What's inside the event horizon

In summary, the conversations discusses the concept of a black hole and its event horizon, which marks the point of no return for anything that enters it, including light. The conversation also touches on the idea of a singularity within the black hole and the limitations of observational evidence in studying it. There is also a discussion about the reliability of using math to understand the event horizon and the possibility of exploring it in the future. The conversation concludes with a discussion about potential sensors for a black hole probe.
  • #71
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.
 
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  • #72
ClamShell said:
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.
 
  • #73
Calimero said:
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!
 
  • #74
Chalnoth said:
That's another good point, but they did have artificial gravity on board!

Ah yes, good old artificial gravity, interstellar traveler's best friend.
 
  • #75
Chalnoth said:
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! :tongue:

Ironically, I just finished watching all five seasons of it. So pretty sharp on it's content at the moment.
 
  • #76
Calimero said:
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.
 
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  • #77
jarednjames said:
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! :tongue:

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.
 
  • #78
ClamShell said:
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.
 
  • #79
Calimero said:
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 [itex]10^{40}[/itex] years, while the lifetime of a solar-mass black hole is about [itex]10^{66}[/itex] 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 [itex]10^{26}[/itex] times as long...
 
  • #80
Chalnoth said:
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 [itex]10^{40}[/itex] years, while the lifetime of a solar-mass black hole is about [itex]10^{66}[/itex] 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 [itex]10^{26}[/itex] 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.
 
  • #81
Chalnoth said:
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 [itex]10^{40}[/itex] years, while the lifetime of a solar-mass black hole is about [itex]10^{66}[/itex] 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 [itex]10^{26}[/itex] 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.
 
  • #82
Calimero said:
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?
 
  • #83
ClamShell said:
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.
 
  • #84
Calimero said:
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.
 
  • #85
qraal said:
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?
 
  • #86
ClamShell said:
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.

ClamShell said:
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.
 
  • #87
Chalnoth said:
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?

Chalnoth said:
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?

Chalnoth said:
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?
 
  • #88
qraal said:
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.
 
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  • #89
ClamShell said:
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:

[tex]ds = \sqrt{dx^2 + dy^2 + dz^2}[/tex]

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):

[tex]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[/tex]

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.

ClamShell said:
The laws of physics are the same everywhere for different observers?
Yes.

ClamShell said:
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.
 
  • #90
Chalnoth said:
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:

[tex]ds = \sqrt{dx^2 + dy^2 + dz^2}[/tex]

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):

[tex]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[/tex]

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. :cry:
 
  • #91
Chalnoth said:
...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.
 
  • #92
nismaratwork said:
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. :cry:

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.
 
  • #93
ClamShell said:
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.
 
  • #94
Chalnoth said:
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...
 
  • #95
Chalnoth said:
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.
 
  • #96
ClamShell said:
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.
 
  • #97
ClamShell said:
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).
 
  • #98
nismaratwork said:
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.
 
  • #99
ClamShell said:
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?
 
  • #100
nismaratwork said:
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?
 
  • #101
ClamShell said:
what's mine have to do with it?

What?! Nothing! I'm not trying to criticize your grammar; there are a number of people here for whom English isn't their first language. I'm wondering, your grammar completely aside, whether you're one such person. I'm trying to understand your behaviour, and it occurred to me that a fundamental series of miscommunications due to language could be at fault. I'm going to assume given your response that I was wrong, and you think I'm saying that you're posts are grammatically incorrect... I'm not... I'm saying the CONTENT sometimes makes no sense.

Lets put aside the language issue then, if I'm wrong... how about the REST of my post?

Nismaratwork said:
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.
 
  • #103
nismaratwork said:
...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.

Got me there...(these 3 dots mean I have paused to ponder your idea)...competent?...
Incompetent is why I'm here...wiki's(wikipedia) are all I've got. I read about metrics
and I was still unclear...a minute with Chalnoth and a glimmer appears...
(edit) it's a hiku (sp)
(you know, that Edison's bulb that appears over your head). After all, we're
BSing here...you have no idea what's over the horizon(of the black hole).
Next, I suspect, you'll be "red herring"ing my spelling errors. You know your
stuff up to just above the horizon(of the black hole)...and this gives you
the ability to predict an empty universe?...ballderdash.
 
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  • #105
OK clamshell, this is ridiculous. I can make little sense of your posts. You tell us you come here for help and so far everything we have told you you have disagreed with and whenever we give you resources you just tell us you don't accept them. What more do you want? Any discussion here is based on the articles we provide you and so will be no different to you reading them.

You don't seem to want to listen to anything people say and have constantly argued anything that disagrees with your own viewpoint. A viewpoint I would like to add, that has changed constantly throughout this thread so it is now way off your original thoughts.

We're "BSing here"? That says all we need to know. You don't care what we say, you ask for us to discuss your opinions, but when we tell you they are wrong you're just going to ignore us and say nobody actually knows anything about black holes.

How do you 'red herring' spelling mistakes? Do you even know what a red herring is? Certainly nothing to do with how you're using it.

We know what we do, because it is based on current theories and what the maths says. You can deny it all you like, but this is not the forum for it. We deal with published and accepted mainstream science, not 'things that go on in your head', and no matter how much you may argue it, so far everything you have proposed is your own overly speculative personal theories which conform very little to mainstream science. The maths gives us the ability to predict an empty universe, and given how we have that on our side and all you have is your own speculation, how can you shoot people down and tell them they know nothing?

I've been nice so far and accommodated you, as have many others here, but now you really are taking the p*** out of what is supposed to be a place for learning and discussion.
 
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<h2>What is the event horizon?</h2><p>The event horizon is the boundary around a black hole from which nothing, including light, can escape.</p><h2>What happens inside the event horizon?</h2><p>Inside the event horizon, the gravitational pull of the black hole is so strong that it causes space and time to become infinitely distorted. This is known as the singularity.</p><h2>Can anything survive inside the event horizon?</h2><p>No, it is believed that anything that crosses the event horizon will be pulled into the singularity and destroyed.</p><h2>What is the size of the event horizon?</h2><p>The size of the event horizon depends on the mass of the black hole. The more massive the black hole, the larger the event horizon.</p><h2>Is there any way to see inside the event horizon?</h2><p>No, since nothing can escape the event horizon, it is impossible to see or gather any information about what is happening inside.</p>

What is the event horizon?

The event horizon is the boundary around a black hole from which nothing, including light, can escape.

What happens inside the event horizon?

Inside the event horizon, the gravitational pull of the black hole is so strong that it causes space and time to become infinitely distorted. This is known as the singularity.

Can anything survive inside the event horizon?

No, it is believed that anything that crosses the event horizon will be pulled into the singularity and destroyed.

What is the size of the event horizon?

The size of the event horizon depends on the mass of the black hole. The more massive the black hole, the larger the event horizon.

Is there any way to see inside the event horizon?

No, since nothing can escape the event horizon, it is impossible to see or gather any information about what is happening inside.

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