What Happens Inside a Black Hole's Event Horizon?

[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.