Jump to content

What difference does it make if the Schwarzschild radii touch?


Robittybob1

Recommended Posts

 

To understand my post two, The formula here best correlates.

 

Then look at the SI base units for energy/density.

 

I think the equation that may suit best on energy/density to curvature relations may be better seen here.

 

[latex]R_{ab}-\frac{1}{2}Rg_{ab}=8\pi GT_{ab}[/latex]

 

[latex]R_{ab}[/latex] is the ricci tensor,[latex] g_{ab}[/latex] is the curved by the presence of energy via the stress tensor [latex]T_{ab}[/latex]. G is the gravitational constant. R is the ricci scalar.

 

Conceptually this equation means curvature=energy...

Equation 1.1 the curvature equals energy statement is footnote 1.2.

 

http://www.google.ca/url?q=http://www.physics.usyd.edu.au/~luke/research/masters-geodesics.pdf&sa=U&ved=0ahUKEwj25uq8qOzLAhVM4WMKHWm4Ca0QFggRMAA&usg=AFQjCNEr4WEHhcvoL-LVhqBLVIcgBRFdkQ

Now as I understand the Schwartzchild metric it assumes the background vacuum =0. However I don't believe it states the curvature has zero energy/density.

Without going into energy being a property lets avoid gravitons and use gravitational field energy.

 

 

(Of course we could get into Komar,Bondi and ADM mass but I don't we really need to.) As it would be off the OP topic

 

https://en.m.wikipedia.org/wiki/Mass_in_general_relativity

They don't instantly combine. They merge (pretty quickly). The fastest the event horizon can grow is c.

Correct but again this doesn't mean that gravitational waves are only emitted during the event horizons merging.

Which is the point I was trying to explain in the first place lol

Link to comment
Share on other sites

To understand my post two, The formula here best correlates.

 

Then look at the SI base units for energy/density.

 

 

 

Correct but again this doesn't mean that gravitational waves are only emitted during the event horizons merging.

Which is the point I was trying to explain in the first place lol

If there is mass at the EH is it going to move at c?

They don't instantly combine. They merge (pretty quickly). The fastest the event horizon can grow is c.

if there is mass there as Mordred says can the EH move at that speed. What does it do with the mass?

Link to comment
Share on other sites

If there is mass at the EH is it going to move at c?

 

All you know about the mass is that the black hole has that mass.

 

 

if there is mass there as Mordred says can the EH move at that speed. What does it do with the mass?

 

The event horizon is not a "thing". It is a location.

 

WE DON"T KNOW WHAT HAPPENS TO THE MASS.

http://arxiv.org/pdf/1506.00560.pdf "Possible golden events for ringdown gravitational waves" This was the paper making predictions about the internal structure of the BH.

 

I skimmed that (most of it is way over my head) and didn't immediately see anything related to that. Can you point out where it says that?

Link to comment
Share on other sites

 

All you know about the mass is that the black hole has that mass.

 

 

The event horizon is not a "thing". It is a location.

 

WE DON"T KNOW WHAT HAPPENS TO THE MASS.

 

I skimmed that (most of it is way over my head) and didn't immediately see anything related to that. Can you point out where it says that?

Let Mordred answer it. If there is mass outside the EH and the EH expands or reshapes how is that mass moved about?

There is no issue if the EH is only a location and not a thing with mass around it.

But then if it is not a thing why does it matter if they touch when the merger is inspiraling?

 

To me it is a tricky situation.

Link to comment
Share on other sites

To be honest you would have a far easier time understanding how gravitational waves work if you focus on the energy/density not mass.

Your question is better described in what are the energy/density distribution changes.

Link to comment
Share on other sites

To be honest you would have a far easier time understanding how gravitational waves work if you focus on the energy/density not mass.

Your question is better described in what are the energy/density distribution changes.

Can you move energy density around at the speed of light?

Link to comment
Share on other sites

Yes all interactions are limitted to the speed of light. The rate depends on the field your describing. You know that both the gravitational field and the electromagnetic field change at the speed of light.

 

For modelling gravitational waves that's sufficient.

 

As changes to spacetime geometry also propogates at maximum c.

Edited by Mordred
Link to comment
Share on other sites

Yes all interactions are limitted to the speed of light. The rate depends on the field your describing. You know that both the gravitational field and the electromagnetic field change at the speed of light.

 

For modelling gravitational waves that's sufficient.

 

As changes to spacetime geometry also propogates at maximum c.

I'm starting to get your picture. So none of the energy density is actual particles of matter so it all can be moved at the speed of light, is that what you are implying?

Link to comment
Share on other sites

We are modelling changes in geometry.

 

Think of the particles as attached to the geometry. Curvature means a higher energy/density.

 

When your modelling changes in spacetime or gravity waves your modelling the changes in geometry.

Granted you need a change in energy/density to change the geometry.

 

Ie change in angular momentum or acceleration etc.

 

Remember the equivalence principle inertia mass is the same as gravitational mass.

Energy is simply the ability to do work. Mass is resistance to inertia.

Edited by Mordred
Link to comment
Share on other sites

We are modelling changes in geometry.

 

Think of the particles as attached to the geometry. Curvature means a higher energy/density.

OK but once you have particles you are not going to get them to move at the speed of light, so you get a separation of the geometry with the radiation and fields able to move at the speed of light but not the particles, They will move but not as far and as fast.

If the EH had a wave in it, will the particles move in and out of the EH? Like a ship in a wave. The boat will remain practically in the same place but the wave will pass along the ship. The wave could even go over the sides of the ship (resistance to inertia).

Edited by Robittybob1
Link to comment
Share on other sites

Particles with mass will always move slower than the speed of light.

 

Particles moving in and out of the EH is kind of nonsense. We're talking geometry change. It's the geometry that's changing. The event horizon is a specific geometry condition.

 

Once a particle crosses the event horizon it's never going to exit.

 

The event horizon is a point of no return. It doesn't matter if the EH moves or not. There is no spacetime path for a particle to ever exit an event horizon. Doesn't matter if the EH gets warped, stretched etc.

 

It still represents the boundary of point of no return.

Edited by Mordred
Link to comment
Share on other sites

 

...

I skimmed that (most of it is way over my head) and didn't immediately see anything related to that. Can you point out where it says that?

 

 

The other method shown in this paper is even simpler. When we focus on the dominant QNM, there is a forbidden

parameter region in GR. Just using the ringdown GWs, we can directly discuss whether the QNM from the remnant

compact object is consistent with the one from a BH predicted by GR or not.

There was another section of the results in http://arxiv.org/pdf/1506.00560.pdf

 

Page 4/9

 

 

A. Only ringdown

First, using only the ringdown GWs, we propose a simple method to test whether the compact object emitting the

ringdown GWs is a BH predicted by GR or not. Figure 2 shows the QNM frequencies for the dominant (` = 2, m = 2)

least-damped (n = 0) mode in the (fR, fI ) plane. In GR, the top-left side of the thick black line is prohibited. The

boundary thick black line corresponds to the Schwarzschild limit, which is obtained by setting αrem = 0, i.e.,

|fI |

fR

≈ 0.236 , (15)

in Eqs. (7), (8) and (9). In principle, if we obtain the parameters in the forbidden region from GW observations, we

can conclude that the compact object is not the one predicted by GR.

I took that to mean that the inside of a BH "is not as predicted by GR" in other words it is not a singularity.

Particles with mass will always move slower than the speed of light.

 

Particles moving in and out of the EH is kind of nonsense. We're talking geometry change. It's the geometry that's changing. The event horizon is a specific geometry condition.

 

Once a particle crosses the event horizon it's never going to exit.

 

The event horizon is a point of no return. It doesn't matter if the EH moves or not. There is no spacetime path for a particle to ever exit an event horizon. Doesn't matter if the EH gets warped, stretched etc.

 

It still represents the boundary of point of no return.

Could it be something new? Particles above the EH moving with extreme velocity around a spinning BH and the EH changes shape. What happens to the particles?

 

There is a Caltech/Cornell simulation YT "Merging black holes, falling spacetime, and gravitational waves" starts at 58 secs.

 

 

Part 2: Head-on collision of two black holes. The individual apparent horizons (blue) move together, and eventually a common apparent horizon (green) pops up discontinuously in time. The event horizon (gray) evolves continuously in time and is always outside or coincident with the apparent horizons.

The simulation and rendering were performed by members of the SXS collaboration (http://www.black-holes.org), using the SpEC code (http://www.black-holes.org/SpEC/) and the open-source visualization application ParaView (http://www.paraview.org/).

.

Edited by Robittybob1
Link to comment
Share on other sites

Any particles that cross the event horizon for whatever reason is essentially lost to the blackhole.

 

We cannot get them back ever.

That's pretty straightforward.

 

A point of no return is just that. We cannot ever get any information beyond the event horizon.

 

That's why I stated the mass or energy that causes gravity waves do not emit from inside the event horizon.

The energy for gravity waves originate from the spacetime geometry outside the event horizon. Not from within.

Edited by Mordred
Link to comment
Share on other sites

Don't read too much into that paper.

 

Lol at least not in your stage of understanding. The paper is theoretical. Ie has plausible mathematics. Yer it states it doesn't conform to GR.

 

Let's stick to how GR describes a BH until evidence shows it is wrong. (Which hasn't happened yet)

The reason I posted that article in your other thread was to show you the ringdown formula itself.

Edited by Mordred
Link to comment
Share on other sites

....

 

That's why I stated the mass or energy that causes gravity waves do not emit from inside the event horizon.

The energy for gravity waves originate from the spacetime geometry outside the event horizon. Not from within.

We might have to look at that in another thread. It doesn't seem right for there would have to be some mechanism to replenish the energy on the outside if that was the case.

Link to comment
Share on other sites

Why would you think that? Does the change in angular momentum of the BINARY system somehow elude you?

 

It is the changes in angular momentum that causes the changes in the spacetime curvature.

 

 

"certain circumstances, accelerating objects generate changes in this curvature, which propagate outwards at the speed of light in a wave-like manner. These propagating phenomena are known as gravitational waves.

 

https://en.m.wikipedia.org/wiki/Gravitational_wave

 

you must have looked at this wiki page 100s of times.

 

Did the quoted section somehow elude you?

 

It specifically states changes to the curvature. It does not state change to the mass of the rotating objects.

Here is another appropriate line.

 

"In theory, the loss of energy through gravitational radiation could eventually drop the Earth into the Sun. However, the total energy of the Earth orbiting the Sun (kinetic energy + gravitational potential energy) is about 1.14×1036 joules of which only 200 joules per second is lost through gravitational radiation, leading to a decay in the orbit"

In other words the Earth Sun system is technically emitting gravity waves.

 

They are simply too weak for us to detect.

 

The Earth and Sun certainly aren't touching each other. Lol.

 

 

Take a look at the equation for radiated power of a gravitational wave.

 

[latex] P=\frac{de}{dt}=-\frac{32}{5}\frac{G^4}{c^5}\frac{(m_1m_2)^2(m_1+m_2)}{r^5}[/latex]

 

You can keep the mass m1 and m2 constant but just change the radius and you emit a gravity wave.

(PS I wish I had remembered this formula dozens of posts ago. Would have saved a lot of effort)

Edited by Mordred
Link to comment
Share on other sites

I took that to mean that the inside of a BH "is not as predicted by GR" in other words it is not a singularity.

 

It doesn't say anything about "inside" or "singularity". It seems to me to be just testing whether the (merged) black hole behaves as predicted by GR. If it doesn't, that will be pretty exciting because that may provide the information needed to work out how GR needs to be modified.

 

It is possible that a version of GR that includes quantum effects might mean that event horizons are not as impenetrable as we currently think. But that is all unknown at the moment. We can only work with the theory we have.

OK but once you have particles you are not going to get them to move at the speed of light,

 

Forget about particles. We are dealing with black holes that have mass but no structure (in current theory).

Link to comment
Share on other sites

quote #1

As you approach a gravitational body the average mass [of spacetime] will steadily increased depending on the energy/density change as a function of radius and the changes in the stress/energy tensor.

 

quote #2

The term vacuum doesn't mean zero energy/density.

Thankfully we can set the background metric as zero even if the background metric is a positive energy/density.

 

Not all the mass is inside the Schwartzchild radius. Any spacetime region of higher gravitational potential has mass.

 

 

Robittybob1:

I regret saying I was done with this thread.

I've found it difficult to work out the conventions used above, especially as my understanding of GR and especially black holes is limited. You seem to have a similar problem.

I'll go back to my Apollo example and use a pseudoclassical qualitative (ie mostly accurate) approach using what I think are Mordred's conventions. I'm sure he'll quickly correct me as necessary.

 

 

 

In the Apollo capsule special relatvityish reference frame:

 

As the capsule falls towards earth tidal forces (from the nearer part being increasingly attracted to earth more than the more distant part) increase; the increasing stress gives the capsule more energy - analogous to quote #1.

 

If the tidal tidal forces were sufficiently large, the capsule would be torn apart and form an accretion disc.

 

Setting the capsule's gravitational potential energy at all times to 0 (quotes #1 and #2), the earth/moon's gravitational potential energy increases enormously as the capsule falls.

[Mordred may be using an event horizon or similar as a reference, with similar effect.]

 

With this approach, the capsule's kinetic energy, transformed to heat in the atmosphere, has been created from nothing as has the increase in gravitational potential energy of everything but the capsule.

 

All this for a special relativityish approach when general relativity would be more helpful - specifically using a distant place as an unchanging reference for gravitational potential energy.

 

 

 

A rehash of a conventional energy conservation approach using distant gravitational potential as a zero reference:

 

As the capsule falls its gravitational potential energy reduces [becomes more negative] and is transformed to kinetic energy, a little gravitational radiation, heat etc.

The increased (tidal) energy in space around the capsule is far less than the reduction due to lower gravitational potential energy.

 

Dealing with two orbiting BHs is much more complex but a similar approach can be used.

 

 

 

That's why I stated the mass or energy that causes gravity waves do not emit from inside the event horizon.

The energy for gravity waves originate from the spacetime geometry outside the event horizon. Not from within.

We might have to look at that in another thread. It doesn't seem right for there would have to be some mechanism to replenish the energy on the outside if that was the case.

 

Using an energy conservation approach:

The energy for gravity waves does not originate from the spacetime geometry outside the event horizon. Nor from within.

Set gravitational P.E. at zero at great distance from the BH.

As a mass (or another BH) approaches the BH its gravitational P.E. becomes increasingly negative due to the BH's gravity and this P.E. is transformed to kinetic energy, heat, gravitational waves etc.

As above, with limited exceptions due to extreme conditions, the locally higher 'tidal' energy aka 'spacetime geometry' is less than the negative P.E.

Spacetime geometry does not originate energy for gravity waves.

All of the energy produced is from the incoming mass interacting with the BH's gravity and other incoming masses.

Black holes and their environs do not create energy.

Link to comment
Share on other sites

I don't recall ever stating that energy is being created. Other than that your explanation is good. However you can see from the curvature statement I quoted from the wikilink that the curvature changes occur. If you look at the curvature to stress/tensor relations. You can see the curvature equates to energy momentum...

Via the stress/energy tensor. The formulas involved are posted in this thread.

Here is the stress/energy tensor relations to the energy density and pressure.

 

[latex]T^{\mu\nu}=(\rho+p)U^{\mu}U^{\nu}+p\eta^{\mu\nu}[/latex]

 

 

what this equation means is that as the gravitational potential increases

 

So does the energy/mass density

I think the equation that may suit best on energy/density to curvature relations may be better seen here.

 

[latex]R_{ab}-\frac{1}{2}Rg_{ab}=8\pi GT_{ab}[/latex]

 

[latex]R_{ab}[/latex] is the ricci tensor,[latex] g_{ab}[/latex] is the curved by the presence of energy via the stress tensor [latex]T_{ab}[/latex]. G is the gravitational constant. R is the ricci scalar.

 

Conceptually this equation means curvature=energy...

Equation 1.1 the curvature equals energy statement is footnote 1.2.

 

http://www.google.ca/url?q=http://www.physics.usyd.edu.au/~luke/research/masters-geodesics.pdf&sa=U&ved=0ahUKEwj25uq8qOzLAhVM4WMKHWm4Ca0QFggRMAA&usg=AFQjCNEr4WEHhcvoL-LVhqBLVIcgBRFdkQ

Now as I understand the Schwartzchild metric it assumes the background vacuum =0. However I don't believe it states the curvature has zero energy/density.

Without going into energy being a property lets avoid gravitons and use gravitational field energy.

 

 

(Of course we could get into Komar,Bondi and ADM mass but I don't we really need to.) As it would be off the OP topic

 

https://en.m.wikipedia.org/wiki/Mass_in_general_relativity

You can see the first equation equates gravitational potential energy to the energy/density and pressure relations.

The second equation relates those relations to the curvature.

That's the part I think your missing.

Edited by Mordred
Link to comment
Share on other sites

I don't recall ever stating that energy is being created. Other than that your explanation is good. However you can see from the curvature statement I quoted from the wikilink that the curvature changes occur. If you look at the curvature to stress/tensor relations. You can see the curvature equates to energy momentum...

Via the stress/energy tensor. The formulas involved are posted in this thread.

 

You can see the first equation equates gravitational potential energy to the energy/density and pressure relations.

The second equation relates those relations to the curvature.

That's the part I think your missing.

Is it possible to put values on the GR terms so we can check those statements?

Link to comment
Share on other sites

I have no idea why you would have difficulty in understanding that the tensors involve scalar and vector field strength at each coordinate.

 

A vector field has both direction and magnitude. How do you define a magnitude without an energy value.

 

Why would you have a problem with a gravitational potential energy at a given location on a spacetime curvature as having an energy/density relation?

 

Does the term gravitational potential energy not mean a specific type of energy ?

 

 

At a given radius from a mass the force of gravity has a field strength. That field strength can be calculated via f=ma.

 

Those equations above equate this to the curvature and stress/energy momentum

Edited by Mordred
Link to comment
Share on other sites

Mordred are you talking to me when you say "I have no idea why you would have difficulty in understanding that the tensors involve scalar and vector field strength at each coordinate."?

 

Who is that "you"?

 

Personally I do have difficulty. Let's put some numbers on these symbols.

Edited by Robittybob1
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.