LHC Black Holes

No, they are different things.

Before relativity, the concept of "Newtonian black holes" was considered, with a horizon which had an escape velocity equal to the speed of light. A /powered/ spaceship could still escape.

What you told me - repeatedly - was that you can't escape from a black hole because you can't accelerate to faster than the speed of light. That is simply an incorrect justification.

A correct justification is that you would need an infinite amount of energy to keep up the thrust needed to escape. Alternatively, you would need your thrust to be pushed out with an infinite total momentum.

And of course, you can say that space-time is so warped that every direction is "down", but that is much harder to understand.

I understand now that there is no escape for the spaceship once it has passed through the event horizon - as I noted, I am not an expert in this subject, and I have learned a little more during this thread. But I don't think your "escape velocity" explanation is any better than my one just because it happens to match the correct end result.

As a stand-alone reason, this is not enough. From the viewpoint of the spaceship, the inside of the black hole is not that special - it sees no discontinuity or jump as it passes through the event horizon.

Reply to
David Brown
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One of the links I followed says that, mathematically, space itself is falling into the BH. At the event horizon, you have to moving at the speed of light just to stay where you are. Beyond, you can't even do that.

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John Devereux
Reply to
John Devereux

All this reasoning still uses classical (even Special Relativity). But to describe a BH GR (general relativity) applies. To talk about speed you need a frame of reference. But GR does not allow to construct a global frame of reference, only local ones covering a region of space where its curvature is small. But in the vicinity of a BH the curvature is large even on small scales. It gets infinite at the event horizon.

Inside of the Schwarzschild region (event horizon) all usual calculations just do not work. We can only describe the physics outside of the event horizon. We do not have real theory to describe what happens inside. So any arguments based on more or less classical physics and even GR can only give meaningless results.

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Reinhardt
Reply to
Reinhardt Behm

Completely wrong stands.

That much is true. Since they are at radically different gravitational potentials. The Mossbauer nuclear resonance provided an exquisite test of the GR predictions of clocks, frequency and time.

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The latest atomic clocks are now so good that moving them up or down by a couple inches is a detectable systematic error. Time ticks more slowly in a deep gravitational well when viewed from outside.

Not really. They are a consequence of the infinite energy required to propel any finite mass at the speed of light.

Laplace first considered this possibility of a massive object so compact that its surface escape velocity was greater than the speed of light. Back then they believed that Newtonian dynamics held good. That is that acceleration adds the same delta_V no matter how fast you are going. We now know that assumption to be false.

It is just another way of looking at it. Inside the black hole by any reasonable definition (made more difficult by the lack of any inertial frames of reference) you are potentially travelling faster than the speed of light to your doom at the central singularity.

Spacetime itself is quite literally falling into the singularity.

That is that you need an infinite amount of energy to reach the speed of light. You know you are in trouble when you pass through the light sphere where some photons despite travelling at the speed of light can remain trapped in circular orbits. Inside that the only way is down.

As a handwaving explanation that it probably the easiest to understand.

Whenever you appeal to classical physics arguments in GR you have to be very careful how you do it. It is very easy to be a factor of 2 out in the actual prediction unless the full GR treatment is used.

Newtonian physics and the Galilean transform will give you incorrect results. You are in strong gravitational fields and at relativistic speeds here where common sense will inevitably lead you astray.

It isn't that special in terms that they notice nothing special about going into the BH provided that it is big enough not to shred them. However, they are in for a big shock when they do try to leave - it is like the Hotel California in that respect. You can never leave.

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Regards, 
Martin Brown
Reply to
Martin Brown

Maybe the entire universe is a gigantic black hole and everything is "falling in", and that is what causes the passage of time?

you can't reverse the passage of time m

Reply to
makolber

Spacetime has had lots of singularities to fall into and a long time to do it. Why do we still have any Spacetime?

Reply to
John S

You keep saying that without understanding what it means in the context of a black hole.

No, you need to understand what it means. The gravity of the black hole is equivalent to acceleration. To move outward from the black hole is like accelerating at a much higher rate than in earth's gravity, even if you are barely moving. Given the extreme gravity moving outward at all is like exceeding the speed of light. Do you understand that? There is

*no* amount of thrust possible that will accelerate you at *any* speed outward from inside a black hole.

I didn't say that. I simply mention that it is impossible to move outward from inside a black hole because of the escape velocity. You seem to be focused solely on the idea that escape velocity implies moving at that speed. No one else here has said that. It is simply a way to quantifying the effort required to escape a gravitational field. Once that velocity reaches the speed of light, you can't even move against it at all.

Yes, I see that. It's because you don't understand what escape velocity means.

No one said it did. But there is a fundamental difference in what happens whether there is a sign post stating "You have entered the Twilight Zone" or not.

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Rick C
Reply to
rickman

Yes, that would make sense if you could somehow show that "in" becomes "out" inside a black hole. Isn't our universe expanding?

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Rick C
Reply to
rickman

Like Jay Leno said in the Tostitos commercial... "Eat all you want, we'll make more!". Is there only a finite amount of space-time?

I don't like the idea of space-time "falling into" a black hole. I'm not even sure what that means.

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Rick C
Reply to
rickman

Not so.

"Travelling at about 17 kilometers per second (11 mi/s) it has the fastest heliocentric recession speed of any spacecraft. As Voyager 1 headed for interstellar space, its instruments continued to study the Solar System."

Also "The orbital velocity needed to maintain a stable low Earth orbit is about 7.8 km/s."

This information is easy to find.

Reply to
John S

No. Curvature doesn't become infinite at the event horizon of a Schwarzchild black hole (at least not for a large galactic mass one).

The apparent singularity at Rs is an illusion due to a poor choice of coordinates and no more significant than the breakdown of latitude and longitude at the North Pole on planet Earth. See for example:

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Only at r=0 is there a hard unavoidable physical singularity.

There are equations that will work across the boundary although their interpretation is somewhat more difficult (and testing them impossible). The usual calculations are easier for all practical purposes but solutions have been found that span the boundary.

Think of the event horizon as a semipermeable membrane that will let anything go through it but only in the inwards direction. Once inside it is inevitable that you will end up crunched at r=0 in finite clock time no matter how hard you try to avoid your fate.

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Regards, 
Martin Brown
Reply to
Martin Brown

Not only expanding but instead of decelerating it is accelerating away from itself hence the need for Dark Energy to balance the books.

There is no way our universe is enclosed inside a BH - the density of mass energy in the observable universe is not sufficiently high. (even including the dark matter contribution)

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Regards, 
Martin Brown
Reply to
Martin Brown

I was thinking something involving frames of reference... not sure exactly what though, none of it makes sense to me now.

"Lag" mainly. Gotta agree with you now.

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This email has not been checked by half-arsed antivirus software
Reply to
Jasen Betts

You can't do that to physics theories. The Hawking radiation fits into and completes a picture we have of a universe. It predicts very faint radiation of largish black holes that we can predict not to be observable by the means we have available now.

Huh? The particle pairs are everywhere. The black hole only makes them visible, but not to the naked eye.

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Albert van der Horst, UTRECHT,THE NETHERLANDS 
Economic growth -- being exponential -- ultimately falters. 
albert@spe&ar&c.xs4all.nl &=n http://home.hccnet.nl/a.w.m.van.der.horst
Reply to
Albert van der Horst

And therefore it has *not* been verified. Simple enough, no?

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Rick C
Reply to
rickman

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Not exactly. The LIGO gravity wave detector has "seen" a couple of relative ly small black holes in the process of merging to make a slightly bigger sm all black hole, so we do know that they exist. Their properties can be dedu ced from regular physics.

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They would be visible to the naked eye, if the eye was close enough to blac k hole. No human eye could survive there, but the effect would be visible a t a greater distance, for a sufficiently small black hole that was emitting Hawking radiation sufficiently rapidly.

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Bill Sloman, Sydney
Reply to
bill.sloman

You mean what we *assume* are black holes... Regardless. Even if they exist, that says nothing about Hawking radiation.

Someday we will create black holes in the lab and may observe Hawking radiation. My understanding is that the strength of the radiation would be proportional to the size of the black hole, so if we created a very small one it would in essence explode immediately. I think we could detect that.

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Rick C
Reply to
rickman

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That is one way of understanding it.

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tively small black holes in the process of merging to make a slightly bigge r small black hole, so we do know that they exist. Their properties can be deduced from regular physics.

We don't "assume" black holes - we postulate them to explain facts that oth erwise seem impossible to explain. Electrons are equally hypothetical.

Hawking radiation is a necessary consequence of Dirac pair production and a n event horizon.

The problem with modern physics is that it is a package deal - you get the counter-intuitive bits as part of the package. Reject them and you are back to flint axes.

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black hole. No human eye could survive there, but the effect would be visib le at a greater distance, for a sufficiently small black hole that was emit ting Hawking radiation sufficiently rapidly.

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Or something considerably larger, and sufficiently long-lived that we could move it off to safe distance before it came apart.

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Bill Sloman, Sydney
Reply to
bill.sloman

"Sonic black holes" have been verified to have Hawking radiation, if I remember correctly. I agree it is not verification for the case of "ordinary" black holes, but it is a strong indication that the theory is correct.

Reply to
David Brown

There is little observational evidence about black holes that require them to be black holes. Mostly it is the lack of other things like radiation, but then the black holes are actually known by all the indirect radiation. We have to postulate, as you say, because we have yet to observe a black hole directly.

Regardless, it is Hawking radiation we were talking about.

Except that we don't fully understand black holes... so we may have this wrong.

It's not a question of what is intuitive, it is a matter of what is proven. We fully expected the Higgs boson, but we looked for it to prove it. We fully believed the theory of relativity, but we looked for proof for a very long time and are still happy when we can find something new that verifies it.

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Rick C
Reply to
rickman

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