View Full Version : Would the EH form first in the center of a star?

tommac

2011-Apr-17, 01:56 PM

From my understanding an EH would first form in the center of a star and actually grow outward to include the rest of star.

As a star collapses at some point the EH is created. It would be created when at some point of the star the gravitational potential energy is great enough so that the EH would be created.

This would first happen at the core as it is the place that has the lowest potential.

Can someone please correct this if it is wrong.

This was a statement that I made on a thread ... which was based on another thread where grant described this in detail.

This is the thread that originally discussed it:

http://www.bautforum.com/showthread.php/100620-How-can-a-Black-Hole-form?highlight=can+an+event+horizon+actually

but now there seems to be debate about it again.

Can someone clarify?

tommac

2011-Apr-17, 02:06 PM

Gravitational potential energy is a scaler ( not a vector ) right?

korjik

2011-Apr-17, 10:09 PM

Energy is a scalar quantity.

Where this is wrong is that you arent even close to using the correct science to figure out how things work in the middle of a type 2 supernova.

A core collapse involves two phase transitions (from electron degenerate to neutron degenerate to collapse), an emission of a rather immense amount of energy, and an extremely turbulent extremely hot environment. The initial event horizon is probably a very chaotic shape based on where matter locally has sufficient density to create one, and then there is an emission of gravitational energy to round it out, all of this taking a very short time, while in the middle of an extremely large exploding star.

The gravitational potential at any point in the core is only one of many factors that goes into the formation of a black hole. Figuring out to any real accuracy what is happening is not really possible right now.

WayneFrancis

2011-Apr-18, 01:14 AM

From my understanding an EH would first form in the center of a star and actually grow outward to include the rest of star.

As a star collapses at some point the EH is created. It would be created when at some point of the star the gravitational potential energy is great enough so that the EH would be created.

This would first happen at the core as it is the place that has the lowest potential.

Can someone please correct this if it is wrong.

This was a statement that I made on a thread ... which was based on another thread where grant described this in detail.

This is the thread that originally discussed it:

http://www.bautforum.com/showthread.php/100620-How-can-a-Black-Hole-form?highlight=can+an+event+horizon+actually

but now there seems to be debate about it again.

Can someone clarify?

As you get smaller and smaller the critical density needs to be higher and higher. So the EH shouldn't start at the very centre as a point and grow outwards. More likely is the fact that as the star is collapsing there will be some point where the EH will form and this will be some distance from the centre of the star. Ie the critical density would occur at some r > 0.

caveman1917

2011-Apr-18, 03:31 AM

As you get smaller and smaller the critical density needs to be higher and higher. So the EH shouldn't start at the very centre as a point and grow outwards. More likely is the fact that as the star is collapsing there will be some point where the EH will form and this will be some distance from the centre of the star. Ie the critical density would occur at some r > 0.

It does form at the center and expands outwards. At least under the assumptions of null pressure, a starting homogeneous mass density and an isotropic collapse.

Solutions that go beyond those assumptions are not generally agreed upon, and are under active research (you'd get naked singularities and other issues).

ETA: this is however not due to the gravitational potential being minimal.

WayneFrancis

2011-Apr-18, 04:18 AM

It does form at the center and expands outwards. At least under the assumptions of null pressure, a starting homogeneous mass density and an isotropic collapse.

Solutions that go beyond those assumptions are not generally agreed upon, and are under active research (you'd get naked singularities and other issues).

ETA: this is however not due to the gravitational potential being minimal.

The pressure in a neutron star....is that homogeneous? Its as tightly packed as possible .....drop more mass onto the neutron star and the mass density of the core doesn't increase until the neutron degeneracy pressure is over come...just working this out in my head...Isn't the gravitational potential the greatest at or near the surface of the neutron star? I REALLY need to crunch some numbers on this because I don't think the answer is very intuitive. Do you have any links or references that talks about the collapse of an object uniform mass density?

My big issue is that the EH can't form at r = 0 because before the density = ∞ at r = 0 there would be a m/(4/3πrsh3) > ?...

Am I getting the maths wrong?

WayneFrancis

2011-Apr-18, 04:19 AM

I am getting the formula wrong...stupid head ache isn't letting me do a simple formula for mass density of a black hole based on rsh

tommac

2011-Apr-18, 02:30 PM

Probably correct ... but lets just take a simple collapse for a second where there is nothing stopping all of the mass of a star from fully collapsing ( lets forget about all of the things that would stop it for a second ).

During the collapse there would be a point where the gravitational potential energy would be sufficient for an EH to form, since the gravitationa potential energy is greatest at the center, the center would be the first place the EH would appear. It would then very quickly grow otwards as additional mass collapsed towards it.

Grant and I discussed this in depth here:

http://www.bautforum.com/showthread.php/100620-How-can-a-Black-Hole-form?highlight=can+an+event+horizon+actually

Energy is a scalar quantity.

Where this is wrong is that you arent even close to using the correct science to figure out how things work in the middle of a type 2 supernova.

A core collapse involves two phase transitions (from electron degenerate to neutron degenerate to collapse), an emission of a rather immense amount of energy, and an extremely turbulent extremely hot environment. The initial event horizon is probably a very chaotic shape based on where matter locally has sufficient density to create one, and then there is an emission of gravitational energy to round it out, all of this taking a very short time, while in the middle of an extremely large exploding star.

The gravitational potential at any point in the core is only one of many factors that goes into the formation of a black hole. Figuring out to any real accuracy what is happening is not really possible right now.

tommac

2011-Apr-18, 02:36 PM

One point to remember is that adding additional mass anywhere will increase the gravitational potential energy at the core. This is partly why the EH extends and grows outwards.

On the other thread:

http://www.bautforum.com/showthread.php/100620-How-can-a-Black-Hole-form?highlight=can+an+event+horizon+actually

my question was how can a star collapse when time stops once the EH has formed. After a long q & a lesson from grant, thanks grant, he pointed out that although nothing more can fall past the EH ... as matter gets close to the EH there is an increase in gravitational potential energy and the EH will expand.

This is similar to when two black holes get close to each other. at the midpoint between the two bhs there is signifigant gravitational potential energy and each holese EH expands in that area even though no additional matter fell into either of the holes.

Jeff Root

2011-Apr-18, 07:18 PM

I'm replying to your OP without having read any posts after it.

I think I read the thread you link to, but that was a year ago.

From my understanding an EH would first form in the center of a

star and actually grow outward to include the rest of star.

As a star collapses at some point the EH is created. It would be

created when at some point of the star the gravitational potential

energy is great enough so that the EH would be created.

This would first happen at the core as it is the place that has the

lowest potential.

The core of a star is a very big thing. The event horizon will be

only a few kilometers in radius after the core has fully collapsed.

The event horizon definitely forms before the star is entirely within

the final Schwarzschild radius, and it definitely grows as matter

from outer parts of the core fall in. So it must start out smaller

than the final size, and grow to that final size. Its size at formation

is what you want to know.

It can't be *very* small, because that would require a density much

greater than the actual density of the collapsing core at that moment.

Assuming that the density of the collapsing core is pretty uniform,

formation of a tiny black hole (say, a centimeter in radius) would

require the density to be such that the entire mass of the core was

*already* compacted into a volume smaller than the Schwarzschild

radius of the core. So it must start out larger than that. With the

assumption that the density of the core is uniform as it collapses,

it should be easy to calculate the density and mass required for an

event horizon to form.

I'll now read the thread and see if anyone has done so yet.

-- Jeff, in Minneapolis

Jeff Root

2011-Apr-18, 07:22 PM

I just said what Wayne said in his first post.

-- Jeff, in Minneapolis

pzkpfw

2011-Apr-18, 07:30 PM

... since the gravitationa potential energy is greatest at the center, the center would be the first place the EH would appear.

It isn't clear to me what you mean by "gravitational potential energy" and what you mean by "greatest at the center". Could you explain your definition?

tommac

2011-Apr-18, 07:33 PM

It can't be *very* small, because that would require a density much

greater than the actual density of the collapsing core at that moment.

Assuming that the density of the collapsing core is pretty uniform,

formation of a tiny black hole (say, a centimeter in radius) would

require the density to be such that the entire mass of the core was

*already* compacted into a volume smaller than the Schwarzschild

radius of the core. So it must start out larger than that. With the

assumption that the density of the core is uniform as it collapses,

it should be easy to calculate the density and mass required for an

event horizon to form.

Can you show an example of this please? I am missing this point. Lets start with an example of 1 cm. vs 1 km.

astromark

2011-Apr-18, 07:45 PM

Talk of the event horizon as if it were a place or a thing is going to get you into trouble.

To say created is wrong. You can create a thing. A event horizon is nothing.. It is a EFFECT. just as gravity is.

From the standpoint of a outside observer a event horizon is where things vanish from view...

the light image can no longer scape the gravity well.

Korjik said in post 3 ... and I find it says it all.

The formation, not creation.. of a event horizon is a effect of the collapse of a large mass star. Its hot and chaotic.

To say the event horizon grows to encapsulate the BH is as near to correct as we can be...

Jeff Root

2011-Apr-18, 07:55 PM

Okay, Tom, I'll work on it.

-- Jeff, in Minneapolis

tommac

2011-Apr-18, 08:10 PM

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

it should be the greatest at the center of a star ( at the core )

It isn't clear to me what you mean by "gravitational potential energy" and what you mean by "greatest at the center". Could you explain your definition?

tommac

2011-Apr-18, 08:12 PM

The way I used it in the other thread was a point where Time stopped for the external observer. The reason that was important to that thread was that if time stopped then how could the rest of the star fall in? It was answered nicely by grant in that other thread.

But in any case that is how I am defining the EH.

Talk of the event horizon as if it were a place or a thing is going to get you into trouble.

To say created is wrong. You can create a thing. A event horizon is nothing.. It is a EFFECT. just as gravity is.

From the standpoint of a outside observer a event horizon is where things vanish from view...

the light image can no longer scape the gravity well.

Korjik said in post 3 ... and I find it says it all.

The formation, not creation.. of a event horizon is a effect of the collapse of a large mass star. Its hot and chaotic.

To say the event horizon grows to encapsulate the BH is as near to correct as we can be...

pzkpfw

2011-Apr-18, 08:57 PM

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

it should be the greatest at the center of a star ( at the core )

(Why do you always type above the quote? It's a little odd. Anyway...)

That page links to gravitational energy, which yes, is related to gravitational potential, but isn't the same. You keep writing gravitational potential energy. That's different. ( http://en.wikipedia.org/wiki/Gravitational_potential_energy#Gravitational_poten tial_energy )

However, I'm not looking for a wiki reference, I want to know what you think it means. Because it seems you are using it differently than others, and until you've explained what you mean by it, it's not clear why you think it's greatest at the centre of the star.

tommac

2011-Apr-18, 11:23 PM

It is the field of energy at a point that is produced from gravity. At any point there is a gravitational effect on the stress energy tensor.

(Why do you always type above the quote? It's a little odd. Anyway...)

That page links to gravitational energy, which yes, is related to gravitational potential, but isn't the same. You keep writing gravitational potential energy. That's different. ( http://en.wikipedia.org/wiki/Gravitational_potential_energy#Gravitational_poten tial_energy )

However, I'm not looking for a wiki reference, I want to know what you think it means. Because it seems you are using it differently than others, and until you've explained what you mean by it, it's not clear why you think it's greatest at the centre of the star.

Jeff Root

2011-Apr-19, 01:24 AM

I put some effort into getting these numbers right, but I won't

be surprised if I've bungled them totally.

When a quantity of matter is packed inside its Schwarzschild

radius, an event horizon forms at that radius. The matter will

continue collapsing without known limit.

If the same quantity of matter is packed with uniform density into

a volume larger than its Schwarzschild radius, it won't have the

required density, so no event horizon forms.

A mass of 6.73 x 10^29 kg has a Schwarzschild radius of 1 km.

When that mass is packed into a sphere of 1 km radius, an event

horizon forms at the surface and we have a black hole.

If that mass is packed with uniform density into a volume with

a radius larger than 1 km, it won't have the required density

to form an event horizon.

A mass of 6.73 x 10^24 kg has a Schwarzschild radius of 1 cm.

When that mass is packed into a sphere of 1 cm radius, an event

horizon forms at the surface and we have a black hole.

If that mass is packed with uniform density into a volume with

a radius larger than 1 cm, it won't have the required density

to form an event horizon.

6.73 x 10^29 kg in a volume of 1 km radius has a density of

1.61 x 10^20 kg/m^3.

6.73 x 10^24 kg in a volume of 1 cm radius has a density of

1.61 x 10^29 kg/m^3.

The collapsing core of a star should have roughly uniform density.

As it collapses, the density should increase uniformly throughout.

For simplicity, I'll further assume that matter outside the core

doesn't contribute to the collapse.

The collapsing core of a large star would reach a density of

1.61 x 10^20 kg/m^3 before it would reach a density of

1.61 x 10^29 kg/m^3. So an event horizon 1 km in radius would

form before an event horizon 1 cm in radius could form. The

collapsing core would never have an event horizon less than

1 km in radius.

Does that explain it?

-- Jeff, in Minneapolis

.

caveman1917

2011-Apr-19, 01:53 AM

Does that explain it?

You're using the equations for a static solution of a black hole to the dynamic situation of the formation of that black hole, thereby getting an incorrect result.

Things turn out to be a lot more difficult than that. An isotropic collapse from a homogeneous density with null pressure will form the event horizon at the center, which will then expand outwards. Using more realistic assumptions gets into a whole range of issues. For example using an initial density gradient that increases towards the center will in some cases not develop an event horizon in the first place, getting you a naked singularity.

See the spacetime diagram i linked to in the other thread, here (http://www.bautforum.com/showthread.php/114712-gravitational-force-vs-gravitational-potential-(-of-a-sphere-)?p=1878663#post1878663).

Jeff Root

2011-Apr-19, 02:35 AM

Exactly what do you mean by "the center"?

How big do you think it is when it forms?

-- Jeff, in Minneapolis

WayneFrancis

2011-Apr-19, 02:39 AM

I put some effort into getting these numbers right, but I won't

be surprised if I've bungled them totally.

When a quanty of matter is packed inside its Schwarzschild radius,

an event horizon forms at that radius. The matter will continue

collapsing without known limit.

If the same quantity of matter is packed with uniform density into

a volume larger than its Schwarzschild radius, it won't have the

required density, so no event horizon forms.

A mass of 6.73 x 10^29 kg has a Schwarzschild radius of 1 km.

When that mass is packed into a sphere of 1 km radius, an event

horizon forms at the surface and we have a black hole.

If that mass is packed with uniform density into a volume with

a radius larger than 1 km, it won't have the required density

to form an event horizon.

A mass of 6.73 x 10^24 kg has a Schwarzschild radius of 1 cm.

When that mass is packed into a sphere of 1 cm radius, an event

horizon forms at the surface and we have a black hole.

If that mass is packed with uniform density into a volume with

a radius larger than 1 cm, it won't have the required density

to form an event horizon.

6.73 x 10^29 kg in a volume of 1 km radius has a density of

1.61 x 10^20 kg/m^3.

6.73 x 10^24 kg in a volume of 1 cm radius has a density of

1.61 x 10^29 kg/m^3.

The collapsing core of a star should have roughly uniform density.

As it collapses, the density should increase uniformly throughout.

For simplicity, I'll further assume that matter outside the core

doesn't contribute to the collapse.

The collapsing core of a large star would reach a density of

1.61 x 10^20 kg/m^3 before it would reach a density of

1.61 x 10^29 kg/m^3. So an event horizon 1 km in radius would

form before an event horizon 1 cm in radius could form. The

collapsing core would never have an event horizon less than

1 km in radius.

Does that explain it?

-- Jeff, in Minneapolis

Sounds good to me

WayneFrancis

2011-Apr-19, 03:06 AM

You're using the equations for a static solution of a black hole to the dynamic situation of the formation of that black hole, thereby getting an incorrect result.

Things turn out to be a lot more difficult than that. An isotropic collapse from a homogeneous density with null pressure will form the event horizon at the center, which will then expand outwards. Using more realistic assumptions gets into a whole range of issues. For example using an initial density gradient that increases towards the center will in some cases not develop an event horizon in the first place, getting you a naked singularity.

See the spacetime diagram i linked to in the other thread, here (http://www.bautforum.com/showthread.php/114712-gravitational-force-vs-gravitational-potential-(-of-a-sphere-)?p=1878663#post1878663).

What that looks like to me is a description of the apparent horizon which is different then an absolute horizon. I've not read through the paper yet.

I think Jeff and I are just pointing out that before you'd get infinite density at the centre of a collapsing star you'd get the critical density at a Schwarzschild radius > 0

Tensor

2011-Apr-19, 04:32 AM

At any point there is a gravitational effect on the stress energy tensor.

That statement makes absolutely no sense at all. The Stress-Energy Tensor, along with some constants, describe the Curvature, which in GR IS gravity. I would seriously suggest you go here (http://xxx.lanl.gov/pdf/gr-qc/9712019v1). Those note were the basis of Carroll's Textbook "Spacetime and Geometry: An Introduction to General Relativity".

WaxRubiks

2011-Apr-19, 05:09 AM

perhaps tommac's OP could be restated as: the center of a star has the greatest time dilation, as seen by a distant observer.

WayneFrancis

2011-Apr-19, 06:12 AM

perhaps tommac's OP could be restated as: the center of a star has the greatest time dilation, as seen by a distant observer.

Yup...can agree with that but I'd point out that time dilation beyond a EH is meaningless.

tommac

2011-Apr-19, 01:34 PM

I sort of get it but not quite ... I may need to read through this 4 or 5 times before it makes sense ... so stick with me ... I will eventually get it.

The problem I am having right now is that say you have a dense core ... but not quite dense enough for an EH ... but then around it there is a collapse so that outside of the core you have more mass closer to the core. All of that mass that is close to the core should also contribute gravitational energy to points within the core.

For example lets take a real EH for example ... when a mass comes close to the EH the EH actually will expand outwards towards that mass because the mass is adding gravitational energy to portions of space that are already very warped ( the area just outside the EH ). As that mass comes close the sum of the gravitational energies from the BH AND the mass will warp the BHs EH until the mass is consumed ... then the EH will just be larger to because of the extra mass consumed.

Another example is the midpoint between two black holes. If you bring two black hole close to each other the midpoint between them will eventually form a EH at a distance I believe of 1.5 x EH outside of each of the BHs EH. This is because the between the BH has gravitational energy from both black holes ( even though the force could be 0 because they are pulling in opposite directions )

going back to the core. You have the core ... PLUS you have the gravitational energy of anything around the core. Once the gravitational energy is high enough an EH will form. Supposing that the gravitational energy doesnt jump from below what is needed for an EH to something for a large EH ... a small event horizon would need to form somewhere ... that place would be at the center.

I put some effort into getting these numbers right, but I won't

be surprised if I've bungled them totally.

When a quantity of matter is packed inside its Schwarzschild

radius, an event horizon forms at that radius. The matter will

continue collapsing without known limit.

If the same quantity of matter is packed with uniform density into

a volume larger than its Schwarzschild radius, it won't have the

required density, so no event horizon forms.

A mass of 6.73 x 10^29 kg has a Schwarzschild radius of 1 km.

When that mass is packed into a sphere of 1 km radius, an event

horizon forms at the surface and we have a black hole.

If that mass is packed with uniform density into a volume with

a radius larger than 1 km, it won't have the required density

to form an event horizon.

A mass of 6.73 x 10^24 kg has a Schwarzschild radius of 1 cm.

When that mass is packed into a sphere of 1 cm radius, an event

horizon forms at the surface and we have a black hole.

If that mass is packed with uniform density into a volume with

a radius larger than 1 cm, it won't have the required density

to form an event horizon.

6.73 x 10^29 kg in a volume of 1 km radius has a density of

1.61 x 10^20 kg/m^3.

6.73 x 10^24 kg in a volume of 1 cm radius has a density of

1.61 x 10^29 kg/m^3.

The collapsing core of a star should have roughly uniform density.

As it collapses, the density should increase uniformly throughout.

For simplicity, I'll further assume that matter outside the core

doesn't contribute to the collapse.

The collapsing core of a large star would reach a density of

1.61 x 10^20 kg/m^3 before it would reach a density of

1.61 x 10^29 kg/m^3. So an event horizon 1 km in radius would

form before an event horizon 1 cm in radius could form. The

collapsing core would never have an event horizon less than

1 km in radius.

Does that explain it?

-- Jeff, in Minneapolis

.

tommac

2011-Apr-19, 01:38 PM

The collapsing core of a star should have roughly uniform density.

As it collapses, the density should increase uniformly throughout.

For simplicity, I'll further assume that matter outside the core

doesn't contribute to the collapse.

.

This is the part I think I am having a problem with. Not only would the matter outside of the core contribute to the collapse of the star ... but wouldnt it also contribute gravitational energy to the core putting points of the core deeper into a gravtiational well than if the core was the only mass?

tommac

2011-Apr-19, 01:41 PM

I thought the Ricci Tensor described the curvature? And that the Stress Energy Tensor destribed the sumation of the energy at an individual point. If the stress energy tensor describes the curvature then what does the ricci tensor describe?

That statement makes absolutely no sense at all. The Stress-Energy Tensor, along with some constants, describe the Curvature, which in GR IS gravity. I would seriously suggest you go here (http://xxx.lanl.gov/pdf/gr-qc/9712019v1). Those note were the basis of Carroll's Textbook "Spacetime and Geometry: An Introduction to General Relativity".

caveman1917

2011-Apr-19, 02:59 PM

Exactly what do you mean by "the center"?

How big do you think it is when it forms?

Zero, it starts out as a point and expands.

What that looks like to me is a description of the apparent horizon which is different then an absolute horizon. I've not read through the paper yet.

No, the absolute event horizon.

I think Jeff and I are just pointing out that before you'd get infinite density at the centre of a collapsing star you'd get the critical density at a Schwarzschild radius > 0

Yes, you'd get the event horizon before the singularity, but it doesn't just pop into existence at rs. It forms as a point in the center, expands, the singularity forms, the EH keeps expanding until all matter has crossed it.

caveman1917

2011-Apr-19, 03:23 PM

Look at it this way: what does an event horizon mean?

It's a boundary that splits spacetime into two sections, an inner part and an outer part.

Loosely speaking, this means that an event that happens in the inner part will not be able to reach the outer part.

In other words, if i emit a photon in the inner part, it will never reach the outer part.

Now consider your (Jeff's and Wayne's) scenario. At some point in time, let's call it t0 you suddenly create an EH at rs.

Now suppose that at some minute time before that, t = t0 - ε, i emit a photon at some minimal distance from the center rε, pointing "outwards".

Track this photon. By the time it reaches rs we are already past t0, so it will reach an EH horizon there. This means that the spacetime event of the emittence of this photon was already inside an event horizon (by what an EH means), so before we had the EH at rs, we already had a smaller EH. The end result is that we had an EH form at the center as a point, which expands at c (though there are things that influence that) until it reaches the standard rs radius.

You just can't use the equations for an EH around a static black hole system in the dynamical situation of the formation of a black hole.

WayneFrancis

2011-Apr-19, 03:25 PM

I thought the Ricci Tensor described the curvature? And that the Stress Energy Tensor destribed the sumation of the energy at an individual point. If the stress energy tensor describes the curvature then what does the ricci tensor describe?

Ok ... someone ask if tommac understands what a tensor is. Point out to him that a tensor it a relationship of a bunch of vector of n dimensions and this is the reason why they are invariant and consequently the EH of a black hole is also invariant... I can just see his next post in my head...it will start out with something like "I know all that....BUT"

WayneFrancis

2011-Apr-19, 03:41 PM

Look at it this way: what does an event horizon mean?

It's a boundary that splits spacetime into two sections, an inner part and an outer part.

Loosely speaking, this means that an event that happens in the inner part will not be able to reach the outer part.

In other words, if i emit a photon in the inner part, it will never reach the outer part.

Now consider your (Jeff's and Wayne's) scenario. At some point in time, let's call it t0 you suddenly create an EH at rs.

Now suppose that at some minute time before that, t = t0 - ε, i emit a photon at some minimal distance from the center rε, pointing "outwards".

Track this photon. By the time it reaches rs we are already past t0, so it will reach an EH horizon there. This means that the spacetime event of the emittence of this photon was already inside an event horizon (by what an EH means), so before we had the EH at rs, we already had a smaller EH. The end result is that we had an EH form at the center as a point, which expands at c (though there are things that influence that) until it reaches the standard rs radius.

You just can't use the equations for an EH around a static black hole system in the dynamical situation of the formation of a black hole.

I understand that you are talking about the apparent horizon and we are talking about the actual schwarzschild horizon.

Lets take your example just a bit further.

At some minute time before the formation of an t0 an EH forms at rs

We can track back and say that the apparent horizon will form at a time t0 - c/rs

Jeff and I are not disagreeing with the apparent horizon....BUT the photon's fate isn't sealed until t0. If something disrupts the critical density from occurring then that photon can still escape because there is a difference between the 2.

So...as I said before...I don't believe Jeff and I are at odds with what you are saying because you are talking about the apparent horizon which should very quickly meet with the absolute horizon and always be inside it and Jeff and I are talking about the absolute horizon.

caveman1917

2011-Apr-19, 03:48 PM

Jeff and I are not disagreeing with the apparent horizon....BUT the photon's fate isn't sealed until t0.

No, i'm talking about the absolute horizon. The photon's fate is sealed from the moment it is emitted (actually from before that). Remember that general relativity plays out on a 4d spacetime manifold, the future is "already there" in the equations that describe this manifold.

caveman1917

2011-Apr-19, 08:24 PM

Perhaps a clarification might be needed.

There is indeed, when using schwarzschild coordinates, an apparent horizon that shields the growing absolute horizon. They will meet up at rs, so from those coordinates it appears you have an apparent horizon growing to rs and there turning into the absolute horizon, which appears to just pop into existence there.

But an event horizon is an event horizon. The difference between an apparent horizon and an absolute horizon is nothing else than wether this horizon appears in all coordinate systems, or just in the used one. So when you're looking at it from a single coordinate system, it doesn't quite make sense to employ the distinction. The boundary you have is an event horizon, period. Every absolute horizon is an horizon in every coordinate system.

We can however use certain coordinates to be able to describe beyond apparent horizons, so the horizons we meet are absolute (not the schwarzschild coordinates). There we find that the absolute horizon also forms at a point, and expands outwards. Every event horizon (either absolute or apparent) must either form from a point or have always been there, in any timeslicing - just track the photons wether they will ever reach infinity given an infinite time in your chosen coordinate system.

So...as I said before...I don't believe Jeff and I are at odds with what you are saying because you are talking about the apparent horizon which should very quickly meet with the absolute horizon and always be inside it and Jeff and I are talking about the absolute horizon.

It will always be "outside" it, in the sense that it will shield the absolute horizon. You cannot have, in a single coordinate system, one horizon shielding another one since the horizon means you can't describe the section of spacetime that it is the boundary of. You can't describe further than the "first" horizon you encounter to see if there's another one "below". You need different coordinates for that.

Jeff Root

2011-Apr-19, 11:38 PM

In another thread, talking about this same subject, I made a

serious error: I said that gravitational potential is the same as

gravitational potential energy. They are NOT the same.

Gravitational potential is a property of a location in space in

relation to a gravitating body. Gravitational potential energy is

a property of a test mass at such a location.

A given location relative to a gravitating body will have some

particular gravitational potential relative to that body. A test mass

at that location will have a particular gravitational potential energy

relative to the body.

Gravitational potential energy has the dimension 'energy'.

Gravitational potential has the dimension 'energy per unit mass'.

-- Jeff, in Minneapolis

WayneFrancis

2011-Apr-20, 02:21 AM

Ok maybe I'm going about this wrong. Instead of talking about the EH lets talk about the curvature of space/time. The EH might appear to grown outward from the centre of the black hole but if one could measure the curvature of space time they would not see that the slope of the curvature radiate outward from a point.

So let me make this perfectly clear. I AM NOT DISAGREEING WITH YOU. I'm pointing out that you don't get ∞ density at r = 0 of a collapsing star before you get a critical density at some r > 0. That is all I've been trying to say. I've never disagreed with your description using light cones and what external observers would see. I've just been pointing out that r = 0 doesn't have ∞ density....actually lets get rid of the singularity problem.

I'll argue that before you get the the Planck mass into the Planck volume within a star you'll get an appropriate amount of mass into a volume where r > lp...that is what I'm saying.

baric

2011-Apr-20, 02:52 AM

Look at the problem in a stepwise fashion..

Let's say you start with a neutron star just on the verge of acquiring enough mass to begin the catastrophic core collapse into a black hole. At this point, there is no event horizon since the core is not dense enough.

Now add just enough mass to trigger the core collapse. In other words, the pressure at the center of the core overcomes neutron degeneracy pressure.

In the instant that the core center begins its collapse, there is no event horizon (just yet) because the maximum density allowed by neutron degeneracy is not enough to create a black hole.

However the center continues to collapse, pulling the outer layers down with it, until it stops at some as-yet-unknown limit. Along the way to this new balance, the core density increased to some point where photons could no longer escape and an event horizon was formed.

As more material fell onto the new BH core and past the EH threshold, the EH expanded to account for the increased core mass.

So yes, the EH begins at the center of the star during its core collapse and expands outward. However, the EH is just an artifact of this collapse, not a cause.

tommac

2011-Apr-20, 01:10 PM

Well worded! One note however as was discussed in the other thread. Once the EH is created time STOPs at the EH for the external observer. At first thought it may seem that the star could never fully collapse because of this time dilation, at least from the point of view of the external observer. However as more mass is pulled towards the EH their is additional gravitational potential energy ( please correct me if I am using the wrong term ) added to the core at the point the EH so that the EH actually expands to then consume the infalling mass. This was an important point as I was not able to figure out how the collapse could continue, from the viewpoint of the external observer, while time was stopped.

Look at the problem in a stepwise fashion..

Let's say you start with a neutron star just on the verge of acquiring enough mass to begin the catastrophic core collapse into a black hole. At this point, there is no event horizon since the core is not dense enough.

Now add just enough mass to trigger the core collapse. In other words, the pressure at the center of the core overcomes neutron degeneracy pressure.

In the instant that the core center begins its collapse, there is no event horizon (just yet) because the maximum density allowed by neutron degeneracy is not enough to create a black hole.

However the center continues to collapse, pulling the outer layers down with it, until it stops at some as-yet-unknown limit. Along the way to this new balance, the core density increased to some point where photons could no longer escape and an event horizon was formed.

As more material fell onto the new BH core and past the EH horizon threshold, the EH expanded to account for the increased core mass.

So yes, the EH begins at the center of the star during its core collapse and expands outward. However, the EH horizon is just an artifact of this collapse, not a cause.

baric

2011-Apr-20, 01:20 PM

Well worded! One note however as was discussed in the other thread. Once the EH is created time STOPs at the EH for the external observer.

Only time as measured by observation from outside the EH. The physics inside the EH continue without regard for the perceptions of any external observers.

Jeff Root

2011-Apr-20, 02:16 PM

The problem I am having right now is that say you have a dense

core ... but not quite dense enough for an EH ... but then around it

there is a collapse so that outside of the core you have more mass

closer to the core. All of that mass that is close to the core should

also contribute gravitational energy to points within the core.

In two ways, you are right, but I'm pretty sure it doesn't have the

impact you think it does.

You are right that material outside the core (or portion of the core

that we are looking at) plays a role in the collapse. The weight of

material above helps compact the material below to the density

needed to become a black hole.

And you are right that the material outside the core contributes to

the total gravitational potential of the object.

But neither of those should affect my argument.

I think I can demolish the second objection.

At the surface of the Earth, we are in a "gravity well". Because

the interior of the Earth is more dense than the outer layers, the

gravitational field strength increases for some depth into the Earth

before it begins to decrease, and becomes zero at the center.

This Wikipedia graph has three curves for Earth's gravitational

field strength. The solid line is for a good model of the real Earth.

The curve made of short dashes is an idealized model with the

same density at the center as the real Earth, but decreasing to

the surface at a linear rate. The straight line made of long dashes

is an idealized model in which the density is uniform throughout

the Earth. All three curves show the same total mass.

http://upload.wikimedia.org/wikipedia/commons/8/86/EarthGravityPREM.jpg

The realistic model shows maximum g at the core/mantle boundary.

The model with linearly-decreasing density is maximum just inside

the lower mantle. The model with uniform density throughout is

maximum at the surface. Above Earth's surface, gravitational field

strength falls with the square of the distance from the center.

Gravitational field strength tells us how hard an object at a given

location is pulled downward by a gravitating body, and it tells us

the object's acceleration at that location.

Gravitational potential is relative, and is confusingly expressed

as a negative number. It has a minimum possible value at Earth's

center. It increases from Earth's center to its theoretical maximum

at infinite distance, where it has a numerical value of zero.

This Wikipedia embedding diagram is a 3-D graph of the gravitational

potential of a sphere with uniform density throughout. The surface

of the sphere isn't shown, but is near the top of the red part. The

scale is arbitrary.

http://upload.wikimedia.org/wikipedia/commons/d/d9/GravityPotential.jpg

Gravitational potential tells us, for example, the speed an object

has in a circular orbit at a given distance from Earth's center (r),

the instantaneous speed an object needs in order to escape Earth

from a given r, and the energy required to lift a test mass from one

given r to another.

Something at Earth's center is deeper in Earth's gravity well than

we are. It has farther to climb to get out. However, the climb is

easier, at first, because the gravitational field strength is low.

Imagine a particle trying to escape the Earth, starting at Earth's

center. At first it has very easy going, starting in zero gravity.

Climbing gets tougher and tougher as it rises through the core,

against gravity that gradually gets stronger until it reaches the

point of maximum gravitational field strength. After that the

going gradually gets easier again. After leaving the Earth, it

gets easier more rapidly as the field strength falls with the

square of the distance from the center. A few million miles from

Earth, and the going is very easy, on out to infinity.

But we are ignoring the Sun's gravity. If the particle climbs from

Earth to infinity, it must also climb out of the Sun's gravity well.

And the gravity well of the Milky Way. Earth is deep in both the

Sun's and the Milky Way's gravity wells.

However, Earth is in orbit around the Sun, and around the Milky

Way. It is in free-fall around them. So the particle doesn't have

to fight their gravity to escape Earth's gravity. It can escape from

Earth's gravity first, then escape the Sun's gravity if it wants to,

and go on to escape the Milky Way's gravity if it wants to.

The presence of the gravity wells of the Sun and the Milky Way

do not make climbing out of Earth's gravity well more difficult.

The event horizon of a black hole is the location where the

gravitational field strength is so great that light headed vertically

outward is unable to make any progress in climbing out of the

well. Insde the event horizon, light and everything else falls

toward the center.

If the black hole is inside the Milky Way, light trying to escape

from it is also deep inside the Milky Way's gravity well. But that

gravity well doesn't make any difference to the light's ability to

climb out of the black hole's gravity well.

Similarly, the gravity well of the matter closely surrounding a

black hole makes no difference to light trying to escape from it,

if that matter is uniformly distributed in a spherical shell around

the black hole. The gravity of the outer matter pulls on the light

from all directions, so even though it puts the light in a deeper

gravity well overall (because the gravitational potential between

the light's location and infinity is greater), it has no effect on

the gravitational field strength at that location.

The matter farther from the center than the light may press down

on the matter below it, increasing the lower matter's density and

triggering a collapse, but it doesn't add directly to the gravity

that can prevent the light from escaping. As long as the matter

below the light is not sufficient to form an event horizon at the

light's position or higher, the light will be able to keep rising

and escape.

-- Jeff, in Minneapolis

Jeff Root

2011-Apr-20, 02:49 PM

Exactly what do you mean by "the center"?

How big do you think it is when it forms?

Zero, it starts out as a point and expands.

Okay. Let's suppose a large star is collapsing to a black hole,

and the event horizon has just started to form. When its radius

is 1 cm, what is the density of the matter in that region? Or in

other words, how much mass is inside the event horizon?

-- Jeff, in Minneapolis

Jeff Root

2011-Apr-20, 03:22 PM

Look at it this way: what does an event horizon mean?

It's a boundary that splits spacetime into two sections, an inner part

and an outer part.

Loosely speaking, this means that an event that happens in the inner

part will not be able to reach the outer part.

In other words, if i emit a photon in the inner part, it will never reach

the outer part.

Now consider your (Jeff's and Wayne's) scenario. At some

point in time, let's call it t0 you suddenly create an EH at rs.

Now suppose that at some minute time before that, t = t0 - ε,

i emit a photon at some minimal distance from the center rε,

pointing "outwards".

Track this photon. By the time it reaches rs we are already

past t0, so it will reach an EH horizon there. This means that

the spacetime event of the emittence of this photon was already

inside an event horizon (by what an EH means), so before we had

the EH at rs, we already had a smaller EH. The end result is

that we had an EH form at the center as a point, which expands at c

(though there are things that influence that) until it reaches the

standard rs radius.

Your account will be debited to pay your bills on the 3rd of the month.

You deposit the money to pay the bills on the 1st. But it takes three

days for your account to be credited. Does that mean you were

overdrawn on the 1st?

The event horizon of a black hole is the location where gravity is so

strong that light cannot move outward. It is even stronger than that

inside the event horizon. If light is moving outward, it is not inside

the event horizon yet, though it may be in the future.

-- Jeff, in Minneapolis

Strange

2011-Apr-20, 03:35 PM

Your account will be debited to pay your bills on the 3rd of the month.

You deposit the money to pay the bills on the 1st. But it takes three

days for your account to be credited. Does that mean you were

overdrawn on the 1st?

Not a good analogy (or not the right analogy). You have some money in the bank now but not enough to pay the bills. Are you overdrawn? In accounting terms, yes. (I worked for a company that had to wind up make everyone redundant because they didn't have enough money to guarantee they could carry on trading and still have enough money to pay all the redundancy money if they had to make people redundant in future!)

The event horizon of a black hole is the location where gravity is so

strong that light cannot move outward.

That is one feature (a side effect) of the the event horizon. The important point is that it is an event horizon. That is why caveperson's explanation makes perfect sense to me.

tommac

2011-Apr-20, 07:30 PM

Only time as measured by observation from outside the EH. The physics inside the EH continue without regard for the perceptions of any external observers.

yes thus the "external observer" part.

tommac

2011-Apr-20, 07:34 PM

Which would have a greater increase of kinetic energy an object dropped from some arbitrary distance above the earths surface to the surface of the earth or an object dropped through a tunnel to the center of the earth?

In two ways, you are right, but I'm pretty sure it doesn't have the

impact you think it does.

You are right that material outside the core (or portion of the core

that we are looking at) plays a role in the collapse. The weight of

material above helps compact the material below to the density

needed to become a black hole.

And you are right that the material outside the core contributes to

the total gravitational potential of the object.

But neither of those should affect my argument.

I think I can demolish the second objection.

At the surface of the Earth, we are in a "gravity well". Because

the interior of the Earth is more dense than the outer layers, the

gravitational field strength increases for some depth into the Earth

before it begins to decrease, and becomes zero at the center.

This Wikipedia graph has three curves for Earth's gravitational

field strength. The solid line is for a good model of the real Earth.

The curve made of short dashes is an idealized model with the

same density at the center as the real Earth, but decreasing to

the surface at a linear rate. The straight line made of long dashes

is an idealized model in which the density is uniform throughout

the Earth. All three curves show the same total mass.

http://upload.wikimedia.org/wikipedia/commons/8/86/EarthGravityPREM.jpg

The realistic model shows maximum g at the core/mantle boundary.

The model with linearly-decreasing density is maximum just inside

the lower mantle. The model with uniform density throughout is

maximum at the surface. Above Earth's surface, gravitational field

strength falls with the square of the distance from the center.

Gravitational field strength tells us how hard an object at a given

location is pulled downward by a gravitating body, and it tells us

the object's acceleration at that location.

Gravitational potential is relative, and is confusingly expressed

as a negative number. It has a minimum possible value at Earth's

center. It increases from Earth's center to its theoretical maximum

at infinite distance, where it has a numerical value of zero.

This Wikipedia embedding diagram is a 3-D graph of the gravitational

potential of a sphere with uniform density throughout. The surface

of the sphere isn't shown, but is near the top of the red part. The

scale is arbitrary.

http://upload.wikimedia.org/wikipedia/commons/d/d9/GravityPotential.jpg

Gravitational potential tells us, for example, the speed an object

has in a circular orbit at a given distance from Earth's center (r),

the instantaneous speed an object needs in order to escape Earth

from a given r, and the energy required to lift a test mass from one

given r to another.

Something at Earth's center is deeper in Earth's gravity well than

we are. It has farther to climb to get out. However, the climb is

easier, at first, because the gravitational field strength is low.

Imagine a particle trying to escape the Earth, starting at Earth's

center. At first it has very easy going, starting in zero gravity.

Climbing gets tougher and tougher as it rises through the core,

against gravity that gradually gets stronger until it reaches the

point of maximum gravitational field strength. After that the

going gradually gets easier again. After leaving the Earth, it

gets easier more rapidly as the field strength falls with the

square of the distance from the center. A few million miles from

Earth, and the going is very easy, on out to infinity.

But we are ignoring the Sun's gravity. If the particle climbs from

Earth to infinity, it must also climb out of the Sun's gravity well.

And the gravity well of the Milky Way. Earth is deep in both the

Sun's and the Milky Way's gravity wells.

However, Earth is in orbit around the Sun, and around the Milky

Way. It is in free-fall around them. So the particle doesn't have

to fight their gravity to escape Earth's gravity. It can escape from

Earth's gravity first, then escape the Sun's gravity if it wants to,

and go on to escape the Milky Way's gravity if it wants to.

The presence of the gravity wells of the Sun and the Milky Way

do not make climbing out of Earth's gravity well more difficult.

The event horizon of a black hole is the location where the

gravitational field strength is so great that light headed vertically

outward is unable to make any progress in climbing out of the

well. Insde the event horizon, light and everything else falls

toward the center.

If the black hole is inside the Milky Way, light trying to escape

from it is also deep inside the Milky Way's gravity well. But that

gravity well doesn't make any difference to the light's ability to

climb out of the black hole's gravity well.

Similarly, the gravity well of the matter closely surrounding a

black hole makes no difference to light trying to escape from it,

if that matter is uniformly distributed in a spherical shell around

the black hole. The gravity of the outer matter pulls on the light

from all directions, so even though it puts the light in a deeper

gravity well overall (because the gravitational potential between

the light's location and infinity is greater), it has no effect on

the gravitational field strength at that location.

The matter farther from the center than the light may press down

on the matter below it, increasing the lower matter's density and

triggering a collapse, but it doesn't add directly to the gravity

that can prevent the light from escaping. As long as the matter

below the light is not sufficient to form an event horizon at the

light's position or higher, the light will be able to keep rising

and escape.

-- Jeff, in Minneapolis

tommac

2011-Apr-20, 07:40 PM

I think the point here ... is that there doesnt need to be a certain amount of mass inside the EH ... there just needs to be enough gravitational potential energy ( maybe I am using the wrong term ... but enough energy ) at a point / volume ... that graviational energy can be from outside of the EH .

In other words lets take an EH of 1 cm ... or roughly 1 earth mass. However you dont need to have that much mass inside the EH when a star is collapsing. Some of the energy can come from outside of the EH. All that matters is that in the region of EH the total energy ( from gravity and everything else ) is sufficient for an EH to form.

Okay. Let's suppose a large star is collapsing to a black hole,

and the event horizon has just started to form. When its radius

is 1 cm, what is the density of the matter in that region? Or in

other words, how much mass is inside the event horizon?

-- Jeff, in Minneapolis

pzkpfw

2011-Apr-20, 08:53 PM

(snip) there just needs to be enough gravitational potential energy ( maybe I am using the wrong term ... but enough energy ) (snip)

This is why I'd been trying to get you to clarify your terms, because it was clear you were using them differently than mainstream science, and it was important to then clarify what it is you think makes a black hole in the first place.

(snip) In other words lets take an EH of 1 cm ... or roughly 1 earth mass. However you dont need to have that much mass inside the EH when a star is collapsing. Some of the energy can come from outside of the EH. All that matters is that in the region of EH the total energy ( from gravity and everything else ) is sufficient for an EH to form.

You know (I assume) that our own Sun would beconme a black hole if it were squeezed down to a small enough size.

So two questions:

1. What is the "the total energy ( from gravity and everything else )" at the centre of our Sun right now?

2. What is the "the total energy ( from gravity and everything else )" at the centre of our Sun if it were squeezed smaller than its Schwarzschild radius?

caveman1917

2011-Apr-20, 08:56 PM

I'm pointing out that you don't get ∞ density at r = 0 of a collapsing star before you get a critical density at some r > 0.

Yes, but it doesn't matter. The formula for using the mass to get the radius of an EH of a black hole (ie density-wise) only works when you have a static situation, it's a mathematical 'shortcut'. It doesn't work when you have the dynamical situation of the formation of the black hole. You have to go back to the actual definition of an event horizon, which is a sectioning of a 4-manifold depending on the intersection of light cones emitted from events (=points on that manifold).

caveman1917

2011-Apr-20, 09:01 PM

However the center continues to collapse, pulling the outer layers down with it, until it stops at some as-yet-unknown limit. Along the way to this new balance, the core density increased to some point where photons could no longer escape and an event horizon was formed.

Please stop thinking about the density along certain points in time :)

Think about it in terms of painting lines on a surface depending on a logical property of the points you're considering.

As more material fell onto the new BH core and past the EH threshold, the EH expanded to account for the increased core mass.

Actually, as material falls into the EH, the rate of expansion of the EH slows down.

caveman1917

2011-Apr-20, 09:04 PM

When its radius

is 1 cm, what is the density of the matter in that region? Or in

other words, how much mass is inside the event horizon?

That's a really difficult question to answer exactly, but i can tell you that the density is a lot lower than the "critical density" you'd get by dividing the schwarzschild radius of a black hole by its mass. And that's because it doesn't matter (well, it does of course, but not in the way you're thinking about it :)).

caveman1917

2011-Apr-20, 09:34 PM

The event horizon of a black hole is the location where gravity is so

strong that light cannot move outward.

Strictly speaking it is the boundary of the region where events (=points on a 4d manifold) will never have their future lightcones intersect with the worldline of an external observer in finite time. Or, more appropiately to the "observer at infinity", a point for which holds that no null geodesic starting at it will asymptotically approach infinity in any spatial coordinate as t \rightarrow \infty, in any coordinate system.

If light is moving outward, it is not inside

the event horizon yet, though it may be in the future.

Don't think about the future as if it's indetermined (we're in GR here). The location and existence of an event horizon (at any point in time) depends on the entire future of the system you're considering.

It's not the light that's moving outwards that tells you something about itself, the photon you are tracking tells you something about the event/point of its emittence.

If that photon never reaches an external observer, put a "yes" on the event of that photon's emittence (it is just a point on your manifold). Do this check for every point on your surface. Draw a boundary around those points where you put "yes". That is your event horizon. Then choose a coordinate system, and slice up your spacetime. This will allow you to talk about space and time seperately. Then you will see how that event horizon "grows over time".

Let's do this in 3-space. Suppose you have drawn the complete future lightcones of every point in 3-space, and found that for all points that hold to the condition follows x^2 + y^2 + z^2 \leq r. You take the boundary, which is a sphere. Only now will you slice it up. Suppose that we take the z-coordinate to be time, and the x and y to be space. So we are taking, "slicing", an x-y plane through our geometry where we let z go from some point to another point, giving us a series of 2d 'pictures' one after the other, somewhat akin to the frames of a movie. What you will see is at first nothing, then suddenly a point that immediately expands in a circle, that keeps expanding (let's stop half way when our circle is biggest, the "top half" is not realistic here). That's how it plays out in a given coordinate system, the event horizon starts at a point and "grows in time".

The event horizon is nothing more than a boundary on a 4-manifold, giving rise to some geometrical figure that divides our manifold in an "inside" and an "outside". The logical decision of wether a point should be inside or outside depends on the entire manifold, there is no distinction between "past" and "future" at this point. That's only when we get to describing it in some coordinate system. Which for the problem at hand of a collapsing star gives us an EH starting from a point and growing over time.

caveman1917

2011-Apr-20, 10:13 PM

Which would have a greater increase of kinetic energy an object dropped from some arbitrary distance above the earths surface to the surface of the earth or an object dropped through a tunnel to the center of the earth?

Arbitrary distance. Kinetic energy depends on velocity, which in turn depends on the total amount of acceleration the object has endured between dropping it and measuring its kinetic energy. You'll be able to have this sum of acceleration (which is the force on the object at each point) be greater when dropping it from some distance as long as you choose your distance to be high enough.

profloater

2011-Apr-20, 10:57 PM

OK out of my depth, but if this collapse is slower than the speed of light and you have densities with big numbers then there will be shock waves and vibrations, talk about a singularity! In ordinary explosive forming or shaped charge explosives these effects cause gigantic pressure waves which can be focussed and would I imagine be focussed at the centre in a sphere. To imagine this as an orderly compaction is to simplify too far I would think. I would expect the local pressures to exceed the binding forces in an oscillating way, and this is without getting to the point about the EH which surely could not be anything orderly during this process.?

WayneFrancis

2011-Apr-20, 11:47 PM

Well worded! One note however as was discussed in the other thread. Once the EH is created time STOPs at the EH for the external observer. At first thought it may seem that the star could never fully collapse because of this time dilation, at least from the point of view of the external observer. However as more mass is pulled towards the EH their is additional gravitational potential energy ( please correct me if I am using the wrong term ) added to the core at the point the EH so that the EH actually expands to then consume the infalling mass. This was an important point as I was not able to figure out how the collapse could continue, from the viewpoint of the external observer, while time was stopped.

This is partly my point ... the actual event is a extreme curvature of space...that happens at a particular r and that is at some point where r > 0. The effect that an apparent horizon is formed at the centre and spreads out is different.

An EH does NOT always mean time stops. Take cosmic expansion. 2 points that recede from each other would view the other as time slowing down just like 2 objects that are moving through space. But unlike the the movement through space there is no actual time dilation. The slowing down is is just an illusion.

Like wise light that is emitted just before that apparent horizon forms is NOT red shifted until it actually passes rs. If the absolute EH forms at t0 at rs then an observer at r < rs at a time < t0 will see that photon and that photon will not be red shifted in the same manner as if it was emitted from a location that was some very small distance from an absolute horizon.

The I'll crunch the the numbers and post up the results but I'm 99% sure that if you where at a point inside of rs before the EH formed at rs then the optical effect would be different because even though you'd still be in a gravity well and even though you'd have to keep accelerated to sty at what ever r you are at you COULD do this but in a real black hole situation you can't because you'd have to then be moving at a speed > c and that is the key difference between the 2 situations. Remember no one can observe the EH form because it also travels at the speed of light. You'd see everything normal and in fact that apparent horizon would pass you and everything would still appear normal until the absolute horizon formed further out then your massive rockets would no longer be able to hold you at your r and you'd start to fall towards the singularity that to you would probably be forming.

caveman1917

2011-Apr-21, 12:47 AM

This is partly my point ... the actual event is a extreme curvature of space...that happens at a particular r and that is at some point where r > 0. The effect that an apparent horizon is formed at the centre and spreads out is different.

You state a couple of posts back that you're not disagreeing, but your description seems to counterindicate that.

Do you agree that the existence of an apparent horizon depends on the coordinate system chosen?

Do you agree that, as long as we're considering things in a single coordinate system, there is no difference between an absolute and an apparent horizon?

An EH does NOT always mean time stops.

*Prove it. (assuming the obvious that you mean coordinate time).

Take cosmic expansion. 2 points that recede from each other would view the other as time slowing down just like 2 objects that are moving through space. But unlike the the movement through space there is no actual time dilation. The slowing down is is just an illusion.

What qualifier makes time dilation "actual" versus "just an illusion"?

Like wise light that is emitted just before that apparent horizon forms is NOT red shifted until it actually passes rs.

Why not?

If the absolute EH forms at t0 at rs then an observer at r < rs at a time < t0 will see that photon and that photon will not be red shifted in the same manner as if it was emitted from a location that was some very small distance from an absolute horizon.

Why?

If there is such a difference between absolute and apparent horizons, why is rindler space a good approximation for the schwarzschild metric near the EH?

The I'll crunch the the numbers and post up the results but I'm 99% sure that if you where at a point inside of rs before the EH formed at rs then the optical effect would be different because even though you'd still be in a gravity well and even though you'd have to keep accelerated to sty at what ever r you are at you COULD do this

Prove that you can keep yourself at constant r until the EH is at rs.

You'd see everything normal and in fact that apparent horizon would pass you

Prove that an apparent horizon can pass you.

and everything would still appear normal until the absolute horizon formed further out, then your massive rockets would no longer be able to hold you at your r and you'd start to fall towards the singularity that to you would probably be forming.

Prove it.

tommac

2011-Apr-21, 01:33 AM

yes so which one would be greater dropping it from an arbitrary distance to the surface of the earth or dropping it from the same distance above the surface of the earth to the center of the earth, assuming there is a tunnel.

Arbitrary distance. Kinetic energy depends on velocity, which in turn depends on the total amount of acceleration the object has endured between dropping it and measuring its kinetic energy. You'll be able to have this sum of acceleration (which is the force on the object at each point) be greater when dropping it from some distance as long as you choose your distance to be high enough.

caveman1917

2011-Apr-21, 01:40 AM

yes so which one would be greater dropping it from an arbitrary distance to the surface of the earth or dropping it from the same distance above the surface of the earth to the center of the earth, assuming there is a tunnel.

You're dropping both from some distance above the earth, but allowing one to continue towards the center?

You can solve this, since both will have had the same trajectory until reaching the surface of the earth, they will have the same velocity.

If one falls through, does it pick up even more speed on its way to the center or is it slowed down?

tommac

2011-Apr-21, 01:43 AM

yes that is my question to you.

You're dropping both from some distance above the earth, but allowing one to continue towards the center?

You can solve this, since both will have had the same trajectory until reaching the surface of the earth, they will have the same velocity.

If one falls through, does it pick up even more speed on its way to the center or is it slowed down?

baric

2011-Apr-21, 03:50 AM

Please stop thinking about the density along certain points in time :)

Think about it in terms of painting lines on a surface depending on a logical property of the points you're considering.

Fair enough, although I'm not sure how to parse that :P

Actually, as material falls into the EH, the rate of expansion of the EH slows down.

I never meant to suggest that the rate of expansion for the EH speeds up, only that it increases in size as the mass within it increases. Obviously mass is one-dimensional and the EH is 3-dimensional, so the rate of increase necessarily slows down.

WayneFrancis

2011-Apr-21, 04:57 AM

You state a couple of posts back that you're not disagreeing, but your description seems to counterindicate that.

No I don't. You keep harping on about the same thing over and over and I don't disagree that there is an EH as you describe. Now here is what I'm pointing out.

Different event horizons can have different characteristics. The the EH at the schwarzschild has some different implications then the apparent horizon that comes out from the centre and that is what I'm doing at the today...I'm either going to show that there are some differences in what is observed and I'll admit that they are very subtle differences or I'll prove to myself that I'm completely wrong but what you have shown me does NOT address the effect I'm talking about.

Do you agree that the existence of an apparent horizon depends on the coordinate system chosen?

Where do I not agree to this? You might as well ask me if I agree that 2 + 2 = 4 when I'm talking about granny smith apples are green.

Do you agree that, as long as we're considering things in a single coordinate system, there is no difference between an absolute and an apparent horizon?

Again I'm not saying that there isn't an EH. I'm pointing out that there are subtle differences just like there are differences in the expansion of the universe having an object recede at c and an object actually traveling THROUGH space at c even though the redshift LOOKS the same.

*Prove it. (assuming the obvious that you mean coordinate time).

Again don't twist my words and don't try to make it look like I've said anything I haven't. When you say "time stops" to a layperson they seem to think that the proper time for that person stops. It doesn't. A better description then "time stops", in my opinion, is that comparing clocks is no longer a valid option. You can't see "time stop" of a free falling object. You see time slow down then the object will disappear from your view in a finite amount of time. There will be a very definite last photon emitted by the falling object. So an external observer will see time slow then the object disappear. An analogy : I have a watch that for what ever reason is slowing down and I seal it in box did time stop because I can't observe it? You might like to think of it that way and for certain things that is fine BUT here is my problem. In the past tommac has made the statement that he believed that objects receding from us at c because of the expansion of the universe would have time stop and that is NOT the case. He mixed up SR with a GR effect. Even at the cosmic horizon those objects aren't going slower then us. That EH causes an illusion that the object is slowing down relative to us, which is not what happens when something falls deeper into a gravity well and it actually does experience less time.

What qualifier makes time dilation "actual" versus "just an illusion"?

See above & below

Take the following thought experiments and tell me if the results would be the same

Thought Experiment 1

Object A & B are next to each other with clocks synced.

Have object A accelerate away from B at .9c for n amount of time.

Have object A stop.

Have object A & B travel towards each other, at any speed, and then compare the clocks.

Object A will have lesser time on it's clock then B does

Thought Experiment 2

Object A & B are next to each other with clocks synced.

Space expands at a rate that separates A & B at .9c for n amount of time.

Space stops expanding.

Have object A & B travel towards each other, at any speed, and then compare the clocks.

Object A & B will have the same time on their clocks.

During the test though observations would appear the same. The red shifts from the objects would appear the same.

Same initial observation but the final comparison of clocks will indicate an entirely different cause to the observed effect.

Why not?

Because the topology of that part of the universe is not the same as rs

Einstein's General Theory of Relativity | Lecture 12 (http://www.youtube.com/watch?v=fVqYlSNqSQk)

explains much of what I'm trying to get across.

Why?

If there is such a difference between absolute and apparent horizons, why is rindler space a good approximation for the schwarzschild metric near the EH?

Again I don't think that really has anything to do with what I'm claiming.

...I'll hold off because your next few questions will be answered by the same explanation

Prove that you can keep yourself at constant r until the EH is at rs.

Prove that an apparent horizon can pass you.

Prove it.

Ok We both agree that the EH you are talking about isn't from rs = 2GM/c2

So lets look at it this way

using

http://latex.codecogs.com/gif.latex?\frac{1}{\sqrt{1-\frac{r_{s}}{r}}}%20-%201

Lets look at the problem

A neutron star has density of 3.7×1017 to 5.9×1017 kg/m3 with core density of up to 8×1017 kg/m3

lets take the core density at 1km out

volume of the core at 1km r = 4/3πr3

that gives us

~4,188,790,205m3 or 4.188790205x109m3

with a total mass of about

3.3510321638291127876934862755x1027kg

What is the rs for that mass?

well 2GM / c2

2 x 6.6742867-11 x 3.3510321638291127876934862755x1027 / 2997924582

44.73149880463373710340511825040416/8.9875517873681764x1016

That gives us ~4.977050465m. IE that amount of mass needs to be squished down by another 995m or ~199x more then it is to get to its rs

So we take a look at a photon at emitted via the

Lets take a the red shift of a photon from r = 1001m and we see its red shift ofhttp://latex.codecogs.com/gif.latex?\frac{1}{\sqrt{1-\frac{4.977050465}{1000}}}%20-%201

or a red shift of

0.002497853

compare this to the actual red shift of a photon being released at 1m above a rs of 999m

which would have a red shift of 30.6227766

THIS is my point it isn't a rs starting from r = 0 and expanding out. THAT is my only point so please don't try to make my argument into on of there isn't an EH propagation outward from r=0 at a rate of c because I agree with that. What I don't agree with is that you'll get rs satisfied at a r ≈ 0 before you get a rs satisfied at an r >> 0.

So for your question/request

Prove that you can keep yourself at constant r until the EH is at rs.

So the speed you need to maintain your position at just above the rs is just under c but since rs of 3.3510321638291127876934862755x1027kg of mass is at about ~5m and your at position is at r = 1000m your speed is well below c. Even as that EH goes past you, note you don't see anything different at this point either because at this point in time the neutron star still exists in its neutron star form. The EH has to travel up another 12+km and this is where our physics get fuzzy because we don't know the mass it takes to form a black hole very well in this situation. So don't expect me to model it out when the best scientist can't tell you.

What an observer inside the rs will observe is something very different then having a rs propagate out from r = 0 at c. IE the first observer might have about 4.00277x10-05 seconds left of proper time and the second would have ~12x less time. Might seem silly but it is a difference. Both cases and external observer would never get to hear about it.

Another thing to note is that tidal forces between the 2 is quite different.

and this doesn't make sense

Prove that an apparent horizon can pass you.

if you are at a r < rs at a distance where you would have to travel > c to get rs then that apparent horizon, which is observer dependent, will pass you because to "see" it you'd have to get out of the black hole before it forms which means you'd have to travel > c which is not possible so you won't. We already have a problem with some poster(s) on here thinking that the Schwarzschild EH isn't invariant.

tommac

2011-Apr-23, 10:57 PM

3d? isnt it 2d or at least 2 spacial + 1 time? BUt wouldnt then also mass be 2 d ( 1 energy + 1 t ) ? or am I confusing what is being said here?

Obviously mass is one-dimensional and the EH is 3-dimensional, so the rate of increase necessarily slows down.

caveman1917

2011-Apr-24, 02:36 AM

I never meant to suggest that the rate of expansion for the EH speeds up, only that it increases in size as the mass within it increases.

We really need to make a proper distinction between the apparent horizon found in schwarzschild coordinates and the absolute horizon.

The two will only coincide in a static spacetime, meaning that no more mass will fall into the black hole, ever.

Before that happens, as is necessarily true when considering what happens when mass falls in, the two will have vastly different behaviour.

The apparent horizon will discontinuously "jump". Take a black hole with mass M and schwarschild radius R. Suppose we let some mass m fall in. We can calculate the new schwarzschild radius r for the black hole M+m. The apparent horizon at R will sit there, until our mass has fallen beyond r, and then this horizon will "jump" to r.

The absolute horizon on the other hand will continuously grow in anticipation of the infalling mass, will decrease its rate of expansion when more mass falls in, will increase its rate of expansion when it's 'in vacuo', and will settle at the schwarzschild radius when the last bit of mass has fallen in.

You might think of an apparent horizon as "reactive" and an absolute horizon as "anticipatory". This is because an apparent horizon depends on the local properties of the metric (the most outer surface with all lightlike normal vectors pointing inwards), where the absolute horizon depends on the global causal structure, the behaviour at future null infinity. The "actual" black hole (enclosed by the absolute horizon) will behave in very counterintuitive ways as long as the spacetime is dynamic, after which it doesn't really behave at all :)

Obviously mass is one-dimensional and the EH is 3-dimensional, so the rate of increase necessarily slows down.

I'm not sure what you mean by this.

grav

2011-Apr-24, 04:18 AM

This is partly my point ... the actual event is a extreme curvature of space...that happens at a particular r and that is at some point where r > 0. The effect that an apparent horizon is formed at the centre and spreads out is different.

An EH does NOT always mean time stops. Take cosmic expansion. 2 points that recede from each other would view the other as time slowing down just like 2 objects that are moving through space. But unlike the the movement through space there is no actual time dilation. The slowing down is is just an illusion.If the event horizon were a direct result of the acceleration of gravity, then since the acceleration of gravity reduces below the surface of a spherical body, only applying to the mass that is contained within any particular spherical shell to the center, then no actual event horizon could form until the radius of any spherical shells within the body becomes greater than the Schwarzschild radius for the mass that lies within that spherical shell. Depending upon the density profile of the body, the event horizon might suddenly come into existence wherever this occurs. Since the acceleration of gravity is typically greatest at the surface, it might not form at all until all of the mass falls below the Schwarzschild radius. But with gravitational time dilation causing the infall of matter to slow as it approaches the Schwarzschild radius, a true event horizon would never fully form in this manner

However, although it might seem that the equivalence principle of Relativity should relate the acceleration of gravity with accelerating observers in flat space-time as it does inertial observers and freefallers and find time dilation and length contraction accordingly, for whatever reason I am not quite sure about, with black holes, it relates instead with the gravitational potential. Although the acceleration of gravity reduces below the surface of a body until it reaches zero at the center, the gravitational potential increases and is greatest at the center (although negative). It is the potential that determines the gravitational time dilation, zero at the event horizon, which will occur first at the center of a body when the gravitational potential,

V{r internal} = G M (r^2 - 3 r_body^2) / (2 r_body^3)

so V = - 3 G M / (2 r_body) at the center, reaches a boundary where - V ≥ c^2 / 2.

As the infall continues and the potential increases, the time dilation at the center becomes complex, space-like, showing that the center now lies within the event horizon, and the potential further away from the center will steadily climb so that the boundary lies further and further from the center as the event horizon grows as the infall continues. When all of the mass falls below the Schwarzschild radius, the boundary then becomes external and remain steady since the external gravitational potential will then be constant at that radius, although it will still take an eternity for all of the mass to fall below this final boundary according to a distant observer.

Another important issue is that the gravitational time dilation is a real effect according to the distant observer's coordinate system, not an optical illusion. The Doppler effect for the redshifting of light coming from a massive body is a direct result of the time dilation. Let's say that the time dilation at some point in a gravitational filed is 1/10 that of a distant observer, so if the local observer emits one photon per second according to local observer's clock, the distant observer receives one photon per ten seconds, so says that the local observer's clock is really ticking ten times slower than the distant observer's own clock. The distant observer says that the first photon is emitted at T0=0 on the local observer's clock. That photon takes a time of t_ph to reach the distant observer. Ten seconds after the first photon was emitted, the distant observer says the local observer's clock now reads T1 = 1 sec and a second photon is emitted, taking the same amount of time to travel to the distant observer. So since both photons took the same amount of time to travel the same distance along identical paths, the only difference in times that will explain this is the times that they were received is the difference in times they were emitted. In other words, the first photon was received in a time of T0 + t_ph and the second after a time of T1 + t_ph, so

(T1 + t_ph) - (T0 + t_ph) = T1 - T0

The local observer says the photons were emitted every second while the distant observer says they must have been emitted every ten seconds according to the distant observer's own clock since they were received every ten seconds, so the distant observer's clock must actually be ticking ten times faster than the local clock, demonstrating that the Doppler shift is a direct consequence of gravitational time dilation.

caveman1917

2011-Apr-24, 04:21 AM

Where do I not agree to this? [that the existence of an apparent horizon depends on the coordinate system used]

(my [..] for reference)

Here

You'd see everything normal and in fact that apparent horizon would pass you

And again here

if you are at a r < rs at a distance where you would have to travel > c to get rs then that apparent horizon, which is observer dependent, will pass you

Think about it.

I'm pointing out that there are subtle differences just like there are differences in the expansion of the universe having an object recede at c and an object actually traveling THROUGH space at c even though the redshift LOOKS the same.

Does it? What would the doppler shift be for an object travelling at c? What would the cosmological redshift be for an object receding at c? Is "receding at c" enough information to calculate cosmological redshift in the first place? Note that we can in fact see those objects.

Again don't twist my words and don't try to make it look like I've said anything I haven't.

This is what you said (my bold)

An EH does NOT always mean time stops.

Either that means that an EH not always has coordinate time stop, or that sometimes an EH does have proper time stop.

Which one did you mean?

That EH causes an illusion that the object is slowing down relative to us, which is not what happens when something falls deeper into a gravity well and it actually does experience less time.

Are you saying proper time slows down for an object deeper in a gravity well?

If yes, please prove it.

If no, how exactly is this different than the time dilation with a cosmological event horizon?

Take the following thought experiments and tell me if the results would be the same {snip}

During the test though observations would appear the same. The red shifts from the objects would appear the same.

What is the formula for cosmological redshift? Will the redshifts appear the same?

So lets look at it this way

{snip}

a red shift of

0.002497853

compare this to the actual red shift of a photon being released at 1m above a rs of 999m

which would have a red shift of 30.6227766

A photon released a bit above a tiny black hole is less redshifted than one above a big black hole..

How is this relevant?

THIS is my point it isn't a rs starting from r = 0 and expanding out.

rs is a value for a certain radius, how can it expand?

That's like saying "my point is that 5 doesn't start at 0 and grows".

What I don't agree with is that you'll get rs satisfied at a r ≈ 0 before you get a rs satisfied at an r >> 0.

What does this even mean? Replace rs with some value (for example 5) in your sentence, and read it again.

So the speed you need to maintain your position at just above the rs is just under c but since rs of 3.3510321638291127876934862755x1027kg of mass is at about ~5m and your at position is at r = 1000m your speed is well below c. Even as that EH goes past you, note you don't see anything different at this point either because at this point in time the neutron star still exists in its neutron star form. The EH has to travel up another 12+km and this is where our physics get fuzzy because we don't know the mass it takes to form a black hole very well in this situation. So don't expect me to model it out when the best scientist can't tell you.

This is wrong at every single step.

What an observer inside the rs will observe is something very different then having a rs propagate out from r = 0 at c. IE the first observer might have about 4.00277x10-05 seconds left of proper time and the second would have ~12x less time. Might seem silly but it is a difference. Both cases and external observer would never get to hear about it.

Which two observers?

Besides, your claim was that an observer stationary at some r < rs will only have his acceleration diverge when the EH reaches rs.

caveman1917

2011-Apr-24, 04:25 AM

yes that is my question to you.

I'm sure you can solve this yourself. The question boils down to: does an object pick up speed going from the surface of the earth to the center?

WaxRubiks

2011-Apr-24, 06:35 AM

I'm sure you can solve this yourself. The question boils down to: does an object pick up speed going from the surface of the earth to the center?

the answer is yes.

although the acceleration falls the closer you get to the center, until there is no acceleration(at the center).

grav

2011-Apr-24, 06:59 AM

As for an apparent horizon as compared to a real horizon, I suppose one could say that every body has an apparent horizon. A spherical mass will have an apparent horizon of R = 2 G M / c^2 even if that horizon cannot physically be reached because the mass has not condensed small enough, as with the sun, Earth, and even a tennis ball. The apparent horizon exists because a distant observer will still observe a time dilation for a local observer at r of sqrt(1 - R/r), where R is the Schwarzschild radius of the body. However, technically the time dilation is really sqrt(1 + 2 V / c^2) externally, where V is the gravitational potential (negative). Outside of the body the potential decreases with the distance, where V = - G M / r. Internally, for a uniformly dense body, it is V{internal} = G M (r^2 - 3 r_body^2) / (2 r_body^3). At the center of a massive body, then, the time dilation is sqrt(1 - 3 G M / (r c^2)) = sqrt(1 - 3 R / (2 r)).

So even though there is still greater gravitational time dilation with greater potential, an actual event horizon will not exist until a potential is produced where - V ≥ c^2. Any points that have a potential where - V > c^2 will lie within an event horizon, with complex time dilation, and - V = c^2 will describe the boundary with zero time dilation. This will hold for any shape or density. For instance, two black holes placed next to each other will have twice the potential halfway between them than they otherwise would have at that distance. If the potential at that point becomes greater than - V > c^2, then that point now lies within the event horizon. The event horizons might now overlap and lie around both black holes, producing one overall event horizon, where the boundary can be determined by finding everywhere that the potential is now - V = c^2. The potential at any point can be found by simply summing the potentials, V = - G M / d for every point mass in the bodies, where d is the distance from a point mass to the point we are measuring for.

Jeff Root

2011-Apr-24, 03:00 PM

No horizon exists if "the mass has not condensed small enough".

The Schwarzschild radius of a body which is not compact enough

to form an event horizon is a purely theoretical thing, with zero

physical consequences.

-- Jeff, in Minneapolis

.

grav

2011-Apr-24, 03:34 PM

No horizon exists if "the mass has not condened small enough".

The Schwarzschild radius of a body which is not compact enough

to form an event horizon is a purely theoretical thing, with zero

physical consequences.

-- Jeff, in Minneapolis

Right, an actual event horizon will not form until the potential within the body becomes great enough. As for the physical consequences, however, the gravitational time dilation can still be worked out for local external observers in terms of the Schwarzschild radius of the body according to its mass, where R = 2 G M / c^2, and the time dilation is sqrt(1 - R/r) even without the event horizon, although technically it should just be found in terms of the potential and there would be less confusion.

caveman1917

2011-Apr-24, 03:49 PM

If the event horizon were a direct result of the acceleration of gravity, then since the acceleration of gravity reduces below the surface of a spherical body, only applying to the mass that is contained within any particular spherical shell to the center, then no actual event horizon could form until the radius of any spherical shells within the body becomes greater than the Schwarzschild radius for the mass that lies within that spherical shell. Depending upon the density profile of the body, the event horizon might suddenly come into existence wherever this occurs. Since the acceleration of gravity is typically greatest at the surface, it might not form at all until all of the mass falls below the Schwarzschild radius.

Have you read any of the posts in this thread?

caveman1917

2011-Apr-24, 03:58 PM

As for an apparent horizon as compared to a real horizon, I suppose one could say that every body has an apparent horizon.

One could in ATM, not in Q&A.

A spherical mass will have an apparent horizon of R = 2 G M / c^2 even if that horizon cannot physically be reached because the mass has not condensed small enough, as with the sun, Earth, and even a tennis ball.

No it doesn't. Jeff has the right of it, the schwarzschild radius is nothing more than a meaningless number until the black hole has fully formed (ie no more mass will fall into it, ever, in its entire future).

The apparent horizon exists because a distant observer will still observe a time dilation for a local observer at r of sqrt(1 - R/r), where R is the Schwarzschild radius of the body.

The interior geometry of a spherical mass is not the schwarzschild geometry. Only the exterior geometry is schwarzschild, please review Birkhoff's theorem.

grav

2011-Apr-24, 04:03 PM

the answer is yes.

although the acceleration falls the closer you get to the center, until there is no acceleration(at the center).Right, and this is where the equivalence principle comes in. Stationary observers within a gravitational field are considered to be accelerating, with the same local effects that would be experienced upon an accelerating platform (with minor gradient differences). Freefalling observers are considered inertial because those experiences are the same also. In other words, each set of observers would gain the same results for experiments performed locally.

So to find the gravitational time dilation as compared to a distant observer at infinity, we just let the distant observer freefall to the same place as the local stationary observer and measure the instantaneous time dilation at that point. SR is valid locally, so all we need is the kinetic time dilation when the freefaller and stationary observer are side by side. The local speed of the freefaller is measured at

v = sqrt(2 G M / r) = c sqrt(R/r), so the time dilation is

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

So the gravitational time dilation is sqrt(1 - R/r) on a local stationary observer's clock according to a distant observer. That is for points external to the body. After passing the surface of the body, the freefaller continues to accelerate, and so measures greater instantaneous relative speeds to stationary observers within the body, and so with greater time dilation, until the local speed reaches a maximum at the center of the body, so this is where the maximum time dilation occurs.

grav

2011-Apr-24, 04:13 PM

Have you read any of the posts in this thread?Right, I am saying that is not the way the event horizon operates, but is probably the way WayneFrancis is thinking of it, in terms of the acceleration of gravity, not the potential, which is common.

No it doesn't. Jeff has the right of it, the schwarzschild radius is nothing more than a meaningless number until the black hole has fully formed (ie no more mass will fall into it, ever, in its entire future).Right, it is just a convenient numerical value, although far from meaningless. An event horizon doesn't actually exist until the potential is great enough (or small since it is negative), but physical consequences such as time dilation can still be found in terms of the Schwarzschild radius externally without the event horizon.

The interior geometry of a spherical mass is not the schwarzschild geometry. Only the exterior geometry is schwarzschild, please review Birkhoff's theorem.Right, externally only.

WayneFrancis

2011-Apr-25, 12:38 PM

I'm done caveman...your warping things I've said. If I've done the maths wrong the point it out and tell me my mistake...Do I have the density of the core of a neutron star wrong? Do I mess up compact that amount of mass needs to be to be within its schwarzschild radius. I'm not above being shown I'm wrong but all you've done is say "Your wrong". You twisted my words....I can not say proper time slows down because that doesn't make any sense you always experience time as 1s/s.

Anyway I've made my point as clear as I care to try to at this point.

WayneFrancis

2011-Apr-25, 12:45 PM

Right, I am saying that is not the way the event horizon operates, but is probably the way WayneFrancis is thinking of it....

Please tell me how you think I'm thinking of it.

I'm saying ONE thing and one thing only. That the schwarzschild radius isn't satisfied at r = 0 then grows from there. That is ALL i'm saying ... that is IT. NOTHING MORE.

If someone cares to show me how the rs for a star forming into a black hole starts a 0 and goes out then feel free.

and by rs I mean where that = 2GM/c2 is satisfied. Saying "What does this even mean? Replace rs with some value (for example 5) in your sentence, and read it again"

5m? well plug rs = 5m into the formula rs = 2GM/c2 and you get a very definite answer for the only variable in that equation which is M. If you have less mass then M in that radius then you don't have a black hole.

tommac

2011-Apr-25, 01:46 PM

I'm sure you can solve this yourself. The question boils down to: does an object pick up speed going from the surface of the earth to the center?

Yes I have my answer ... it was meant as a question to you.

tommac

2011-Apr-25, 01:49 PM

the answer is yes.

although the acceleration falls the closer you get to the center, until there is no acceleration(at the center).

Agreed, so wouldnt that mean that the gravitational potential is less at the center than at the surface? Or that the gravitational potential energy is the greatest at the center?

This is the point that I am confused about. I always thought that the gravitational potential energy at the center was the greatest. if it is not then why would a falling object be able to gain more kinetic energy by falling to the center?

tommac

2011-Apr-25, 01:52 PM

You could in theory create a black hole by just using a laser or a series of lasers. All that you need is enough energy at one particular point for an EH to form.

No horizon exists if "the mass has not condensed small enough".

The Schwarzschild radius of a body which is not compact enough

to form an event horizon is a purely theoretical thing, with zero

physical consequences.

-- Jeff, in Minneapolis

.

EDIT ... I think I misinterpreted was was being said in Jeffs Quote ... please ignore this post.

caveman1917

2011-Apr-25, 02:39 PM

I'm done caveman...your warping things I've said. If I've done the maths wrong the point it out and tell me my mistake...

The problem is, as i've said a couple of times, that you're using a formula where it doesn't apply.

The geometry is not schwarzschild until the black hole is fully formed. That's what makes your math wrong, and the conclusions about redshift and other things through the formation of the black hole.

Suppose we have a cube that's slowly warping into a sphere. We want to know the volume as it relates to the radius throughout that transformation.

What you're doing is using \frac{4}{3}\pi r^3 all the time. Which by itself is the correct calculation for the volume of a sphere, but it simply doesn't apply until the transformation is complete and we actually do have a sphere.

caveman1917

2011-Apr-25, 02:46 PM

Please tell me how you think I'm thinking of it.

I'm saying ONE thing and one thing only. That the schwarzschild radius isn't satisfied at r = 0 then grows from there. That is ALL i'm saying ... that is IT. NOTHING MORE.

If someone cares to show me how the rs for a star forming into a black hole starts a 0 and goes out then feel free.

Strictly speaking that is correct, but it is irrelevant, that's the issue.

Anyway, it's been brought to my attention that i've only linked to a spacetime diagram which used an analytic approximation for the formation.

I have to go now, but i'll see if i can dig up some references later today or tomorrow with the full correct treatment of the formation of a black hole.

caveman1917

2011-Apr-25, 02:50 PM

Btw, an apparent horizon cannot pass you because if you're using an observer inside the black hole, there is going to be no apparent horizon (or it'll be in some other place, depending on details). It's observer dependent, if you're using an observer that would pass the apparent horizon, this means that from this observer there won't be any apparent horizon there in the first place.

grav

2011-Apr-25, 03:19 PM

I'm saying ONE thing and one thing only. That the schwarzschild radius isn't satisfied at r = 0 then grows from there. That is ALL i'm saying ... that is IT. NOTHING MORE.

Okay, but in this thread we are discussing the formation of the event horizon, not that of the Schwarzschild radius which is just a fixed numerical value based upon the body's mass.

5m? well plug rs = 5m into the formula rs = 2GM/c2 and you get a very definite answer for the only variable in that equation which is M. If you have less mass then M in that radius then you don't have a black hole.

Since the surface of the mass never quite falls below the event horizon or the Schwarzschild radius according to a distant observer, a black hole could never fully form with that definition. The event horizon, however, is defined as the boundary beyond which light cannot escape. Even if surrounded by mass outside of the Schwarzschild radius, if the potential is large enough at the center, an event horizon will form there. If we add more mass to a black hole where the event horizon lies just below the Schwarzschild radius, then the event horizon will then grow beyond the Schwarzschild radius for the previous mass, but not now for the new Schwarzschild radius with the combined mass. The Schwarzschild radius defines a limit for the event horizon of a particular mass, not the actual boundary.

pzkpfw

2011-Apr-25, 08:17 PM

This is the point that I am confused about. I always thought that the gravitational potential energy at the center was the greatest. if it is not then why would a falling object be able to gain more kinetic energy by falling to the center?

Because when you are not at the centre, there is still net gravitational attraction to the centre. That will cause acceleration. However, that acceleration will get less and less, the closer you get to the centre, because the net gravitational attraction goes to zero at the centre, because some of the mass is now "above " you.

The maximum gravitational potential energy is at infinite distance away from an object. It is zero (least) at the centre of mass of the object.

Edit to add:

If you let go of a ball a million miles from Earth, it will gain quite a bit of kinetic energy by the time it got to Earth.

If you let go of a ball a half a million miles from Earth, it will gain less kinetic energy than before.

If you let go of a ball at the centre of Earth, it won't gain any kinetic energy at all.

caveman1917

2011-Apr-25, 11:32 PM

Booth 2005 (paper here (http://arxiv.org/abs/gr-qc/0508107)) seems to give an accessible overview of the differences between different horizons and how they behave, including quite some references to specific issues.

macaw

2011-Apr-26, 12:00 AM

V{r internal} = G M (r^2 - 3 r_body^2) / (2 r_body^3)

so V = - 3 G M / (2 r_body) at the center, reaches a boundary where - V ≥ c^2 / 2.

But this makes no sense since your -V>c^2/2 is equivalent to:

r_s/r_{body}>2/3

which is clearly false. For example, in the case of the Earth r_s=9mm and r_{body}=6400km. In general, r_s<<r_{body}

grav

2011-Apr-26, 12:53 AM

But this makes no sense since your -V>c^2/2 is equivalent to:

r_s/r_{body}>2/3

which is clearly false. For example, in the case of the Earth r_s=9mm and r_{body}=6400km. In general, r_s<<r_{body}That is why the Earth does not contain an event horizon, the potential isn't great enough. For a uniformly dense body, when R / r_body > 2/3, an event horizon will begin to form at the center of the body. As R / r_body approaches 1, the event horizon will have expanded to nearly the surface of the body as the surface also infalls to increase the potential.

macaw

2011-Apr-26, 01:05 AM

That is why the Earth does not contain an event horizon,

Sure it does, the event horizon is the sphere of Schwarzschild radius, i.e. 9mm .

For a uniformly dense body, when R / r_body > 2/3, an event horizon will begin to form at the center of the body. As R / r_body approaches 1, the event horizon will expand toward the surface of the body as the surface also infalls to increase the potential.

Where did you get this? Reference?

tommac

2011-Apr-26, 01:24 PM

Sure it does, the event horizon is the sphere of Schwarzschild radius, i.e. 9mm .

UGGGG that is the Schwarzschild radius NOT the event Horizon ... the earth does not have an event horizon ...

In order for your numbers to work the entire mass of the earth would need to be contained inside the SR ... which is not the case ... therefore no EH.

What we are talking about in this thread is slightly different. As you can have an EH without having mass inside teh EH. but you cant just use the calculations to calculate SR ... it is different ( the mass is above my skill set but I realize that it is not the same calculation ). The EH at the core of an object is based on the amount of gravitational energy at the core. Once that is high enough ( and no the SR is not enough ) a EH will form in the center of the object.

tommac

2011-Apr-26, 01:25 PM

Macaw,

Do you understand the difference between an Event Horizon and the Schwarzchild Radius?

Sure it does, the event horizon is the sphere of Schwarzschild radius, i.e. 9mm .

Where did you get this? Reference?

macaw

2011-Apr-26, 02:50 PM

UGGGG that is the Schwarzschild radius NOT the event Horizon ... the earth does not have an event horizon ...

In order for your numbers to work the entire mass of the earth would need to be contained inside the SR ... which is not the case ... therefore no EH.

What we are talking about in this thread is slightly different. As you can have an EH without having mass inside teh EH. but you cant just use the calculations to calculate SR ... it is different ( the mass is above my skill set but I realize that it is not the same calculation ). The EH at the core of an object is based on the amount of gravitational energy at the core. Once that is high enough ( and no the SR is not enough ) a EH will form in the center of the object.

It is very simple , really, I explained it before. The sphere of radius r_s where r_s is the Schwarzschild radius forms the event horizon.

I asked grav for a reference on his claims, would you care to provide it yourself?

tommac

2011-Apr-26, 03:49 PM

It is very simple , really, I explained it before. The sphere of radius r_s where r_s is the Schwarzschild radius forms the event horizon.

Again this is slightly wrong. It doesnt always create an Event Horizon only if there is enough energy. And your formula is ONLY for the case where all of that mass fits inside the SR.

Using YOUR formula if a mass is not located within that volume of space ( the earth in 9 mm ) then NO EVENT HORIZON forms. An Event horizon doesnt always exist for every mass. While an SR does exist for every mass. Do you see the difference?

tommac

2011-Apr-26, 03:50 PM

Are you claiming that the earth does in fact have an Event Horizon at its core? If so then this portion of the discussion should be ATM.

It is very simple , really, I explained it before. The sphere of radius r_s where r_s is the Schwarzschild radius forms the event horizon.

I asked grav for a reference on his claims, would you care to provide it yourself?

macaw

2011-Apr-26, 04:14 PM

Are you claiming that the earth does in fact have an Event Horizon at its core?

No, you need to read the definition carefully: the event horizon is simply the sphere of radius r_s. Since all gravitational bodies have a Schwarzshild radius, an easy conclusion follows.

If so then this portion of the discussion should be ATM.

Can you provide the references requested to grav' claims? Yes or no? That was a direct question and that was the subject of discussion with grav. If you want to step in for him, then I would like you to answer the question I asked him.

Swift

2011-Apr-26, 04:43 PM

This thread is closed. After almost 100 posts, the original question has not only been answered but beaten to death and it would appear this thread has evolved into yet another of the endless series of the same two or three people talking past each other. No wonder they are called blackhole threads.

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