[Kryptid]

Okay, now for some calculations.

Since the total mass of an electron-positron pair is 1.022 MeV, a photon must also have at least this much energy in order to be converted into such a pair (by interacting with an atomic nucleus, for example).

Hawking provided us with equations that allow us to predict the properties of Hawking radiation based on a black hole’s mass. The following calculator does this: https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator

When we put in a peak photon energy of 1.022 MeV, we get an associated black hole mass of around 4 x 10¹³ kilograms. This is somewhat smaller than the previous number I gave of 10¹⁴ kilograms, but the mechanism of producing the electron-positron pairs is also different. The electrons and positrons are not being produced directly as a part of Hawking radiation, but rather as the interaction of photon-based Hawking radiation interacting with some form of matter surrounding the black hole (either hydrogen atoms in the vacuum or perhaps accretion disk material).

Theoretically, electron-positron pairs can be produced from photon-photon interactions. This would require half the photon energy, since it involves two photons instead of one (with a resulting black hole twice as heavy as the previous calculation). The following paper discusses this phenomenon around black holes: https://pdfs.semanticscholar.org/f4bc/24655948d538fdca3804a41cbf812e39cf34.pdf

Of interest, the results of calculations done by Don Page can be seen in the image here: https://i.stack.imgur.com/vmSpB.jpg

To summarize the image, a black hole with a mass above of 10¹⁷ grams emits Hawking radiation with a consistency of 81.4% neutrinos and antineutrinos, 16.7% photons and 1.9 % gravitons, whereas one with a mass between 10¹⁷ grams and 10¹⁴ grams emits 45% positrons and electrons, 45% neutrinos and antineutrinos, 9% photons and 1% gravitons. Between 10¹⁴ grams and 3.16 x 10¹³ grams, the hole emits 12% neutrons and antineutrons, 28% electrons and positrons, 48% neutrinos and antineutrinos, 11% photons and 1% gravitons.

The image was found here: https://physics.stackexchange.com/questions/89983/how-many-of-which-particles-are-in-hawking-radiation

[Dave Lev (OP)]

With regards to the Tidal VS Lorentz force

The fatal flaw in your calculations is that you are considering the overall gravitational force, not the tidal force. Redo the calculations, but for tidal forces, and you'll be on the right track.

We have already agreed that Lorentz force under the magnetic field can do that job. Therefore, there is no need for tidal.

The Lorentz force would be there, but it wouldn't be "positive goes out and negative goes in". The Lorentz force would deflect the path of particles at a right angle to the field lines (assuming that they were already on a path perpendicular to the field lines. If they are parallel to the field lines, there is no force).

Therefore, the only open question is about the feasibility of a SMBH to create new pair of charged particles as a BH can do.
Let's start with a BH:

Yes. I don't know what it is off the top of my head, but the calculations have already been done by physicists (Don N. Page, for example). I do know that, according to those calculations, a black hole with a mass much larger than 1014 kilograms (about one-tenth the mass of Mar's satellite Deimos) doesn't produce electron-positron pairs.

However, in all the articles that you have offered they specifically focus on the black body radiation:
https://physics.stackexchange.com/questions/89983/how-many-of-which-particles-are-in-hawking-radiation
"The black body radiation (semiclassical form) description of Hawking radiation is realistic for the low temperatures of black holes acting as black bodies, because the energy needed to create a detectable particle from the vacuum fluctuation framework is large, 1 MeV for e+and e- , one of them falling back."
So, If I understand it correctly, they have calculated the requested energy that is needed to create those charged particles, extract the requested mass (or gravity) and then verify if the BH will can set the radiation.
So, if we will try to work with the same formulas of energy/gravity at SMBH, we can easily find that SMBH can also create those kinds of particles (at the relevant radius/energy). However, due to it's massive size, those particles would not be ejected and therefore, there will be no radiation. That is very clear to me.
However, in all of those articles, I couldn't find even one word about magnetism.
So, they have totally ignored the great impact of Lorenz force.
That was a severe mistake.
If our scientists will add the impact of the Lorentz force (especially at the SMBH) they should find how easy it is for the SMBH to eject those new born positive charged particles as a BH can do without the need for Lorentz force.
Therefore, the following statement is fully correct:

Super-massive black holes are many, many orders of magnitude more massive than that. So you don't get charged particles from them.

However, it is correct as long as we ignore the magnetism and Lorentz force.
Never the less, once we add Lorentz force, we should find that it can easily extract the radiation of the Positive charged particles from the SMBH.

[Kryptid]

We have already agreed that Lorentz force under the magnetic field can do that job. Therefore, there is no need for tidal.

Who is "we"? I never agreed that Hawking radiation can be created by a magnetic field.

However, in all of those articles, I couldn't find even one word about magnetism.
So, they have totally ignored the great impact of Lorenz force.
That was a severe mistake.
If our scientists will add the impact of the Lorentz force (especially at the SMBH) they should find how easy it is for the SMBH to eject those new born positive charged particles as a BH can do without the need for Lorentz force.

Absolutely not! I have explained this to you several times already: you can't use magnetism to pull particles out of an event horizon! The swapping of space-time coordinates inside of a black hole do not allow matter that is moving below the speed of light to travel away from the singularity. It's a consequence of the geometry of space inside of the horizon. It doesn't work. It can't be done. Period. The one and only way to escape from a black hole is by traveling faster than light, which is not a velocity that a magnetic field can provide.

Again, here is the link that speaks of space and time "swapping" such that escape is impossible for any particles travelling slower than light: http://www.jimhaldenwang.com/black_hole.htm

Notice how the minus sign has moved from the t coordinate to the r coordinate.  This means that inside the event horizon, r is the timelike coordinate, not t.  In relativity, the paths of material particles are restricted to timelike world lines.  Recall the discussion of timelike separation earlier in this paper (2).  It is the coordinate with the minus sign that determines the meaning of "timelike."  According to relativity, inside a black hole time is defined by the r coordinate, not the t coordinate.  It follows that the inevitability of moving forward in time becomes, inside a black hole, the inevitability of moving toward r = 0.  This swapping of space and time occurs at r = 2M.  Thus, r = 2M marks a boundary, the point where space and time change roles.  For the observer inside this boundary, the inevitability of moving forward in time means that he must always move inward toward the center of the black hole at r = 0.  All timelike and lightlike world lines inside r = 2M lead inevitably to r = 0 (the end of time!)  Because it is not possible for any particle or photon inside r = 2M to take a path where r remains constant or increases, the boundary r = 2M is called the event horizon of the black hole.  No observer inside the event horizon can communicate with any observer outside the event horizon.  The event horizon can be thought of as a one-way boundary.

Here is a video explaining it:

Another analogy for this is to visualize space as "flowing" into the black hole. Outside of the event horizon, this flow is moving below the speed of light. This allows particles that are moving quickly enough to escape. Inside the event horizon, however, space is flowing faster than light. Since no material object can move faster than light, it is impossible for anything to move against this flow. It must inevitably be pulled into the singularity. Even infinite force would be insufficient (as infinite force can only get you up to the speed of light, not beyond it).

Here is yet another explanation of space and time swapping roles (I want to be thorough with this because it's a strange concept that can be hard to understand): http://www.einstein-online.info/spotlights/changing_places.html

By the way, your agreement that black holes have positive mass means that the infalling negative mass particles formed from the Hawking process decrease the mass of the hole (exactly as modern physics says). Give it long enough and the hole eventually disappears.

[evan_au]

Some news: A recent outburst was seen in Sgr A*: It was suddenly 75x brighter than normal, when observed over a 2-hour period.
- It was observed in the infra-red, by the Keck telescope
- Normally, the black hole is hardly visible in the infra-red, but on this occasion, Sgr A* was brighter than S2, confusing the astronomers somewhat.
- During the observation period, Sgr A* was declining in brightness, so it was probably even brighter before they observed it.

It sounds like a big clump of matter got added to the accretion disk and/or swallowed by the black hole, causing a burst of infra-red.
See: https://www.universetoday.com/143150/milky-ways-black-hole-just-flared-growing-75-times-as-bright-for-a-few-hours/

[Dave Lev (OP)]

Some news: A recent outburst was seen in Sgr A*: It was suddenly 75x brighter than normal, when observed over a 2-hour period.
- It was observed in the infra-red, by the Keck telescope
- Normally, the black hole is hardly visible in the infra-red, but on this occasion, Sgr A* was brighter than S2, confusing the astronomers somewhat.
- During the observation period, Sgr A* was declining in brightness, so it was probably even brighter before they observed it.
See: https://www.universetoday.com/143150/milky-ways-black-hole-just-flared-growing-75-times-as-bright-for-a-few-hours/

The orbital period cycle of S2 is about 15.2 Years
https://www-istp.gsfc.nasa.gov/stargaze/Kep3laws.htm
In 2002 and in 2017 it was very close to SMBH.
The next time should be in 2032.
So, how could it be that in 2019 it comes back again to the SMBH?
Is it just to make the impossible – possible??

I have explained this to you several times already: you can't use magnetism to pull particles out of an event horizon!

The magnetism can pull particles out of an event horizon.
There is no need for a unrealistic high velocity. Lorentz force can do it!!!
This is very clear to me and I thought that it was also clear for you.
However, are you the same person that sent me the following reply #635?

Dave - 1. New created particles - New pair of particles are created constantly around the SMBH (at the event of horizon or below).
Kryptid - Right.
Dave- If one particle carry a positive charge, the other one gets a negative charge.
Kryptid - For very small black holes, this would be true. This would not be true for super-massive black holes as they have insufficient tidal forces to produce anything other than photons, gravitons and maybe neutrinos. Charged particles like electrons, protons or muons have too much mass-energy to be generated by the (relatively) weak tidal forces present.
Dave - 2. Magnetics field - Around the SMBH there is magnetic field. This magnetic field is quite strong at the event of horizon (or deeper?...)
Kryptid- Although true black holes cannot have such a magnetic field, I will submit to the possibility that something like MECOs (magnetospheric eternally collapsing objects) could, maybe, be what "black holes" actually are. So I will tentatively agree that "black holes" could be MECOs and as such could have magnetic fields. I'll consider this plausible for the sake of discussion: https://en.wikipedia.org/wiki/Magnetospheric_eternally_collapsing_object
Dave - 3. Lorentz force - Based on Lorentz force, the magnetic fields deflects differently the path of the orbital new born particles pair:
Hence, if the positive charged particle will be deflected outwards, the negative charged particle will be deflected inwards.
Therefore, while the negative is pushed inwards into the center of the SMBH, the positive is pulled outwards and get's eventually into the accretion disc.
Kryptid - The Lorentz force would be there, but it wouldn't be "positive goes out and negative goes in". The Lorentz force would deflect the path of particles at a right angle to the field lines (assuming that they were already on a path perpendicular to the field lines. If they are parallel to the field lines, there is no force).

So, you had agreed that the positive particles can be ejected outwards due to Lorentz force.
However, you were sure that the negative particles that fall in could decrease the size of the BH:

The contradiction is your claim that this process causes the black hole to grow in mass. The mass of the black hole has to shrink, not grow, as the negative mass (not negative charge, an important difference) particle is invariably the one that passes into the hole (because the swapping of time and space coordinates inside of the event horizon is what makes that particular particle have a negative mass in the first place). That negative mass subtracts from the positive overall mass of the hole, causing it to become smaller. If you are willing to accept this point, then I will agree that your model no longer violates the first law of thermodynamics.

I wonder why did you agree in reply 635 to accept the impact of the magnetism and Lorentz force, while now you have decided to withdraw from your willing to accept it.
I'm not going to force you to do it and I have no more questions about magnetism at this phase.

So, would you kindly answer the following question?

I have proved that whole your idea about the Virial theorem is totally wrong.
In that theory we must OBSERVED the orbital objects. As we can't observe the dark matter we can't know its orbital velocity (and if it has any sort of velocity)
Therefore our understanding of how the galaxy works must also be updated.
As we have based the gravity force that holds the Sun around the galaxy on the Virial theorem - than we must look for better theory.
How are you going to address this key issue?

[Kryptid]

The orbital period cycle of S2 is about 15.2 Years
https://www-istp.gsfc.nasa.gov/stargaze/Kep3laws.htm
In 2002 and in 2017 it was very close to SMBH.
The next time should be in 2032.
So, how could it be that in 2019 it comes back again to the SMBH?
Is it just to make the impossible – possible??

Nobody said that it did. Its closest approach was "in the middle of 2018" according to the article. The hypothesis that the article is talking about is the following:

It’s possible that SO-2’s close approach disrupted the way that material flows into Sgr. A*. That would generate the kind of variability and bright flaring that astronomers saw in May, about one year after the star’s close approach.

Since the star's closest approach was 17 light hours from the hole (18.7 billion kilometers), any disruption of the material orbiting at that distance would not have had any kind of immediate impact on the flow of material into the hole. It would have taken time for such a disruption in an accretion disk to propagate to the hole.

The magnetism can pull particles out of an event horizon.

No it can't. That would require the laws of physics to be broken. Magnetism could only move particles away from the hole if they are already outside of the horizon.

There is no need for a unrealistic high velocity. Lorentz force can do it!!!
This is very clear to me and I thought that it was also clear for you.

I feel as if you completely and utterly ignored everything in my last post in regards to space and time swapping inside of an event horizon. Almost the entirety of post 650 was about that subject, so it isn't possible that you merely missed it. I posted a video from PBS about it. I posted an analogy about space flowing into the hole faster than light. I posted two links explaining it in text form (one of which used visual figures to aid in the explanation). Go back and actually look at them. Nothing can get out of an event horizon because the inevitable flow of space into the singularity will not let it. It isn't the same as pushing yourself away from a conventional gravity field with rocket engines (or, in this case, with a magnetic field).

Right now, you are unavoidably travelling into the future. You can't stop that. It's the same with traveling towards the singularity inside of a black hole because space has taken on the unstoppable movement property of time. You can't stop yourself from going into the future outside the hole, and you can't stop yourself from moving towards the singularity inside the hole. Magnetism, even if it is literally infinitely strong, will make no difference.

So, you had agreed that the positive particles can be ejected outwards due to Lorentz force.

You are lying. What is truly bizarre about this is that you even quoted where I stated, "The Lorentz force would be there, but it wouldn't be "positive goes out and negative goes in"." So I stated the opposite of what you claim. What were you trying to gain by lying about what I said? I am genuinely beginning to wonder if you are a troll. I don't see how a serious person could continuously cite one thing and then say the exact opposite in the same post the way you do. I refuse to believe that your reading comprehension skills are really that bad (unless, perhaps, you have some kind of language disability akin to dyslexia that I have never heard of before). If you did, then I would be more understanding.

However, you were sure that the negative particles that fall in could decrease the size of the BH:

Negative mass particles, not negatively-charged particles. Don't make that mistake again.

I wonder why did you agree in reply 635 to accept the impact of the magnetism and Lorentz force, while now you have decided to withdraw from your willing to accept it.

You are being deceptive. I agreed that magnetism would have an effect on electrically-charged particles generated as Hawking radiation (which are outside of the horizon), but at no point did I agree that magnetism can pull them out of an event horizon. It can't.

So, would you kindly answer the following question?

I already said, "Since when have I ever said anything about virial theorem, let alone say that it was "my idea"? Or are you talking to Halc here?"

I don't know much about virial theorem or what it has to do with super-massive black holes, so I don't even know how to answer your question.

[Kryptid]

At unequal velocities, one has momentum that is not the negative of the other. The sum of the momentum of the system is not zero.

If they started at unequal velocities, then shouldn't the system have started with a momentum sum that wasn't zero anyway?

[Dave Lev (OP)]

Dear Kryptid
Sorry as I didn't understand you correctly.

Magnetism could only move particles away from the hole if they are already outside of the horizon.

I agreed that magnetism would have an effect on electrically-charged particles generated as Hawking radiation (which are outside of the horizon), but at no point did I agree that magnetism can pull them out of an event horizon. It can't.

So, do you mean that if there is a positive particle out of the event of horizon of a SMBH (But very close to it), based on magnetism and Lorentz force this particle could be ejected outwards. However, if it is located inwards into the event of horizon, than there is no way to push it outwards even if the magnetism is infinity?

you can't stop yourself from moving towards the singularity inside the hole. Magnetism, even if it is literally infinitely strong, will make no difference.

In order to get better understanding:
Let assume that one positive particle is located one meter inwards from the Event of horizon and the other one is located one meter outwards from the event of horizon
Both particles orbits almost at the same velocity around the SMBH.
Therefore:
Do you agree that the gravity force on both two positive particles is finite and almost identical?
If the magnetism is infinite do you agree that Lorentz force should also be infinite?
If so, I can't understand why infinite Lorentz force can only extract the particle that orbits outwards from the event of horizon but can't do it with the one that orbits two meter inwards?
Let me offer a solution:
1. A particle that orbits inside the event of horizon (even one meter inwards) gets infinite gravity force in order to overcome the infinite Lorentz force. Is it feasible?
2.  A particle that orbits inside the event of horizon doesn't get infinite gravity force. However, the Lorentz force there is zero. So, could it be that once we cross the event of horizon the magnetic force goes to zero?

[Kryptid]

So, do you mean that if there is a positive particle out of the event of horizon of a SMBH (But very close to it), based on magnetism and Lorentz force this particle could be ejected outwards.

Possibly. It depends on the direction the particle is travelling in and the relative orientation of the magnetic field.

However, if it is located inwards into the event of horizon, than there is no way to push it outwards even if the magnetism is infinity?

Yes, that's exactly what I'm saying. In order to escape an event horizon, you have to travel faster than the speed of light. Infinite force, whether it is produced by a magnetic field or otherwise, can only get you up to the speed of light. That isn't enough.

In order to get better understanding:
Let assume that one positive particle is located one meter inwards from the Event of horizon and the other one is located one meter outwards from the event of horizon
Both particles orbits almost at the same velocity around the SMBH.

There are no stable orbits inside of an event horizon. In fact, there are no stable orbits within a radius of 1.5 times that of the event horizon (called the photon sphere). At this distance, only objects moving at the speed of light can remain in orbit. Inside of this radius, not even light can stay in orbit.

Do you agree that the gravity force on both two positive particles is finite and almost identical?

The gravitational force is finite, but that is not what inevitably draws particles towards the singularity once they get inside the horizon. The particles are driven by the swapping of space and time. Time travels forwards outside the horizon whether you want it to or not. Space travels inwards towards the singularity whether you want it to or not. It's like trying to swim upstream when the current is moving too fast for even an infinite force to swim against.

If the magnetism is infinite do you agree that Lorentz force should also be infinite?

Yes, but infinite force won't help. Infinite force can only get you up to the speed of light, not beyond it.

If so, I can't understand why infinite Lorentz force can only extract the particle that orbits outwards from the event of horizon but can't do it with the one that orbits two meter inwards?

It's for the same reason that you can't stop getting older. Space behaves like time inside the horizon.

1. A particle that orbits inside the event of horizon (even one meter inwards) gets infinite gravity force in order to overcome the infinite Lorentz force. Is it feasible?

Orbits don't exist inside the horizon nor is the gravity there infinite in strength.

2.  A particle that orbits inside the event of horizon doesn't get infinite gravity force. However, the Lorentz force there is zero. So, could it be that once we cross the event of horizon the magnetic force goes to zero?

Magnetism probably doesn't go to zero. I think I've seen papers about magnetic field lines (from an outside source) threading the event horizons of spinning black holes as a way to draw angular momentum away from them.

[Dave Lev (OP)]

Thanks Kryptid
So, the info is as follow:
1. A positive particle out of the event of horizon will be ejected outwards due to magnetism and Lorentz force
2. A positive particle in the event of horizon won't be ejected outwards due to space time.
3. Magnetism in the event of horizon doesn't go to zero.

If that is correct, let me ask the following:

1. Space time.
If I understand it correctly, Minkowski had developed the Space time module for the Universe.
Based on this module our scientists have found that there must be a curvature in our universe.
However, so far they didn't find any curvature in our universe.
Therefore, my personal understanding is that the space time is just a module which doesn't necessarily represents the real Universe  - especially, once we get into the edge of the module.
Now our scientists have decided to use this model also for the aria inwards the event horizon.
Why is it? Based on what data?
What is the chance that this model is not relevant for the event horizon?
Can you please prove that space time module works also at the event of horizon?
How the extreme conditions at that aria could affect the space time?
If we use this module for the event of horizon, why we can't use it for outside that aria?
Why not using it for the center Bulge or even for the whole galaxy?
Why do we insist to fix it only for that limited aria - Event horizon?


2. Speed limit -
We all know that the maximal speed limit is the speed of light.
We also know that if we go in one direct line, we will never come back to the starting point.
However, based on the space time module, we have discovered that if we go in one direct line, we might come back to the same starting point.
Therefore in the same token - if the space time can set a curvature in space, why it can't eliminate the speed limit?
Why under the space time module, the speed can't get to infinite.
Did we try to verify this issue in our space time modeling?

3. Creation of new particle
You have already confirmed that new particle can be created below the event of horizon in a BH.
This particle can even be ejected from the event horizon if the BH is small enough.
That proves that particle can orbits below the event horizon.
Now you claim that:

Orbits don't exist inside the horizon nor is the gravity there infinite in strength.

How could it be?

[Halc]

Thanks Kryptid
So, the info is as follow:
1. A positive particle out of the event of horizon will be ejected outwards due to magnetism and Lorentz force

Even a regular stellar black hole is far too large to create charged particles.  Photons are unaffected by magnetism and such.

1. Space time.
If I understand it correctly, Minkowski had developed the Space time module for the Universe.
Based on this module our scientists have found that there must be a curvature in our universe.
However, so far they didn't find any curvature in our universe.

It is found wherever mass is nearby, like here on Earth.

We also know that if we go in one direct line, we will never come back to the starting point.

True.  One cannot cross the event horizon, and you would need to cross it to go all the way around if there was an 'around' to go.

However, based on the space time module, we have discovered that if we go in one direct line, we might come back to the same starting point.

I don't know what 'the space time module' is.  Minkowski didn't develop a module.  A model perhaps, but that model is a local model, not one of the universe.

Therefore in the same token - if the space time can set a curvature in space, why it can't eliminate the speed limit?
Why under the space time module, the speed can't get to infinite.

Maximum speed of c follows from frame independent speed of light, not from curved spacetime.

You have already confirmed that new particle can be created below the event of horizon in a BH.

We have not. We're just not in denial of the possibility.

This particle can even be ejected from the event horizon if the BH is small enough.

No it cannot.

That proves that particle can orbits below the event horizon.

Again, no it cannot. Asserting nonsense is not proof of anything. You can't even orbit anywhere close to the event horizon on the outside.

[Kryptid]

1. A positive particle out of the event of horizon will be ejected outwards due to magnetism and Lorentz force

This is no more likely than a negatively-charged particle being ejected.

1. Space time.
If I understand it correctly, Minkowski had developed the Space time module for the Universe.
Based on this module our scientists have found that there must be a curvature in our universe.
However, so far they didn't find any curvature in our universe.
Therefore, my personal understanding is that the space time is just a module which doesn't necessarily represents the real Universe  - especially, once we get into the edge of the module.
Now our scientists have decided to use this model also for the aria inwards the event horizon.
Why is it? Based on what data?

You are talking about a universe with overall curvature. That is a different matter than the curvature of space-time around a massive body due to gravity. A massive body will produce space-time curvature regardless of whether the Universe as a whole has any curvature or is flat. Don't confuse local curvature with universal curvature.

What is the chance that this model is not relevant for the event horizon?
Can you please prove that space time module works also at the event of horizon?
How the extreme conditions at that aria could affect the space time?
If we use this module for the event of horizon, why we can't use it for outside that aria?
Why not using it for the center Bulge or even for the whole galaxy?
Why do we insist to fix it only for that limited aria - Event horizon?

It's only a matter of degree. Space-time mixing happens in every day life as well, but the effects are too subtle to notice unless you have sufficiently-sensitive equipment (the phenomena of length contraction and time dilation are manifestations of this mixing). Here is an excerpt from "Black Holes & Time Warps" about this:


spacetimemix.jpg (622.89 kB . 711x1198 - viewed 3819 times)

It's just that, at the event horizon, the conditions become so extreme that the mixing becomes a complete swapping so that space becomes fully time-like. This is what defines the event horizon and what makes it different from other locations in a galaxy.

Therefore in the same token - if the space time can set a curvature in space, why it can't eliminate the speed limit?
Why under the space time module, the speed can't get to infinite.
Did we try to verify this issue in our space time modeling?

There is no reason for this to be the case. The speed of light in a vacuum has consistently been measured as finite and constant regardless of reference frame.

This particle can even be ejected from the event horizon if the BH is small enough.

Only if it is outside the horizon.

That proves that particle can orbits below the event horizon.

I have no idea how you came to that conclusion, but it's wrong. The closer you get to a gravitational source, the faster your orbit has to be. At the photon sphere, that required orbital speed is the speed of light. Inside of that radius, you would have to move faster than light to remain in orbit, which cannot be done with either light or material objects. So merely orbiting is not an option to prevent you from falling in. Please keep in mind that the inability to orbit is not the same as the inability to move away from the black hole. You can do that for so long as you are still outside the horizon.

[Dave Lev (OP)]

You are talking about a universe with overall curvature. That is a different matter than the curvature of space-time around a massive body due to gravity. A massive body will produce space-time curvature regardless of whether the Universe as a whole has any curvature or is flat. Don't confuse local curvature with universal curvature.

How do we know that  "A massive body will produce space-time curvature regardless of whether the Universe as a whole has any curvature or is flat?

spacetimemix.jpg (622.89 kB . 711x1198 - viewed 3 times)
It's just that, at the event horizon, the conditions become so extreme that the mixing becomes a complete swapping so that space becomes fully time-like. This is what defines the event horizon and what makes it different from other locations in a galaxy.

In the articale which you have offered there is not even one word about event horizon.
So, why did you decide to set the space time only below the event horizon?
There must be a solid proof for that. Would you kindly offer it?
However, if there is no mathematical roof for that, why the accretion disc can't be also part of the space time?
It is also around a massive Body.  The conditions there are also extreme. The orbital velocity could be above 0.3c.
If we can claim that the accretion disc could also be part of the space time, than nothing should also be ejected from it.
Why can't we look at the whole central bulge as space time?
Can you please prove it?

[Kryptid]

How do we know that  "A massive body will produce space-time curvature regardless of whether the Universe as a whole has any curvature or is flat?

The consequences of that curvature (time dilation, gravitational lensing, relativistic orbital precession, frame dragging and the geodetic effect) have been measured.

In the articale which you have offered there is not even one word about event horizon.
So, why did you decide to set the space time only below the event horizon?

What do you mean I decided to "set the space time"? Space-time is everywhere, not only below the horizon.

There must be a solid proof for that. Would you kindly offer it?
However, if there is no mathematical roof for that,

Go back and watch that PBS video I posted. They go into the math of the behavior of space-time below the horizon.

why the accretion disc can't be also part of the space time?

It is a part of space-time. Everything we know of is. Why wouldn't it be? The gravity in that area isn't strong enough to swap space and time, though.

It is also around a massive Body.  The conditions there are also extreme. The orbital velocity could be above 0.3c.

Not extreme enough. The escape velocity has to exceed the speed of light for this space-time swapping to take place.

If we can claim that the accretion disc could also be part of the space time, than nothing should also be ejected from it.

It's not inside of an event horizon.

Why can't we look at the whole central bulge as space time?
Can you please prove it?

Again, space-time is everywhere so I don't know what you are talking about.

[Dave Lev (OP)]

Thanks Halc and Kryptid

where the accretion disc is is part of spacetime, as is Earth.

What do you mean I decided to "set the space time"? Space-time is everywhere, not only below the horizon.

space-time is everywhere so I don't know what you are talking about.

So now you both agree that even the accretion disc is under the space time.
However - we clearly see that 99% of the matter in the accretion disc is ejected outwards.
Hence
How could it be that the accretion disc which is fully under the impact of space time can eject most of its mass, while the event of horizon can't do so just because of the same space time.
You have stated that based on space-time, nothing could be ejected from inside the horizon:
Please see the following:

The gravitational force is finite, but that is not what inevitably draws particles towards the singularity once they get inside the horizon. The particles are driven by the swapping of space and time. Time travels forwards outside the horizon whether you want it to or not. Space travels inwards towards the singularity whether you want it to or not. It's like trying to swim upstream when the current is moving too fast for even an infinite force to swim against.

Why space time that works so nicely at the event of horizon and prevents from any particle to be ejected outwards, works so badly at the accretion disc and has no resistance for the massive ejection at that aria?
In other words -
If the space time has no objection that most of the particle can be ejected from the accretion disc, could it be that it also has no objection that particles can be ejected also below the horizon?
Why it prevents from particles to be ejected from the horizon, while it has no resistance that most of the particles can be ejected from the accretion disc?
Do you have better idea than space-time to show the difference between the horizon and the accretion?

[Halc]

So now you both agree that even the accretion disc is under the space time.

Spacetime is simply where and when events occur.

the accretion disc which is fully under the impact of space time

Spacetime has no impact.  As I said, it is simply where and when events occur.  It is a geometric 4D space in which a coordinate system can be mapped.

You have stated that based on space-time, nothing could be ejected from inside the horizon:

For the same reason I cannot give an apple to a hungry mammoth, no matter what force I put on the apple.  I am in the future of said mammoth and cannot push the apple back in time.  It isn't a matter of lacking the energy or force to do it.

Why space time that works so nicely at the event of horizon

Spacetime does not work so nicely at the event horizon.  There is a singularity there, meaning all the laws of physics break down.  Lacking a unified theory, we cannot meaningfully describe what goes on at the event horizon.

[Kryptid]

So now you both agree that even the accretion disc is under the space time.

"Under the space time" makes no sense. How can you be "under" space-time?

How could it be that the accretion disc which is fully under the impact of space time can eject most of its mass, while the event of horizon can't do so just because of the same space time.
You have stated that based on space-time, nothing could be ejected from inside the horizon:
Please see the following:

What you are missing is that space-time behaves differently inside of the horizon than it does outside of the horizon.

Why space time that works so nicely at the event of horizon and prevents from any particle to be ejected outwards, works so badly at the accretion disc and has no resistance for the massive ejection at that aria?
In other words -
If the space time has no objection that most of the particle can be ejected from the accretion disc, could it be that it also has no objection that particles can be ejected also below the horizon?
Why it prevents from particles to be ejected from the horizon, while it has no resistance that most of the particles can be ejected from the accretion disc?

Because space is fully time-like inside the horizon but it isn't outside the horizon.

Do you have better idea than space-time to show the difference between the horizon and the accretion?

The differing behavior of space-time at those two locations is what makes all of the difference. Space-time behaves differently in different reference frames. It behaves differently in strong gravitational fields than it does in weak ones. It behaves differently at high accelerations than at low ones. This "all or nothing" thinking process of yours is wrong.

[Dave Lev (OP)]

What you are missing is that space-time behaves differently inside of the horizon than it does outside of the horizon.

Because space is fully time-like inside the horizon but it isn't outside the horizon.

Thanks
So, there are two types of space-time
A. Fully time-like Space-time which works exactly up to the event Horizon - Nothing can escape from it.
B. Not fully time-like space time which works in the accretion disc - Everything can escape from it.
Questions:
1. How could it be that the two types of space-time behave so differently?
2. What is so unique in the "fully time-like" that prevents from any particle to escape, while his brother - the "None fully time-like" has no problem that everything can escape? Can you please show the different formulas for each one?
3. How the space-time knows exactly where is the border of the "fully time-like"?
4. Why the "fully time-like" is exactly located at the event horizon?
5. Why not 5% or 50% outwards or inwards from the horizon.
6. Would you kindly offer real proof for the border of the "fully time-like" space-time version (If possible -mathematical calculation)

[Kryptid]

So, there are two types of space-time
A. Fully time-like Space-time which works exactly up to the event Horizon - Nothing can escape from it.
B. Not fully time-like space time which works in the accretion disc - Everything can escape from it.

This is overly simplistic. There is actually a gradual transition from one to the other. Any time there is a gravitational field present, it causes some degree of space-time mixing. This causes objects in space to move in a preferred direction: towards the source of the field. In weak fields like on Earth, this direction is merely preferred but not mandatory. As the gravitational field becomes stronger, space becomes increasingly time-like and the preference for movement into the field becomes stronger. Once the field becomes so strong that the escape velocity reaches the speed of light space has become fully time-like and what was merely a preferred direction becomes a mandatory direction instead. It's like travelling down a river that goes faster and faster until it eventually becomes impossible to resist its flow.

Questions:
1. How the space-time knows exactly where is the border of the "fully time-like"?

Space-time doesn't "know" anything. It simply "is" fully time-like at the event horizon. This border automatically exists at any location where the escape velocity reaches the speed of light. So it is defined by gravity.

2. Why the "fully time-like" is exactly located at the event horizon?

Because that is the very thing that defines it. An event horizon is what it is specifically because nothing can get out of it and the reason nothing can get out is because space is fully time-like there.

3. Why not 5% or 50% outwards or inwards from the horizon.

Because then the event horizon wouldn't be an event horizon.

4. Would you kindly offer real proof for the border of the "fully time-like" space-time version (If possible -mathematical calculation)

The math is discussed in the PBS video, as well as here: http://www.jimhaldenwang.com/black_hole.htm

This page explains it using light cones: https://www.quora.com/If-the-pulling-force-inside-the-event-horizon-is-not-infinite-why-cant-I-push-myself-out-with-a-greater-force

For more information on light cones: https://en.wikipedia.org/wiki/Light_cone

[Kryptid]

Don't understand this.  There are 4 dimensions, and they don't seem to 'mix'.  Nothing moves through spacetime. They move through space, but have worldlines in spacetime. No worldline can be angled so much that its events become separated in a space-like manner.  That's the speed of light restriction. It's the same as saying that my worldline must be entirely contained in my own causal cones.
All this is no different inside the event horizon.  There's still 1 time and 3 spatial dimensions, but the event horizon is in the past light cone of any event in there.  None of it is in the future light cone.

I'm just using the terminology that Kip Thorne uses in his book (see the page I scanned above). The main thrust of what I'm saying is that the light cone tilting is not something that happens suddenly at the event horizon. It becomes more and more tilted as you approach it.

Not necessarily true inside the event horizon.  There are still the 3+1 dimensions with no preference for direction of movement through space. In theory, masses could form with stuff orbit them and such. There would be no obvious tidal force tearing such systems apart. The spacetime is reasonably normal and not different in a way that one is termed 'time-like' and the other not.

Perhaps my terminology is poor then. I figured there were degrees of "time-like"-ness.

I find the river a poor analogy because sufficient force would allow one to resist any arbitrarily large flow.  The river is still space. Time is not something that flows in a spacetime model.

Perhaps you are right. Perhaps I could instead say that the river flows at the speed of light. Then it becomes more apparent that force alone won't allow you to resist it.