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Mike Holland
2011-May-12, 01:38 AM
I am posting this under ATM, but it is pure Mainstream. I am not proposing any new theory, just pointing out aspects of existing theory which many people are unaware of.

When discussing black holes, there are basically two points of view, that of a remote observer and that of a poor spaceman who falls into one. The difference is caused by gravitational time dilation. From the remote viewer’s point of view (or in his time frame, if you prefer), the passage of time is retarded near the black hole, and comes to a complete stop at the Schwarzschild radius. So as far as he is concerned, the falling spaceman never reaches the Schwarzschild radius (which for convenience I will abbreviate to R0), but hovers just outside it gradually edging closer and closer. The spaceman, in turn, would have a very different experience, falling through R0 in a very short period of time.

The consequence of this is that as far as the outside universe is concerned, the spaceman never enters the black hole. And neither does any other falling matter. Nothing has ever fallen into a black hole as far as our clocks are concerned!

But you don’t need a black hole to get time dilation. We have it right here due to Earth’s gravitational field, as has been measured using atomic clocks in orbit. We ourselves are time-delayed relative to a remote observer in space. So a super-massive collapsing object which is nearly a black hole would itself be highly time dilated (by our clocks), and the collapse process itself would slow down and come to a complete stop just as it reaches black hole status - when our clocks read infinity.

What we end up with is an object collapsing more and more slowly as it tries to fit within its Schwarzschild radius, and this almost EH area becomes extended as more material falls onto it. The almost-EH is not a surface, but a whole volume of the collapsing mass, with never enough mass within that radius to actually form an event horizon.


Here are some big guns to back up this picture of events:

“What would happen if you fall in? As seen from the outside, you would take an infinite amount of time to fall in, because all your clocks – mechanical and biological – would be perceived as having stopped’”
- Carl Sagan “Cosmos”, 1981

“ .. a critical radius, now called the “Schwarzschild radius,” at which time is infinitely dilated.”
- Paul Davies “About Time”, 1995

“From the standpoint of an outside observer, time grinds to a halt at the event horizon.”
- Timothy Ferris “The Whole Shebang”, 1997

“The closer we are to the event horizon, the slower time ticks away for the external observer. The tempo dies down completely on the boundary of the black hole.”
- Igor Novikov “The River of Time”, 1998

“When all thermonuclear sources of energy are exhausted a sufficiently heavy star will collapse. This contraction will continue indefinitely till the radius of the star approaches asymptotically its gravitational radius.”
- Oppenheimer and Snyder “Phys.Rev. 56,455” 1939

“According to the clocks of a distant observer the radius of the contracting body only approaches the gravitational radius as t -> infinity.”
- Landau and Lifschitz “The Classical Theory of Fields”, 1971

“What looks like a black hole is “in reality” a star frozen in the very late stages of collapse.”
- Paul Davies “About Time”, 1995


“At the stage of becoming a black hole, time dilatation reaches infinity.”
- Jayant Narlikar

In all his writings, Fred Hoyle referred to them as “near black holes”.


The problem, as I see it, is that everyone has become obsessed with the time frame of a falling observer, because it is mathematically so much more interesting. Frozen stars are very dull. And so many people do not realize that all the black hole phenomena do not occur in the universe we live in, but only beyond the far end of time when the universe is infinitely old.

At this point I would like to mention different time dilation processes which occur. For a static object, suspended near R0 (not possible, I know!), we would only observe the gravitational time dilation. For a falling object we have, added to this, the Doppler effect due to the speed of the object away from us, and another apparent dilation due to the longer and longer time it takes for light signals to escape the gravity as the object moves in. This last light signal delay is often confused with the delay effect of gravity itself.


Now, just out is interest, I will add some figures to the time dilation process.

Einstein’s General Relativity states that relative to a remote viewer, time is dilated in a gravitational field, and in the Schwarzschild metric the dilation in the vicinity of a non-rotating massive spherically-symmetric object is governed by the formula
dT/dt = √(1 - r0/r)
where dT = proper time interval of the falling object
dt= time interval observed by remote viewer
r = distance from the centre of the black hole
r0 = Schwarzschild radius of the black hole.


To simplify calculations, I will replace r with r0+dr, so that dr is the distance from the Schwarzschild event horizon.
Then (1 – r0/r) = ((r0 + dr)/(r0+dr) – r0/(r0+dr)) = (dr/(r0+dr)) = dr/r0 for dr<<r0.
And so, close in to the Schwarzschild radius,
dT/dt= √(dr/r0)

Next I will consider a black hole with mass about 3x the mass of the sun, and radius 10 km.

Then at a distance of 1 meter, we have dT/dt = √ 1/10,000 = 1/100, so one minute of local time at the falling object would correspond to 100 minutes in our time frame. Using this method, we can easily find the distance from the event horizon for any degree of time dilation we choose. For a dilation of 1:10^6 we need dr/r0 = 10^-12.

Obviously we need to get very close to the EH to achieve a respectable amount of dilation. At the surface of a neutron star it is only 20%! How close can we go? Well, the smallest distance we can consider is the Planck length, 1.6x10^-35m. Then
Dr/r0 = 1.6x10^-39, and dT/dt= 4x10^-19. A year is just over 3x10^10 seconds, and so a one second interval at this distance would pass in about 100,000,000 years in our timeframe. If two objects fell passed this point on two successive days in local time, then in our time frame we would see them pass this point about 10,000,000,000,000,000 years apart (IF we could see anything this close in).

But because the time dilation effects occur so close in to the EH, it would be impossible to detect them. So no observation would show the difference between a black hole and an eternally collapsing object.


To avoid repetition, here are some objections posted by WayneFrancis in another topic:

“If stuff really didn’t ever cross the EH of a black hole then they’d be far from black. They’d just have very long radio wavelengths. This is where it gets into what really happens. Because take the 3600x time dilatation. If I had a green shirt on then that green shirt would no longer be green to an external observer.

I’ll go through the maths.
Green wavelength – 545 nanometers or 5.45x10-7
At a time dilatation of 3600x its wavelength would now be 1.962x10-3m
This is on the border between infrared and microwave radiation. VERY easily picked up.
Now work out how close an object would have to be to be time dilated to something like 10Hz. I’ll give you a clue … you’d have to have something time dilated by
~1.000.000.000.000.000 times.

So if things didn’t actually fall through the EH then in theory we would still be receiving photons from them that we could pick up for like 10 million years or so.

The objects DO pass the event horizon and the proof is not only in the maths but in our observations and even if they didn’t pass through the EH, as I and others have pointed out, the EH would bubble out past that shell anyway. End effect is the same. The stuff is lost inside the EH and the only way its getting back out is the slow process of HR.”


Yes, Wayne, the maths show that objects DO pass the event horizon, but the maths also shows that it happens when our clocks read infinity. As for observations? Do you have a photograph of an object falling through? I think I have shown that our observations would not show the difference.

I have explained what happens regarding “the EH would bubble out”. There IS no EH, only a volume extremely close to being one.



I have just been re-reading Kip Thorne’s book “Black Holes and Time Warps”, an excellent description of the development of the subject. On page 218 he refers to Oppenheimer and Snyder’s view that the implosion freezes forever as measured in the static external frame but continues rapidly on past the freezing point as measured in the frame of the star’s surface. In the early 1960’s Wheeler was converted to this view. But then Astronomers became obsessed with the falling mass, and the “space” inside the event horizon, and stopped looking at the outside picture. Thorne writes “because of the enormous difficulty light has escaping gravity’s grip, as seen from afar the implosion seems to take forever; the star’s surface seems never quite to reach the critical circumference, and the horizon never forms.” But the horizon never forms because of the gravitatiional effect on time. The light delay is an optical effect, not a real time delay.

WayneFrancis
2011-May-12, 02:48 AM
So you are trying to say a black hole can never grow in size to an external observer?
Do you think the EH will never bubble out? Again I'm agnostic to the question of "does something pass through the EH or does the EH pass through the falling object" end result is the same for what most people are talking about.
Also when quoting people try using the QUOTE tag. Others can see my post in full context of the thread

WaxRubiks
2011-May-12, 03:37 AM
or when an supposed event horizon expands, is it maybe just gravitational self lensing that increases?

Mike Holland
2011-May-12, 03:55 AM
WayneFrancis, I am not saying that not that the EH will not bubble out. I am saying that there IS NO EH - not until our clocks trick over infinity! A Schwarzschild radius exists for the mass (as for any mass), but an EH does not come into existence until all the mass is within that radius, and that contraction is slowed by gravity, just as a falling object is.

The near-black hole will expand, as the original collapsing mass gets nearer and nearer to being a black hole, and further mass slowly accumulates around it adding to it.

Sorry about not using the QUOTE option. I developed my post over several days and then copied and pasted it. Didn't know how QUOTE works. Ill look it up.

Tensor
2011-May-12, 04:23 AM
WayneFrancis, I am not saying that not that the EH will not bubble out. I am saying that there IS NO EH - not until our clocks trick over infinity! A Schwarzschild radius exists for the mass (as for any mass), but an EH does not come into existence until all the mass is within that radius, and that contraction is slowed by gravity, just as a falling object is.

You do realize that the event horizon forms from the inside of the star and works it way outward. The EH breaks through the surface of the star at the same instant as the remaining mass falls below the EH. I'm also wondering how you explain those frames in GR that show what the outside observer and infalling observer are seeing at the same time in that combined frame.


The problem, as I see it, is that everyone has become obsessed with the time frame of a falling observer, because it is mathematically so much more interesting. Frozen stars are very dull. And so many people do not realize that all the black hole phenomena do not occur in the universe we live in, but only beyond the far end of time when the universe is infinitely old.

It's not just that the infalling observer is so much more interesting, it's the reason it's so much more interesting. The EH is not freezing, and there is no time dilation in the frame of the observer, so we can explore what happens near, at and below the EH. Not only that, but even if the time dilation is infinite at the EH, it simply doesn't matter as far as the black hole phenomena that we can observer. Those phenomena (accretion disks, jet formation, observation of external orbits, disruption of orbits, effects on pulsars, etc) take place far enough away that time dilation simply isn't a factor.

Let's face it, if someone is going to explore a blackhole, they need to know what to expect when they get there. Not what they can see from a distance.

Mike Holland
2011-May-12, 05:46 AM
Tensor, a neutron star has a Schwarzschild radius somewhere inside it, but no event horizon. If it collapsed further, then at some depth inside it the amount of mass within that depth would come close to satisfying the equation r = 2GM/c^2. In fact, a whole range of radii might come close to satisfying it. But then over this whole range of radii extreme time dilation would be in effect, and in our timeframe the collapse would "freeze" and proceed extremely slowly, coming to a stop after an infinite time.

Time dilation doesn't snap into existence when an event horizon appears. It is present around any mass and becomes extreme as a mass collapses towards becoming a black hole.

I agree, it doersn't matter as far as the phenomena that we can observe is concerned. But we should still get our facts straight.

What happens to the spaceman exploring a black hole may be interesting, but he can never tell us whether our theories are correct. Not just because of the event horizon which will have come into existence when he gets there, but because he will have travelled into our infinite future and effctively left our universe.

As the spaceman approaches the frozen star, he would see our clocks ticking more quickly, just as we see ones in orbit ticking more quickly. But he wouldn't see the whole universe grow old - his fall would take place too quickly. Some say he would see a bright flash as all the photons following him over millenia catch up. Personally, I think he would catch up with other infalling matter and go splat, but I'm not certain about that.

WaxRubiks
2011-May-12, 05:58 AM
Personally, I think he would catch up with other infalling matter and go splat, but I'm not certain about that.
or evaporate via pre-Hawking radiation, maybe.

....They find that the gravity of the collapsing mass starts to disrupt the quantum vacuum, generating what they call "pre-Hawking" radiation. Losing that radiation reduces the total mass-energy of the object - so that it never gets dense enough to form an event horizon and a true black hole. "There are no such things", Vachaspati told New Scientist. "There are only stars going toward being a black hole but not getting there."http://www.newscientist.com/article/dn12089-do-black-holes-really-exist.html

Mike Holland
2011-May-12, 06:47 AM
Hi Froggy,
Yes, I'm aware of that theory, which gave rise to the term ECO which I have borrowed. I can't comment on its validity, but what the hell, we have ECOs anyway just with Einstein's theory.

baskerbosse
2011-May-12, 06:50 AM
As the spaceman approaches the frozen star, he would see our clocks ticking more quickly, just as we see ones in orbit ticking more quickly. But he wouldn't see the whole universe grow old - his fall would take place too quickly. Some say he would see a bright flash as all the photons following him over millenia catch up. Personally, I think he would catch up with other infalling matter and go splat, but I'm not certain about that.

I don't think this is correct.
Surely it is the other way around. Photons from the surrounding universe would find it increasingly hard to catch up with him as he accelerates, no?

Peter

WayneFrancis
2011-May-12, 07:53 AM
WayneFrancis, I am not saying that not that the EH will not bubble out. I am saying that there IS NO EH - not until our clocks trick over infinity! A Schwarzschild radius exists for the mass (as for any mass), but an EH does not come into existence until all the mass is within that radius, and that contraction is slowed by gravity, just as a falling object is.

The near-black hole will expand, as the original collapsing mass gets nearer and nearer to being a black hole, and further mass slowly accumulates around it adding to it.

Sorry about not using the QUOTE option. I developed my post over several days and then copied and pasted it. Didn't know how QUOTE works. Ill look it up.

Then sounds like you are changing the definition of what an event horizon is a boundary in spacetime, beyond which events cannot affect an outside observer. There is most definitely a region which event can not affect outside observers.

WayneFrancis
2011-May-12, 08:01 AM
Tensor, a neutron star has a Schwarzschild radius somewhere inside it, but no event horizon. If it collapsed further, then at some depth inside it the amount of mass within that depth would come close to satisfying the equation r = 2GM/c^2. In fact, a whole range of radii might come close to satisfying it. But then over this whole range of radii extreme time dilation would be in effect, and in our timeframe the collapse would "freeze" and proceed extremely slowly, coming to a stop after an infinite time.

Time dilation doesn't snap into existence when an event horizon appears. It is present around any mass and becomes extreme as a mass collapses towards becoming a black hole.

I agree, it doersn't matter as far as the phenomena that we can observe is concerned. But we should still get our facts straight.

What happens to the spaceman exploring a black hole may be interesting, but he can never tell us whether our theories are correct. Not just because of the event horizon which will have come into existence when he gets there, but because he will have travelled into our infinite future and effctively left our universe.

As the spaceman approaches the frozen star, he would see our clocks ticking more quickly, just as we see ones in orbit ticking more quickly. But he wouldn't see the whole universe grow old - his fall would take place too quickly. Some say he would see a bright flash as all the photons following him over millenia catch up. Personally, I think he would catch up with other infalling matter and go splat, but I'm not certain about that.

The free faller can not look back, while falling, and watch the universe age. To do that the free faller would have to stop their journey above the EH then they could watch the universe age at a rate dependant on their time dilation at that location according to GR. Once below the EH they have no options and will hit the singularity in a finite amount of proper time. Otherwise, if they continue free falling, the picture of the universe they see is dependent on how long in their proper time they are falling into the black hole....well that is my understanding until I'm shown in detail what I've got wrong

Mike Holland
2011-May-12, 08:04 AM
I don't think this is correct.
Surely it is the other way around. Photons from the surrounding universe would find it increasingly hard to catch up with him as he accelerates, no?

Peter

Peter, he would be accelerating towards c, if he was in free fall from a long way off. But the photons would be at c and therefore faster. In his timeframe they would still be at c relative to him (I don't know the maths on this, but I'm pretty certain c = c in any timeframe).

Mike Holland
2011-May-12, 08:22 AM
The free faller can not look back, while falling, and watch the universe age. To do that the free faller would have to stop their journey above the EH then they could watch the universe age at a rate dependant on their time dilation at that location according to GR. Once below the EH they have no options and will hit the singularity in a finite amount of proper time. Otherwise, if they continue free falling, the picture of the universe they see is dependent on how long in their proper time they are falling into the black hole....well that is my understanding until I'm shown in detail what I've got wrong

I'm afraid you are wrong on this one. If I start falling towards a black hole and am still a million miles away from it and moving slowly, what is to stop me looking back at the universe behind me? Why can I not look back at any stage until I actually pass through the EH? If I am falling off a tall building, why can I not compare my clock with one in orbit? (OK! a VERY tall building!).

Once passed the EH, I agree with you. But while falling towards it I am just another observer outside the EH, and am only time dilated relative to observers and clocks further out.

Mike Holland
2011-May-12, 08:25 AM
Something I forgot to mention in my original post is that thanks to the elegance of Einstein's theory, time dilation means we live in a universe with no singularities, no zeros or infinities, no wormholes or other such science fiction stuff, and no places where physics breaks down. A very reasonable universe! (Apart from quantum mechanics).

WayneFrancis
2011-May-12, 10:06 AM
I'm afraid you are wrong on this one. If I start falling towards a black hole and am still a million miles away from it and moving slowly, what is to stop me looking back at the universe behind me? Why can I not look back at any stage until I actually pass through the EH? If I am falling off a tall building, why can I not compare my clock with one in orbit? (OK! a VERY tall building!).

Once passed the EH, I agree with you. But while falling towards it I am just another observer outside the EH, and am only time dilated relative to observers and clocks further out.

Let me read my post and see if I messed up what I meant.

Ok I the first line there should have been.

"The free faller can not look back, while falling, and watch the universe age to infinity."

IE the free fallers proper time doesn't let them see the end of the universe. They'll only see light from a period of time that is relative to the amount of proper time they are falling.

WayneFrancis
2011-May-12, 10:09 AM
Something I forgot to mention in my original post is that thanks to the elegance of Einstein's theory, time dilation means we live in a universe with no singularities, no zeros or infinities, no wormholes or other such science fiction stuff, and no places where physics breaks down. A very reasonable universe! (Apart from quantum mechanics).

You are contradicting yourself there. I agree there is no place where physics breaks down...just our models of the physics. I'm not sure if I agree with the no infinities. Please point to where Einstein says the universe can't be infinite in size?

Mike Holland
2011-May-12, 10:30 AM
You are contradicting yourself there. I agree there is no place where physics breaks down...just our models of the physics. I'm not sure if I agree with the no infinities. Please point to where Einstein says the universe can't be infinite in size?

OK, I should have been more specific. I mean no infinite forces or gravitational fields and such. Obviously there can be zero space between two objects in contact. I was thinking of things like relativistic mass, which approaches infinity as velocity approaches c,m but never gets there.

Mike Holland
2011-May-12, 10:39 AM
Ok I the first line there should have been.

"The free faller can not look back, while falling, and watch the universe age to infinity."

IE the free fallers proper time doesn't let them see the end of the universe. They'll only see light from a period of time that is relative to the amount of proper time they are falling.

Yes, as I stated above,
"As the spaceman approaches the frozen star, he would see our clocks ticking more quickly, just as we see ones in orbit ticking more quickly. But he wouldn't see the whole universe grow old - his fall would take place too quickly."

But he would certainly have time to see the remote universe speed up.

NB. we have several options regarding his speed, whether he is free-falling, accelerating into the BH, or using his engines to slow his fall. This obviously affects the time he has for making his observations. Up to a certain point he could hover on his engines while observing us and our clocks.

grav
2011-May-12, 11:40 AM
As for what a freefaller would literally see of the light coming from a distant observer when looking back, a stationary observer at r would see the light gravitationally blueshifted by a factor of 1 / sqrt(1 - R/r), and this is directly tied to the gravitational time dilation, so the stationary observer at r says that the distant observer's clock is ticking 1 / sqrt(1 - R/r) faster. However, a freefaller instantaneously falling past the stationary observer at r with a relative speed of v will see the same light Doppler redshifted by a factor of sqrt[(1 - v/c) / (1 + v/c)], whereas a freefaller falling from rest at infinity will have a relative speed to the stationary observer of v = c sqrt(R/r), so the freefaller sees an overall redshift for the distant observer of

D = [1 / sqrt(1 - R/r)] sqrt[(1 - v/c) / (1 + v/c)]

= [1 / sqrt(1 - R/r)] [sqrt(1 - (v/c)^2) / (1 + v/c)]

= [1 / sqrt(1 - R/r)] [sqrt(1 - R/r) / (1 + v/c)]

= 1 / (1 + v/c)

= 1 / (1 + sqrt(R/r))

Due to the acceleration, the freefaller will actually see an increasing redshift of the distant observer, so in regards to what is directly observed by the light that falls upon the freefaller's eyes, the freefaller will physically see things occurring that much slower for the distant observer as well. Upon reaching the event horizon, where r = R, the redshift would be 1/2 and the freefaller would still physically see things occurring at half the rate for the distant observer.

Mike Holland
2011-May-12, 12:24 PM
Thanks for that, Grav. So while an observer suspended close to the EH would see the distant observer blue-shifted, the freefalling one would see the opposite. As he reaches c at the EH, the gravitational time dilation -> 0 and the relativistic time dilation -> infinity, and in this case the limit of the product of the two is 1/2!

But if he doesn't fall from infinity, his speed will be less than c, and the gravitational dilation will dominate as he nears the EH.

I hope that settles that, because it is all getting away from my original position, that in our remote observer timeframe, no black holes have ever quite formed, and that falling objects slow down and "freeze on the ege of the near-black hole.

macaw
2011-May-12, 01:39 PM
However, a freefaller instantaneously falling past the stationary observer at r with a relative speed of v will see the same light Doppler redshifted by a factor of sqrt[(1 - v/c) / (1 + v/c)],


whereas a freefaller falling from rest at infinity will have a relative speed to the stationary observer of v = c sqrt(R/r), so the freefaller sees an overall redshift for the distant observer of

D = [1 / sqrt(1 - R/r)] sqrt[(1 - v/c) / (1 + v/c)]

= [1 / sqrt(1 - R/r)] [sqrt(1 - (v/c)^2) / (1 + v/c)]

= [1 / sqrt(1 - R/r)] [sqrt(1 - R/r) / (1 + v/c)]

= 1 / (1 + v/c)

= 1 / (1 + sqrt(R/r))


The above is incorrect. The correct solution is obtained as follows:

(c d\tau)^2=(1-r_s/r)(c dt)^2-dr^2/(1-r_s/r)

(d \tau/dt)^2=(1-r_s/r)(1-(dr/cdt)^2/(1-r_s/r)^2)


Since the coordinate speed for an object falling from infinity dr/dt is known:

dr/dt=c (1-r_s/r) \sqrt{r_s/r}

it follows that

d\tau/dt=1-r_s/r

Your error stems from your gratuitous multiplication of the gravitational redshift of a STATIONARY (hovering) observer with the SR Doppler effect of a FREEFALLING observer as in your equation

D = [1 / sqrt(1 - R/r)] sqrt[(1 - v/c) / (1 + v/c)]

I have pointed out this error several times before, yet you continue to propagate it.


Upon reaching the event horizon, where r = R, the redshift would be 1/2 and the freefaller would still physically see things occurring at half the rate for the distant observer.

This is also false, when r->r_s , d\tau/dt->0 . See above for the correct derivation.

macaw
2011-May-12, 03:49 PM
the limit of the product of the two is 1/2!

No, it isn't. The correct answer is 0, see previous post.



I hope that settles that, because it is all getting away from my original position, that in our remote observer timeframe, no black holes have ever quite formed, and that falling objects slow down and "freeze on the ege of the near-black hole.

Actually, grav's error reinforces your errors.

kzb
2011-May-12, 05:56 PM
This topic of discussion has come up several times on these forums over the past few years.

I must admit to big conceptual problems: if it takes infinite time to fall throught the EH (to a remote observer), but to the person falling, he does fall through, we have two different outcomes depending on the reference frame.

However, the general relativity experts say that the EH actually pops up to engulf the infalling observer (because he has mass too), so he IS observed to pass through the EH in real time to a remote observer.

This means that the long list of august references in the OP are in fact wrong.

Observationally, it seems there are difficulties with black holes. They appear to have strong magnetic fields, which in theory they cannot have. That is more in line with the MECO theory.


Schild and Leiter 2010:
Black Hole or Meco: Decided by a thin Luminous Ring Structure Deep Within Quasar Q0957+561

Mike Holland
2011-May-12, 11:09 PM
I must admit to big conceptual problems: if it takes infinite time to fall throught the EH (to a remote observer), but to the person falling, he does fall through, we have two different outcomes depending on the reference frame.

However, the general relativity experts say that the EH actually pops up to engulf the infalling observer (because he has mass too), so he IS observed to pass through the EH in real time to a remote observer.

This means that the long list of august references in the OP are in fact wrong.

Yes, your conceptual problem has been around since Special Relativity hit the scene. Observers in different reference frames see things differently.

However your unnamed GR experts seem to forget that time dilation is already in play long before the collapse reaches the BH stage, and in fact slows the collapse process to zero as the EH forms, after an infinite time. So the EH does not pop up to engulf the infalling observer. Rather, he becomes part of the slowly expanding almost-EH region.

All my reasoning is based on the simple fact that time dilation becomes infinite at an EH, in the reference frame of an outside observer (which I expand to include all of our universe that is outside the BH). And I think this is an accepted consequence of Einstein's theory. So until you can name one of your experts who disproved this, and give a reference, I shall stick with my list of august references.

NB. I think the existence of MECOs re-inforces the view of my august references.

WayneFrancis
2011-May-13, 12:07 AM
Yes, your conceptual problem has been around since Special Relativity hit the scene. Observers in different reference frames see things differently.

However your unnamed GR experts seem to forget that time dilation is already in play long before the collapse reaches the BH stage, and in fact slows the collapse process to zero as the EH forms, after an infinite time. So the EH does not pop up to engulf the infalling observer. Rather, he becomes part of the slowly expanding almost-EH region.

All my reasoning is based on the simple fact that time dilation becomes infinite at an EH, in the reference frame of an outside observer (which I expand to include all of our universe that is outside the BH). And I think this is an accepted consequence of Einstein's theory. So until you can name one of your experts who disproved this, and give a reference, I shall stick with my list of august references.

NB. I think the existence of MECOs re-inforces the view of my august references.

So according to you a free faller falling into a 20km radius black hole will forever hover just above the 20km "almost event horizon".

Tell me then what happens to that free faller if you dump 1 solar mass into that "almost" black hole?
If you now calculate that "almost black hole" object it has a SR of ~23km yet the free faller was at 20km. Did the free faller get pushed outwards? from the centre?

I mean the way your describing it no EH can ever form and nothing can ever be inside it yet their is a SR for a given mass so what happens in your idea? Does all that mass some how get pushed outward from the centre?

WayneFrancis
2011-May-13, 12:13 AM
No, it isn't. The correct answer is 0, see previous post.



Actually, grav's error reinforces your errors.

Macaw, just to be sure I have the right picture in my head, a free faller doesn't see any real shift via time or spectrum?

grav
2011-May-13, 12:25 AM
Since the coordinate speed for an object falling from infinity dr/dt is known:

dr/dt=c (1-r_s/r) \sqrt{r_s/r}

it follows that

d\tau/dt=1-r_s/rIt also follows from

dτ / dt = gravitational time dilation * kinetic time dilation

= sqrt(1 - R/r) * sqrt(1 - (v_loc/c)^2)

For a body falling from rest at infinity, whereas v_loc = c sqrt(R/r), so that sqrt(1 - (v_loc/c)^2) = sqrt(1 - R/r), which gives

dτ / dt = sqrt(1 - R/r) * sqrt(1 - (v_loc/c)^2)

= sqrt(1 - R/r) * sqrt(1 - R/r)

= 1 - R/r

But we are discussing the Doppler shift that the freefaller directly observes of the distant observer's light. If the distant observer emits light with a frequency of f, then an observer that is stationary at r will receive it blueshifted with a frequency of f' = f / sqrt(1 - R/r). If a third observer is in the same immediate vicinity of the stationary observer at r but is travelling away from the stationary observer with a relative speed of v_loc, then if the stationary observer emits light with a frequency of f', the observer that is travelling away will measure it redshifted as f" = f' sqrt[(1 - v_loc/c)/(1 + v_loc/c)]. So if the distant observer emits the light at f and the stationary observer receives it at f', then the stationary observer re-emits it as f' and the freefaller receives it as f", then the freefaller is measuring the light to be redshifted overall by a factor of

D = [1 / sqrt(1 - R/r)] * sqrt[(1 - v_loc/c) / (1 + v_loc/c)]

= [1 / sqrt(1 - R/r)] * [sqrt(1 - (v_loc/c)^2) / (1 + v_loc/c)]

where v_loc = c sqrt(R/r) for a freefaller falling from rest at infinity, so

= [1 / sqrt(1 - R/r)] * [sqrt(1 - R/r) / (1 + v_loc/c)]

= 1 / (1 + v_loc/c)

= 1 / (1 + sqrt(R/r))

Of course, having the stationary observer at r receive and then re-emit the light is the same as having the light simply pass the stationary observer on its way to the freefaller. Here is a reference (http://www.actaphys.uj.edu.pl/vol39/pdf/v39p1357.pdf) also.

macaw
2011-May-13, 12:37 AM
It also follows from

dτ / dt = gravitational time dilation * kinetic time dilation

= sqrt(1 - R/r) * sqrt(1 - (v_loc/c)^2)

For a body falling from rest at infinity, whereas v_loc = c sqrt(R/r), so that sqrt(1 - (v_loc/c)^2) = sqrt(1 - R/r), which gives

dτ / dt = sqrt(1 - R/r) * sqrt(1 - (v_loc/c)^2)

= sqrt(1 - R/r) * sqrt(1 - R/r)

= 1 - R/r

So, are you admitting that the correct result is 1-r_s/r? Then why are you insisting in pushing your incorrect derivation?

[
But we are discussing the Doppler shift that the freefaller directly observes of the distant observer's light.

...which is exactly what I have derived. Were you able to follow the derivation?





If the distant observer emits light with a frequency of f, then an observer that is stationary at r will receive it blueshifted with a frequency of f' = f / sqrt(1 - R/r).

There is no reason to introduces a stationary observer since, according to your admission above, you were trying to calculate the time dilation for the FREEFALLING observer. This is one of the errors in your derivation I pointed out in the earlier post.



If a third observer is in the same immediate vicinity of the stationary observer at r but is travelling away from the stationary observer with a relative speed of v_loc, then if the stationary observer emits light with a frequency of f', the observer that is travelling away will measure it redshifted as f" = f' sqrt[(1 - v_loc/c)/(1 + v_loc/c)].

No, this is a repeat of your original error in your original post. First off , this hack is unnecessary since I showed you how to obtain the result directly.
Second off, nothing allows you to multiply the gravitational effect of the stationary observer with the SR effect (kinematic) of the frefalling observer.
Thirdly, you end up with the wrong result.



So if the distant observer emits the light at f and the stationary observer receives it at f', then the stationary observer re-emits it as f' and the freefaller receives it as f", then the freefaller is measuring the light to be redshifted overall by a factor of

D = [1 / sqrt(1 - R/r)] * sqrt[(1 - v_loc/c) / (1 + v_loc/c)]

= [1 / sqrt(1 - R/r)] * [sqrt(1 - (v_loc/c)^2) / (1 + v_loc/c)]

where v_loc = c sqrt(R/r) for a freefaller falling from rest at infinity, so

= [1 / sqrt(1 - R/r)] * [sqrt(1 - R/r) / (1 + v_loc/c)]

= 1 / (1 + v_loc/c)

= 1 / (1 + sqrt(R/r))

Yet, this result is false, it even contradicts the result you calculated at the beginning of your post. The correct result , derived in a few simple steps is D=1-r_s/r
A quick examination of your method shows that , at the event horizon r=r_s you obtain an absurd value for the time dilation , D=1/2 whereas the correct value should be D=0. The error is even more glaring since , if the observer is stationary, the result is the well-known D=\sqrt {1-r_s/r} that also zeroes out at the EH.

macaw
2011-May-13, 12:46 AM
Macaw, just to be sure I have the right picture in my head, a free faller doesn't see any real shift via time or spectrum?

The freefaller records a blueshift:

d\tau/dt=1-r_s/r

At the EH, r=r_s so d\tau/dt=0

Conversely, the observer at infinity records a redshift:

dt/d\tau=1/(1-r_s/r)

At the EH, r=r_s so dt/d\tau=oo

The result is consistent with what a stationary (hovering) observer records:

d\tau/dt=\sqrt{1-r_s/r}

At the EH, r=r_s so d\tau/dt=0

baskerbosse
2011-May-13, 01:05 AM
Peter, he would be accelerating towards c, if he was in free fall from a long way off. But the photons would be at c and therefore faster. In his timeframe they would still be at c relative to him (I don't know the maths on this, but I'm pretty certain c = c in any timeframe).

Since he is accelerating away from the emitter of the photons, they would surely redshift to him, no?

Obviously c is always c, but shouldn't the redshift of the photons approach infinite wavelength as he approaches the event horizon?
The photons energy therfore approaching 0 as he approaches the event horizon?

Once he passes the event horizon, the photon energy would be 0, wouldn't it?
Can there be a photon that does not have positive energy?
I'd say it just can't hit you anymore.


Peter

macaw
2011-May-13, 01:13 AM
Since he is accelerating away from the emitter of the photons, they would surely redshift to him, no?

Se post above yours for the exact math.

John Mendenhall
2011-May-13, 01:33 AM
Gentlemen, keep in mind the very artificial boundary conditions. In our real universe, all collapsed objects will rotate - and very rapidly, at that.

John Mendenhall
2011-May-13, 02:04 AM
Speaking of boundary conditions, I should have said 'initially'.

grav
2011-May-13, 02:45 AM
Yet, this result is false, it even contradicts the result you calculated at the beginning of your post. The correct result , derived in a few simple steps is D=1-r_s/r
A quick examination of your method shows that , at the event horizon r=r_s you obtain an absurd value for the time dilation , D=1/2 whereas the correct value should be D=0. The error is even more glaring since , if the observer is stationary, the result is the well-known D=\sqrt {1-r_s/r} that also zeroes out at the EH.That is incorrect. Did you even read the link? From what you have there, if the stationary observer at r measures a blueshift of D = sqrt(1 - R/r) and the freefaller measures D = 1 - R/r {although it should be the inverse of those}, then the freefaller sees an even greater blueshift than the stationary observer according to you, but the freefaller is travelling away from the light coming from the distant observer in respect to the stationary observer, so should measure a lesser shift than the stationary observer measures, so redshifted from that value, not more blueshifted.

macaw
2011-May-13, 02:52 AM
That is incorrect. Did you even read the link?

yes, i read it, it is pretty mundane paper, has nothing to do with your repeated errors.


From what you have there, if the stationary observer at r measures a blueshift of D = sqrt(1 - R/r) and the freefaller measures D = 1 - R/r {although it should be the inverse of those},

The math says you are wrong. Interestingly enough, with a little bit of prodding I managed to get you to obtain the same (http://www.bautforum.com/showthread.php/115485-Black-Holes-or-Eternally-Collapsing-Objects?p=1887756#post1887756) result. Yet, for some unexplained reason you still cling to your original incorrect results. Why?


but the freefaller is travelling away from the light coming from the distant observer in respect to the stationary observer, so should measure a lesser shift than the stationary observer measures, so redshifted from that value, not more blueshifted.

You are wrong again:

d\tau_{free}/dt=1-r_s/r

so:

T_{free}=T_{distant}(1-r_s/r)

d\tau_{stationary}/dt=\sqrt{1-r_s/r}

so:

T_{stationary}=T_{distant}\sqrt{1-r_s/r}

Therefore:

T_{free}<T_{stationary}

making your above claim false (again).

so:

f_{free}>f_{stationary}

The freefaller sees the frequency more blueshifted, so my derivation, contrary to your claims, is correct. This is not SR, it isn't Doppler shift, the speed plays no role (the radial distance does), so you need to know how to apply the equations of GR. You need to abandon your intuitions (since they are wrong) and you need to use correct derivations based on the Schwarzschild metric, not some canned formulas.

macaw
2011-May-13, 03:15 AM
As for what a freefaller would literally see of the light coming from a distant observer when looking back, a stationary observer at r would see the light gravitationally blueshifted by a factor of 1 / sqrt(1 - R/r), and this is directly tied to the gravitational time dilation, so the stationary observer at r says that the distant observer's clock is ticking 1 / sqrt(1 - R/r) faster. However, a freefaller instantaneously falling past the stationary observer at r with a relative speed of v will see the same light Doppler redshifted by a factor of sqrt[(1 - v/c) / (1 + v/c)], whereas a freefaller falling from rest at infinity will have a relative speed to the stationary observer of v = c sqrt(R/r), so the freefaller sees an overall redshift for the distant observer of

D = [1 / sqrt(1 - R/r)] sqrt[(1 - v/c) / (1 + v/c)]

= [1 / sqrt(1 - R/r)] [sqrt(1 - (v/c)^2) / (1 + v/c)]

= [1 / sqrt(1 - R/r)] [sqrt(1 - R/r) / (1 + v/c)]

= 1 / (1 + v/c)

= 1 / (1 + sqrt(R/r))

This is impossible since at the event horizon (r=R in your notation) you would be getting:

d\tau/dt=1/2

or, even worse:

dt/d\tau=2

when, in reality, dt/d\tau->oo when r->R

grav
2011-May-13, 03:23 AM
yes, i read it, it is pretty mundane paper, has nothing to do with your repeated errors.equ.14 - 20, u_t = y_s y_g, time dilation factorizes into gravitational and kinematical components. Last equation, w_IFO / w_IO = [1 / sqrt(g_tt)) [sqrt(1 - V_IFO^2) / (1 + V_IFO)] {redshift}.


The math says you are wrong. Interestingly enough, with a little bit of prodding I managed to get you to obtain the same result. Yet, for some unexplained reason you still cling to your original incorrect results. Why?

You are wrong again:

d\tau_{free}/dt=1-r_s/r

so:

T_{free}=T_{distant}(1-r_s/r)

d\tau_{stationary}/dt=\sqrt{1-r_s/r}

so:

T_{stationary}=T_{distant}\sqrt{1-r_s/r}

Therefore:

T_{free}<T_{stationary}

making your above claim false (again).

so:

f_{free}>f_{stationary}

The freefaller sees the frequency more blueshifted, so my derivation, contrary to your claims, is correct. This is not SR, it isn't Doppler shift, the speed plays no role (the radial distance does), so you need to know how to apply the equations of GR. You need to abandon your intuitions (since they are wrong) and you need to use correct derivations based on the Schwarzschild metric, not some canned formulas.That is not the Doppler shift the freefaller sees of the distant observer's light, but the time dilation of the freefaller's clock according to the distant observer.

macaw
2011-May-13, 03:25 AM
equ.14 - 20, u_t = y_s y_g, time dilation factorizes into gravitational and kinematical components. Last equation, w_IFO / w_IO = [1 / sqrt(g_tt)) [sqrt(1 - V_IFO^2) / (1 + V_IFO)] {redshift}.

That is not the Doppler shift the freefaller sees

There is no "Doppler shift" in GR, I just explained that to you.



of the distant observer's light, but the time dilation of the freefaller's clock according to the distant observer.

f=1/T, have you forgotten?

Now, if you are serious about this discussion, you would need to find the error in my derivation, the way I found the errors in yours.
As an aside, you would be very hard pressed to explain why your method produces dt/d\tau=2 at the EH when the correct answer is infinity. this should have clued you in that your method is wrong.

macaw
2011-May-13, 03:39 AM
yes, i read it, it is pretty mundane paper, has nothing to do with your repeated errors.



The math says you are wrong. Interestingly enough, with a little bit of prodding I managed to get you to obtain the same (http://www.bautforum.com/showthread.php/115485-Black-Holes-or-Eternally-Collapsing-Objects?p=1887756#post1887756) result. Yet, for some unexplained reason you still cling to your original incorrect results. Why?



You are wrong again:

d\tau_{free}/dt=1-r_s/r

so:

T_{free}=T_{distant}(1-r_s/r)

d\tau_{stationary}/dt=\sqrt{1-r_s/r}

so:

T_{stationary}=T_{distant}\sqrt{1-r_s/r}

Therefore:

T_{free}<T_{stationary}

making your above claim false (again).

so:

f_{free}>f_{stationary}

The freefaller sees the frequency more blueshifted, so my derivation, contrary to your claims, is correct. This is not SR, it isn't Doppler shift, the speed plays no role (the radial distance does), so you need to know how to apply the equations of GR. You need to abandon your intuitions (since they are wrong) and you need to use correct derivations based on the Schwarzschild metric, not some canned formulas.

OK, since you couldn't understand the above, let's try a different route.

The period of a freefalling emitter, as seen by a distant observer is:

T_{distant-free}=T_{free}/(1-r_s/r) (redshift)

The period of a stationary emitter, as seen by a distant observer is:

T_{distant-stationary}=T_{stationary}/\sqrt{1-r_s/r} (less redshift)

So:

T_{distant-free}>T_{distant-stationary}

or, expressed in frequencies:

f_{distant-free}<f_{distant-stationary}

i.e., the freefalling emitter appears more redshifted than the stationary emitter to the same observe placed at infinity.

In GR, as opposed to SR, the effects are not symmetrical, the observer higher up in the gravity well sees the sources below redshifted. The observers lower in the gravity well see the sources above blueshifted. This is one more reason to abandon any "intuition" tying the effect to SR.

baskerbosse
2011-May-13, 04:41 AM
Se post above yours for the exact math.

Sorry, I missed that one..
So basically, time dilation more than compensates for doppler shift caused by accelerating away from the light source?


Thanks,
Peter

baskerbosse
2011-May-13, 04:43 AM
There is no "Doppler shift" in GR, I just explained that to you.


-Doh!

I should hit refresh more often..

-Excuse me while I go and lurk for a while..


Peter

Mike Holland
2011-May-13, 04:47 AM
So according to you a free faller falling into a 20km radius black hole will forever hover just above the 20km "almost event horizon".

Tell me then what happens to that free faller if you dump 1 solar mass into that "almost" black hole?
If you now calculate that "almost black hole" object it has a SR of ~23km yet the free faller was at 20km. Did the free faller get pushed outwards? from the centre?

I mean the way your describing it no EH can ever form and nothing can ever be inside it yet their is a SR for a given mass so what happens in your idea? Does all that mass some how get pushed outward from the centre?

OK, picture a sphere and choose a point inside it at distance d from the centre. Then the gravity acting on it depends on the mass contained within the sphere of radius d. The hollow shell of matter outside the surface of that sphere is pulling equally in every direction and doesn't add to the gravity of the contained sphere. Now think of the near-black hole as a sequence of such spheres, each just a little larger than its SR and trying desperately to squeeze into its SR. This is the condition right through the collapsing mass as it tries to become a BH. We don't have an EH; we have a whole region which is nearly an EH, and additional mass falling in just extends this region.

I saw some discussion elsewhere about whether there is any pressure if the mass is in freefall. I don't know the answer. Pressure might slightly increase the rate of infall, but this is a problem for quantum mechanics, and our ideas of pressure might not apply in this medium.

Anyway, our poor falling spaceman would become another layer trying to join the forming BH, and a star collapsing on top of him would add additional layers, all slowly moving in. I don't know whether this would speed up his movement, because I don't know the answer regarding pressure. But additiional mass outside his radius does not add to the gravity where he is (as long as it is spread uniformly around the sphere), and so does not affect the near-EH conditions inside the ECO.

macaw
2011-May-13, 05:00 AM
Sorry, I missed that one..
So basically, time dilation more than compensates for doppler shift caused by accelerating away from the light source?


Thanks,
Peter

Yes and no, things are very DIFFERENT in GR. As I was explaining to grav, what counts is the position in the gravitational well. The observer above sees the source below as redshifted and the observer below sees the source above as blueshifted REGARDLESS of speed. Post 39 gives a good summary.

grav
2011-May-13, 05:23 AM
OK, since you couldn't understand the above, let's try a different route.

The period of a freefalling emitter, as seen by a distant observer is:

T_{distant-free}=T_{free}/(1-r_s/r) (redshift)

The period of a stationary emitter, as seen by a distant observer is:

T_{distant-stationary}=T_{stationary}/\sqrt{1-r_s/r} (less redshift)

So:

T_{distant-free}>T_{distant-stationary}

or, expressed in frequencies:

f_{distant-free}<f_{distant-stationary}

i.e., the freefalling emitter appears more redshifted than the stationary emitter to the same observe placed at infinity.

In GR, as opposed to SR, the effects are not symmetrical, the observer higher up in the gravity well sees the sources below redshifted. The observers lower in the gravity well see the sources above blueshifted. This is one more reason to abandon any "intuition" tying the effect to SR.Again, that is the time dilation of the freefaller according to the distant observer, not the redshift. The distant observer will measure a gravitational redshift of sqrt(1 - R/r) for the light coming from a stationary observer at r because they are stationary to each other, so only the gravitational time dilation comes into effect. But since a freefaller has a relative speed, we must take into account the classical Doppler shift due to the freefaller's motion and the kinetic time dilation. According to a stationary observer at r, the light coming from the freefaller is Doppler shifted due to the freefaller's motion by 1 / (1 + v_loc/c). It is also time dilated by sqrt(1 - (v_loc/c)^2) due to the relative speed, so giving an overall kinematic redshift of sqrt(1 - (v_loc/c)^2) / (1 + v_loc/c) = sqrt[(1 - v_loc/c) / (1 + v_loc/c)]. If the stationary observer were to then re-emit the same light to the distant observer, the distant observer would measure a gravitational time dilation for that same already kinetically redshifted light, so measures

D = [sqrt(1 - (v_loc/c)^2) / (1 + v_loc/c)] sqrt(1 - R/r)

= [sqrt(1 - R/r) / (1 + v_loc/c)] sqrt(1 - R/r)

= (1 - R/r) / (1 + v_loc/c)

= (1 - R/r) / (1 + sqrt(R/r))

So while the time dilation the distant observer measures of the freefaller is just

dτ / dt = gravitational time dilation * kinematic time dilation = (1 - R/r),

the redshift the distant observer measures of the light coming from the freefaller is

D = gravitational time dilation * kinematic time dilation * component of motion

= (1 - R/r) / (1 + v_loc/c)

The component of motion is due to the freefaller emitting one pulse and then moving further away before emitting the next, then moving further still before emitting the next, and so on, so that the light takes a longer time to travel a further distance between each pulse.

macaw
2011-May-13, 06:09 AM
Again, that is the time dilation of the freefaller according to the distant observer, not the redshift.

Q1: What is the difference?



The distant observer will measure a gravitational redshift of sqrt(1 - R/r) for the light coming from a stationary observer at r because they are stationary to each other, so only the gravitational time dilation comes into effect. But since a freefaller has a relative speed, we must take into account the classical Doppler shift due to the freefaller's motion and the kinetic time dilation.

Yes, you need to but not through the hack that you keep pulling.
It is very simple, really, you need to start from the Schwarzschild solution, as I showed you earlier.

(cd\tau)^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)

Divide by dt^2 and out comes the correct result.




According to a stationary observer at r, the light coming from the freefaller is Doppler shifted due to the freefaller's motion by 1 / (1 + v_loc/c).

Q2: Prove it. By deriving it, not by asserting it.





It is also time dilated by sqrt(1 - (v_loc/c)^2) due to the relative speed,

Q3: Why would it be? This is not SR, this is GR. Doppler effect is derived for inertial motion, not for accelerated one.

Q4: How did you come up with the factor sqrt(1 - (v_loc/c)^2)

Q5: In an earlier attempt, you used sqrt((1 - v_loc/c)/(1+v_loc/c)). What entitles you to keep changing the multiplier factors?



so giving an overall kinematic redshift of sqrt(1 - (v_loc/c)^2) / (1 + v_loc/c) = sqrt[(1 - v_loc/c) / (1 + v_loc/c)]. If the stationary observer were to then re-emit the same light to the distant observer, the distant observer would measure a gravitational time dilation for that same already kinetically redshifted light, so measures

D = [sqrt(1 - (v_loc/c)^2) / (1 + v_loc/c)] sqrt(1 - R/r)

Q6: But you claimed in the beginning of this post that the dilation between the distant observer and the stationary one is (1-R/r). Yet, you have surreptitiously replaced the (1-R/r) with sqrt(1-R/r). Is this an error or just a sleigh of hand?


= [sqrt(1 - R/r) / (1 + v_loc/c)] sqrt(1 - R/r)

= (1 - R/r) / (1 + v_loc/c)

= (1 - R/r) / (1 + sqrt(R/r))

Q7: So, you changed your final answer? So, the answer is no longer 1/ (1 + sqrt(R/r) but (1 - R/r) / (1 + sqrt(R/r))? Now you arranged to have everything multiplied by (1-R/r) because you realized that it doesn't go to 0 when r =R, right? It is still wrong. Differently wrong but still wrong. What entitles you to make the above claims? There is no derivation, you are just cobbling expressions together.


So while the time dilation the distant observer measures of the freefaller is just

dτ / dt = gravitational time dilation * kinematic time dilation = (1 - R/r),

the redshift the distant observer measures of the light coming from the freefaller is

D = gravitational time dilation * kinematic time dilation * component of motion

Q8: Where did the "component of motion" come from?
Q9: How is it different from "kinematic time dilation"?


= (1 - R/r) / (1 + v_loc/c)

You are developing your own ATM inside someone else's ATM. The standard derivation says that you are wrong.

Q10: If you think that you are right, you will need to derive (not to assert) your expressions.




The component of motion is due to the freefaller emitting one pulse and then moving further away before emitting the next, then moving further still before emitting the next, and so on, so that the light takes a longer time to travel a further distance between each pulse.

Q11: But the math describing the phenomenon says otherwise, there is no "v" in the formula. If you think otherwise, you'll need to do a derivation from base principles. This means that you can't multiply things together , you need to start from the Schwarzschild solution and do the derivation correctly. No sleigh of hands as in showing first a dilation by 1-r/R only to replace it midway with sqrt(1-r/R). Such basic errors are easy to spot.

Mike Holland
2011-May-13, 06:21 AM
Grav, that's pretty much what I had assumed about the different factors involved in the observed redshift and time dilation, except that I had forgotten to apply the kinematic and doppler redshifts/dilations back on the faller's view of the distant observer. Thanks for sorting out all the maths.

macaw
2011-May-13, 06:24 AM
Grav, that's pretty much what I had assumed about the different factors involved in the observed redshift and time dilation, except that I had forgotten to apply the kinematic and doppler redshifts/dilations back on the faller's view of the distant observer. Thanks for sorting out all the maths.

grav's math is still wrong. Different from before but just as wrong.

Mike Holland
2011-May-13, 06:44 AM
OK, I guess its pistols at dawn, then. I will go along with the victor. Just as long as the gravitational time dilation at an EH is infinity, I am happy.

kzb
2011-May-13, 12:30 PM
Yes, your conceptual problem has been around since Special Relativity hit the scene. Observers in different reference frames see things differently.

However your unnamed GR experts seem to forget that time dilation is already in play long before the collapse reaches the BH stage, and in fact slows the collapse process to zero as the EH forms, after an infinite time. So the EH does not pop up to engulf the infalling observer. Rather, he becomes part of the slowly expanding almost-EH region.

All my reasoning is based on the simple fact that time dilation becomes infinite at an EH, in the reference frame of an outside observer (which I expand to include all of our universe that is outside the BH). And I think this is an accepted consequence of Einstein's theory. So until you can name one of your experts who disproved this, and give a reference, I shall stick with my list of august references.

NB. I think the existence of MECOs re-inforces the view of my august references.

I'm afraid I can't use the Search function on here ! But believe me this very same question has come up at least twice before. One of the "unnamed experts" was Ken G.

My gut feeling is that BH's will turn out to be nothing more than a GR thought-experiment that neglects quantum effects. Something else will happen, probably based on the quantum world, that prevents the EH from forming.

Does Sgr A* Have an Intrinsic Magnetic Moment Instead of an Event Horizon?
Quote: .....magnetic fields and their synchrotron radiations are ubiquitous among the black hole
candidates.

http://arxiv.org/abs/astro-ph/0603746

grav
2011-May-13, 01:29 PM
What is the difference?The difference between time dilation and redshift is the factor of the component of motion, or classical Doppler, 1 / (1 + v/c), due to the increasing distance with the relative motion between the source and receiver between emitting pulses. For instance, classical Doppler is multiplied by the kinematic time dilation to get relativistic Doppler, but kinematic time dilation and relativistic Doppler are not the same thing any more than the time dilation and redshift in a gravitational field are.


Why would it be? This is not SR, this is GR. Doppler effect is derived for inertial motion, not for accelerated one.SR is valid locally, briefly between observers at the same place in a gravitational field. We've been over this before and I provided you several references.


How did you come up with the factor sqrt(1 - (v_loc/c)^2)That is the kinematic time dilation due to the relative speed that is measured locally, briefly at the same place in the gravitational field. We have been over this as well.


In an earlier attempt, you used sqrt((1 - v_loc/c)/(1+v_loc/c)). What entitles you to keep changing the multiplier factors?The factors are classical Doppler and relativistic time dilation, so we get

[1 / (1 + v_loc/c)] * [sqrt(1 - (v_loc/c)^2)] = sqrt[(1 - v_loc/c) / (1 + v_loc/c)]


But you claimed in the beginning of this post that the dilation between the distant observer and the stationary one is (1-R/r). Yet, you have surreptitiously replaced the (1-R/r) with sqrt(1-R/r). Is this an error or just a sleigh of hand?The gravitational time dilation that the distant observer measures of a stationary observer at r is just sqrt(1 - R/r). The time dilation that the distant observer measures of a freefaller at r is sqrt(1 - R/r) * sqrt(1 - (v_loc/c)^2), which for a freefaller falling from rest at inifinity, where v_loc = c sqrt(R/r), becomes 1 - R/r.


So, you changed your final answer? So, the answer is no longer 1/ (1 + sqrt(R/r) but (1 - R/r) / (1 + sqrt(R/r))?1 / (1 + sqrt(R/r)) is the redshift the freefaller sees for light coming from the distant observer, blueshifting within the gravity well but redshifting due to the relative speed. (1 - R/r) / (1 + sqrt(R/r)) is the redshift the distant observer sees of the light coming from the freefaller, redshifting from both the gravity well and the relative speed.


Where did the "component of motion" come from?It is the same as classical Doppler. It must be multiplied by the relativistic time dilation to get relativistic Doppler, and multiplied by the gravitational time dilation to get the redshift in a gravitational field.


How is it different from "kinematic time dilation"?Kinematic time dilation is sqrt(1 - v/c)^2). It is multiplied by classical Doppler 1 / (1 + v/c) to get relativistic Doppler with sqrt[(1 - v/c) / (1 + v/c)]. They are not the same thing.

grav
2011-May-13, 01:37 PM
Grav, that's pretty much what I had assumed about the different factors involved in the observed redshift and time dilation, except that I had forgotten to apply the kinematic and doppler redshifts/dilations back on the faller's view of the distant observer. Thanks for sorting out all the maths.No problem. You're welcome. :)

macaw
2011-May-13, 02:06 PM
The difference between time dilation and redshift is the factor of the component of motion, or classical Doppler, 1 / (1 + v/c),

Q11: Is this from a book or you just made it up? Can you give a reference?
Q12: What entitles you to mix in relativity with classical Doppler effect?
Q13: What entitles you to multiply?



due to the increasing distance with the relative motion between the source and receiver between emitting pulses. For instance, classical Doppler is multiplied by the kinematic time dilation to get relativistic Doppler,

Q14: Do you really think that this is how the relativistic Doppler effect was derived? Through a hack?




SR is valid locally, briefly between observers at the same place in a gravitational field. We've been over this before and I provided you several references.

Yes, we've been over this before but you are clearly missaplying this concept by hacking together incorrect formulas.

Q15: Can you derive the correct formula starting from the Schwarzschild metric without using any hack (like multiplication by various factors)?




That is the kinematic time dilation due to the relative speed that is measured locally, briefly at the same place in the gravitational field. We have been over this as well.

Yet, no textbook ever uses such a hack.

Q16: Can you give one reference where you saw this?




The factors are classical Doppler and relativistic time dilation, so we get

[1 / (1 + v_loc/c)] * [sqrt(1 - (v_loc/c)^2)] = sqrt[(1 - v_loc/c) / (1 + v_loc/c)]

Q17: What entitles you to mix classical with relativity?


The gravitational time dilation that the distant observer measures of a stationary observer at r is just sqrt(1 - R/r). The time dilation that the distant observer measures of a freefaller at r is sqrt(1 - R/r) * sqrt(1 - (v_loc/c)^2),


Q18: But you still have not explained why you switched surreptitiously in your "derivation" from 1-R/r to sqrt(1-R/r). To make your new results look like mine?

Q18a: Can you derive this instead of asserting it? You have asserted it (in different forms) multiple times but you never derived it, so, please try to do so.



which for a freefaller falling from rest at inifinity, where v_loc = c sqrt(R/r), becomes 1 - R/r.

Q19: If it is 1-R/r, where does your extra factor 1/\sqrt{1+R/r} comes from?




1 / (1 + sqrt(R/r)) is the redshift the freefaller sees for light coming from the distant observer, blueshifting within the gravity well but redshifting due to the relative speed.

Q20: Can you derive this rather than asserting it?


(1 - R/r) / (1 + sqrt(R/r)) is the redshift the distant observer sees of the light coming from the freefaller, redshifting from both the gravity well and the relative speed.

Q20: But earlier in the thread you asserted that the answer was 1/\sqrt{1+R/r}, after I pointed out that that answer was wrong you have changed your claim that it is (1-R/r)/\sqrt{1+R/r}. Which one is it?


It is the same as classical Doppler.

Q21: How can relativity be the same as classical physics? There are countless examples where relativity contradicts classical physics, what entitles you to mix classical physics with relativity?




It must be multiplied by the relativistic time dilation to get relativistic Doppler, and multiplied by the gravitational time dilation to get the redshift in a gravitational field.

Q22: Why does it must? . Try deriving formulas from base principles, without using these erroneous "it must".


Kinematic time dilation is sqrt(1 - v/c)^2). It is multiplied by classical Doppler 1 / (1 + v/c) to get relativistic Doppler with sqrt[(1 - v/c) / (1 + v/c)].

Q23: Is this how Einstein derived the formulas for the relativistic Doppler effect? By hacking? Can you point to where in his paper he resorts to your hack?

macaw
2011-May-13, 02:21 PM
OK, I guess its pistols at dawn, then. I will go along with the victor. Just as long as the gravitational time dilation at an EH is infinity, I am happy.

Well, grav has recently switched surreptitiously from claiming that it is 2 to claiming that it is infinity. So, he's making (slow) progress.

grav
2011-May-13, 03:12 PM
Well, grav has recently switched surreptitiously from claiming that it is 2 to claiming that it is infinity. So, he's making (slow) progress.The freefaller approaching r = R will observe half the frequency for the light that came from the distant observer than the frequency that the photons were originally emitted by the distant observer, blueshifted by the gravitational field and redshifted due to the relative motion. If the freefaller emits a frequency of light while approaching r = R, then when the distant observer receives it, it will be redshifted to almost zero, due to the redshift of the gravitational field and compounded by a redshift due to the relative motion. Here also is a link to relavistic Doppler (http://en.wikipedia.org/wiki/Relativistic_Doppler_effect), composed of classical Doppler and relativistic time dilation, so having to do with more than just taking the inverse of time dilation by itself.

macaw
2011-May-13, 04:13 PM
The freefaller approaching r = R will observe half the frequency for the light that came from the distant observer than the frequency that the photons were originally emitted by the distant observer,

Q24: The presence of the factor (1-r/R) contradicts your above claim. It shows that the frequency observed by the freefaller is actually infinite. So, it cannot be "double" ANYTHING. If you think otherwise, you'll have to prove it.



blueshifted by the gravitational field and redshifted due to the relative motion.

Q25: the math describing the phenomenon shows that BOTH effects are blueshift, if you think otherwise , please prove your claim.

As a starting point, feel free to use the Schwarzschild solution:

(c d\tau)^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)=(1-r_s/r)((cdt)^2-(dr/(1-r_s/r))^2)

resulting into:

(d\tau/dt)^2=(1-r_s/r)(1-(dr/cdt(1-r_s/r))^2)

You can easily see that BOTH factors are smaller than 1. This clearly contradicts your claim above.





If the freefaller emits a frequency of light while approaching r = R, then when the distant observer receives it, it will be redshifted to almost zero, due to the redshift of the gravitational field and compounded by a redshift due to the relative motion. Here also is a link to relavistic Doppler (http://en.wikipedia.org/wiki/Relativistic_Doppler_effect), composed of classical Doppler and relativistic time dilation,

Q26: Using a terrible wiki page as justification for your claims is not going to cut it. Once again, where did Einstein derive the relativistic Doppler effect as being "composed of Classical Doppler effect and relativistic time dilation"? Please go back to his paper (http://www.fourmilab.ch/etexts/einstein/specrel/www/) and point out where is this hack being used.

Q27: What entitles you to expand the SR claim to GR without any derivation from base principles?


so having to do with more than just taking the inverse of time dilation by itself.


Q28: Once again, you are mixing theories, this time GR with SR. What entitles you to multiply the effects of the two different theories?

WayneFrancis
2011-May-13, 04:31 PM
OK, picture a sphere and choose a point inside it at distance d from the centre. Then the gravity acting on it depends on the mass contained within the sphere of radius d. The hollow shell of matter outside the surface of that sphere is pulling equally in every direction and doesn't add to the gravity of the contained sphere. Now think of the near-black hole as a sequence of such spheres, each just a little larger than its SR and trying desperately to squeeze into its SR. This is the condition right through the collapsing mass as it tries to become a BH. We don't have an EH; we have a whole region which is nearly an EH, and additional mass falling in just extends this region.

I saw some discussion elsewhere about whether there is any pressure if the mass is in freefall. I don't know the answer. Pressure might slightly increase the rate of infall, but this is a problem for quantum mechanics, and our ideas of pressure might not apply in this medium.

Anyway, our poor falling spaceman would become another layer trying to join the forming BH, and a star collapsing on top of him would add additional layers, all slowly moving in. I don't know whether this would speed up his movement, because I don't know the answer regarding pressure. But additiional mass outside his radius does not add to the gravity where he is (as long as it is spread uniformly around the sphere), and so does not affect the near-EH conditions inside the ECO.

Ummm I don't think you understand. If I dump 1 solar mass on to a black hole then the EH will be 2.95km further out then where it started from Doesn't matter that it was outside of the EH at one point. What matters is that there is enough mass within a given volume to an external observer.

OK, let me explain it another way without even dumping mass onto a black hole

take this scenario

http://www.users.on.net/~waynefrancis/bh_s1a.png

I'm falling into a black hole and even if I never reach the EH like you claim what happens when another black hole gets close

http://www.users.on.net/~waynefrancis/bh_s1b.png

I haven't moved...still above the EH of BH #1 but there is a new EH formed from the combined mass of the 2.

Even if you think that they are some solid froozen star there still comes a point where there is enough mass within a given volume to produce a EH before the EH's even touch. What happens in your model of 2 black holes orbiting each other? They just freeze and stop orbiting because the closest points keep slowing down?


That said If I take a 10 solar mass object object it would have a SR of ~29.5km. If I dump another 1 solar mass onto it then even if it stayed just above the 29.5km that mass adds to the 10 solar masses for a total of 11 solar masses. This means that the EH would now have to be 32.45km but you say that the 1 solar mass of material will hover just above the EH @ 29.5km...will it hover just 3km above the EH? Tell me what is the time dilation for a point 3km above the EH of a 10 solar mass frozen star. Tell me why that mass can not fall below the 32.45km radius because that is what it would need to do so that a EH would not be formed.

According to you the Earths crust doesn't add to the gravity well of the Earths core because it is

...hollow shell of matter outside the surface of that sphere is pulling equally in every direction and doesn't add to the gravity of the contained sphere...
It isn't about the effect it has on the mass inside of that shell but the effect it has to an external observer that is important.

Mike Holland
2011-May-14, 12:20 AM
If the freefaller emits a frequency of light while approaching r = R, then when the distant observer receives it, it will be redshifted to almost zero, due to the redshift of the gravitational field and compounded by a redshift due to the relative motion. Here also is a link to relavistic Doppler (http://en.wikipedia.org/wiki/Relativistic_Doppler_effect), composed of classical Doppler and relativistic time dilation, so having to do with more than just taking the inverse of time dilation by itself.

Yes, but you missed out one additional apparent dilation. As the faller moves deeper and deeper into the gravity field, light signals that he emits are slowed more and more, and take longer and longer to escape the gravity, adding a further redshift. Your article only describes Special Relativity Doppler.

So while the slowing down of the faller's clock is described by GR time dilation, the time dilation that we see is further redshifted but these several optical effects.

Mike Holland
2011-May-14, 12:27 AM
OK, WayneFrancis, you've given me a poser there. My initial take is that your diagram is incorrect because you show a blue black hole with two lumps of red black hole sticking out on the sides, and as a black hole has no hair, that is impossible. By definition a black hole is spherical and featureless (alright, a spinning one would be flattened). I will have to think about it, but perhaps you should think a bit deeper too. What happens to "no hair" when two black holes merge? Haven't read anything about that yet!

Edit: OK, I see the error in your diagram. If the two BHs have mass M and radius R, then the combined BH would have mass 2M and radius 2R. So for the merger to take place their event horizons would have to come into contact, and they would both be fully inside the new EH. Does the new EH just snap into existence when the two original ones touch? Doesn't sound likely! Doesn't make any difference to your argument, 'though.

grav
2011-May-14, 01:27 AM
Yes, but you missed out one additional apparent dilation. As the faller moves deeper and deeper into the gravity field, light signals that he emits are slowed more and more, and take longer and longer to escape the gravity, adding a further redshift. Your article only describes Special Relativity Doppler.

So while the slowing down of the faller's clock is described by GR time dilation, the time dilation that we see is further redshifted but these several optical effects.Right. The article was in reply to macaw, asking where the classical Doppler part comes in, rather than just taking the inverse of the time dilation itself directly for the redshift with f = 1 / t. Relativistic Doppler is in reference to what is measured locally, which includes a component of classical Doppler and the kinematic time dilation. Between observers at different locations in the gravitational field, one freefalling and one stationary, there are three factors involved, the component of classical Doppler 1 / (1 + v_loc/c), the kinematic time dilation sqrt(1 - (v_loc/c)^2), and the gravitational time dilation. All three are multiplied to find the total redshift or blueshift observed.

macaw
2011-May-14, 01:54 AM
Right. The article was in reply to macaw, asking where the classical Doppler part comes in, rather than just taking the inverse of the time dilation itself directly for the redshift with f = 1 / t.

Q29: So, you don't think that f=1/T? Is this yet another ATM that you are developing within this thread?



Relativistic Doppler is in reference to what is measured locally, which includes a component of classical Doppler and the kinematic time dilation. Between observers at different locations in the gravitational field, one freefalling and one stationary, there are three factors involved, the component of classical Doppler 1 / (1 + v_loc/c),

Q30: Once again, what entitles you to mix "classical" mechanics into relativity? Especially after I have shown the correct derivation that uses GR only?






the kinematic time dilation sqrt(1 - (v_loc/c)^2), and the gravitational time dilation. All three are multiplied to find the total redshift or blueshift observed.

Q31: Once again, please show a mainstream reference to the above. If you cannot, then you should admit it and split your ATM from the current thread.

WayneFrancis
2011-May-14, 02:07 AM
OK, WayneFrancis, you've given me a poser there. My initial take is that your diagram is incorrect because you show a blue black hole with two lumps of red black hole sticking out on the sides, and as a black hole has no hair, that is impossible. By definition a black hole is spherical and featureless (alright, a spinning one would be flattened). I will have to think about it, but perhaps you should think a bit deeper too. What happens to "no hair" when two black holes merge? Haven't read anything about that yet!

Edit: OK, I see the error in your diagram. If the two BHs have mass M and radius R, then the combined BH would have mass 2M and radius 2R. So for the merger to take place their event horizons would have to come into contact, and they would both be fully inside the new EH. Does the new EH just snap into existence when the two original ones touch? Doesn't sound likely! Doesn't make any difference to your argument, 'though.

I covered that already in my post.


...
Even if you think that they are some solid froozen star there still comes a point where there is enough mass within a given volume to produce a EH before the EH's even touch. ...


Sure their EH would have to come into contact to get an EH that completely encompasses both but the EH would actually form before that as shown by this simulation (http://www.youtube.com/watch?v=50aTv0R460w)Grey pointed us to in the Black Holes near miss? (http://www.bautforum.com/showthread.php/115455-Black-Holes-near-miss?p=1887369#post1887369) thread.

So since this is ATM I ask you to show with maths what happens to 2 of your "frozen stars" almost make contact.

Tensor
2011-May-14, 05:00 AM
If it collapsed further, then at some depth inside it the amount of mass within that depth would come close to satisfying the equation r = 2GM/c^2. In fact, a whole range of radii might come close to satisfying it.

That may be, but the Event Horizon starts from the center of the collapsing star and moves outward from the center. If you don't agree, please demonstrate the math showing this isn't the case.


But then over this whole range of radii extreme time dilation would be in effect, and in our timeframe the collapse would "freeze" and proceed extremely slowly, coming to a stop after an infinite time.

Interesting. You do realize that we would not be able to see the freeze or the slowing past a certain point, right?


Time dilation doesn't snap into existence when an event horizon appears. It is present around any mass and becomes extreme as a mass collapses towards becoming a black hole.

Of course. But, you're still missing the point. Where does the horizon first form? Please show us, mathematically, that it doesn't start in the center of the collapse.


I agree, it doersn't matter as far as the phenomena that we can observe is concerned. But we should still get our facts straight.

Of course. So you can show us where the horizon first starts, right. Just so we can get our facts straight?

Mike Holland
2011-May-16, 07:10 AM
Tensor, I don't know where the EH starts, and I don't care much. Can you prove mathematically that it does start in the centre?

Seems to me it all depends on the mass distribution at the time. If the collapsing superstar had a uniform density, like a neutron star, then the EH would form on the outside as it sank below its SR. Any contained sphere within the mass would not be within its SR. But normally we would be dealing with the aftermath of a nova or supernova, so all bets are off as far as I am concerned.

But the facts that I would like to get straight are that Oppenheimer, Snyder, Landau, Lifschitz, Novikov & co have proved mathematically that in our outside reference frame, it takes an infinite time for a black hole to form, and as our universe is only 13.7 billion years old, there aren't any in our universe.

If anyone wishes to prove me wrong, just disprove that last sentence. That is my whole case in a nutshell.

And yes, I know that an observer falling into a collapsing supermassive object would encounter a black hole and event horizon in his time-frame, but he would have left our universe at the far end of time.

WayneFrancis, I am still working on a reply to you. It is taking a bit of research, but my ideas are becoming clearer. Please be patient.

WaxRubiks
2011-May-16, 09:37 AM
Tensor, I don't know where the EH starts, and I don't care much. Can you prove mathematically that it does start in the centre?

Seems to me it all depends on the mass distribution at the time. If the collapsing superstar had a uniform density, like a neutron star, then the EH would form on the outside as it sank below its SR.

surely the density will always be greatest in the center? And, conventionally, that would be the first place that would satisfy the conditions for a black hole.....where the mass would be dense enough to be inside its schwarzschild radius(conventionally)....even if it was a 1kilo piece of star matter(a very small BH).

It would at least be where the time dilation was greatest, and thus closest to where an event horizon would (conventionally) form, wouldn't it?

Mike Holland
2011-May-16, 10:40 AM
Yes, Frog March, that seems reasonable in our everyday Newtonian way of thinking, but I think this is a problem for quantum mechanics, when the matter has collapsed way beyond potons and electrons, and gravity waves are shooting around in every direction. I am too ignorant of the physics here to have an opinion.

kzb
2011-May-16, 11:47 AM
There was a paper from the LHC physicists some time back, around the time people were worried about the LHC producing micro-BH's.

They had exactly the same issue. How do you detect if a micro-BH has been produced, when, to a remote observer, they cannot form in finite time? I can't find this paper anymore -my old references have gone into their own black hole. Along with previous threads on here on this very same question !

Anyway, it was a paper written by some high ranking particle physicists, who you would think would have a better understanding than most. Altho they are not GR experts of course.

Personally I can't follow the maths anyway. I have just about got my head around the idea that the EH expands to surround incoming extra mass, so that mass can be observed to pass through an EH in finite time. However that is not the same thing as an EH forming in the first place. I still have difficulties with that one.

Mike Holland
2011-May-16, 01:15 PM
They had exactly the same issue. How do you detect if a micro-BH has been produced, when, to a remote observer, they cannot form in finite time? Personally I can't follow the maths anyway.

I have just about got my head around the idea that the EH expands to surround incoming extra mass, so that mass can be observed to pass through an EH in finite time. However that is not the same thing as an EH forming in the first place. I still have difficulties with that one.

And there is the whole problem I am up against. You say that for us remote observers they cannot form in a finite time, and then you talk of an existing EH expanding. The whole point is that there is no existing EH. Not until our clocks read infinity.

Of course if you switch to the time frame of the Black Hole (in our infinite future), then the EH does exist, but you are no longer talking about anything in our universe.

Mike Holland
2011-May-16, 01:31 PM
WayneFrancis, I have finally got my head around the conundrum you posed with merging Black Holes.

Your picture shows an event horizon forming around the two black holes with a radius of 2R when the two equal black holes get within 2R of each other and their singularities are within that sphere. But this means that light moving away from the area near the black holes would suddenly change direction and start moving inwards as the surrounding EH springs into existence. This doesn't sound right!

This AIP Conference Proceedings report states that the holes merge like a pair of pants – the two holes making contact and then merging into a single hole.

http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=APCPCS000493000001000035000001&idtype=cvips&gifs=yes&ref=no

Here is the link Grav provided in the “Black holes near miss?” topic. Note that it says on the “Black hole collisions: Ring down” page that after merger the merged black hole will be highly distorted, and this is illustrated on the “Coordinate orbits” page to which Grav refers.
http://www.wfu.edu/~cookgb/NCSU_2006.pdf

These two references show that a surrounding spherical black hole does not spring into existence the moment that the two event horizons get within 2R of each other, but rather that the event horizons distort and stretch towards each other until they make contact and join in a dumbbell shape, and then this rapidly smooths out into a sphere. So my hovering spaceman would not suddenly find himself inside a black hole.

Pity no-one has done the simulation or calculation from a remote observer's point of view, taking time dilation into account.

You can prove me wrong very easily, as I said a few posts back. Just prove that the time-dilated "freezing" of a contracting mass as observed by a remote viewer, described by Oppenheimer et al, doesn't happen. If you can't, then you need to rethink your own ideas.

Strange
2011-May-16, 01:44 PM
And there is the whole problem I am up against. You say that for us remote observers they cannot form in a finite time, and then you talk of an existing EH expanding. The whole point is that there is no existing EH. Not until our clocks read infinity.

But there is an existing event horizon. It was formed at the center of the collapsing star and accelerated out at the speed of light.

On the other hand, if there is no event horizon, then time doesn't slow to zero and it doesn't take an infinite time for the event horizon to form. Which sounds like a reductio ad absurdum demonstration that your idea is wrong. But even if you are right, and we can only see stuff falling into a black hole form an (asymptotically) infinitely thin layer, infinitely close to the event horizon - until we can no longer see it because the last photon before it fell through the event horizon has been emitted - and nothing can emerge from that point, then that is, as far as I can see, indistinguishable from an event horizon around a black hole.



Of course if you switch to the time frame of the Black Hole (in our infinite future), then the EH does exist, but you are no longer talking about anything in our universe. It could be argued that a black hole (ie. the "inside" of the event horizon) doesn't exist in our universe anyway.

macaw
2011-May-16, 01:58 PM
But the facts that I would like to get straight are that Oppenheimer, Snyder, Landau, Lifschitz, Novikov & co have proved mathematically that in our outside reference frame, it takes an infinite time for a black hole to form,

This is COORDINATE time, not PROPER Time. PROPER time to BH formation is finite.



and as our universe is only 13.7 billion years old, there aren't any in our universe


It is the COORDINATE time that is infinite, not the PROPER time. Coordinate time is irrelevant, so your conclusion is false.

WaxRubiks
2011-May-16, 02:00 PM
On the other hand, if there is no event horizon, then time doesn't slow to zero and it doesn't take an infinite time for the event horizon to form.

but wouldn't the high density of collapsing matter mean that time dilation would be huge, and gradually increasing slower and slower, approaching infinite time dilation, but never reaching it?

Strange
2011-May-16, 02:05 PM
but wouldn't the high density of collapsing matter mean that time dilation would be huge, and gradually increasing slower and slower, approaching infinite time dilation, but never reaching it?

I would refer you to macaw's answer above. The event horizon exists.

macaw
2011-May-16, 02:35 PM
but wouldn't the high density of collapsing matter mean that time dilation would be huge, and gradually increasing slower and slower, approaching infinite time dilation, but never reaching it?

Expressed in terms of coordinate time, yes. Expressed in terms of proper time, no. It is the proper time that counts, coordinate time is an artifact of our choice of ...coordinates.

Tensor
2011-May-16, 03:21 PM
Tensor, I don't know where the EH starts, and I don't care much. Can you prove mathematically that it does start in the centre?

I would suggest you check out rule #13. Here, it is up to the ATM proponent to show the proof. Since having the EH beginning in the center is mainstream Science, it is up to you to show otherwise.


Seems to me it all depends on the mass distribution at the time.

Yes, the proper time as macaw and strange(referring to macaw) have pointed out.


If the collapsing superstar had a uniform density, like a neutron star, then the EH would form on the outside as it sank below its SR. Any contained sphere within the mass would not be within its SR. But normally we would be dealing with the aftermath of a nova or supernova, so all bets are off as far as I am concerned.

Please show this mathematically. Since you seem to have a inaccurate idea of how Black Holes form, it would appear you have an inaccurate idea of the nature of the idea of a frozen horizon.



But the facts that I would like to get straight are that Oppenheimer, Snyder, Landau, Lifschitz, Novikov & co have proved mathematically that in our outside reference frame, it takes an infinite time for a black hole to form, and as our universe is only 13.7 billion years old, there aren't any in our universe.

And I would say you are taking the word explanation of the collapse into a black hole too far. It's a common problem for people who can't follow the math and have to rely on popular science explanation. It happens here all the time.


If anyone wishes to prove me wrong, just disprove that last sentence. That is my whole case in a nutshell.

Again, not how it works here. Check rule #13. Please show us mathematically, that it takes an infinite proper time for black holes to form. Relying on a popular science explanation can give a lot of people the wrong idea of what is going on. The actual situation is a bit more complicated and requires specific knowledge of how GR works. Which is why they use such simplified examples to explain it.

General Relativity is a Local Theory. Trying to show that two proper times are the same in General Relativity is a GLOBAL Problem and can't be done.


And yes, I know that an observer falling into a collapsing supermassive object would encounter a black hole and event horizon in his time-frame, but he would have left our universe at the far end of time.

And yet, in this very post said this:

But the facts that I would like to get straight are that Oppenheimer, Snyder, Landau, Lifschitz, Novikov & co have proved mathematically that in our outside reference frame, it takes an infinite time for a black hole to form, and as our universe is only 13.7 billion years old, there aren't any in our universe.

Care to explain how there can't be any black holes, but an observer can fall into one?

kzb
2011-May-16, 05:29 PM
And there is the whole problem I am up against. You say that for us remote observers they cannot form in a finite time, and then you talk of an existing EH expanding. The whole point is that there is no existing EH. Not until our clocks read infinity.

Of course if you switch to the time frame of the Black Hole (in our infinite future), then the EH does exist, but you are no longer talking about anything in our universe.

That last paragraph is another difficulty I have. The final outcome is different in the two reference frames. Note it is not just the timing that is different -I can accept that- it is that to one observer the final outcome is to fall through the EH, and to the other it is to hover over the EH for ever. There are two different final outcomes.

It's also worth noting as I said above, that if the EH does form in finite time, and things can be observed to go beyond it in finite time, then the long list of respected authorities you give are wrong in this respect. So we are in good company.

WayneFrancis
2011-May-17, 12:33 AM
But there is an existing event horizon. It was formed at the center of the collapsing star and accelerated out at the speed of light.

On the other hand, if there is no event horizon, then time doesn't slow to zero and it doesn't take an infinite time for the event horizon to form. Which sounds like a reductio ad absurdum demonstration that your idea is wrong. But even if you are right, and we can only see stuff falling into a black hole form an (asymptotically) infinitely thin layer, infinitely close to the event horizon - until we can no longer see it because the last photon before it fell through the event horizon has been emitted - and nothing can emerge from that point, then that is, as far as I can see, indistinguishable from an event horizon around a black hole.


It could be argued that a black hole (ie. the "inside" of the event horizon) doesn't exist in our universe anyway.

Well put.

Strange
2011-May-17, 12:36 AM
The final outcome is different in the two reference frames.and things can be observed to go beyond it in finite time

Nothing can be observed to pass the event horizon anyway (it's an event horizon) which is one reason I think the whole discussion is moot.

Strange
2011-May-17, 12:37 AM
Well put.

I seem to have a fan club...

WayneFrancis
2011-May-17, 12:44 AM
WayneFrancis, I have finally got my head around the conundrum you posed with merging Black Holes.

Your picture shows an event horizon forming around the two black holes with a radius of 2R when the two equal black holes get within 2R of each other and their singularities are within that sphere. But this means that light moving away from the area near the black holes would suddenly change direction and start moving inwards as the surrounding EH springs into existence. This doesn't sound right!

This AIP Conference Proceedings report states that the holes merge like a pair of pants – the two holes making contact and then merging into a single hole.

http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=APCPCS000493000001000035000001&idtype=cvips&gifs=yes&ref=no

Here is the link Grav provided in the “Black holes near miss?” topic. Note that it says on the “Black hole collisions: Ring down” page that after merger the merged black hole will be highly distorted, and this is illustrated on the “Coordinate orbits” page to which Grav refers.
http://www.wfu.edu/~cookgb/NCSU_2006.pdf

These two references show that a surrounding spherical black hole does not spring into existence the moment that the two event horizons get within 2R of each other, but rather that the event horizons distort and stretch towards each other until they make contact and join in a dumbbell shape, and then this rapidly smooths out into a sphere. So my hovering spaceman would not suddenly find himself inside a black hole.

Pity no-one has done the simulation or calculation from a remote observer's point of view, taking time dilation into account.

You can prove me wrong very easily, as I said a few posts back. Just prove that the time-dilated "freezing" of a contracting mass as observed by a remote viewer, described by Oppenheimer et al, doesn't happen. If you can't, then you need to rethink your own ideas.

Much smarter people in here have put it much better then I can. At this point I'd be saying the same thing over and over to which you seem to hand wave away so I see no point to try to continue to show you what you are getting confused.

baskerbosse
2011-May-17, 03:06 AM
That last paragraph is another difficulty I have. The final outcome is different in the two reference frames. Note it is not just the timing that is different -I can accept that- it is that to one observer the final outcome is to fall through the EH, and to the other it is to hover over the EH for ever. There are two different final outcomes.

I'd say that's not really two outcomes. It's the same outcome from two different perspectives.
Spacetime is curved to such an extent that as you fall past the event horizon, you are 'rotated' (4D-wise) so far that space becomes time.
Your forward movement is now 'forward' in time instead (for external observers).
Since approaching the event horizon happens in the future, the only way to observe it, is to go there.

Is the inside of a black hole part of the universe? I would think so, but that depends on your definition of universe I guess..

What I can not get my head around is how Hawking Radiation works with respect to the inside of a black hole.
According to this, a black hole evaporates and therefore ceases to exist in a finite time.
I do not understand how anything can pass the event horizon if this is the case..


Peter

kzb
2011-May-17, 11:45 AM
I'd say that's not really two outcomes. It's the same outcome from two different perspectives.
Spacetime is curved to such an extent that as you fall past the event horizon, you are 'rotated' (4D-wise) so far that space becomes time.
Your forward movement is now 'forward' in time instead (for external observers).
Since approaching the event horizon happens in the future, the only way to observe it, is to go there.

Is the inside of a black hole part of the universe? I would think so, but that depends on your definition of universe I guess..

What I can not get my head around is how Hawking Radiation works with respect to the inside of a black hole.
According to this, a black hole evaporates and therefore ceases to exist in a finite time.
I do not understand how anything can pass the event horizon if this is the case..


Peter

For the "two outcomes" issue, I don't think there are two outcomes: this is because we are told that the infalling observer IS observed to pass beyond the EH in finite time. This is because the EH expands to accommodate the added mass. If this is correct, then there are not two different final outcomes. This also means the list of authors in the OP also misunderstand the issue.

Hawking radiation: I believe it is not a dead cert that this even happens. There are controversies in the published material on this.

We need to remember GR neglects quantum effects. In GR, the EH will have a precise location, the precision only limited by our measurement abilities. In reality, the precision is fundamentally limited by the uncertainty principle. The EH is actually fuzzy, and particles will be able to quantum tunnel in and out.

Same goes for the singularity I guess. It can't really be an infinitely small point.

Mike Holland
2011-May-18, 12:41 AM
For the "two outcomes" issue, I don't think there are two outcomes: this is because we are told that the infalling observer IS observed to pass beyond the EH in finite time. This is because the EH expands to accommodate the added mass. If this is correct, then there are not two different final outcomes. This also means the list of authors in the OP also misunderstand the issue.

Same goes for the singularity I guess. It can't really be an infinitely small point.

KZB, this IS the whole issue. If, as you say, the authors I quoted are wrong, then PPLLEEAASSEE give me some references to show this. My whole case rests on time freezing around a supermassive object as it collapses so that it never quite becomes a black hole - in our reference frame (ie. as measured by our calendars and wristwatches).

You say "we are told", but by whom?

With all the responses I have had to my originial post, not one has presented evidence challenging this. WayneFrancis at least had a damn good try, and had me worried for a while, until I did some studying on merging black holes.

baskerbosse
2011-May-18, 01:47 AM
My whole case rests on time freezing around a supermassive object as it collapses so that it never quite becomes a black hole - in our reference frame (ie. as measured by our calendars and wristwatches).

But isn't that what a black hole IS?
A whole chunk of space essentially doesn't exist in our reference frame. Space itself has turned into time and is literally falling into the future..

The hole exists. It's black because it can not be observed.

Peter

Cougar
2011-May-18, 02:39 AM
But the facts that I would like to get straight are that Oppenheimer, Snyder, Landau, Lifschitz, Novikov & co have proved mathematically that in our outside reference frame, it takes an infinite time for a black hole to form, and as our universe is only 13.7 billion years old, there aren't any in our universe... If anyone wishes to prove me wrong, just disprove that last sentence. That is my whole case in a nutshell.

Of course, "in our outside reference frame..." is the key assumption here. As macaw and I'm sure others have pointed out, that effects a coordinate choice. Other coordinate choices "in our universe" do allow "time for a black hole to form." I expect it is correct that from our planet's point of view, we cannot see the "final formation" of a black hole. But then, we don't expect to "see" a black hole very well as it is. We do know that spacetime near the event horizon is going to be extremely warped. When we detect signals traversing that warped spacetime, it's like looking at a funhouse mirror. The "oddity" of the image is illusory. The black hole did collapse. It's just the "mirror" you're looking through that shows something different. GR gives us a method to "correct" for the curvature, except when it yields singularities, such as within black holes.

Mike Holland
2011-May-18, 05:00 AM
Baskerbosse, that is correct, except that technically it is not quite a black hole, and there is no event horizon, in an outside reference frame. But observationally we can't tell the difference.

Cougar, I cannot recall anything in GR that says it only applies to observers on Earth. All the references I have simply refer to a distant outside observer - ANY observer who is a respectable distance away from the forming BH, be they on Mars or somewhere in the Andromeda galaxy. I agree this includes an infinity of different reference frames, but the calculations still apply to them all. Please inform me if you can think up an outside reference frame relative to which the collapse would not take an infinite time, except for the observer who is falling in.

macaw
2011-May-18, 05:37 AM
Please inform me if you can think up an outside reference frame relative to which the collapse would not take an infinite time, except for the observer who is falling in.

What I am going to tell you is mainstream physics, you can find it in any book:

1. The coordinate time necessary for a freefalling observer to reach the EH (r=r_s) when falling from a radial distance r_0>r_s is infinite. Indeed

dr/dt=(1-r_s/r)\sqrt{r_s/r-r_s/r_0}

meaning that :

dt/dr=1/(\sqrt{r_s/r-r_s/r_0}(1-r_s/r))

Integrate from r_s to r_0 and you get oo

2. 1. The proper time necessary for a freefalling observer to reach the EH (r=r_s) when falling from a radial distance r_0>r_s is finite. Indeed:

dr/d\tau=\sqrt{r_s/r-r_s/r_0}

Integrate from r_s to r_0 and you get a finite number.

The above proves your statements false.

WayneFrancis
2011-May-18, 06:06 AM
↑↑↑↑↑See. What smarter people then I have said ↑↑↑↑↑

Mike Holland
2011-May-18, 07:03 AM
I'm confused. Is that proper time ours or the free fallers? If it is the free faller's, then that is what I have been saying all along. I have never said that the free faller takes an infinite time according to his own clock!

Strange
2011-May-18, 07:05 AM
there is no event horizon

If there were no event horizon, that would mean that events "inside" the black hole are observable in one frame of reference but not another. Which is obviously not true.


But observationally we can't tell the difference.

Nor in any other way. Quacks like a duck, etc.

kzb
2011-May-18, 11:53 AM
You CAN tell the difference. One, a BH cannot have an intrinsic magnetic field, whereas a MECO does. Two, an EH should generate Hawking radiation.

baskerbosse
2011-May-18, 11:54 AM
Baskerbosse, that is correct, except that technically it is not quite a black hole, and there is no event horizon, in an outside reference frame. But observationally we can't tell the difference.


I understand what you are getting at, -that you can not observe the event horizon externally in a finite timeframe. But that's one of the characteristics of a black hole.
The event horizon still exists. It has a radius.
You have to go there to observe it (not that there is anything special to see there), but then you can not return. Going there is going into the future.

If you call it 'Eternally Collapsing Object' you are picking an external frame of reference only for describing it. Why limit the description?
Is it really a better name? It's not half as catchy as 'Black Hole' IMHO.

Peter

Mike Holland
2011-May-18, 12:32 PM
You CAN tell the difference. One, a BH cannot have an intrinsic magnetic field, whereas a MECO does. Two, an EH should generate Hawking radiation.

Yes, kzb, I stand corrected. But a number of the supermassive objects observed in the centre of other galaxies do have magnetic fields, which supports the ECO/MECO view.

Mike Holland
2011-May-18, 12:40 PM
The event horizon still exists. It has a radius.
If you call it 'Eternally Collapsing Object' you are picking an external frame of reference only for describing it. Why limit the description?
Is it really a better name? It's not half as catchy as 'Black Hole' IMHO.

Peter

Every object has a Schwarzschild radius. But it has to fit within that radius to have an event horizon.

I agree the EH exists when you get there, but not until then. And as you get there, you leave our universe at the far end of time.

As kzb has just remarked, it does make a difference whether we have ECOs or BHs. But mostly I like the idea of not having any singularities around with infinite density. It goes against the grain. Or, for that matter, infinite time dilations or wormholes. Rather leave them to the realm of science fiction.

WaxRubiks
2011-May-18, 12:51 PM
but might not a black hole evaporate(via some sort of Hawking type radiation) before any event horizon formed, even in the distant future?

If pre-Hawking radiation is right, you wouldn't need an event horizon, so a free faller would just get crushed into the collapsing mass, and then evaporate, in the distant future.

macaw
2011-May-18, 01:54 PM
I'm confused. Is that proper time ours or the free fallers? If it is the free faller's, then that is what I have been saying all along. I have never said that the free faller takes an infinite time according to his own clock!

One last time: the only "time" that counts is proper time. Coordinate time doesn't count.

Tensor
2011-May-18, 02:08 PM
Yes, kzb, I stand corrected. But a number of the supermassive objects observed in the centre of other galaxies do have magnetic fields, which supports the ECO/MECO view.

And your proof of this statement?

ETA: You do know that ECO/MECO view has been thoroughly refitted mathematically, right? Those object cannot exist, based on the math.

Tensor
2011-May-18, 02:17 PM
I agree the EH exists when you get there, but not until then.

So you're claiming the EH just pops in and out of existence? It only appears when something comes along to fall in?


And as you get there, you leave our universe at the far end of time.

But, wait, aren't you still in our universe, before you fall in, no matter how close you get to the EH?


As kzb has just remarked, it does make a difference whether we have ECOs or BHs. But mostly I like the idea of not having any singularities around with infinite density.

And the popular science explanations show. There are not infinite densities in GR. Densities approach infinite asymptotically, but they never reach it. If you think otherwise, by all means, please show us.


It goes against the grain. Or, for that matter, infinite time dilations or wormholes. Rather leave them to the realm of science fiction.

Ahhhhh, yes. Argument by "I don't like it".

grav
2011-May-18, 03:13 PM
Every object has a Schwarzschild radius. But it has to fit within that radius to have an event horizon.

I agree the EH exists when you get there, but not until then. And as you get there, you leave our universe at the far end of time.

As kzb has just remarked, it does make a difference whether we have ECOs or BHs. But mostly I like the idea of not having any singularities around with infinite density. It goes against the grain. Or, for that matter, infinite time dilations or wormholes. Rather leave them to the realm of science fiction.According to a distant observer, a black hole never "fully" forms in that, due to the gravitational time dilation, some of its matter will be frozen above the Schwarzschild radius for the mass of the star involved. The thing is, though, the Schwarzschild radius is not the same as the event horizon unless all of the mass falls below it, so the Schwarzschild radius represents an upper limit for the largest event horizon that can form which includes all of the mass below its limit, but an event horizon that is smaller than the Schwarzschild radius can still be present.

As the matter of the star infalls, the potential increases. The time dilation is based upon the potential and is greatest at the center. At some point the potential will become great enough at the center that an event horizon forms there. The time dilation becomes infinite at the center and matter is frozen there, but the overall potential continues to increase as matter which is further away and is not so time dilated continues to infall, causing the event horizon to expand. This produces a region within the infalling star where light cannot escape and the matter there no longer lies within real space. The event horizon will take an infinite time to grow to reach all the way to the Schwarzschild radius limit according to a distant observer, but it still exists.

That is the mainstream view anyway. Although the surface of the star is forever frozen beyond the limit of the Schwarzschild radius to the distant observer, most of its matter has already fallen past its already formed event horizon that has grown outward from the center. If you don't like the idea of objects falling into non-real space, then you would have to re-define the kinematics of a black hole. Others have tried it by re-evaluating the equivalence principle. But then, upon doing so, you would be giving up expressing gravity in terms of Schwarzschild coordinates, so would also have to demonstrate how, with essentially a new hypothesis about gravity altogether, one still gains gravitational lensing and the Shapiro delay, as well as the precession of Mercury and the decay of the orbits of binary star systems.

kzb
2011-May-18, 05:36 PM
KZB, this IS the whole issue. If, as you say, the authors I quoted are wrong, then PPLLEEAASSEE give me some references to show this. My whole case rests on time freezing around a supermassive object as it collapses so that it never quite becomes a black hole - in our reference frame (ie. as measured by our calendars and wristwatches).

You say "we are told", but by whom?

With all the responses I have had to my originial post, not one has presented evidence challenging this. WayneFrancis at least had a damn good try, and had me worried for a while, until I did some studying on merging black holes.

When I say the quoted authors are wrong, I am simply saying that if the mainstream-BH theory supporters on this thread are correct, it means that not just you, but also all those published authors are wrong. I don't have any references, and I myself am sympathetic to your position but I am also sitting on the fence.

"We are told" refers to the pro-mainstream BH supporters. This is my non-mathematical understanding of their position, and that is the EH and infalling observer become one system, with joint mass and gravity. The EH expands and engulfs infalling mass like an ameoba.

I believe it is also possible to find yourself inside a BH almost by accident. If you have sufficiently large mass, the average density to form a BH is not that great. A large enough cluster of galaxies that approached each other sufficiently closely could find themselves within a large black hole. Pinched off from the rest of the universe in their own bubble.

However, please could someone explain to us dumbos precisely what is the definition of proper time in this context? Do I, as a remote observer, observe an EH forming in finite time (as measured by my wristwatch) ?

If I do, do I then observe matter being engulfed by it, again in finite time as measured by me?

caveman1917
2011-May-18, 05:50 PM
Mike Holland, what exactly do you mean by "observe an event horizon"?

If you agree, as you seem to do, that in our outside coordinates we have a 2-surface with diverging time dilation, how is this not "observing an event horizon"?

caveman1917
2011-May-18, 05:57 PM
According to a distant observer, a black hole never "fully" forms in that, due to the gravitational time dilation, some of its matter will be frozen above the Schwarzschild radius for the mass of the star involved.

It will not just sit there frozen. Once the matter is within the new schwarzschild radius (ie the one of the mass of the star combined with the infalling stuff), the apparent horizon will jump over it and everything will be normal again. Note that this happens in finite proper time for the outside observer.
The absolute horizon is another matter, but nobody "observes" any absolute horizon as it's a global property, not local (ie there's no divergence or any other way to locate it for any observer).

macaw
2011-May-18, 06:54 PM
Do I, as a remote observer, observe an EH forming in finite time (as measured by my wristwatch) ?

If I do, do I then observe matter being engulfed by it, again in finite time as measured by me?

Interestingly, an observer situated at radial coordinate r_0 will need infinite amount of coordinate time to see light arriving from the EH. Indeed, start with the Schwarzschild metric (again):

ds^2=(1-r_s/r)dt^2-dr^2/(1-r_s/r)

The equation of light geodesic is ds^2=0 so:

dt/dr=1/(1-r_s/r)

Integrate :

t=r_s log(r-r_s)-r

between the limits r=r_s and r=r_0. The answer is infinity. The coordinate time is the only time in this case, there is no proper time.

caveman1917
2011-May-18, 07:14 PM
Indeed, it diverges at r = rs, as it should. The event horizon is, by definition, a surface with all lightlike normal vectors pointing inwards. That's why the entire concept of "observing an event horizon" is problematic.

macaw
2011-May-18, 07:21 PM
Indeed, it diverges at r = rs, as it should. The event horizon is, by definition, a surface with all lightlike normal vectors pointing inwards. That's why the entire concept of "observing an event horizon" is problematic.

Yes, absolutely. Light coming from within the sphere cannot escape. Light coming from the sphere surface, reaches us in infinite time. Either way, we can never "see" the EH sphere. Math is wonderful.

caveman1917
2011-May-18, 07:39 PM
Either way, we can never "see" the EH sphere.

Which is why i have trouble seeing the entire point in the OP. We know of an event horizon because that's the place where we get certain divergences. In fact we're just defining "well, let's call that place where we get all those divergences an 'event horizon'". And in this specific case we add "and let's call the stuff behind that event horizon 'black hole'".

The OP seems to state there is a surface of diverging time dilation, so in essence that there's an event horizon there - and by extrapolation a black hole. And from that making the statement that "this means there's no black hole".

Mike Holland
2011-May-18, 11:21 PM
What I am going to tell you is mainstream physics, you can find it in any book:

1. The coordinate time necessary for a freefalling observer to reach the EH (r=r_s) when falling from a radial distance r_0>r_s is infinite. Indeed

dr/dt=(1-r_s/r)\sqrt{r_s/r-r_s/r_0}

meaning that :

dt/dr=1/(\sqrt{r_s/r-r_s/r_0}(1-r_s/r))

Integrate from r_s to r_0 and you get oo

2. 1. The proper time necessary for a freefalling observer to reach the EH (r=r_s) when falling from a radial distance r_0>r_s is finite. Indeed:

dr/d\tau=\sqrt{r_s/r-r_s/r_0}

Integrate from r_s to r_0 and you get a finite number.

The above proves your statements false.

Amazing! Did you just copy these formulae out of some textbook without examining them, or have you not read any of my previous posts?

Lets start with the second equation. The proper time, ie. the time taken on his clock, will be finite when he encounters the EH. That's what I have been saying all the time, and everyone agrees with this. No problem, and no disproof of anything I have said.

The first equation is far more interesting. Firstly, it uses the expression "coordinate time". This is technically more corrrect, and it differs from what I have called "our timeframe" by a small amount due to Earth's gravity, rotation and orbital movement. Not enough to affect the final infinity. So the equation states that within this remote coordinate time, it will take an infinite time for the falling observer to reach the EH. This again is exactly what I have been saying all along.

So by what topsy-turvey logic do you conclude that the equations prove my statements false?

But I have gone further than these equations by claiming that the original collapse towards a BH will itself be retarded by time dilation, which is what Oppenheimer and others proved, and no-one has come up with a disproof of this yet.

Even if a BH did form in a finite time, your first equation tells us that any further mass falling in would be "frozen" as it neared the EH, and it would form an ECO around the BH anyway. The equation tells us that nothing has fallen into a BH yet. The falling observer may do it by his clock, but not by any external clock.

If you think you have proved me false, please quote a particular statement of mine and show how it conflicts with your equations.

Mike Holland
2011-May-18, 11:27 PM
The OP seems to state there is a surface of diverging time dilation, so in essence that there's an event horizon there - and by extrapolation a black hole. And from that making the statement that "this means there's no black hole".

There is a Schwarzschild radius there, where the EH would form after an infinite time. But remember that it takes an infinite time for the time dilation to become infinite. The whole situation is approached asymptotically, the nearer you get the slower you move.

WayneFrancis
2011-May-18, 11:32 PM
You CAN tell the difference. One, a BH cannot have an intrinsic magnetic field, whereas a MECO does. Two, an EH should generate Hawking radiation.

Why can't a BH have a magnetic field? As I understand it the force particles that carry the electromagnetic and possibly gravitational forces don't have the same limit as non virtual photons.

macaw
2011-May-18, 11:43 PM
Amazing! Did you just copy these formulae out of some textbook without examining them, or have you not read any of my previous posts?

I did not copy , I derived them.




The first equation is far more interesting. Firstly, it uses the expression "coordinate time". This is technically more corrrect, and it differs from what I have called "our timeframe" by a small amount due to Earth's gravity, rotation and orbital movement. Not enough to affect the final infinity. So the equation states that within this remote coordinate time, it will take an infinite time for the falling observer to reach the EH. This again is exactly what I have been saying all along.


And what you have been told all along is that coordinate time is meaningless.


If you think you have proved me false, please quote a particular statement of mine and show how it conflicts with your equations.

See post 70 where I have already corrected your false statements.

macaw
2011-May-18, 11:53 PM
There is a Schwarzschild radius there, where the EH would form after an infinite time.

Q1: What proof do you have that "the EH will form after an infinite time"? Citation, please.



But remember that it takes an infinite time for the time dilation to become infinite.

Q2: Prove it. Mathematically.



The whole situation is approached asymptotically, the nearer you get the slower you move.

This is also false, both proper and coordinate speed increase as you approach EH.

Q3: If you think otherwise, you'll have to prove it. Feel free to use some of the equations that I have already posted in this thread, they are all mainstream.

baskerbosse
2011-May-19, 12:29 AM
There is a Schwarzschild radius there, where the EH would form after an infinite time. But remember that it takes an infinite time for the time dilation to become infinite. The whole situation is approached asymptotically, the nearer you get the slower you move.

To quote Doc Emmet Brown -"You are not thinking four dimensionally" ;-)

What if the external observer falls into another black hole?
;-)
He can then observe the first faller, from a distance, passing the event horizon.
Shouldn't they then see each other disappear into the respective event horizon?


Peter

grav
2011-May-19, 12:42 AM
The equation tells us that nothing has fallen into a BH yet.On the contrary, as the star infalls, the potential at the center increases until an event horizon forms. As the star continues to infall, the event horizon expands until most of the matter of the star lies within it. The equations only tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius for the mass of the star, and for the event horizon to expand to the Schwarzschild radius also, but that's all. That is because there is no more mass to infall and increase the potential further, but if outside matter were to fall in as well, that would increase the potential more and expand the event horizon great enough to pass the surface the star, but then the new infalling matter would take an infinite amount of time for all of it to reach the new Schwarschild radius for the original mass of the star plus the new mass.

baskerbosse
2011-May-19, 12:57 AM
On the contrary, as the star infalls, the potential at the center increases until an event horizon forms. As the star continues to infall, the event horizon expands until most of the matter of the star lies within it. The equations only tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius for the mass of the star, and for the event horizon to expand to the Schwarzschild radius also, but that's all. That is because there is no more mass to infall and increase the potential further, but if outside matter were to fall in as well, that would increase the potential more and expand the event horizon great enough to pass the surface the star, but then the new infalling matter would take an infinite amount of time for all of it to reach the new Schwarschild radius for the original mass of the star plus the new mass.

Or to rephrase it as a question;
Where is the original star, if not inside the event horizon?


Peter

macaw
2011-May-19, 01:12 AM
The equations only tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius for the mass of the star, and for the event horizon to expand to the Schwarzschild radius also, but that's all.

What equations "tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius"? Can you cite a mainstream reference to your claim?

grav
2011-May-19, 03:24 AM
What equations "tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius"? Can you cite a mainstream reference to your claim?Here (http://apod.nasa.gov/htmltest/gifcity/bh_pub_faq.html) is one, halfway through question #3, but you can also google many others. It is the reason they were called "frozen stars" before John Wheeler coined the term "black hole".

macaw
2011-May-19, 04:11 AM
Here (http://apod.nasa.gov/htmltest/gifcity/bh_pub_faq.html) is one, halfway through question #3, but you can also google many others. It is the reason they were called "frozen stars" before John Wheeler coined the term "black hole".

It doesn't show any "The equations only tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius for the mass of the star".
There are no equations and there aren't any claims remotely close to what you claimed.
Why are you making up things and why are you giving false references?

Mike Holland
2011-May-19, 06:33 AM
And what you have been told all along is that coordinate time is meaningless.

Macaw, I am getting exasperated. You keep repeating that coordinate time is meaningless. Yet you produced a formula for calculatiing coordinate time. So is your formula meaningless? And does this in turn imply that GR is meaningless? This "meaningless" term is used throughout the literature!

Or are you just trying to "disown" the infinity that came out of your equation because it embarasses you?

Please explain why you say it is meaningless. Repeating it over and over does not explain anything or refute anything. And also please explain how you interpret that infinity.
Thanks.

grav
2011-May-19, 06:40 AM
It doesn't show any "The equations only tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius for the mass of the star".
There are no equations and there aren't any claims remotely close to what you claimed.
Why are you making up things and why are you giving false references?It says,


If you attempt to witness the black hole's formation, you'll see the star collapse more and more slowly, never precisely reaching the Schwarzschild radius.

although it also says it is an optical effect caused by the paths of the light rays, but it is not, it is a coordinate effect. Anyway, that is because particles falling toward the Schwarzschild radius will become gravitationally time dilated so that it appears to slow toward zero coordinate speed according to a distant observer. The outermost particles that lie upon the surface of the infalling star would appear to "freeze" at the Schwarzschild radius. Here is a reference from Wiki (http://en.wikipedia.org/wiki/Black_hole),


...the collapsed stars were called "frozen stars,"[13] because an outside observer would see the surface of the star frozen in time at the instant where its collapse takes it inside the Schwarzschild radius.

Mike Holland
2011-May-19, 06:48 AM
On the contrary, as the star infalls, the potential at the center increases until an event horizon forms. As the star continues to infall, the event horizon expands until most of the matter of the star lies within it.

Grav, I enjoy your posts because they are all so well reasoned out and presented, and I don't normally have any quibbles with what you say.

But I have a small quibble this time. An EH does not sdimply snap into existence out of nowhere. It follows on from a steadily (?) increasing gravitational field and corresponding degree of time dilation. Of course in local time this happens extremely swiftly, but in a remote timeframe this dilation would reach infinity in the centre just as the EH came into existence, and this means that just before that moment the dilation was extreme. And so I question your idea of how long it would take for the event horizon to form in the centre. Please give your view on this, or let me know whether someone has done the calculations regarding a remote viewers take on the scene. Perhaps it is covered by Oppenheimer, Wheeler and co. in their caculations. I haven't read them, only their conclusions!

Strange
2011-May-19, 06:59 AM
If you agree, as you seem to do, that in our outside coordinates we have a 2-surface with diverging time dilation, how is this not "observing an event horizon"?

:clap:

This reminds me of the "does 0.999... = 1?" thread. Mike's description has something that behaves infinitely similarly to event horizon that forms infinitely closely to the event horizon .... but is not quite an event horizon.

WaxRubiks
2011-May-19, 07:16 AM
nah, 0.999... does equal 1, where as collapsing objects are not quite at the point of forming any event horizon.

Mike Holland
2011-May-19, 07:17 AM
:clap:

This reminds me of the "does 0.999... = 1?" thread. Mike's description has something that behaves infinitely similarly to event horizon that forms infinitely closely to the event horizon .... but is not quite an event horizon.

.... and becomes an event horizon after an infinite time!

I think of it in terms of "do parallel lines never meet or do they meet at infinity?", and also the graph of y = 1/x, shooting up to infinity as x -> zero.

Strange
2011-May-19, 07:29 AM
Please explain why you say it is meaningless.

If you see a ship sailing away towards the horizon, is it really getting smaller? Or is it just an artefact of the Euclidean coordinate system?

If Einstein on the platform sees a train go through at 90% of c, has that train really got shorter? Or is it just an artefact of the Minkowski coordinate system?

If you see matter slow, redshift and disappear from sight before it can fall through the event horizon, does that mean the event horizon doesn't reallyexist? Or is it just an artefact of the Schwarzschild coordinate system?

Whose ruler should we use to measure what is really happening, yours or theirs?

WayneFrancis
2011-May-19, 07:40 AM
Ok Mike, do the following. This is a direct set question against your ATM claim.


Show us what the radius of a 100 solar mass frozen star would be.

Next calculate the surface area of the 100 solar mass frozen star

Next calculate how thick 1 solar mass of neutron star material would be on the surface of the 100 solar mass frozen star.

Next show how this now 101 solar mass frozen star doesn't have enough mass within a radius that would form a 101 solar mass black hole.

Your only solution that I can see is that as mass is added onto the surfaces all mass on the surface some how gets pushed outward. This means the 100 solar mass object would some how have to be hollow as to not have a EH anywhere.

grav
2011-May-19, 08:01 AM
Grav, I enjoy your posts because they are all so well reasoned out and presented, and I don't normally have any quibbles with what you say.

But I have a small quibble this time. An EH does not sdimply snap into existence out of nowhere. It follows on from a steadily (?) increasing gravitational field and corresponding degree of time dilation. Of course in local time this happens extremely swiftly, but in a remote timeframe this dilation would reach infinity in the centre just as the EH came into existence, and this means that just before that moment the dilation was extreme. And so I question your idea of how long it would take for the event horizon to form in the centre. Please give your view on this, or let me know whether someone has done the calculations regarding a remote viewers take on the scene. Perhaps it is covered by Oppenheimer, Wheeler and co. in their caculations. I haven't read them, only their conclusions!That looks right to me, so I'm not sure what the quibble is. :)

grav
2011-May-19, 08:07 AM
.... and becomes an event horizon after an infinite time!Don't you mean that it takes an infinite time to produce a fully formed black hole, whereas the forever falling surface of the star meets the forever growing event horizon?

Mike Holland
2011-May-19, 08:59 AM
Don't you mean that it takes an infinite time to produce a fully formed black hole, whereas the forever falling surface of the star meets the forever growing event horizon?

HaHa! No, I believe that the event horizon never ever forms in the centre of the almost-black hole, because of the time dilation. This was the afore-mentioned quibble. But I can see that it makes absolutely no difference to the outcome, except the universe as I see it has no singularities.

We'll have to send a Star Trek mission into one to find out. Need to invent a time machine first so they can come back to tell us.

Mike Holland
2011-May-19, 09:02 AM
OK, WayneFrancis, I shall do the calculations and then think about the results. Give me a day or two, 'though. I do have a life!

kzb
2011-May-19, 11:39 AM
If you see a ship sailing away towards the horizon, is it really getting smaller? Or is it just an artefact of the Euclidean coordinate system?

If Einstein on the platform sees a train go through at 90% of c, has that train really got shorter? Or is it just an artefact of the Minkowski coordinate system?

If you see matter slow, redshift and disappear from sight before it can fall through the event horizon, does that mean the event horizon doesn't reallyexist? Or is it just an artefact of the Schwarzschild coordinate system?

Whose ruler should we use to measure what is really happening, yours or theirs?

Hang on though, if an EH is formed in the remote observer's time frame, that remote observer should observe Hawking radiation. He should also observe (assuming that the original collapsing matter was a star), that the magnetic field has disappeared.

So there should be real physical outcomes in the remote observer's time frame. Would these outcomes actually be observed in finite time?

macaw
2011-May-19, 01:25 PM
Macaw, I am getting exasperated. You keep repeating that coordinate time is meaningless.

This is your problem, not mine. Physically, coordinate time has no meaning, any coordinate - dependent answer in physics is meaningless.



Yet you produced a formula for calculatiing coordinate time.

Yes, only for the purpose of giving a mathematical foundation to the discussion.



So is your formula meaningless?

No, the formulas are quite meaningful from a mathematical standpoint. They have no meaning from a physical POV.





And does this in turn imply that GR is meaningless? This "meaningless" term is used throughout the literature!

I have already explained that it is the coordinate-dependent answers that are meaningless. Because you get adifferent answer for each different system of coordinate that you choose.


Or are you just trying to "disown" the infinity that came out of your equation because it embarasses you?

Please explain why you say it is meaningless. Repeating it over and over does not explain anything or refute anything. And also please explain how you interpret that infinity.
Thanks.

I have patiently explained my statements. Now, please answer the questions Q1-Q3. Thank you.

macaw
2011-May-19, 01:31 PM
It says,


If you attempt to witness the black hole's formation, you'll see the star collapse more and more slowly, never precisely reaching the Schwarzschild radius.



Yet, you claimed it says something totally different:


The equations only tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius for the mass of the star, and for the event horizon to expand to the Schwarzschild radius

Mike Holland
2011-May-19, 01:44 PM
I have patiently explained my statements. Now, please answer the questions Q1-Q3. Thank you.

Macaw, I have searched through your previous posts, and eventually found your questions Q1-Q3, in post 47 addressed to Grav. Which questions do you want me to answer?

macaw
2011-May-19, 02:18 PM
Macaw, I have searched through your previous posts, and eventually found your questions Q1-Q3, in post 47 addressed to Grav. Which questions do you want me to answer?

Post 47 addresses the ATM developed by grav. He has been using your thread to develop his own ATM.
The questions I'd like you to answer are posted much later, in post 110.
Are you now clear on why coordinate dependent answers are not used in GR?

grav
2011-May-19, 02:50 PM
HaHa! No, I believe that the event horizon never ever forms in the centre of the almost-black hole, because of the time dilation. This was the afore-mentioned quibble. But I can see that it makes absolutely no difference to the outcome, except the universe as I see it has no singularities.

We'll have to send a Star Trek mission into one to find out. Need to invent a time machine first so they can come back to tell us.No event horizon forms at the center because of the time dilation? I'm not sure I follow. We don't need an expedition to a black hole to find out, though. We can start with the gravitational time dilation of Earth. Under your hypothesis, what is the gravitational time dilation at the surface of the Earth according to a distant observer (along a radial path and ignoring rotation)? What is the gravitational time dilation at the center of the Earth? What equations would you use to find the time dilations at those points?

grav
2011-May-19, 03:29 PM
Yet, you claimed it says something totally different:Well, the first part of my statement is the same as what is quoted, so do you now mean the second part about the event horizon not quite growing enough to expand all the way to the Schwarzschild radius according to a distant observer? Before you only asked about the first part, so are you now accepting that the surface of the star never quite falls past the Schwarzschild radius according to a distant observer?

As for the second part, if all of the matter were within the Schwarzschild radius of the black hole, the event horizon would lie at the Schwarzschild radius with infinite time dilation there. Since to the distant observer, the surface of the star never quite reaches the Schwarzschild radius, then the potential is slightly smaller because some of the mass is further out, so the time dilation at the Schwarzschild radius is smaller also, less than infinite, so finite, and the event horizon instead lies just below the Schwarzschild radius.

macaw
2011-May-19, 03:36 PM
Do you now mean the second part about the event horizon not quite growing enough to expand all the way to the Schwarzschild radius according to a distant observer?

Once again, this is what you claimed (post 112):


The equations only tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius for the mass of the star, and for the event horizon to expand to the Schwarzschild radius

Where did you get the above? The website you tried to link in does not support your fringe claim.





Before you only asked about the first part, so are you now accepting that the surface of the star never falls past its event horizon according to a distant observer?

No, three times in a row I challenged you to provide support for the same fringe claim you made. Every time you tried to evade the question a different way.

macaw
2011-May-19, 03:41 PM
On the contrary, as the star infalls, the potential at the center increases

Q35: Prove that "the potential increases", please. Use math, please.


The equations only tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius for the mass of the star, and for the event horizon to expand to the Schwarzschild radius also, but that's all.

Q36: Can you post "the equations" that support your above ATM?

grav
2011-May-19, 04:05 PM
Let's review. You asked


What equations "tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius"? Can you cite a mainstream reference to your claim?

in post #114. I gave a reference and you stated


It doesn't show any "The equations only tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius for the mass of the star".

in post #116. I then quoted the part that reads


If you attempt to witness the black hole's formation, you'll see the star collapse more and more slowly, never precisely reaching the Schwarzschild radius.

as well as Wiki that states


...the collapsed stars were called "frozen stars,"[13] because an outside observer would see the surface of the star frozen in time at the instant where its collapse takes it inside the Schwarzschild radius.

If you are now attempting to change that to asking about the second part of my statement instead, how the event horizon grows, it is based upon the first part, so do you now accept the first part of my statement, that according to a distant observer, the surface of the star takes an infinite time to reach the Schwarzschild radius?

macaw
2011-May-19, 04:25 PM
If you are now attempting to change that to asking about the second part of my statement instead, how the event horizon grows, it is based upon the first part, so do you now accept the first part of my statement, that according to a distant observer, the surface of the star takes an infinite time to reach the Schwarzschild radius?

There is no mention of any observer (distant or otherwise) in your post. Here is your (incorrect) post again:


On the contrary, as the star infalls, the potential at the center increases until an event horizon forms. As the star continues to infall, the event horizon expands until most of the matter of the star lies within it. The equations only tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius for the mass of the star, and for the event horizon to expand to the Schwarzschild radius also, but that's all.

Your post (http://www.bautforum.com/showthread.php/115485-Black-Holes-or-Eternally-Collapsing-Objects?p=1890438#post1890438) is incorrect, hence the challenge.

grav
2011-May-19, 05:10 PM
There is no mention of any observer (distant or otherwise) in your post. Here is your (incorrect) post again:That post is in reply to Mike Holland, who is looking at a black hole from the perspective of a distant observer, so the post is in respect to a distant observer also.


Your post (http://www.bautforum.com/showthread.php/115485-Black-Holes-or-Eternally-Collapsing-Objects?p=1890438#post1890438) is incorrect, hence the challenge.Do you agree that it is correct from the perspective of a distant observer?

macaw
2011-May-19, 05:12 PM
That post is in reply to Mike Holland, who is looking at a black hole from the perspective of a distant observer, so the post is in respect to a distant observer also.

It doesn't matter who you are responding to, the point is that your post is incorrect.
You have made a large number of counter-factual statements in this thread. In fact, you are using this thread to develop your own ATM.

caveman1917
2011-May-19, 05:14 PM
There is a Schwarzschild radius there, where the EH would form after an infinite time. But remember that it takes an infinite time for the time dilation to become infinite. The whole situation is approached asymptotically, the nearer you get the slower you move.

CM1: which metric describes a collapsing shell of dust? (hint: it is not the schwarzschild metric)
CM2: after having done the calculations in the correct metric, what is the proper time until an event horizon forms for an outside observer?

caveman1917
2011-May-19, 05:19 PM
Even if a BH did form in a finite time, your first equation tells us that any further mass falling in would be "frozen" as it neared the EH

No it does not.
CM3: Do you know why?

grav
2011-May-19, 05:25 PM
It doesn't matter who you are responding to, the point is that your post is incorrect.
You have made a large number of counter-factual statements in this thread. In fact, you are using this thread to develop your own ATM.I am using GR as it is presented by mainstream and have given plenty of references to back up everything I have stated throughout the thread, so your rude and baseless accusations are blatantly false. Since you persist, do you or do you not agree with the statement that according to a distant observer, the surface of the (infalling) star takes an infinite amount of time to reach the Schwarzschild radius?

macaw
2011-May-19, 05:30 PM
I am using GR as it is presented by mainstream

your post contradicts your claim above

macaw
2011-May-19, 05:35 PM
Since you persist, do you or do you not agree with the statement that according to a distant observer, the surface of the (infalling) star takes an infinite amount of time to reach the Schwarzschild radius?

But this not what you claimed in your incorrect post. What you claimed is:


On the contrary, as the star infalls, the potential at the center increases until an event horizon forms. As the star continues to infall, the event horizon expands until most of the matter of the star lies within it. The equations only tell us that the surface of the star takes an infinite amount of time to reach the Schwarzschild radius for the mass of the star, and for the event horizon to expand to the Schwarzschild radius also, but that's all.

grav
2011-May-19, 05:40 PM
your post contradicts your claim aboveIn what way? Because I didn't specifically state in that particular post that it is from the perspective of the distant observer like I did with almost every other post? That would be understood in my reply to Mike Holland, just as it is understood that we are only considering a non-rotating collapsing star, with a mass density that is uniformly spread around each spherical shell, and external to the star is vacuum. Should I have included these conditions with every post I make as well?


But this not what you claimed in your incorrect post. What you claimed is:That is the process by which the event horizon forms, which is based upon the increasing potential as the star collapses, forming first at the center and growing outward, with infinite time dilation at its boundary according to a distant observer. The result is that according to the distant observer, the surface of the infalling star never quite reaches the Schwarzschild radius.

macaw
2011-May-19, 05:42 PM
In what way? Because I didn't specifically state in that particular post that it is from the perspective of the distant observer like I did with almost every other post?

because you were sloppy.

Besides, it isn't your only mistake in the post. I challenged you to show that "the potential increases". See questions Q35-6.
BTW, when do you plan to answer Q1-34?

kzb
2011-May-19, 05:42 PM
Yes, absolutely. Light coming from within the sphere cannot escape. Light coming from the sphere surface, reaches us in infinite time. Either way, we can never "see" the EH sphere. Math is wonderful.

But the formation of an EH (in remote observer time frame) would initiate Hawking radiation production . Hawking radiation can be differentiated, observationally, from other types of radiation, if we have a small black hole mass.

Watching a very small star collapse towards a notional black hole state remotely, would I detect Hawking radiation being switched on at any finite time?

grav
2011-May-19, 05:47 PM
because you were sloppy.

Besides, it isn't your only mistake in the post. I challenged you to show that "the potential increases". See questions Q35-6.
BTW, when do you plan to answer Q1-34?Don't you think that as the star collapses, the potential at the center of the star increases?

Even at a small scale, for a spherical mass M of radius r with uniform density, the potential at the surface is - G M / r while the potential at the center is - (3/2) G M / r. As the radius decreases as the mass collapses, the potential at both the center and the surface increase inversely proportional to the radius.

macaw
2011-May-19, 05:51 PM
But the formation of an EH (in remote observer time frame) would initiate Hawking radiation production . Hawking radiation can be differentiated, observationally, from other types of radiation, if we have a small black hole mass.

Watching a very small star collapse towards a notional black hole state remotely, would I detect Hawking radiation being switched on at any finite time?

A correct answer to your question can be formulated only in the context of a quantum gravity theory, we do not have such a theory at the moment. There seems to be (http://en.wikipedia.org/wiki/Hawking_radiation#Experimental_observation_of_Hawk ing_radiation) experimental proof of Hawking radiation detection. The equations I derived are classical GR gravity.

macaw
2011-May-19, 05:52 PM
Don't you think that as the star collapses, the potential at the center of the star increases?

Q37: Prove it. Use math.

When do you plan to answer Q1-36?

grav
2011-May-19, 05:58 PM
Prove it. Use math.I added this to my previous post.

Even at a small scale, for a spherical mass M of radius r with uniform density, the potential at the surface is - G M / r while the potential at the center is - (3/2) G M / r. As the radius decreases as the mass collapses, the potential at both the center and the surface increase inversely proportional to the radius.


When do you plan to answer Q1-36?I don't. I have already explained all of the concepts in detail and have given references, that is enough. Now it is up to you to read through them and attempt to understand. If you have further questions about mainstream GR, Doppler, or whatever else, feel free to ask them in Q&A.

caveman1917
2011-May-19, 06:10 PM
Watching a very small star collapse towards a notional black hole state remotely, would I detect Hawking radiation being switched on at any finite time?

Semi-classical gravity calculations say yes.

macaw
2011-May-19, 06:10 PM
I added this to my previous post.

Even at a small scale, for a spherical mass M of radius r with uniform density, the potential at the surface is - G M / r while the potential at the center is - (3/2) G M / r. As the radius decreases as the mass collapses, the potential at both the center and the surface increase inversely proportional to the radius.

Q38: You are using Newtonian potential. This is a cosmological GR problem. What entitles you to mix in Newtonian and GR theories?

Q39: Even if you were allowed to mix Newtonian into GR, it is well known that the potential increases in the opposite direction, when r varies away from zero , going to infinity. When r->0 the potential tends to minus infinity. So, contrary to your claim, the potential is known to have a minimum, not a maximum at the center of the sphere. But the question to you is subtler than that: you claimed that the potential increases when the star collapses. What does the potential increase have to do with the star collapse?





I don't. I have already explained all of the concepts in detail and have given references, that is enough.

According to ATM rules, you should. You have been developing your own ATM within this thread. So, please , answer the questions.

grav
2011-May-19, 06:25 PM
You are using Newtonian potential. This is a cosmological GR problem. What entitles you to mix in Newtonian and GR theories?GR reduces to Newton on the small scale. Even on the small scale, the (negative) potential increases as the mass infalls. It would do so on a large scale also, or at least there is nothing in GR that says it would do otherwise, if that is what you are suggesting.


Even if you were allowed to mix Newtonian into GR, it is well known that the potential increases in the opposite direction, when r varies away from zero , going to infinity. So, contrary to your claim, the potential is known to have a minimum, not a maximum at the center of the sphere. But the question to you is subtler than that: you claimed that the potential increases when the star collapses.By convention it is negative, but the absolute value increases.


What does the potential increase have to do with the star collapse?As the star collapses, the (negative) potential increases until the time dilation at the center becomes infinite and an event horizon forms there.


According to ATM rules, you should. You have been developing your own ATM within this thread. So, please , answer the questions.Your nitpicking and rather severe inability to grasp basic concepts does not automatically mean that anything you disagree with, which seems to most everything dealing with mainstream Relativity, is ATM.

caveman1917
2011-May-19, 06:30 PM
GR reduces to Newton on the small scale.

The small scale of what? If you put it like that it seems to say on the scale of "small spatial distances", which is definitely not true.


As the star collapses, the (negative) potential increases until the time dilation at the center becomes infinite and an event horizon forms there.

I would also like to see a reference to a mathematical derivation of this statement.

grav
2011-May-19, 07:11 PM
The small scale of what? If you put it like that it seems to say on the scale of "small spatial distances", which is definitely not true.A small scale like the Earth or the sun, where the radius is much, much greater than the Schwarzschild radius for the mass involved. If the radius of the Earth were to shrink with the same mass, the potential at the center would increase (or decrease since it is negative).


I would also like to see a reference to a mathematical derivation of this statement.Here (http://www.physicsforums.com/showthread.php?t=323684) it is derived using the interior Schwarzschild solution. If the event horizon is defined by the boundary where infinite time dilation occurs according to a distant observer, then according to the metric, the event horizon will form as a point at the center when the radius of the mass is at (9/8) r_s. Below that, the event horizon is larger.

caveman1917
2011-May-19, 07:39 PM
A small scale like the Earth or the sun, where the radius is much, much greater than the Schwarzschild radius for the mass involved.

Even if this were correct, it's a contradiction since you're using that as an argument to invoke newtonian approximations when the radius almost equals the schwarzschild radius.


Here (http://www.physicsforums.com/showthread.php?t=323684) it is derived using the interior Schwarzschild solution. If the event horizon is defined by the boundary where infinite time dilation occurs according to a distant observer, then according to the metric, the event horizon will form as a point at the center when the radius of the mass is at (9/8) r_s. Below that, the event horizon is larger.

There is nothing there about the gravitational potential. Nor about the formation of a black hole or event horizon for that matter.

grav
2011-May-19, 08:07 PM
Even if this were correct, it's a contradiction since you're using that as an argument to invoke newtonian approximations when the radius almost equals the schwarzschild radius.No, I'm using it to show that if we decrease the radius of a body with the same mass, the absolute value of the potential at the center increases (although since it is negative, the potential decreases would be the correct terminology).


There is nothing there about the gravitational potential. Nor about the formation of a black hole or event horizon for that matter.What they refer to as the singularity is the boundary of the event horizon where the time dilation is infinite. As for gravitational time dilation being a result of potential, there are many sources on the web that state this, including this one from Wiki (http://en.wikipedia.org/wiki/Gravitational_time_dilation). The very first sentence reads


Gravitational time dilation is the effect of time passing at different rates in regions of different gravitational potential; the lower the gravitational potential, the more slowly time passes.

macaw
2011-May-19, 09:16 PM
GR reduces to Newton on the small scale.

Doesn't entitle you to mix in Newtonian mechanics with GR. You have been repeating this error throughout this thread.



Even on the small scale, the (negative) potential increases as the mass infalls.

Actually, simple math shows exactly the opposite. I have already shown you that this is a mistake, why do you persist in it?





It would do so on a large scale also, or at least there is nothing in GR that says it would do otherwise, if that is what you are suggesting.

Q40: Prove it. Assertions don't cut it.




By convention it is negative, but the absolute value increases.

You weren't talking about absolute value, why change the subject? Because you've been shown wrong? Admit that you were in error.


As the star collapses, the (negative) potential increases until the time dilation at the center becomes infinite and an event horizon forms there.

Q41: What is the "time dilation at the center"?
Q42: Can you calculate "the time dilation at the center" and prove that "it becomes infinite"? (Hint: it doesn't)
Q43: An even horizon forms at the center? Not at the Schwarzschild radius? Can you cite a reference that supports the ATM?




Your nitpicking and rather severe inability to grasp basic concepts does not automatically mean that anything you disagree with, which seems to most everything dealing with mainstream Relativity, is ATM.

Please answer the questions.

macaw
2011-May-19, 09:25 PM
Here (http://www.physicsforums.com/showthread.php?t=323684) it is derived using the interior Schwarzschild solution.

Q44: Where in the thread do you see the derivation for the formation of the EH? (Hint: caveman beat me to pointing out that you have cited again an irrelevant link)



If the event horizon is defined by the boundary where infinite time dilation occurs according to a distant observer,

The EH is defined as the sphere of radius equal to the Schwarzschild radius, there is no need for you to introduce ATM definitions.



then according to the metric, the event horizon will form as a point at the center when the radius of the mass is at (9/8) r_s.

Q45: How can the EH "form at the center" when it is located at r=r_s?




Below that, the event horizon is larger.

Q46: Larger than what?

grav
2011-May-19, 09:52 PM
Doesn't entitle you to mix in Newtonian mechanics with GR. You have been repeating this error throughout this thread.I'm not mixing anything. The small scale reduces to Newtonian.


Actually, simple math shows exactly the opposite. I have already shown you that this is a mistake, why do you persist in it?As the mass infalls, the potential at the center decreases (becomes more negative).


But you weren't talking about absolute value, why change the subject? Because you've been shown wrong? Admit that you were in error.I am used to thinking in terms of absolute value and that is what I was talking about, as I said. You are right, though, since the potential is negative, the correct terminology would be that the potential is least at the center, and decreases with infall, producing a greater time dilation there.

grav
2011-May-19, 09:54 PM
The EH is defined as the sphere of radius equal to the Schwarzschild radius, there is no need for you to introduce ATM definitions.That is not the event horizon. For instance, the Earth has a Schwarzschild radius for its mass but no event horizon.

macaw
2011-May-19, 09:56 PM
I'm not mixing anything. The small scale reduces to Newtonian.

Q47: you mean it allows you to treat black holes as Newtonian? This is definitely ATM.
Q48: you have been mixing GR with Newtonian mechanics throughout the ATM you've been developing in this thread, like you dd in your attempts to calculate gravitational frequency shifts, why would you not extend this approach to the formation of black holes?




As the mass infalls, the potential at the center decreases (becomes more negative).

Good, so you are learning from your previous mistakes.




I am used to thinking in terms of absolute value and that is what I was talking about, as I said. You are right, though, since the potential is negative, the correct terminology would be that the potential is least at the center, and decreases with infall, producing a greater time dilation there.

Q49: But you claimed that the time dilation is "infinite at the center". Are you withdrawing this claim as well?

macaw
2011-May-19, 09:58 PM
That is not the event horizon. For instance, the Earth has a Schwarzschild radius for its mass but no event horizon.

This was not the point, the point was your ATM definition of the EH.

grav
2011-May-19, 10:06 PM
you mean it allows you to treat black holes as Newtonian? This is definitely ATM.It would be if that was what I am saying, but it is not.


But you claimed that the time dilation is "infinite at the center". Are you withdrawing this claim as well?When a star infalls far enough, the potential at the center decreases far enough that an event horizon forms there with infinite time dilation (or zero depending upon how you look at it) and the boundary of the event horizon grows outward as the infall continues.

grav
2011-May-19, 10:07 PM
This was not the point, the point was your ATM definition of the EH.So how would you define the boundary of an event horizon?

macaw
2011-May-19, 10:14 PM
It would be if that was what I am saying, but it is not.

When a star infalls far enough, the potential at the center decreases far enough that an event horizon forms there with infinite time dilation (or zero depending upon how you look at it) and the boundary of the event horizon grows outward as the infall continues.

Q50: Prove it. Use math. Like all your other ATM claims in this thread, you will need to back up your assertion.

macaw
2011-May-19, 10:18 PM
So how would you define the boundary of an event horizon?

This is the fourth time I am explaining this to you, the EH for a BH is the sphere of radius r=r_s. Please read here (http://en.wikipedia.org/wiki/Event_horizon#Event_horizon_of_a_black_hole). Please abstain from coming back with the irrelvant example of bodies that have a Schwarzschild radius (they all do) but have no EH.

grav
2011-May-19, 10:20 PM
This is the fourth time I am explaining this to you, the EH for a BH is the sphere of radius r=r_s. Please read here (http://en.wikipedia.org/wiki/Event_horizon#Event_horizon_of_a_black_hole). Please abstain from coming back with the irrelvant example of bodies that have a Schwarzschild radius (they all do) but have no EH.So according to you, the event horizon is always at r_s, but does not become active until the star becomes a black hole? The event horizon doesn't form at the center and grow outward?

macaw
2011-May-19, 10:23 PM
So according to you, the event horizon is always at r_s, but does not become active until the star becomes a black hole? The event horizon doesn't form at the center and grow outward?

Did you read the link I posted for your education? Where did I say that? (Hint: I did not).
Now, please answer the questions pertaining your ATM.

grav
2011-May-19, 10:27 PM
Did you read the link I posted for your education? Where did I say that?Sure, the first couple of sentences read


In general relativity, an event horizon is a boundary in spacetime beyond which events cannot affect an outside observer. In layman's terms it is defined as "the point of no return" i.e. the point at which the gravitational pull becomes so great as to make escape impossible.

According to a distant observer, that boundary occurs first at the center of the star as it infalls and grows outward from there. Do you disagree?

macaw
2011-May-19, 10:31 PM
Sure, the first couple of sentences read



According to a distant observer, that boundary occurs first at the center of the star as it infalls and grows outward from there. Do you disagree?

Read on:


The surface at the Schwarzschild radius acts as an event horizon in a non-rotating body that fits inside this radius (although a rotating black hole operates slightly differently).

Now, would you please stop the diversions and start answering the questions about your ATM?

grav
2011-May-19, 10:35 PM
Read on:"That fits inside this radius" being the operative words there. Are you saying that the event horizon doesn't form first at the center and grow outward as the star infalls, that it only ever existed at r_s?

macaw
2011-May-19, 10:40 PM
"That fits inside this radius" being the operative words there.

Q51: Do you have difficulty in understanding what that means? It refers to the non-rotating body that fits inside the sphere of radius r=r_s. it is pretty self - explanatory.



Are you saying that the event horizon doesn't form first at the center and grow outward as the star infalls, that it only ever existed at r_s?

No, I am not saying that and I have no idea how you make this inferences. Please stop trying to put words in my mouth and start answering the challenges against your ATM. As you know, I am not obligated to explain mainstream science to you but you are obligated to answer the questions leveled against your ATM.

grav
2011-May-19, 10:57 PM
No, I am not saying that and I have no idea how you make this inferences. Please stop trying to put words in my mouth and start answering the challenges against your ATM. As you know, I am not obligated to explain mainstream science to you but you are obligated to answer the questions leveled against your ATM.I don't have to put words in your mouth, I can read your posts. Since you still persist in claiming that what I am saying is ATM, let me ask you these to find out where you think so.

1. Does an event horizon describe a boundary beyond which events cannot affect an outside observer?

2. What is the gravitational time dilation at that boundary?

3. Is gravitational time dilation a function of gravitational potential?

4. Is the gravitational potential smallest at the center of a star?

5. Where does the greatest time dilation occur in a star?

6. As a star infalls, does the potential at the center decrease?

7. As a star infalls, does the time dilation at the center slow according to a distant observer?

8. When the gravitational potential becomes small enough at the center, will time stop there according to a distance observer?

9. If time stops at the center of an infalling star, would that describe the boundary of an event horizon?

10. Does an event horizon initially form at the center and grow outward as the star infalls?

Mike Holland
2011-May-20, 12:04 AM
Macaw, you keep evading Gravs question, so I will provide it again here.

You wanted a mainstream, accessible reference. I have before me "The Classical Theory of Fields" by Landau and Lifschitz.
After a couple of pages of maths, they report, on page 298

"This means that according to the clocks t of a distant observer the radius of the contracting body only approaches the gravitational radius asymptotically as t -> infinity."

Do you or do you not accept this?

macaw
2011-May-20, 12:50 AM
I don't have to put words in your mouth, I can read your posts. Since you still persist in claiming that what I am saying is ATM, let me ask you these to find out where you think so.

1. Does an event horizon describe a boundary beyond which events cannot affect an outside observer?

2. What is the gravitational time dilation at that boundary?

I have already shown you the correct formula earlier in this thread. Please go back and revisit it.




3. Is gravitational time dilation a function of gravitational potential?

Yes. I have posted the correct formulas earlier on (post 21).




4. Is the gravitational potential smallest at the center of a star?

Yes, I corrected your claim to the contrary, have you forgotten?




5. Where does the greatest time dilation occur in a star?

This is for you to find out. Hint: Contrary to your repeated erroneous claims, it isn't in the center of the star.


6. As a star infalls, does the potential at the center decrease?

This is the homework that you botched repeatedly by mixing Newtonian mechanics with GR, so you will need to do it by yourself.




7. As a star infalls, does the time dilation at the center slow according to a distant observer?

Nope, I will give you a hint, the correct answer is:

d\tau/dt=1/2(3\sqrt{1-r_s/R}-1)

where R is the radius of the star

Now, it is left for you to figure out how I derived this formula.




8. When the gravitational potential becomes small enough at the center, will time stop there according to a distance observer?

Nope, see above.


9. If time stops at the center of an infalling star, would that describe the boundary of an event horizon?

Once again, time doesn't "stop at the center of an infalling star".


10. Does an event horizon initially form at the center and grow outward as the star infalls?

You will need to learn cosmological relativity in order to figure out the answer to this question.

Now, I have answered your questions (though I do not have to), correcting in passing quite a few of your misconceptions. Since it is you who formulated an ATM by hijacking Mike Holland's ATM, it is you who needs to answer the questions I asked you. Please do so.

showboat
2011-May-20, 12:51 AM
Well my bar bell theory [2 space time balls with a connection is not newtonian physics, but a two same space time balls are the same[quantum entaglement]], the connection has matter and energy exists as more of a information/ probability transfer between such.

The quantum entanglement between the same [duplicate photons] is not like studying proton smashing and will be math rich explained and proved to the satisfaction of math guys.thats the model is the best at the time.

Lasers as to opposed proton colliders, both expensive.

I think the lasers will give up more info.

I think the black holes in center of galazys is more of a information transfer as a type of quantum entanglement.

Different from a matter /energy outlook so thats what the physics guy do to prove in math [profession called metaphysics]

Nuclear fusion might be the funding. but I don't think its safe

macaw
2011-May-20, 12:55 AM
Macaw, you keep evading Gravs question, so I will provide it again here.

This is not grav's question, so you have provided the answer to a question no one asked.
Now, you have some questions to answer, would please do so.


You wanted a mainstream, accessible reference. I have before me "The Classical Theory of Fields" by Landau and Lifschitz.
After a couple of pages of maths, they report, on page 298

"This means that according to the clocks t of a distant observer the radius of the contracting body only approaches the gravitational radius asymptotically as t -> infinity."

Do you or do you not accept this?

Good answer, wrong question. Please answer the questions I asked you.

grav
2011-May-20, 02:29 AM
This is for you to find out. Hint: Contrary to your repeated erroneous claims, it isn't in the center of the star.So you are saying that time doesn't slow the most at the center of a star even though you agree that the potential is least there and that the time dilation is a function of the potential? Please explain.


This is the homework that you botched repeatedly by mixing Newtonian mechanics with GR, so you will need to do it by yourself.Okay. Decreased radius means smaller potential at the center. There, done.


Nope, I will give you a hint, the correct answer is:

d\tau/dt=1/2(3\sqrt{1-r_s/R}-1)

where R is the radius of the star

Now, it is left for you to figure out how I derived this formula.Simple. It is the interior Schwarzschild solution that I linked to earlier, using r = 0 for the center of the star. At the point where the event horizon appears at the center with zero time dilation there, so setting dτ / dt = 0, we find that the star will be at a radius of (9/8) r_s.


Nope, see above.

Once again, time doesn't "stop at the center of an infalling star".So what time dilation do you get at the center of the star using the interior Schwarschild solution you posted for R = (9/8) r_s?


You will need to learn cosmological relativity in order to figure out the answer to this question.The answer is yes.

grav
2011-May-20, 02:33 AM
This is not grav's question, so you have provided the answer to a question no one asked.It is the question I asked you in posts 138, 140, and 144.


Good answer, wrong question.So it appears that after all the hype is said and done, you do at least now agree.

macaw
2011-May-20, 02:58 AM
So you are saying that time doesn't slow the most at the center of a star even though you agree that the potential is least there and that the time dilation is a function of the potential? Please explain.

The time dilation dt/d\tau at the center of the sphere is finite, the time dilation at the EH is infinite.


Okay. Decreased radius means smaller potential at the center. There, done.

You aren't done yet, you have a lot of questions pending on your ATM. Besides, you need to stop intermixing Newtonian mechanics and GR.




Simple. It is the interior Schwarzschild solution that I linked to earlier, using r = 0 for the center of the star. At the point where the event horizon appears at the center with zero time dilation there, so setting dτ / dt = 0, we find that the star will be at a radius of (9/8) r_s.

Wrong answer, you can't "set dτ / dt = 0", this is what you are being asked to derive, please try solving the problem correctly, not by hacking it. (Hint: nothing to do with (9/8)r_s)





So what time dilation do you get at the center of the star using the interior Schwarschild solution you posted for R = (9/8) r_s?

You proposed the ATM, you need to answer the questions. I am simply correcting your errors and helping clear your misconceptions, I am not obligated to answer (according to the rules).


The answer is yes.

Then, it is time for you to hit the books and start finding the correct answers to the questions.

macaw
2011-May-20, 03:01 AM
It is the question I asked you in posts 138, 140, and 144.

So it appears that after all the hype is said and done, you do at least now agree.

Much earlier, in post 114, I pointed out the mistake (sloppiness) in your statement made in post 112. It took you about 40 posts to admit to the error (without ever being able to produce the equations asked of you). The difference is that Landau and Lif****z derived and formulated the statement correctly, you did neither. Big difference.

grav
2011-May-20, 03:25 AM
The time dilation dt/d\tau at the center of the sphere is finite, the time dilation at the EH is infinite.What is the time dilation at the center for R = (9/8) r_s?

macaw
2011-May-20, 03:34 AM
What is the time dilation at the center for R = (9/8) r_s?

You can plug R=(9/8)r_s into the formula I gave you, it is a simple algebraic substitution. I derived for you the general formula, valid for any R, why are you fixated on a particular value for R?
I am not quite sure why you are so fixated on R=(9/8)r_s but I think that you are still stuck on the irrelevant link you cited earlier, where R=(9/8)r_s appears and has nothing to do with time dilation.You should concentrate on solving the simple exercise I gave. Try deriving the formula.

grav
2011-May-20, 03:38 AM
You can plug R=(9/8)r_s into the formula I gave you, it is a simple algebraic substitution.
I am not quite sure why you are so fixated on R=(9/8)r_s but I think that you are still stuck on the irrelevant link you cited earlier, where R=(9/8)r_s appears and has nothing to do with time dilation.You should concentrate on solving the simple exercise I gave. Try deriving the formula.Forget the link. Plug R = (9/8) r_s into the formula you had. What time dilation does it give you for the center of the star?

macaw
2011-May-20, 03:53 AM
Forget the link.

Why? You brought it in in your attempt to answer a question you couldn't answer.



Plug R = (9/8) r_s into the formula you had. What time dilation does it give you for the center of the star?

You can do it all by yourself, no? Did you get 0? Is this why you are so fixated? Did you understand that I gave you the general formula, that applies for any R?
Now that I answered so many of your questions, it is time for you to answer mine. You have a lot of work to do, you have Q1-55 to answer.

grav
2011-May-20, 04:15 AM
You can do it all by yourself, no? Did you get 0?Okay, so at R = (9/8) r_s, we get dτ / dt = 0. A couple of posts back, you stated that the time dilation dt / dτ is infinite at the boundary of the event horizon, which inversed becomes dτ / dt = 0. So would this time dilation at the center describe the boundary of an event horizon that is just beginning to form there according to a distant observer when the star infalls to R = (9/8) r_s?

macaw
2011-May-20, 04:36 AM
Okay, so at R = (9/8) r_s, we get dτ / dt = 0.

There is nothing magical about R = (9/8) r_s, dτ / dt ! = 0 for any[/b[] other R.
You need to learn that physics is not a sequence of hacks and random guesses.



A couple of posts back, you stated that the time dilation dt / dτ is infinite at the boundary of the event horizon, which inversed becomes dτ / dt = 0.

Yet, dτ / dt ! = 0 for [b]any R!=(9/8)r_s.

Mike Holland
2011-May-20, 06:39 AM
Grav, just another thought on our disagreement over whether the EH first forms in the centre of the Almost-BH. If it started in the centre, and grew outwards from there, then at some stage when it has grown half-way, and we would have an EH at (final SR)/2 and a black hole within this radius. As I understand it, that means that at this radius we have infinite time dilation, i.e. time stands still relative to our outside view, and therefore more matter cannot fall in.

This is why I have this view of matter right through the collapsing body approaching its SR and freezing, not just the outside layer near the overall SR.

kzb
2011-May-20, 11:27 AM
A correct answer to your question can be formulated only in the context of a quantum gravity theory, we do not have such a theory at the moment. There seems to be (http://en.wikipedia.org/wiki/Hawking_radiation#Experimental_observation_of_Hawk ing_radiation) experimental proof of Hawking radiation detection. The equations I derived are classical GR gravity.

I was asking the question because it seems I cannot observe the EH itself in finite time, therefore it is a moot point as to whether it "really" exists in my time frame. The Hawking radiation would be an observable that would prove the EH has formed in my time frame.

The link to the experiment does not really answer this. They have formed a region where c=0 by toying with refractive indices as far as I can make out. Perhaps this can occur in finite time, because it seems to me a completely different situation to BH formation.

kzb
2011-May-20, 11:36 AM
Semi-classical gravity calculations say yes.

I think this is what people were after. The appearance of Hawking radiation, within a remote oberver's finite time scale, would be proof that an event horizon has formed in that observer's reference frame. So it does happen at least in theory.

Next problem- no Hawking radiation has been observed out there in the galaxy to date ?

WaxRubiks
2011-May-20, 11:39 AM
I don't think that observing radiation from a black hole would be proof that Hawking radiation is real; there could be some other process producing said radiation.

kzb
2011-May-20, 11:40 AM
Grav, just another thought on our disagreement over whether the EH first forms in the centre of the Almost-BH. If it started in the centre, and grew outwards from there, then at some stage when it has grown half-way, and we would have an EH at (final SR)/2 and a black hole within this radius. As I understand it, that means that at this radius we have infinite time dilation, i.e. time stands still relative to our outside view, and therefore more matter cannot fall in.

This is why I have this view of matter right through the collapsing body approaching its SR and freezing, not just the outside layer near the overall SR.

Another angle on this: when the notional EH has first formed, it will have a very small radius (in fact does it start out with r=0 ?), this means its Hawking radiation output will be very high. In fact, evaporation will be fighting against expansion, and I wonder which one wins?. I'm guessing that the theorists must have taken this into account?

Mike Holland
2011-May-20, 12:07 PM
Another angle on this: when the notional EH has first formed, it will have a very small radius (in fact does it start out with r=0 ?), this means its Hawking radiation output will be very high. In fact, evaporation will be fighting against expansion, and I wonder which one wins?. I'm guessing that the theorists must have taken this into account?

[language] Aren't things complicated enough already!

grav
2011-May-20, 02:06 PM
There is nothing magical about R = (9/8) r_s, dτ / dt ! = 0 for any[/b[] other R.
You need to learn that physics is not a sequence of hacks and random guesses.

Yet, dτ / dt ! = 0 for [b]any R!=(9/8)r_s.dτ / dt ≠ 0 for R > (9/8) r_s with any positive r, meaning no event horizon has formed yet. According to the metric, as stated in the link, with a Schwarzschild radius r_s for the mass, when the surface has fallen to R, the boundary of the event horizon lies at r = R^2 (9 - 8 R / r_s). With R > (9/8) r_s, r is negative, so no event horizon exists. At R = (9/8) r_s, an event horizon just begins to form at the center. At R< (9/8) r_s, the event horizon has expanded outward. The limit lies at R = r_s, where the surface stops infalling according to the distant observer, and we have r = R.

grav
2011-May-20, 02:13 PM
Grav, just another thought on our disagreement over whether the EH first forms in the centre of the Almost-BH. If it started in the centre, and grew outwards from there, then at some stage when it has grown half-way, and we would have an EH at (final SR)/2 and a black hole within this radius. As I understand it, that means that at this radius we have infinite time dilation, i.e. time stands still relative to our outside view, and therefore more matter cannot fall in.

This is why I have this view of matter right through the collapsing body approaching its SR and freezing, not just the outside layer near the overall SR.Right, the matter at the boundary of an event horizon at r = r_s/2 cannot fall in because the time dilation there is zero, but matter further out is not time dilated so much and will continue to infall. As the further out matter infalls, the potential still decreases, causing the event horizon to expand past the "frozen" matter that was at its boundary. It will only stop expanding when there is no more matter to infall, which is when the surface of the star reaches r_s and the event horizon has expanded to r_s also. As the surface approaches r_s, it will fall ever more slowly, and since that is the only matter left beyond the horizon that will decrease the potential, the event horizon expands ever more slowly toward r_s as well.

macaw
2011-May-20, 02:17 PM
dτ / dt ≠ 0 for R > (9/8) r_s with any positive r, meaning no event horizon has formed yet.

Huh? If the EH horizon starts forming at the center, it will pass through r_s BEFORE it reaches (9/8)r_s, so you are wrong.
Besides, there is nothing magical about (9/8)r_s, as I explained, dτ / dt ≠ 0 for any R ≠ (9/8)r_s. There is nothing special about R=(9/8)r_s.


According to the metric, as stated in the link, with a Schwarzschild radius r_s for the mass, when the surface has fallen to R, the boundary of the event horizon lies at r = R^2 (9 - 8 R / r_s). With R > (9/8) r_s, r is negative, so no event horizon exists. At R = (9/8) r_s, an event horizon just begins to form at the center. At R< (9/8) r_s, the event horizon has expanded outward. The limit lies at R = r_s, where the surface stops infalling according to the distant observer, and we have r = R.

If that were true, then for all r_s<R<(9/8)r_s dτ / dt ≠ 0, proving your initial statement that dτ / dt = 0 at the center of the BH , FALSE. No matter how you twist and turn, your statements are false.

macaw
2011-May-20, 02:23 PM
Right, the matter at the boundary of r = r_s/2 cannot fall in because the time dilation there is zero,

Q56: This is a general question, what does time dilation, a measuring artifact, have to do with the collapse of a body into a BH (Hint: nothing).





but matter further out is not time dilated so much and will continue to infall.

Q57: "Matter is time dilated"? Where did you learn this? Did you make it up?



As the further out matter infalls, the potential still rises,

Q58: I thought that you admitted to this mistake, the potential does NOT "rise", it has a MINIMUM not a maximum as you keep claiming erroneously.




causing the event horizon to expand past the "frozen" matter that was at its boundary. It will only stop expanding when there is no more matter to infall, which is when the surface of the star reaches r_s and the event horizon has expanded to r_s also.

Q59: But in your previous post you were claiming the opposite, that the EH propagates from r>(9/8)r_s . Which one is it, grav?

macaw
2011-May-20, 02:25 PM
Next problem- no Hawking radiation has been observed out there in the galaxy to date ?

I gave you two references to the contrary, did you click the links?

grav
2011-May-20, 02:30 PM
Huh? If the EH horizon starts forming at the center, it will pass through r_s BEFORE it reaches (9/8)r_s, so you are wrong.The event horizon starts at r = 0 and expands outward toward r = r_s. The surface of the star infalls toward R = r_s, passing R = (9/8) r_s.


Besides, there is nothing magical about (9/8)r_s, as I explained, dτ / dt ≠ 0 for any R ≠ (9/8)r_s. There is nothing special about R=(9/8)r_s.Not for r = 0, no. r = 0 is the center, and the boundary is only present there at R = (9/8) r_s, no other.


If that were true, then for all r_s<R<(9/8)r_s dτ / dt ≠ 0, proving your initial statement that dτ / dt = 0 at the center of the BH , FALSE. No matter how you twist and turn, your statements are false.The event horizon is present for all R < (9/8) r_s, gaining dτ / dt = 0 at the boundary of the event horizon at r.

macaw
2011-May-20, 02:37 PM
The event horizon is present for all R < (9/8) r_s, gaining dτ / dt = 0 at the boundary of the event horizon at r = R^2 (9 - 8 R / r_s).

You don't get it, do you? d\tau/dt>0 for all R>(9/8)r_s and for all R<(9/8)r_s. That is, for an INFINITE number of values for R.


The event horizon is present for all R < (9/8) r_s, gaining dτ / dt = 0 at the boundary of the event horizon at r.

Meaning NONE. Therefore, your claim that started this all is false.


Not for r = 0, no. r = 0 is the center, and the boundary is only present there at R = (9/8) r_s, no other.

This is also false. The formula I gave you, on which you are basing your ATM claims, has been derived for , and ONLY for, get this, r=0. If you knew how to derive it, you would have known this.

grav
2011-May-20, 02:53 PM
You don't get it, do you? d\tau/dt>0 for all R>(9/8)r_s and for all R<(9/8)r_s. That is, for an INFINITE number of values for R.It is dτ / dt > 0 for all R > (9/8) r_s for all positive r. With R = (9/8) r_s, r = 0 for dτ / dt = 0, so the event horizon is just forming at the center. For R < (9/8) r_s, r > 0 for dτ / dt = 0, so the event horizon has expanded outward.


Meaning NONE. Therefore, your claim that started this all is false.The full form of the equation you posted for radial infall is

dτ / dt = (1/2) [3 sqrt(1 - r_s / R) - sqrt(1 - r_s r^2 / R^3)]

The boundary of the event horizon lies at dτ / dt = 0, so we have

0 = (1/2) [3 sqrt(1 - r_s / R) - sqrt(1 - r_s r^2 / R^3)]

3 sqrt(1 - r_s / R) = sqrt(1 - r_s r^2 / R^3)

9 (1 - r_s / R) = 1 - r_s r^2 / R^3

8 - 9 r_s / R = - r_s r^2 / R^3

r^2 = (R^3 / r_s) (9 r_s / R - 8)

r^2 = R^2 (9 - 8 R / r_s)

r = R sqrt(9 - 8 R / r_s)

It appears that a mathematical error was made in the link where r^2 was dropped to just r. Anyway, according to the metric, the boundary of the event horizon lies at r = R sqrt(9 - 8 R / r_s).

macaw
2011-May-20, 02:57 PM
the boundary of the event horizon lies at r = R^2 (9 - 8 R / r_s). .

The r = R^2 (9 - 8 R / r_s) is incorrect . You have r in the LHS and R^2 in the RHS, this cannot be. This is what happens when you copy stuff that you don't understand.

Q60: What is the correct formula and how is it derived, grav? Can you derive the correct formula?

macaw
2011-May-20, 02:58 PM
It is dτ / dt > 0 for all R > (9/8) r_s for all positive r.

Err, no. It is dτ / dt > 0 for all R > (9/8) r_s and for all R< (9/8) r_s. Either way, the point was that dτ / dt > 0 for an INFINITE number of cases.



With R = (9/8) r_s, r = 0 for dτ / dt = 0, so the event horizon is just forming at the center. For R < (9/8) r_s, r > 0 for dτ / dt = 0, so the event horizon has expanded outward.

The full form of the equation you posted for radial infall is

dτ / dt = (1/2) [3 sqrt(1 - r_s / R) - sqrt(1 - r_s r^2 / R^3)]


Q61: How did I derive this?



The boundary of the event horizon lies at dτ / dt = 0, so we have

0 = (1/2) [3 sqrt(1 - r_s / R) - sqrt(1 - r_s r^2 / R^3)]

Nope, this is not the correct line of thinking. You do not KNOW that dτ / dt = 0 at the EH. This is a CONSEQUENCE of the math, not a starting point.


3 sqrt(1 - r_s / R) = sqrt(1 - r_s r^2 / R^3)

9 (1 - r_s / R) = 1 - r_s r^2 / R^3

8 - 9 r_s / R = - r_s r^2 / R^3

r^2 = (R^3 / r_s) (9 r_s / R - 8)

r^2 = R^2 (9 - 8 R / r_s)

r = R sqrt(9 - 8 R / r_s)

It appears that a mathematical error was made in the link where r^2 was dropped to just r. Anyway, according to the metric, the boundary of the event horizon lies at r = R sqrt(9 - 8 R / r_s).

Q62: How could that be, the boundary of the EH is, BY DEFINITION, r=r_s. Does not depend on R.

This is not how the formula came to be. Sorry, you corrected it but it is not the correct derivation.

You are still not admitting to your other errors of judgement wrt to time dilation.

grav
2011-May-20, 03:34 PM
Err, no. It is dτ / dt > 0 for all R > (9/8) r_s and for all R< (9/8) r_s. Either way, the point was that dτ / dt > 0 for an INFINITE number of cases.So what do you get at r=0 for say, R = 101/100 r_s?


How did I derive this?You didn't.


Nope, this is not the correct line of thinking. You do not KNOW that dτ / dt = 0 at the EH. This is a CONSEQUENCE of the math, not a starting point.

How could that be, the boundary of the EH is, BY DEFINITION, r=r_s. Does not depend on R.By definition, the boundary of the event horizon is where dτ / dt = 0, the point of no return, the boundary beyond which events cannot affect outside observers.

macaw
2011-May-20, 03:45 PM
So what do you get at r=0 for say, R = 101/100 r_s?

Q63: Why are you continuing along this line? I have already shown your conclusions to be in error.




You didn't.

I must have since I arrived to the formula. So, how was it done, grav? can you reconstruct the steps?




By definition, the boundary of the event horizon is where dτ / dt = 0, the point of no return, the boundary beyond which events cannot affect outside observers.

We've been over this before. You still don't get it, the EH is the sphere of radius r=r_s.

Now, since you have used this thread to advance your ATM concepts about BH formation, event horizons and gravitational redshift, I would recommend that you invested in "The Classical Theory of Fields", fourth edition, by Landau and that you read chapter 102, "Gravitational collapse of a spherical body".

It will correct all your misconceptions:

1. it will show that d\tau/dt=1-r_s/r, as I derived it early in this thread, so your gravitational time dilation formula is wrong
2. it will show that the internal form of the Schwarzschild metric plays no role in the way they derive the formation of the EH, they use only the external form
3. it will show that, contrary to your claims, the EH is at r_s and that there is no particular meaning attached to your (9/8)r_s.

And so on. If the book doesn't correct your misconceptions and errors of calculation, nothing will. I recommend that you invest some money and buy the book.

grav
2011-May-20, 03:50 PM
Another angle on this: when the notional EH has first formed, it will have a very small radius (in fact does it start out with r=0 ?), this means its Hawking radiation output will be very high. In fact, evaporation will be fighting against expansion, and I wonder which one wins?. I'm guessing that the theorists must have taken this into account?That is an interesting point. I'm not sure how that would work out, but here is a paper (http://arxiv.org/PS_cache/arxiv/pdf/1011/1011.2219v2.pdf) about pre-Hawking radiation from a collapsing shell.

grav
2011-May-20, 04:12 PM
Why are you continuing along this line? I have already shown your conclusions to be in error.So does that mean you are getting dτ / dt > 0 at r = 0 for R = 101 / 100 r_s? I didn't think so. That is because the center now lies within the event horizon.


I must have since I arrived to the formula. So, how was it done, grav? can you reconstruct the steps?Well, I would really like to see that. It's more than just a simple matter of arithmetic. One would really have to sit down and think it through for a fluid model where different points infall at different rates according to a distant observer and determine the time dilations at each of those points during infall while the event horizon is also expanding.


We've been over this before. You still don't get it, the EH is the sphere of radius r=r_s.

Now, since you have used this thread to advance your ATM concepts about BH formation, event horizons and gravitational redshift, I would recommend that you invested in "The Classical Theory of Fields", fourth edition, by Landau and that you read chapter 102, "Gravitational collapse of a spherical body".

It will correct all your misconceptions:

1. it will show that d\tau/dt=1-r_s/r, as I derived it early in this thread, so your gravitational time dilation formula is wrong
2. it will show that the internal form of the Schwarzschild metric plays no role in the way they derive the formation of the EH, they use only the external form
3. it will show that, contrary to your claims, the EH is at r_s and that there is no particular meaning attached to your (9/8)r_s.So we're back to this again. The gravitational time dilation of a freefaller falling from rest at infinity toward an idealized fully formed black hole according to a distant observer is dτ / dt = 1 - r_s / r. I never said otherwise. As for the event horizon, it is the boundary beyond which events cannot affect outside observers, as Wiki defines it, hence the name event horizon, which occurs at dτ / dt = 0 since beyond that, the time dilation is complex so cannot affect us. If you insist on making it the same as the Schwarzschild radius, which would only otherwise be the case if the black hole is fully formed, then we must call it something else. so what would you call the boundary of dτ / dt = 0 that starts at the center and then grows outward after a star has infallen enough?

macaw
2011-May-20, 04:14 PM
dτ / dt ≠ 0 for R > (9/8) r_s with any positive r, meaning no event horizon has formed yet. According to the metric, as stated in the link, with a Schwarzschild radius r_s for the mass, when the surface has fallen to R, the boundary of the event horizon lies at r = R^2 (9 - 8 R / r_s). With R > (9/8) r_s, r is negative, so no event horizon exists. At R = (9/8) r_s, an event horizon just begins to form at the center. At R< (9/8) r_s, the event horizon has expanded outward. The limit lies at R = r_s, where the surface stops infalling according to the distant observer, and we have r = R.

All the above is, of course, false.

1. The internal Schwarzschild metric plays no role in the derivation of the location of the EH, only the external form does (see Landau, "The theory of Fields")

2. The value r=R\sqrt{9-8R/r_s} is definitely not the "boundary of the EH". It is simply the value where the coefficient for dt is null in the internal Schwarzschild metric. As the internal Schwarzschild metric plays no role (see point 1), the whole thing you introduced is a red herring. This is what happens when you copy and paste from other websites. BTW, the participants in the link you cited were not dealing with BH formation, so you cut and paste the wrong results.

Once again, I recommend that you buy the book and that you start reading it.

grav
2011-May-20, 04:19 PM
All the above is, of course, false.

1. The internal Schwarzschild metric plays no role in the derivation of the location of the EH, only the external form does (see Landau, "The theory of Fields")

2. The value r=R\sqrt{9-8R/r_s} is definitely not the "boundary of the EH". It is simply the value where the coefficient for dt is null in the internal Schwarzschild metric. As the internal Schwarzschild metric plays no role (see point 1), the whole thing you introduced is a red herring. This is what happens when you copy and paste from other websites. BTW, the participants in the link you cited were not dealing with BH formation, so you cut and paste the wrong results.

Once again, I recommend that you buy the book and that you start reading it.The internal metric is for the inside of an infalling star, but according to an external distant observer. It only reduces to the classic form after the black hole has fully formed, whereas the event horizon and surface of the star both lie idealistically close to the Schwarzschild radius.

macaw
2011-May-20, 04:21 PM
The internal metric is for the inside of an infalling star, but according to an external distant observer. It only reduces to the classic form after the black hole has fully formed, whereas the event horizon and surface of the star both lie idealistically close to the Schwarschild radius.

There is NO internal metric in the derivation, the value r=R\sqrt{9-8R/r_s} is definitely not the "boundary of the EH", do you get this?

macaw
2011-May-20, 04:22 PM
So we're back to this again. The gravitational time dilation of a freefaller falling from rest at infinity toward an idealized fully formed black hole according to a distant observer is dτ / dt = 1 - r_s / r. I never said otherwise.

Yes, we are back to your incorrect claim from post 19.

grav
2011-May-20, 04:28 PM
There is NO internal metric in the derivation, the value r=R\sqrt{9-8R/r_s} is definitely not the "boundary of the EH", do you get this?So according to you, then, where does the boundary for the "point of no return" start as the star infalls, at the center and expand outward, does it just pop at r_s at some point, or did it exist at r_s all along?

macaw
2011-May-20, 04:31 PM
So according to you, then, where does the boundary for the "point of no return" start as the star infalls, at the center and expand outward, does it just pop at r_s at some point, or did it exist at r_s all along?


you need to buy the book and read chapter 102, it will clear all your misconceptions. I will tell you one thing, it has NOTHING to do with any r=R\sqrt{9-8R/r_s}
I have answered a lot of your misconceptions in an attempt to educate you, you formulated the ATM, you need to answer the questions.

grav
2011-May-20, 04:31 PM
Yes, we are back to your incorrect claim from post 19.You mean post 27. Post 19 is not the gravitational time dilation of a freefaller, it is the redshift of a distant observer as observed by a freefaller, so you are mistaken.

macaw
2011-May-20, 04:32 PM
You mean post 27. Post 19 is not the gravitational time dilation of a freefaller.

same error, the D= 1 / (1 + sqrt(R/r)) incorrect conclusion

WayneFrancis
2011-May-20, 04:34 PM
Grav, just another thought on our disagreement over whether the EH first forms in the centre of the Almost-BH. If it started in the centre, and grew outwards from there, then at some stage when it has grown half-way, and we would have an EH at (final SR)/2 and a black hole within this radius. As I understand it, that means that at this radius we have infinite time dilation, i.e. time stands still relative to our outside view, and therefore more matter cannot fall in.

This is why I have this view of matter right through the collapsing body approaching its SR and freezing, not just the outside layer near the overall SR.

And this view of yours is faulty because this means no object would ever get even close to becoming a black hole. Lets take a 10 solar mass star...lets shed 1/2 of its mass during the explosion of the star leaving 5 solar masses to collapse on itself. The 5 solar mass core has to be larger then about 15km in radius. Any smaller and it would be a black hole. Now in real terms the neutron star like material would be much larger then 15km but your saying that at sr/2 or 7.5km the collapsing would stop because nothing could pass an EH. This means you would have 7.5km crust around that "frozen star". Now calculate the time dilation for the surface that is at probably more like 22-25km from the centre and 15km from your "almost event horizon". You'll see that the time dilation isn't much at all. This means this "frozen star" would be very capable or radiating in at least radio wave lengths. These object would be very noticeable to astronomers but guess what...we don't see the black hole candidates radiating at all. All we see is a gravitational lens or a companion star orbiting a spot where there appears to be nothing but the mass required to orbit such a spot would be have much more mass then our sun but NOTHING is being radiated.

This is one reason why your idea that no EH can ever form breaks down.

grav
2011-May-20, 04:41 PM
you need to buy the book and read chapter 102, it will clear all your misconceptions. I will tell you one thing, it has NOTHING to do with any r=R\sqrt{9-8R/r_s}
I have answered a lot of your misconceptions in an attempt to educate you, you formulated the ATM, you need to answer the questions.I'm asking you, since you are still resisting every explanation. According to you, where does the boundary for the "point of no return" start as the star infalls, at the center and expand outward, does it just pop up at r_s at some point, or did it exist at r_s all along?

macaw
2011-May-20, 04:46 PM
I'm asking you, since you are still resisting every explanation.

I am not resisting explanations, at each post you introduce new misconceptions, this slows down the process because I need to point out the new errors in each of your new posts.



According to you, where does the boundary for the "point of no return" start as the star infalls, at the center and expand outward, does it just pop up at r_s at some point, or did it exist at r_s all along?

None of the above, read Landau, chapter 102. It approaches r_s from r>r_s

Now, that I have answered all your misconceptions, please answer the challenges against your ATM.

grav
2011-May-20, 04:53 PM
None of the above, read Landau, chapter 102. It approaches r_s from r>r_s

Now, that I have answered all your misconceptions, please answer the challenges against your ATM.So you're saying the boundary lies beyond r_s and then decreases in size with infall of the star? How does that work? At what point does the boundary come into existence and at what initial radius?

macaw
2011-May-20, 04:56 PM
So you're saying the boundary lies beyond r_s and then decreases in size?

The "boundary" IS r=r_s. It is called convergence, it doesn't work in the naive way you are formulating your question. Read Landau.



At what point does the boundary come into existence and and what initial radius?

This kind of question shows that you do not understand the concept, there is no meaning to your question. No more free classes, buy the book.
Now, it is time for you to stop asking questions and to start answering them. It is your ATM, so , please answer Q1-62. thank you

grav
2011-May-20, 05:10 PM
It is called convergence, it doesn't work in the naive way you are formulating your question. Read Landau.

This kind of question shows that you do not understand the concept, there is no meaning to your question. No more free classes, buy the book.
Now, it is time for you to stop asking questions and to start answering them. It is your ATM, so , please answer Q1-62. thank youIt's not my ATM, and everything I have presented is mainstream, so stop requesting that. I have explained every concept in more than enough detail, but you still refuse to answer the simplest questions, like this one about the formation of a black hole. That would be essential in a thread about the formation of black holes, don't you think? You claim my posts are wrong, yet you still haven't shown what you believe to be the correct explanation for the formation. It is a simple request, and would help clear up any misconceptions. So according to you, as a star infalls, at what point does the boundary for the "point of no return" come into existence and at what initial radius?

macaw
2011-May-20, 05:12 PM
It's not my ATM, and everything I have presented is mainstream,

No, it isn't since everything you posted is contradicted by mainstream physics. So, please answer the questions.


So according to you, as a star infalls, at what point does the boundary for the "point of no return" come into existence and at what initial radius?

What in the "boundary" IS r=r_s didn't you understand? There is no "boundary...come into existence"? These are fringe concepts that you are generating.

grav
2011-May-20, 05:15 PM
The "boundary" IS r=r_s. It is called convergence, it doesn't work in the naive way you are formulating your question. Read Landau.

This kind of question shows that you do not understand the concept, there is no meaning to your question. No more free classes, buy the book.
Now, it is time for you to stop asking questions and to start answering them. It is your ATM, so , please answer Q1-62. thank youSo now you are saying the boundary for the "point of no return" lies at r = r_s? What happened to r > r_s? So did it always exist at r_s or did it pop into existence at some point as the star infalls?

grav
2011-May-20, 05:17 PM
No, it isn't since everything you posted is contradicted by mainstream physics. So, please answer the questions.



What in the "boundary" IS r=r_s didn't you understand?What I am posting is mainstream. I'm still trying to decipher what you are posting. So far it sounds pretty ATMish to me.

macaw
2011-May-20, 05:19 PM
What I am posting is mainstream.

Well, you won't find the basic mistakes from posts 19, 27, etc. in any book.

grav
2011-May-20, 05:23 PM
Well, you won't find the basic mistakes from posts 19, 27, etc. in any book.What does that have to do with the formation of a black hole, besides changing the subject? There are no mistakes in those posts, but let's stay focused on the topic at hand. Please describe the formation of a black hole in your own words, describing where and when the boundary for the "point of no return" comes into existence as a star infalls.

macaw
2011-May-20, 05:36 PM
What does that have to do with the formation of a black hole, besides changing the subject?.

It points out that none of your formulas are mainstream.

caveman1917
2011-May-20, 08:01 PM
A quick note to everyone, the schwarzschild geometry does not describe collapsing stars. It describes an existing black hole that's fully formed (no more mass will fall into it, ever). Using it to derive conclusions about black hole formation is wrong from the start.

grav, your explanations are ATM. The geometry you linked to is that for a static spherically symmetric perfect fluid. When you use R<=(9/8)M you are trying to describe a perfect fluid sphere (something akin to a toy model star) that's so small (yet static) that it's almost the size of its schwarzschild radius. It is not surprising that pressure singularities show up in that case, they have nothing to do with the event horizon, you're just trying to describe something beyond any reason. This is not how black hole formation is done, it is nothing more than a classic case of "garbage in, garbage out".

Mike Holland, you seemed to have skipped over my direct questions in post 142 and post 143, yet replied to later posts. Could you please answer my questions?

grav
2011-May-20, 08:56 PM
grav, your explanations are ATM. The geometry you linked to is that for a static spherically symmetric perfect fluid. When you use R<=(9/8)M you are trying to describe a perfect fluid sphere (something akin to a toy model star) that's so small (yet static) that it's almost the size of its schwarzschild radius. It is not surprising that pressure singularities show up in that case, they have nothing to do with the event horizon, you're just trying to describe something beyond any reason. This is not how black hole formation is done, it is nothing more than a classic case of "garbage in, garbage out".Yes, you are right, that is for a static sphere. That still doesn't negate the point that with a smaller radius, a lesser potential is produced at the center, and with time dilation being a function of potential and the potential being least at the center, as a star infalls, the event horizon will form there initially and expand outward. That is what I am saying, pure and simple, and it is mainstream. The precise metric for infall doesn't matter unless it shows something different than that.

Now, if you believe what I just stated, which is what I have been saying all along, to be ATM, then I will ask to do the same thing I have been asking macaw. Please, in your own words, describe the formation of a black hole, describing (in general, no math needed) where and when the initial boundary for the "point of no return" comes into existence as a star infalls.

grav
2011-May-20, 09:04 PM
It points out that none of your formulas are mainstream.Please, in your own words, describe the formation of a black hole, describing (in general, no math needed) where and when the initial boundary for the "point of no return" comes into existence as a star infalls.

macaw
2011-May-20, 10:06 PM
Please, in your own words, describe the formation of a black hole, describing (in general, no math needed)

This is BAUT, remember? I gave you a reference to a mainstream book that describes in excruciating detail the mathematical model of the formation of black holes.It is up to you to purchase and to study the book, I am through with giving you free lessons.



where and when the initial boundary for the "point of no return" comes into existence as a star infalls.

This question that you keep repeating doesn't even make sense. There is no answer for such an ill formed question.
Both I and caveman1917 consider your postings as pure ATM. As such, you need to start defending your ATM, so, please, start answering the questions.

grav
2011-May-20, 10:19 PM
This is BAUT, remember? I gave you a reference to a mainstream book that describes in excruciating detail the mathematical model of the formation of black holes.It is up to you to purchase and to study the book, I am through with giving you free lessons.

This question that you keep repeating doesn't even make sense. There is no answer for such an ill formed question.
Both I and caveman1917 consider your postings as pure ATM. As such, you need to start defending your ATM, so, please, start answering the questions.So are you saying you have no idea how a black hole and its event horizon form as a star infalls, that you can't even give a general description?

macaw
2011-May-20, 10:25 PM
So are you saying you have no idea how a black hole and its event horizon form as a star infalls, that you can't even give a general description?

What I am saying is that we have mainstream mathematical models about the formation of black holes.
What I am saying is that I gave you a reference to a book that explains these mathematical models in excruciating detail.
What I am saying is that no one has ever been able to observe the formation of a BH and never will (Q64: do you know why?)
What I am saying is that I am through giving you free lessons, that you will need to buy Landau's book and study it.
What I am saying is that, contrary to what permeates from your nonsense question, the EH is not a physical entity, it is simply a component of a mathematical mode. As such, an EH does not "form". I think I've told you this repeatedly.
What I am saying is that you hijacked this thread with your own ATM, trying to disguise it as "mainstream" and that your errors have been pointed out to you as you have been posting them
What I am saying is that, since your developed your own ATM in this thread, you are bound to defend it by answering the questions leveled by me and caveman1917 against your ATM. So, once again, stop asking (nonsensical) questions and start answering the questions we asked you.

grav
2011-May-20, 10:59 PM
What I am saying is that we have mainstream mathematical models about the formation of black holes.
What I am saying is that I gave you a reference to a book that explains these mathematical models in excruciating detail.
What I am saying is that no one has ever been able to observe the formation of a BH and never will (Q64: do you know why?)
What I am saying is that I am through giving you free lessons, that you will need to buy Landau's book and study it.
What I am saying is that, contrary to what permeates from your nonsense question, the EH is not a physical entity, it is simply a component of a mathematical mode. As such, an EH does not "form". I think I've told you this repeatedly.
What I am saying is that you hijacked this thread with your own ATM, trying to disguise it as "mainstream" and that your errors have been pointed out to you as you have been posting them
What I am saying is that, since your developed your own ATM in this thread, you are bound to defend it by answering the questions leveled by me and caveman1917 against your ATM. So, once again, stop asking (nonsensical) questions and start answering the questions we asked you.So you have no clue. That's what I thought. Then how can you claim my posts are ATM if you don't know?

Here (http://www.bautforum.com/showthread.php/114713-Would-the-EH-form-first-in-the-center-of-a-star) is a recent thread in Q&A asking about the formation of an event horizon. Notice post 5 where caveman states that the event horizon forms at the center and expands outward. In fact, it is that post that originally got me thinking about it more. I had forgotten it was caveman, ironically, that had posted it, but it is the same thing I have been repeating word for word in this thread, so I'm not sure why caveman would have a problem with that here, except that he also adds that it is not due to the potential being minimum, although I'm not sure why he states that.

PetersCreek
2011-May-20, 11:19 PM
Thread closed to preclude further combativeness. Moderator discussion to follow concerning behavior on both sides of this discussion.