What difference does it make if the Schwarzschild radii touch?

[Mordred]

Forget the term mass. Just think of the curvature the term mass isn't particularly well defined in GR.

 

Yes mass causes curvature but this doesn't strictly mean rest mass, or inertial mass, Komar mass, Bondi mass, ADM mass etc etc etc.

 

How we measure mass or energy is also subjective to the observer.

Edited April 3, 2016 by Mordred

[Robittybob1]

Forget the term mass. Just think of the curvature the term mass isn't particularly well defined in GR.

 

Yes mass causes curvature but this doesn't strictly mean rest mass, or inertial mass, Komar mass, Bondi mass, ADM mass etc etc etc.

I would love to be able to do that but what is the image? I don't want to think in terms of curved rubber sheets. I have wanted to put some values on the curvature so I can begin. How do you think of curvature?

Do you just say "curvature" but have no mental image? Is it just the same as saying gravity for no one has an image of gravity either?

Edited April 3, 2016 by Robittybob1

[Mordred]

The rubber sheet analogy is useful if you think of the observer being on the sheet itself. In this regard it doesn't do a bad job of mapping light paths.

 

The problem is when you think in terms of Mass and orbits there is the tendency to ignore observer affects.

 

It tends to lead to a more Newtonian

way of drawing conclusions.

 

This is fine if you remember to include relativity effects.

 

For example some of the furmulas used for gravity waves are fairly Newtonian like. Those work fine but you have to be careful what conclusions you draw from them.

 

 

This formula for example is a good one.

 

[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]

 

In essence it's a Newtonian formula.

 

The problem with formula comes into play is the measured mass of M_1 and M_2. Do you use the rest mass or do you also include the inertial mass?

Edited April 3, 2016 by Mordred

[Robittybob1]

 

Inside (or maybe at) the event horizon. If we knew any more than that it would violate the "no hair" theorem.

 

 

Yes, because I have watched simulations of the merger of black holes.

 

 

Why would the speed change its shape?

No hair theorem - a BH can be characterised by 3 parameters:

1. Mass

2. Momentum

3. Electric charge

 

In the list of where the mass could be. If it was a solid core then overlapping of the EH would be a big deal.

The rubber sheet analogy is useful if you think of the observer being on the sheet itself. In this regard it doesn't do a bad job of mapping light paths.

 

The problem is when you think in terms of Mass and orbits there is the tendency to ignore observer affects.

 

It tends to lead to a more Newtonian

way of drawing conclusions.

 

This is fine if you remember to include relativity effects.

 

For example some of the furmulas used for gravity waves are fairly Newtonian like. Those work fine but you have to be careful what conclusions you draw from them.

 

 

This formula for example is a good one.

 

[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]

 

In essence it's a Newtonian formula.

 

The problem with formula comes into play is the measured mass of M_1 and M_2. Do you use the rest mass or do you also include the inertial mass?

Is there such a thing as gravitational mass so you don't need to worry about that? "P" what was that formula calculating?

[Mordred]

The formula calculates the radiated power of a gravity wave.

 

No it's not so simple for gravitational mass. That's what Newtonian theory uses.

 

However this theory ran into a problem with Mercury.

 

Gravitational mass works fine until you add significant relativity effects.

 

Say you use that formula on two objects moving slowly, with a decent separation distance. Extremely accurate in this case.

 

However if both objects are moving at 0.5 c you would get the wrong radiated power.

( This is hinting at the radiated mass value in the Ligo paper disceptancy you asked about)

 

 

Now let's really blow your mind away.

 

Let's compare the Schwartzchild (uncharged, non rotating)

metric to the Kerr metric (uncharted, rotating.

 

Schwartzchild case. Event horizon is

 

[latex]r_s=\frac{2GM}{c^2}[/latex]

 

Kerr metric. (You have not one but two event horizons,one inner and one outer).

Outer.

 

[latex]r_o=m+\sqrt{m^2-a^2}[/latex]

 

Inner

 

[latex]r_i=m-\sqrt{m^2-a^2}[/latex]

In the Kerr metric form the last equations are given differently.

 

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

 

they use the Schwartzchild radius to calculate the inner/outer event horizons.

The formulas on the wikilink are more accurate as it will show that the event horizons are slightly flattened. Wider on the equator.

( The first two equations are a simplified form ignoring frame dragging and working with the singularity mass.)

Edited April 3, 2016 by Mordred

[Strange]

Just confirm for me please: a BH has mass beyond the EH and that mass produces/has gravity?

 

The black hole has mass (ignoring the complexities). That is all you know / need to know. You can't know "where" that mass is (if it is anywhere) with our current theories and it makes no difference.

 

And gravity isn't like light. It doesn't have to "get out". The mass of the black hole causes (or *is*) the curvature that we call gravity.

[Robittybob1]

 

The black hole has mass (ignoring the complexities). That is all you know / need to know. You can't know "where" that mass is (if it is anywhere) with our current theories and it makes no difference.

 

And gravity isn't like light. It doesn't have to "get out". The mass of the black hole causes (or *is*) the curvature that we call gravity.

Gosh. So a larger BH (has more mass) and has a wider area (volume) of curvature. For would you say they all have the same curvature at the EH. Have you got a favourite image of the spacetime curvature round a black hole?

Do you think of space itself falling into the BH?

Edited April 3, 2016 by Robittybob1

[Mordred]

Does a ball of one metre in radius have the same curvature as a ball 10 metres in radius ?

 

No spacetime coordinates do not fall into a BH.

[Robittybob1]

The formula calculates the radiated power of a gravity wave.

 

No it's not so simple for gravitational mass. That's what Newtonian theory uses.

 

However this theory ran into a problem with Mercury.

 

Gravitational mass works fine until you add significant relativity effects.

 

Say you use that formula on two objects moving slowly, with a decent separation distance. Extremely accurate in this case.

 

However if both objects are moving at 0.5 c you would get the wrong radiated power.

( This is hinting at the radiated mass value in the Ligo paper discrepancy you asked about)

 

 

Now let's really blow your mind away.

 

....

 

Kerr metric. (You have not one but two event horizons,one inner and one outer).

....

As an aside: I was impressed with what New Zealander Roy Kerr had calculated I phoned him up and thanked him (about 16 years ago). He said "it was nothing", but recently it was suggested he might receive the Nobel prize for his work on rotating black holes.

Does a ball of one metre in radius have the same curvature as a ball 10 metres in radius ?

 

No spacetime coordinates do not fall into a BH.

Do all circles have the same curvature? I'm tempted to say "no" but I'll accept correction. http://www.math.washington.edu/~lee/Books/Riemannian/c1.pdf

 

 

Theorem 1.5 says that two circles are congruent if and only if they have the same curvature,

That seems to suggest different curvatures are possible.

[Mordred]

[latex]k=\frac{1}{r}[/latex]

 

So correct the curvature varies with the radius.

[Robittybob1]

[latex]k=\frac{1}{r}[/latex]

 

So correct the curvature varies with the radius.

But at the EH there would always be the same curvature (gradient) at that point only. It is the point where the curvature equals the escape velocity. So the tangent to the curvature will have the same slope I presume. Was that the point you were making?

[Mordred]

The event horizon in 3D for a Schwartzchild blackhole is a sphere with all the same mathematics that can be applied to a sphere.

 

The radius of that sphere is given by the Schwartzchild radius formula posted previously.

 

The curvature of that sphere is the last formula I posted. That curvature is at any point along the circumference of that radius.

 

 

The 2d hyperslice curvature as you move away from the Schwartzchild radius would follow the Flamms parabloid

 

https://en.m.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid.

 

 

describing curvature as you approach a BH in 3d is rather complex. So you will have to often use slices.

 

Nearly impossible in 4d.

 

No particle will follow the Flamms parabloid as the time coordinate is set as constant.

Well here is a handy list of representations corresponding to some of the different geometry changes.

 

http://www.google.ca/url?q=http://www.rpi.edu/dept/phys/Courses/Astronomy/CurvedSpacetimeAJP.pdf&sa=U&ved=0ahUKEwiGorW_1fPLAhVkn4MKHX_jBK8QFgglMAY&usg=AFQjCNEQurpyze2nnu2lCkZ7nnuVmeW5Fg

 

has a good math break of the different relations involved depending on which coordinates are changing.

(You can see from it some of the shapes can get extremely complex.)

Edited April 4, 2016 by Mordred

[Robittybob1]

The event horizon in 3D for a Schwartzchild blackhole is a sphere with all the same mathematics that can be applied to a sphere.

 

The radius of that sphere is given by the Schwartzchild radius formula posted previously.

 

The curvature of that sphere is the last formula I posted. That curvature is at any point along the circumference of that radius.

 

 

The 2d hyperslice curvature as you move away from the Schwartzchild radius would follow the Flamms parabloid

 

https://en.m.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid.

 

 

describing curvature as you approach a BH in 3d is rather complex. So you will have to often use slices.

 

Nearly impossible in 4d.

 

No particle will follow the Flamms parabloid as the time coordinate is set as constant.

Well here is a handy list of representations corresponding to some of the different geometry changes.

 

http://www.google.ca/url?q=http://www.rpi.edu/dept/phys/Courses/Astronomy/CurvedSpacetimeAJP.pdf&sa=U&ved=0ahUKEwiGorW_1fPLAhVkn4MKHX_jBK8QFgglMAY&usg=AFQjCNEQurpyze2nnu2lCkZ7nnuVmeW5Fg

 

has a good math break of the different relations involved depending on which coordinates are changing.

(You can see from it some of the shapes can get extremely complex.)

So does the "curvature" of a circle (the EH) relate to the curvature of spacetime? I've never heard that before.

Have you got some reference to that?

You might have thrown me into the deep end before I know how to swim. "Schwarzschild metric"!

Edited April 4, 2016 by Robittybob1

[Mordred]

You already know the radius of the event horizon. Which gives you the point of no return hyperslice.

 

That is the equation I posted earlier.

 

[latex]r_s=\frac{2GM}{c^2}[/latex]

 

that is a specific coordinate hyperslice. It does not represent the hyperslice further away from a BH

 

Of the same mass.

 

Yes this is complex. Welcome to relativity. There is no easy way to understand all the relations without intensive mathematical study.

 

When you ask for a curvature relations you must specify by which coordinates are you referring to.

 

If you set the time coordinate at a constant value say at rest.you get a different curvature than an observer at a different time reference frame. One that is moving.

Edited April 4, 2016 by Mordred

[Robittybob1]

You already know the radius of the event horizon. Which gives you the point of no return hyperslice.

 

That is the equation I posted earlier.

 

[latex]r_s=\frac{2GM}{c^2}[/latex]

 

that is a specific coordinate hyperslice. It does not represent the hyperslice further away from a BH

 

Of the same mass.

 

...

You are right.

[Robittybob1]

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

That was #76

Is the event horizon spinning in the case of a spinning BH? If the spin is in the same direction as the orbital motion would there need to be the need for the addition of velocities using the Lorentz factor? https://en.wikipedia.org/wiki/Lorentz_factor

Edited April 4, 2016 by Robittybob1

[Strange]

Gosh. So a larger BH (has more mass) and has a wider area (volume) of curvature.

 

I don't know what you mean by area or volume of curvature. The curvature caused mass-energy extends infinitely (in principle).

 

Do you think of space itself falling into the BH?

 

The Gullstrand-Painlevé coordinates can be thought of in that way.

http://www.physicspages.com/2013/11/24/painleve-gullstrand-global-rain-coordinates/

http://jila.colorado.edu/~ajsh/insidebh/waterfall.html

Is the event horizon spinning in the case of a spinning BH?

 

I'm not sure you can say the event horizon is spinning, after all it isn't a "thing". The black hole has angular momentum, but I'm not sure you can say where that "is", any more than you can say where the mass is.

 

If the spin is in the same direction as the orbital motion would there need to be the need for the addition of velocities using the Lorentz factor? https://en.wikipedia.org/wiki/Lorentz_factor

 

It clearly has an effect (as seen in the calculation of gravitational waves) but it won't be that simply because we are dealing with something that can only be described using GR (the Lorentz transform only applies in locally flat [Minkowski] space-time).

[Robittybob1]

 

I don't know what you mean by area or volume of curvature. The curvature caused mass-energy extends infinitely (in principle).

 

 

The Gullstrand-Painlevé coordinates can be thought of in that way.

http://www.physicspages.com/2013/11/24/painleve-gullstrand-global-rain-coordinates/

http://jila.colorado.edu/~ajsh/insidebh/waterfall.html

 

I'm not sure you can say the event horizon is spinning, after all it isn't a "thing". The black hole has angular momentum, but I'm not sure you can say where that "is", any more than you can say where the mass is.

 

 

It clearly has an effect (as seen in the calculation of gravitational waves) but it won't be that simply because we are dealing with something that can only be described using GR (the Lorentz transform only applies in locally flat [Minkowski] space-time).

Thanks Strange. I like the images of 3D spacetime as opposed to the rubber sheets. Like this http://cdnimg.visualizeus.com/thumbs/df/49/grid,planet,space,spacetime-df4956ec0ccee7ecf4f84ea9439b57d0_h.jpg?ts=93246

 

There seems to be the idea of frame dragging so space rotates with the spinning BH.

OK (the Lorentz transform only applies in locally flat [Minkowski] space-time).

[Strange]

There seems to be the idea of frame dragging so space rotates with the spinning BH.

 

True. Although I don't think it is quite as simple as "space rotating" (I don't really understand the maths around it) and I'm not sure how it relates to the angular momentum of the black hole. (It applies to all rotating masses.)

 

I keep thinking of it as the Len Stirring effect:

https://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precession