
a few minutes of fuel. Bob is way off track about the fuel issue.

 

[Chalnoth]

Which "amplified the effects of the time dilation from the black holes gravity".

From the point of view of those on board the ship, no time passed, but for those outside 300 years went by.

But yes, the gravity would pancake everyone on board and you would need some wicked engine power to get away from the event horizon.

In the TV show they were 'towed' out with what I can only describe as the worlds greatest tow rope!

Ironically, I just finished watching all five seasons of it. So pretty sharp on it's content at the moment.

Hehe, we've certainly gone off on a bit of a tangent here, haven't we? But it's a fun tangent!

As for the tow rope, it doesn't necessarily have to have been absurdly strong, because the ship was supposedly keeping itself from falling into the black hole under its own power. They only need to give it a little extra pull to get it out.

However, what should have happened then is the ship rocketing off under its own power away from the black hole, after that initial bit of outward pull was provided.

 

[Calimero]

If Alice tried to hover longer by turning on her
retrorockets it would be in vain; better to conserve fuel for finding an earth-
like planet after the BH has evaporated.

There are no earth-like planets, after bh finishes evaporating. Universe is almost empty, with few galaxies full of dying stars.

 

[Chalnoth]

There are no earth-like planets, after bh finishes evaporating. Universe is almost empty, with few galaxies full of dying stars.

I was pretty sure that black holes don't evaporate until all the stars are already dead. A quick glance at Wikipedia indicates that all of the stars will be gone after about 10^{40} years, while the lifetime of a solar-mass black hole is about 10^{66} years.

You might be able to get stars to last a bit longer with exotic physics, but I doubt you can get them to last 10^{26} times as long...

 

[Calimero]

I was pretty sure that black holes don't evaporate until all the stars are already dead. A quick glance at Wikipedia indicates that all of the stars will be gone after about 10^{40} years, while the lifetime of a solar-mass black hole is about 10^{66} years.

You might be able to get stars to last a bit longer with exotic physics, but I doubt you can get them to last 10^{26} times as long...

If protons decay I was off for a few gazillion years. It turns out that when present day stellar mass black hole evaporates, there will not be anything other then black holes (black hole era), and scarce radiation.

 

[nismaratwork]

I was pretty sure that black holes don't evaporate until all the stars are already dead. A quick glance at Wikipedia indicates that all of the stars will be gone after about 10^{40} years, while the lifetime of a solar-mass black hole is about 10^{66} years.

You might be able to get stars to last a bit longer with exotic physics, but I doubt you can get them to last 10^{26} times as long...

That was part of my point to George Jones: you have no HR until long after anything coherent exists in the universe, stars included. I think it's safe to say that evaporating black holes is pretty much one of the last stages of heat death for the universe, to be followed by ever more even distribution of radiation. The universe has to greatly "cool" before HR is emitted.

 

[ClamShell]

There are no earth-like planets, after bh finishes evaporating. Universe is almost empty, with few galaxies full of dying stars.

That is bad news...I'm zero for two, in my attempt to find
large-scale features of the black hole that could be symmetrical.

Do you(plural) know of any possible candidates that might be
symmetrical on the event horizon?

 

[qraal]

Yes, when I "probe" the event horizon with Newton's equation for orbital
velocity:

V = square root[GM/R] and plug in R = 2.95 Kilometers and M = our sun,
I get 212000 km/s, not 300000 km/s as I expected. I'm not telling you
my values for G and M because I am now thinking that I've got them
wrong...do you get 300000 km/s for the orbital velocity near the event
horizon? Does Newton's equation need more terms when relativity
is accounted for?

Newton gets left behind long before the Event Horizon is reached because of the curvature of space around the black hole. At a radius of 1.5 times the schwarzschild radius, photons have their paths deflected by 90 degrees i.e. they're bent into a circular path around the hole if they graze that distance tangentially. Any other orbit within that distance, like a hyperbolic flyby, becomes very hard to plot and the Newtonian/Keplerian approximations are useless. As gets mentioned in most introductions to the Schwarzschild metric, the radius at which the Newtonian escape velocity is equal to lightspeed is only coincidentally equal to the radius of the event horizon - they're derived via different assumptions.

 

[ClamShell]

There are no earth-like planets, after bh finishes evaporating. Universe is almost empty, with few galaxies full of dying stars.

OK, let me be more specific...am I correct in assuming
that a *metric* is just another name for what I call a
*coordinate system*? say a Cartesian coordinate system,
(x, y, z) or a Spherical coordinate system,
(R, theta, phi)? And, if so, is the coordinate system on
our side of the event horizon in any way symmetrical with
the coordinate system chosen for the other side of the
event horizon? I am looking for symmetries that imply
conservation of entropy, as always.

 

[ClamShell]

Newton gets left behind long before the Event Horizon is reached because of the curvature of space around the black hole. At a radius of 1.5 times the schwarzschild radius, photons have their paths deflected by 90 degrees i.e. they're bent into a circular path around the hole if they graze that distance tangentially. Any other orbit within that distance, like a hyperbolic flyby, becomes very hard to plot and the Newtonian/Keplerian approximations are useless. As gets mentioned in most introductions to the Schwarzschild metric, the radius at which the Newtonian escape velocity is equal to lightspeed is only coincidentally equal to the radius of the event horizon - they're derived via different assumptions.

That 90 degree deflection seems analogous to the deflection
(left or right-hand rule) of electrons in a charge field due to
the magnetic field created by the motion of the charge around
the nucleus.

As per other posts on this thread V_escape = root(2)*V_orbital and
V_orbital = root(GM/R) for Newton and Einstein (but derived via
different assumptions). "Only coincidental" seems to imply that they
came together randomly, but I suspect they arise due to identities
in each of the derivations(not between the derivations). Can you
enlighten me a bit?

 

[Chalnoth]

OK, let me be more specific...am I correct in assuming
that a *metric* is just another name for what I call a
*coordinate system*?

Well, no, not really. A metric describes how to calculate lengths between points in a specific coordinate system. Yes, it is true that metrics do appear different in different coordinate systems. However, there is more to it than just this, because the metric also encodes the curvature of the underlying manifold.

One way to think about it is that you can perform a wide variety of coordinate transformations, and the metric changes as you change coordinates so that any lengths you calculate stay the same.

And, if so, is the coordinate system on
our side of the event horizon in any way symmetrical with
the coordinate system chosen for the other side of the
event horizon? I am looking for symmetries that imply
conservation of entropy, as always.

No, it really isn't, because the important quantities here are ones that are independent of the choice of coordinate system. For instance, if we take a path of an object that comes out from infinity and strikes the black hole, it spends an infinite amount of proper time outside the black hole traveling towards it, but once it reaches the event horizon, it takes a finite amount of proper time to strike the singularity. That sort of behavior is about as asymmetric as you can get.

 

[ClamShell]

Well, no, not really. A metric describes how to calculate lengths between points in a specific coordinate system. Yes, it is true that metrics do appear different in different coordinate systems. However, there is more to it than just this, because the metric also encodes the curvature of the underlying manifold.

I guess then that root(x^2 + y^2 + z^2) might be a metric?

One way to think about it is that you can perform a wide variety of coordinate transformations, and the metric changes as you change coordinates so that any lengths you calculate stay the same.

The laws of physics are the same everywhere for different observers?

No, it really isn't, because the important quantities here are ones that are independent of the choice of coordinate system. For instance, if we take a path of an object that comes out from infinity and strikes the black hole, it spends an infinite amount of proper time outside the black hole traveling towards it, but once it reaches the event horizon, it takes a finite amount of proper time to strike the singularity. That sort of behavior is about as asymmetric as you can get.

Could an asymmetry in one metric be a symmetry in another metric?

 

[qraal]

Newton gets left behind long before the Event Horizon is reached because of the curvature of space around the black hole. At a radius of 1.5 times the schwarzschild radius, photons have their paths deflected by 90 degrees i.e. they're bent into a circular path around the hole if they graze that distance tangentially. Any other orbit within that distance, like a hyperbolic flyby, becomes very hard to plot and the Newtonian/Keplerian approximations are useless. As gets mentioned in most introductions to the Schwarzschild metric, the radius at which the Newtonian escape velocity is equal to lightspeed is only coincidentally equal to the radius of the event horizon - they're derived via different assumptions.

There's a simple derivation of the photon sphere radius on Wikipedia...

http://en.wikipedia.org/wiki/Schwar...vistic_circular_orbits_and_the_photon_sphere"

...which aids understanding - a bit more anyway.

 

[Chalnoth]

I guess then that root(x^2 + y^2 + z^2) might be a metric?

Almost. Metrics are differential, so that you can calculate distances over curves of any shape, not just straight lines. The metric for Euclidean space, then, is:

ds = \sqrt{dx^2 + dy^2 + dz^2}

You calculate distances by doing a line integral, typically parameterizing the line by some parameter like so (here I'll use time as the parameter for convenience):

s = \int_{t_1}^{t_2} \sqrt{\left({dx \over dt}\right)^2+\left({dy \over dt}\right)^2+\left({dz \over dt}\right)^2}dt

If you have some path that includes the information x(t), y(t), and z(t), you can calculate the path length by performing the above integral over time. You should discover that if you take the path to be a straight line, you'll end up with the distance you mentioned above.

The laws of physics are the same everywhere for different observers?

Yes.

Could an asymmetry in one metric be a symmetry in another metric?

Stated this way, it makes little sense to me. First, for the example I gave, I listed a parameter which is completely independent of the choice of coordinates. If two metrics aren't related to one another by a change in coordinates, then those two metrics aren't describing the same system. So coordinate-independent constructions, such as the proper time of an infalling particle, can be taken as real properties of the system, independent of whatever arbitrary choice of coordinates we make.

Second, properties can be divided into two sorts: symmetric properties and every other way things can be. In general, when you have a symmetric property of a system, a change of coordinates won't necessarily show the same symmetry. However, if you have an asymmetric system, there is no guarantee that you can make the system look symmetric, no matter how you try to change the coordinates.

 

[nismaratwork]

Almost. Metrics are differential, so that you can calculate distances over curves of any shape, not just straight lines. The metric for Euclidean space, then, is:

ds = \sqrt{dx^2 + dy^2 + dz^2}

You calculate distances by doing a line integral, typically parameterizing the line by some parameter like so (here I'll use time as the parameter for convenience):

s = \int_{t_1}^{t_2} \sqrt{\left({dx \over dt}\right)^2+\left({dy \over dt}\right)^2+\left({dz \over dt}\right)^2}dt

If you have some path that includes the information x(t), y(t), and z(t), you can calculate the path length by performing the above integral over time. You should discover that if you take the path to be a straight line, you'll end up with the distance you mentioned above.


Yes.


Stated this way, it makes little sense to me. First, for the example I gave, I listed a parameter which is completely independent of the choice of coordinates. If two metrics aren't related to one another by a change in coordinates, then those two metrics aren't describing the same system. So coordinate-independent constructions, such as the proper time of an infalling particle, can be taken as real properties of the system, independent of whatever arbitrary choice of coordinates we make.

Second, properties can be divided into two sorts: symmetric properties and every other way things can be. In general, when you have a symmetric property of a system, a change of coordinates won't necessarily show the same symmetry. However, if you have an asymmetric system, there is no guarantee that you can make the system look symmetric, no matter how you try to change the coordinates.

Oh lord, I've been here for too long, I'm understanding the math and thinking, "well that's just simple algebra and no matrices or operators involved!" *rubs temples*. If I forget what I need to do my job because visions of sugarplums and tensors dancing in my head, I'm blaming this site, and the books of math and physics it inspired me to read.

 

[ClamShell]

...because the important quantities here are ones that are independent of the choice of coordinate system. For instance, if we take a path of an object that comes out from infinity and strikes the black hole, it spends an infinite amount of proper time outside the black hole traveling towards it, but once it reaches the event horizon, it takes a finite amount of proper time to strike the singularity. That sort of behavior is about as asymmetric as you can get.

My question was:

"...is the coordinate system on
our side of the event horizon in any way symmetrical with
the coordinate system chosen for the other side of the
event horizon? I am looking for symmetries that imply
conservation of entropy, as always."

My question is now:

Then it would seem to me that the Schwarzschild metric

(a metric now means to me; a tool for modeling a system,
on some underlying manifold, containing many particles,
many motions, and many observers)

...that the Schwarzschild metric transforms infinities
into finities as the BH horizon is crossed. Would it
not be more elegant for it to conserve (cleverly), the
time and space infinities on both sides of the horizon?
IE, if the horizon was just a regular curtain, the above
object would be free to continue its journey to another
(possibly different) infinity? Or, if the BH horizon is
really very exotic, to transform the above infinities
into into an infinite trajectory on the inside of the
event horizon?

No personal theory here...I just want entropies to balance.

 

[ClamShell]

Oh lord, I've been here for too long, I'm understanding the math and thinking, "well that's just simple algebra and no matrices or operators involved!" *rubs temples*. If I forget what I need to do my job because visions of sugarplums and tensors dancing in my head, I'm blaming this site, and the books of math and physics it inspired me to read.

There is nothing threatening about the details, they are easy for
daydreamers just to down right ignore. The general principles
are enough for some of us.

 

[Chalnoth]

My question was:

"...is the coordinate system on
our side of the event horizon in any way symmetrical with
the coordinate system chosen for the other side of the
event horizon? I am looking for symmetries that imply
conservation of entropy, as always."

My question is now:

Then it would seem to me that the Schwarzschild metric

(a metric now means to me; a tool for modeling a system,
on some underlying manifold, containing many particles,
many motions, and many observers)

...that the Schwarzschild metric transforms infinities
into finities as the BH horizon is crossed. Would it
not be more elegant for it to conserve (cleverly), the
time and space infinities on both sides of the horizon?
IE, if the horizon was just a regular curtain, the above
object would be free to continue its journey to another
(possibly different) infinity? Or, if the BH horizon is
really very exotic, to transform the above infinities
into into an infinite trajectory on the inside of the
event horizon?

No personal theory here...I just want entropies to balance.

This doesn't make any sense to me. Again, as I said, when you deal with coordinate-independent quantities, the inside of the black hole is qualitatively different from the outside. No amount of fiddling with coordinates can possibly change this.

As for "getting the entropy to balance", I have no idea why you would want to do this, or what it would even mean if you did.

 

[ClamShell]

This doesn't make any sense to me. Again, as I said, when you deal with coordinate-independent quantities, the inside of the black hole is qualitatively different from the outside. No amount of fiddling with coordinates can possibly change this.

As for "getting the entropy to balance", I have no idea why you would want to do this, or what it would even mean if you did.

Party-Pooper...

 

[ClamShell]

This doesn't make any sense to me. Again, as I said, when you deal with coordinate-independent quantities, the inside of the black hole is qualitatively different from the outside. No amount of fiddling with coordinates can possibly change this.

As for "getting the entropy to balance", I have no idea why you would want to do this, or what it would even mean if you did.

Seriously (my previous post said "party-pooper"), when
I analyze electronic circuits that have amplifiers and filters
(when I'm doing it right, of course), the Shannon entropy
of the input equals the entropy of the interesting output
plus the entropy of the losses. The entropy of the
interesting output is lower (more information) than the
input entropy (hopefully). Normally, when I discuss this
stuff it's with Shannon folks(I'm a fish out-of-water here).
My trouble was that there is really only one(favored) observer
when electronics display information. I needed to assume
that all observers would see the same thing. Now I have
extended my concept to have the ability to calculate
what any of the other observers is seeing. But I can't now
see how this would change things enough to make information
not be conserved in a real universe. Information is the
negative of entropy, but cannot go below 0, since you know
everything at entropy equal to 0.

My electronic circuits are real things and their "blackhole"
is related fabrication problems.

What I suggest is that when you admit no knowledge of
what is inside the event horizon, don't be happy with
extrapolating a model into it. Put what you know
already, behind the horizon...a universe. Please do
not hit me with that justification that you're here
only to teach mainstream dogma. Your dogma has a
black hole...real dogma's only have fleas. There
I go again with the metaphores and riddles.

 

[Chalnoth]

There I go again with the metaphores and riddles.

Yeah, I have no idea what you're talking about here.

I will just add a couple of notes on black holes and entropy. First, in the real world, entropy can and does genuinely increase. Entropy is not a conserved quantity. If you have a universe with some amount of mass, the highest-entropy configuration for that universe is for all of the mass to be in a single black hole. Thus we should be completely unsurprised at the calculations which suggest that as time goes forward, eventually all of the matter in the visible universe will become black holes.

Then, if you consider a universe with a single evaporating black hole, the entropy of the universe as a whole is higher after the black hole evaporates, so that the highest-entropy configuration of a region of space-time is for that region to be completely empty. And this is, in fact, where our universe appears to be headed.

So to me, the entropy considerations with respect to a black hole make perfect sense.

 

[nismaratwork]

There is nothing threatening about the details, they are easy for
daydreamers just to down right ignore. The general principles
are enough for some of us.

I was kidding Clam... I already have a decent grasp of basic algebra (which this is). I was just playing around with Chalnoth. I have to ask you... is English your first language? I don't mean that as an insult, but you seem to be speaking in a somewhat odd fashion. I'm wondering if maybe this just a language barrier issue... if so there may be someone here who speaks your first language well enough to get past it. If not, then I have to say it seems to me you're more interested in word games than physics or cosmology. If you're genuinely interested in the physics of what you're asking, then the riddles and metaphors really don't do you or anyone else any good.

For instance, you say if it was "headed there, likely it would already be there." That's genuinely nonsensical, no two ways about it. The dissipation of radiation and the life-cycles of stars, black holes, and the "evening" out of radiation in a given space takes time. The simple answer is that it is precisely where the universe is headed, but we're not at that time yet, or anywhere near it.

If language isn't the issue, then I have no idea where you're getting this. From what I gather you're a computer or electrical engineer, so frankly you should realize that it's important to understand basics before moving on to more complex issues. You can't read a bit of wikipedia and expect to be competent in a debate about the fate of the universe, the nature of black holes as they're described by GR, and might be in a unified framework of quantum gravity. Reading this thread, I could almost believe you've been possessed by the ghost of Lewis Carrol (that's a joke btw).

 

[ClamShell]

I was kidding Clam... . Reading this thread, I could almost believe you've been possessed by the ghost of Lewis Carrol (that's a joke btw).

Now I must think you are always kidding.

I must admit that your grammar is very good, and your use
of the "red herring" ploy is excellent.

 

[nismaratwork]

Now I must think you are always kidding.

I must admit that your grammar is very good, and your use
of the "red herring" ploy is excellent.

What red herring?! Other than the Lewis Carroll bit I'm completely serious, and what does my grammar have to do with anything?

 

[ClamShell]

What red herring?! Other than the Lewis Carroll bit I'm completely serious, and what does my grammar have to do with anything?

what's mine have to do with it?