[Dave Lev (OP)]

Space-time:
https://en.wikipedia.org/wiki/Spacetime
Space-time interval = (Δs)
(Δs)^2 = (Δx)^2 - (Δt)^2
As: t represents the time, and x represents the space.
I was under the impression that in space-time the space is orthogonal to time.
Therefore, we should get the following:
(Δs)^2 =  (Δx)^2 + (Δt)^2
That Minus sign in the (Δt) really confused me.
So, I have tried to understand more about the source for this Minus sign.
I have found the following great explanation about space time:
https://www.quora.com/Is-time-perpendicular-to-the-other-three-spatial-dimensions
"The basis vectors of the four dimensional Minkowski space (which is used in relativity) are orthogonal. But, you need to remember that a visualization of these four dimensions works different than just adding one extra dimension to our common three dimensional Euclidean space.
The time dimension is something special. Distances between two points in our three dimensional Euclidean space are always real positive numbers (the distance from New York to Chicago is 790 miles). Minskowski space allows for imaginary distances that need to be interpreted as timelike distances. These are events that can have a causal relation, they are in our future or past. Minskowski allows also for zero distances between two points that are spatially separated. These need to be interpreted as the paths of light signals send between two points. Positive distances in a Minkowski space are interpreted as events that don't have a causal relationship. Some observers may see these events happening at the same time, some don't. But everyone agrees they happen at different locations.
So yes, time is "perpendicular" to the other spacial dimensions in Minkowski space.  No, there is an important twist in the mathematics, interpretation, making the time dimension not the same as the spatial dimension."

So, "Minkowski space allows for imaginary distances that need to be interpreted as timelike distances."
Hence, this imaginary distances (or Imaginary time), sets the Minus sign.
However, once we start with imaginary distance, the time could also be imaginary and the whole outcome of the space time formula could also be imaginary.
Let's assume that Minkowski formula for the two dimensions was as follow:
(Δc)^2 = (Δa)^2 - (Δb)^2
So, does it mean that this formula is correct?
What Mr. Pythagoras would say about that?
In the same token, I wonder what Mr. Einstein (who was the student of Minkowski) had stated about that imaginary formula.
I really can't understand how our scientists can get any real information from that imaginary formula..
In my opinion, the outcome of imaginary formula of Minkowski must also be imaginary/unreal.
Therefore, this imaginary formula sets the imaginary/unreal curvature in our Universe and imaginary/unreal curvature/singularity/outcome in a BH.
Hence, If Mr. Einstein didn't approve the use of that formula by Minkowski, I will also not going to accept it and we all should reject it.
Any idea about Mr. Einstein approach to this formula?

Let's continue our discussion about

You have it backwards. The reason nothing can escape is because space is time-like within the horizon. This, in turn, is what causes it to be an event horizon.

So, what comes first?
As we know, Space time is not about a BH or a Universe.
However, due to the Imaginary space, there is a situation where
(Δs)^2 = (Δx)^2 - (Δt)^2 = 0
I assume that this set the border of the "fully time-like" space time.
But there is no real data about the exact radius in this space-time formula.
So, how do we know what is the value of the radius?  Where is the boarder?

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.

I agree it's defined by gravity.
Space time doesn't give any information about Gravity at the BH.
If we could extract the gravity and the radius from the space-time, than we could clearly say that it is he reason for the event horizon.
However, we must use different formula to find the boarders.
So, the Space time can't have any influence on the formula which sets the event horizon formula.
Therefore - the "fully time-like" of the space time can't be used to set the event horizon.
This Event Horizon is set by the following (Schwarzschild) Radius Calculation:
http://www.eguruchela.com/physics/calculator/Black-Hole-Schwarzschild-Radius-Calculator.php
"The calculator calculates the Schwarzschild Radius for given gravitational constant, light speed and body mass. A region in space which does not allow anything to pass out with such a gravity pull is called as the black hole. The radius of the boundary of such an event horizon (hole) is called as the schwarzschild or the gravitational radius.
Formula: rs = 2GM / c2"
So, Schwarzschild tells Minkowski where are the  boarders and not the other way.
Therefore, Minkowski space time has no impact on those boarder and therefore, it doesn't prove anything about those boarders.
Hence, I really don't understand why you insist that:

The reason nothing can escape is because space is time-like within the horizon

[Halc]

Space-time:
https://en.wikipedia.org/wiki/Spacetime
Space-time interval = (Δs)
(Δs)^2 = (Δx)^2 - (Δt)^2
As: t represents the time, and x represents the space.
I was under the impression that in space-time the space is orthogonal to time.

Yes, it's like the Pythagorean theorem that works only for triangles with two orthogonal sides.  The formula is more complicated if non-orthogonal axes are chosen.

Therefore, we should get the following:
(Δs)^2 =  (Δx)^2 + (Δt)^2
That Minus sign in the (Δt) really confused me.
So, I have tried to understand more about the source for this Minus sign.
I have found the following great explanation about space time:
https://www.quora.com/Is-time-perpendicular-to-the-other-three-spatial-dimensions

Be very cautious about anything you get from quora. The don't have a mechanism to eliminate the crap answers, and there are a lot of them.

"The basis vectors of the four dimensional Minkowski space (which is used in relativity) are orthogonal. But, you need to remember that a visualization of these four dimensions works different than just adding one extra dimension to our common three dimensional Euclidean space.
The time dimension is something special. Distances between two points in our three dimensional Euclidean space are always real positive numbers (the distance from New York to Chicago is 790 miles). Minskowski space allows for imaginary distances that need to be interpreted as timelike distances.

This seems nonsense.  Distance along the time dimension is measured in real numbers as well.  It doesn't take me 2i-.5 hours to do something.

Minskowski allows also for zero distances between two points that are spatially separated. These need to be interpreted as the paths of light signals send between two points.

Blatantly wrong.  The statement is self contradictory. Light cannot travel between two spatially separated spacetime points, nor can it travel between temporally separated points (at least not in a vacuum).  Only light-like separated points (events) can have light travel between them.
The quora answer also speaks of points, not events, a terminology error.

Positive distances in a Minkowski space are interpreted as events that don't have a causal relationship.

Here at least they use 'events' correctly, but it should read 'Positive intervals', not 'Positive distances' for the statement to be correct.  Intervals are not distances.  You might observe a star exploding, and the event of the explosion and the event of you seeing it might be separated by a (frame dependent) distance of 500 light years, but a frame independent interval of zero.  Distance and interval are very different things.

So, "Minkowski space allows for imaginary distances that need to be interpreted as timelike distances."
Hence, this imaginary distances (or Imaginary time), sets the Minus sign.

You have a troll's talent for spotting the crap in a web site and quoting it as gospel, but questioning all the correct sites.
Imaginary numbers don't set a minus sign.  They set an imaginary component i.  If you want to make sense of time being measured as 5i or something, be my guest.  Last I looked, my clock didn't have an i on it.

However, once we start with imaginary distance, the time could also be imaginary and the whole outcome of the space time formula could also be imaginary.
Let's assume that Minkowski formula for the two dimensions was as follow:
(Δc)^2 = (Δa)^2 - (Δb)^2
So, does it mean that this formula is correct?
What Mr. Pythagoras would say about that?

The interval is not a measurement of a hypotenuse, something that would be frame dependent.  Pythagoras was describing a spatial relation, not a spacetime one.

Therefore, this imaginary formula sets the imaginary/unreal curvature in our Universe and imaginary/unreal curvature/singularity/outcome in a BH.

The formula is not imaginary and does not describe curvature.  It works in Euclidean space, and is thus more of an SR thing than a GR thing.

Hence, If Mr. Einstein didn't approve the use of that formula by Minkowski,

Where do you get this assertion?

As we know, Space time is not about a BH or a Universe.
However, due to the Imaginary space, there is a situation where
(Δs)^2 = (Δx)^2 - (Δt)^2 = 0
I assume that this set the border of the "fully time-like" space time. But there is no real data about the exact radius in this space-time formula.

The above is word salad and has nothing to do with event horizons.  Read my prior posts for description of how the orientation of causal light cones relates to the Schwarzschild radius and hence an event horizon (a singularity) must exist there.

So, how do we know what is the value of the radius?

Approximately 2GM/c˛.  This isn't a real radius (an actual length), but rather a measure of the circumference divided by 2π.

Where is the boarder?

In the spare room.

This Event Horizon is set by the following (Schwarzschild) Radius Calculation:
http://www.eguruchela.com/physics/calculator/Black-Hole-Schwarzschild-Radius-Calculator.php
"The calculator calculates the Schwarzschild Radius for given gravitational constant, light speed and body mass. A region in space which does not allow anything to pass out with such a gravity pull is called as the black hole. The radius of the boundary of such an event horizon (hole) is called as the schwarzschild or the gravitational radius.
Formula: rs = 2GM / c2"

There, see?  You answered your own question.
The site is a calculator and lets you specify a gravitational constant that is in units of m/sec˛, which is wrong.  Mistakes (including grammatical ones) on the site aside, the quote above is correct.

So, Schwarzschild tells Minkowski where are the  boarders and not the other way.
Therefore, Minkowski space time has no impact on those boarder and therefore, it doesn't prove anything about those boarders.

And we're back to word salad. Minkowski never published anything about black holes.  Schwarzschild didn't tell Minkowski anything since the latter died before making that contribution.

[Kryptid]

In the same token, I wonder what Mr. Einstein (who was the student of Minkowski) had stated about that imaginary formula.
I really can't understand how our scientists can get any real information from that imaginary formula..
In my opinion, the outcome of imaginary formula of Minkowski must also be imaginary/unreal.

Don't mistake the word "imaginary" for "fictional" in the context of mathematics.

Any idea about Mr. Einstein approach to this formula?

He was initially unenthusiastic, but later realized its great importance. To quote from "Black Holes & Time Warps":

When Einstein learned of Minkowski's discovery, he was not impressed. Minkowski was merely rewriting the laws of special relativity in a new, more mathematical language; and, to Einstein, the mathematics obscured the physical ideas that underlie the laws. As Minkowski continued to extol the beauties of his spacetime viewpoint, Einstein began to make jokes about Gottingen mathematicians describing relativity in such complicated language that physicists wouldn't be able to understand it.

The joke, in fact, was on Einstein. Four years later, in 1912, he would realize that Minkowski's absolute spacetime is an essential foundation for incorporating gravity into special relativity.

The idea of warpage of both time and space was rather daunting. Since the Universe admits an infinite number of different reference frames, each moving with a different velocity, there would have to be an infinity of warped times and an infinity of warped spaces! Fortunately, Einstein realized, Hermann Minkowski had provided a powerful tool for simplifying such complexity: "Henceforth, space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality." There is just one, unique, absolute, four-dimensional spacetime in our Universe; and a warpage of everyone's time and everyone's space must show up as a warpage of Minkowski's single, unique, absolute spacetime.

This was the conclusion to which Einstein was driven in the summer of 1912 (though he preferred to use the word "curvature" rather than "warpage"). After four years of ridiculing Minkowski's idea of absolute spacetime, Einstein had finally been driven to embrace is, and warp it.

So yes, Einstein did accept the Minkowski metric.

So, what comes first?

This is like asking, "which comes first, the water or the wetness?" Neither "comes first".

However, due to the Imaginary space, there is a situation where
(Δs)^2 = (Δx)^2 - (Δt)^2 = 0
I assume that this set the border of the "fully time-like" space time.
But there is no real data about the exact radius in this space-time formula.
So, how do we know what is the value of the radius?  Where is the boarder?

That's easy. Anywhere that (Δx)^2 equals (Δt)^2, you will get zero as your answer. That's basic algebra.

I agree it's defined by gravity.
Space time doesn't give any information about Gravity at the BH.
If we could extract the gravity and the radius from the space-time, than we could clearly say that it is he reason for the event horizon.
However, we must use different formula to find the boarders.
So, the Space time can't have any influence on the formula which sets the event horizon formula.
Therefore - the "fully time-like" of the space time can't be used to set the event horizon.
This Event Horizon is set by the following (Schwarzschild) Radius Calculation:
http://www.eguruchela.com/physics/calculator/Black-Hole-Schwarzschild-Radius-Calculator.php
"The calculator calculates the Schwarzschild Radius for given gravitational constant, light speed and body mass. A region in space which does not allow anything to pass out with such a gravity pull is called as the black hole. The radius of the boundary of such an event horizon (hole) is called as the schwarzschild or the gravitational radius.
Formula: rs = 2GM / c2"
So, Schwarzschild tells Minkowski where are the  boarders and not the other way.
Therefore, Minkowski space time has no impact on those boarder and therefore, it doesn't prove anything about those boarders.
Hence, I really don't understand why you insist that:

The reason nothing can escape is because space is time-like within the horizon

This is a bunch of nonsense. You know that gravity is a distortion in space-time, right? So obviously, the local curvature of space-time is going to tell you everything about the gravity there.

Also, no one is pretending that a single equation is going to tell you everything you want to know about something. You can't use one equation to tell you both what the surface gravity on Earth is and what its orbital period around the Sun is. Those two equations, however, are closely related and rely on the same basic physics of gravity to work. They are consistent with each other. Same thing for Minkowski's equations and those of Schwarzschild.

[Dave Lev (OP)]

Therefore, we should get the following:
(Δs)^2 =  (Δx)^2 + (Δt)^2
That Minus sign in the (Δt) really confused me.
So, I have tried to understand more about the source for this Minus sign.
I have found the following great explanation about space time:
https://www.quora.com/Is-time-perpendicular-to-the-other-three-spatial-dimensions

Be very cautious about anything you get from quora. The don't have a mechanism to eliminate the crap answers, and there are a lot of them.

Thanks Halc
So, you  don't like the explanation by " quora", however, you didn't offer any real explanation for the following formula by Minkowski:
(Δs)^2 =  (Δx)^2 - (Δt)^2
How could it be that the (Δt)^2 comes with Minus sign???
Don't forget that even if the original sign of (Δt) was minus, than after the following calculation:
(-Δt)^2 = (Δt)^2
We should get it as a positive value.
So, even if the (-Δt) represents a time in the past, the outcome of the (-Δt)^2 calculation MUST  be positive.
I have never ever found any possibility that
(-Δt)^2 = -(Δt)^2

Don't mistake the word "imaginary" for "fictional" in the context of mathematics.

I'm not sure that I understand this answer.
Do you agree that the Δt is "imaginary" as stated in " quora"?
Would you kindly explain how could it be that:
(+/-Δt)^2 = -(Δt)^2
I really want to understand this calculation.

[Kryptid]

How could it be that the (Δt)^2 comes with Minus sign???

It doesn't. What the equation is doing is subtracting (Δt)² from (Δx)².

I have never ever found any possibility that
(-Δt)^2 = -(Δt)^2

Good, because that's not what the equation says.

I'm not sure that I understand this answer.
Do you agree that the Δt is "imaginary" as stated in " quora"?

https://en.wikipedia.org/wiki/Imaginary_time

Would you kindly explain how could it be that:
(+/-Δt)^2 = -(Δt)^2

No one is saying that is the case.

[Halc]

Therefore, we should get the following:
(Δs)^2 =  (Δx)^2 + (Δt)^2
That Minus sign in the (Δt) really confused me.
So, I have tried to understand more about the source for this Minus sign.
I have found the following great explanation about space time:
https://www.quora.com/Is-time-perpendicular-to-the-other-three-spatial-dimensions

Be very cautious about anything you get from quora. The don't have a mechanism to eliminate the crap answers, and there are a lot of them.

Thanks Halc
So, you  don't like the explanation by " quora", however, you didn't offer any real explanation for the following formula by Minkowski:
(Δs)^2 =  (Δx)^2 - (Δt)^2
How could it be that the (Δt)^2 comes with Minus sign???

Because it only works that way.  There would be no Lortentz transform if it were a plus sign.  Light speed would be frame dependent.  It would not be the universe we observe.

Don't forget that even if the original sign of (Δt) was minus, than after the following calculation:
(-Δt)^2 = (Δt)^2
We should get it as a positive value.
So, even if the (-Δt) represents a time in the past, the outcome of the (-Δt)^2 calculation MUST  be positive.

Correct again.  That means that the spacetime interval is imaginary for any pair of events separated in a time-like manner.  That's how you know the separation is time-like.  Real for space-like (causally separated), and zero for light-like.

Do you agree that the Δt is "imaginary" as stated in " quora"?

Quora is wrong more than it is right.  Don't get your science from there.
More specifically, if Δt was imaginary, then squaring it would yield a potentially negative number and the interval calculation would not be frame independent. That makes it wrong if he said that, but he didn't say time was imaginary. He called the interval an 'imaginary distance', but the word 'distance' is misleading.  An interval is not a distance.

I really want to understand this calculation.

Why are you trying to understand the concept of spacetime interval when the discussion was concerning event horizons?  There isn't a defined interval between events on opposite sides of a gravitational event horizon.

I'm not sure that I understand this answer.
Do you agree that the Δt is "imaginary" as stated in " quora"?

https://en.wikipedia.org/wiki/Imaginary_time

Oh that's interesting.  It is a different interpretation that assumes t itself is imaginary in an effort to make the interval a real number. I don't think the quora answer is assuming this interpretation. It requires a change of Minkowski's formula to: d˛=x˛+y˛+z˛+(it)˛ which always yields a real distance (Δd) instead of an imaginary interval Δs for causally connected events.  But can time have a nonzero real component then?  If not, the formula above seems incorrect in the general case.  Does it make sense to say time imaginary like that?  Kinetic energy would then always be negative, which seems to make no sense.

[Dave Lev (OP)]

Why are you trying to understand the concept of spacetime interval when the discussion was concerning event horizons?

Because the Space-time is used as a proof for the idea that nothing can escape from the radius below the event horizon.

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.

If you give up on that, I won't ask any more questions about the space-time.

[Halc]

Why are you trying to understand the concept of spacetime interval when the discussion was concerning event horizons?

Because the Space-time is used as a proof for the idea that nothing can escape from the radius below the event horizon.

1) Spacetime is not used as such a proof, and 2) you didn't answer the question, which was why you're suddenly interested in an interval computation. I don't see its relevancy to the inability of anything to escape an event horizon.

[Dave Lev (OP)]

1) Spacetime is not used as such a proof,

Perfect.
So, particle pair that had just been created below the event of horizon, are not forced to stay there due to space-time.
However, the event horizon radius is very clear and it is calculated by the following (Schwarzschild) Radius formula:
http://www.eguruchela.com/physics/calculator/Black-Hole-Schwarzschild-Radius-Calculator.php
"The calculator calculates the Schwarzschild Radius for given gravitational constant, light speed and body mass. A region in space which does not allow anything to pass out with such a gravity pull is called as the black hole. The radius of the boundary of such an event horizon (hole) is called as the Schwarzschild or the gravitational radius.
Formula: rs = 2GM / c2"
We all know that the maximal velocity of particle is the speed of light.
Therefore, even  if a particle orbits at the maximal speed (speed of light) and it is located below that radius, it won't be able to be ejected from that radius due to gravity force.
Therefore, the simple conclusion is that nothing can be ejected from the event of horizon.
So far so good.
However, if we add the impact of the Lorentz force due to magnetic field, than now the outcome is quite different.
Let's assume that one of the new particle pair carry positive charge, while the other one carry negative charge.
So, please don't say the following message again:

It absolutely is about negative and positive mass, since we are talking about super-massive black holes (which cannot produce charged particles because those particles are too massive).

Let's assume that one has a negative charge while the other carry a positive charge.
Let's also assume that there is a magnetic field around that BH (let's assume that it is very high (infinite).

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

Let's also assume that the positive charged particle is deflected outwards due to Lorentz force (assuming that it is already on a path perpendicular to the field lines):

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).

Therefore, as the Magnetic field is almost infinite, let's assume that the Lorentz force on the positive charged particle might is also infinite.
So, do you agree that this infinite Lorentz force should deflect the path of the new born positive particle?
Actually, this particle will face two forces (assuming that it orbits at ultra high velocity - almost the speed of light).
One - Gravity force - that pulls it inwards.
Two - Lorentz force - That pushes it outwards
So based on all above assumptions, do you agree that If Lorentz force is stronger than the gravity force, the positive particle should eventually be ejected outwards from the event horizon even while its velocity is still below the speed of light?

[Kryptid]

Perfect.
So, particle pair that had just been created below the event of horizon, are not forced to stay there due to space-time.

It's forced to move inward because space is time-like inside the horizon.

However, if we add the impact of the Lorentz force due to magnetic field, than now the outcome is quite different.

No it isn't, because space is still time-like below the horizon.

Let's assume that one of the new particle pair carry positive charge, while the other one carry negative charge.
So, please don't say the following message again:

Sorry, but if you are going to insist that super-massive black holes can produce charged particles, then I am going to have to keep repeating myself: they can't.

Let's also assume that the positive charged particle is deflected outwards due to Lorentz force (assuming that it is already on a path perpendicular to the field lines):

That's not how the Lorentz force works.

So, do you agree that this infinite Lorentz force should deflect the path of the new born positive particle?

Only if it is already outside of the horizon and even then it won't deflect it away from the black hole. It will deflect it at a right angle to its path instead: https://www.khanacademy.org/test-prep/mcat/physical-processes/magnetism-mcat/a/using-the-right-hand-rule

Let's say you are sitting at the north pole of the black hole and you are watching positively-charged particles stream away from the equator. As they move away, they encounter magnetic field lines running perpendicular to their path (the field lines running downwards relative to you). You can use the right-hand rule to tell you what reaction these charged particles will have to the field lines. If the particle is moving away from the equator while the direction of the magnetic field line is downward, then the force acting on a positively-charged particle is to the left.

However, since this left-moving particle is still interacting with magnetic field lines, it will want to turn left (relative to its current direction) yet again. This results in the particle travelling into a counter-clockwise circle (from your vantage point atop of the black hole). So the magnetic field neither pushes the particle away from the horizon nor pulls it in. The opposite is true for a negatively-charged particle: it will be deflected right and end up circling around the field lines in a clockwise circle.

Actually, this particle will face two forces (assuming that it orbits at ultra high velocity - almost the speed of light).
One - Gravity force - that pulls it inwards.
Two - Lorentz force - That pushes it outwards
So based on all above assumptions, do you agree that If Lorentz force is stronger than the gravity force, the positive particle should eventually be ejected outwards from the event horizon even while its velocity is still below the speed of light?

No, for the reasons we have repeated over and over and over...

[Halc]

1) Spacetime is not used as such a proof,

Perfect.
So, particle pair that had just been created below the event of horizon, are not forced to stay there due to space-time.

Since 'due to space-time' doesn't even make syntactic sense, this statement is not even wrong.

However, the event horizon radius is very clear and it is calculated by the following (Schwarzschild) Radius formula:
http://www.eguruchela.com/physics/calculator/Black-Hole-Schwarzschild-Radius-Calculator.php

The calculator there yielded incorrect answers when I attempted to use it, but the formula is correct for a Schwarzschild black hole.  Sgr-A is not such a black hole, but close enough.

"The calculator calculates the Schwarzschild Radius for given gravitational constant, light speed and body mass. A region in space which does not allow anything to pass out with such a gravity pull is called as the black hole. The radius of the boundary of such an event horizon (hole) is called as the Schwarzschild or the gravitational radius.
Formula: rs = 2GM / c2"
We all know that the maximal velocity of particle is the speed of light.

Neither particles nor light  travel to the past. That's the part that matters.

Therefore, even  if a particle orbits at the maximal speed (speed of light) and it is located below that radius, it won't be able to be ejected from that radius due to gravity force.
Therefore, the simple conclusion is that nothing can be ejected from the event of horizon.

No, this does not follow from your statement.

However, if we add the impact of the Lorentz force due to magnetic field, than now the outcome is quite different.

Nothing can get out.  Applying force makes zero difference since no force can push an object into the past nor pull it from the future.  So no, the outcome is no different when arbitrary force is considered.

Let's also assume that the positive charged particle is deflected outwards due to Lorentz force

It cannot be.  That direction is not a spatial direction.  Force cannot be applied in a non-spatial direction.  So lets not assume this at all. Such assumptions seem to be the base of your problems.

[Dave Lev (OP)]

Thanks Halc
It is clear to me that you don't agree with (almost) any idea that I offer.
Let me understand if you agree with the observation about the Milky Way
1. More than 99% of the matter in the accretion disc is ejected outwards. So out of 100 particles - more than 99 are ejected outwards.
2. So far our scientists couldn't find even one particle that is falling into the accretion disc from outside, or from the accretion disc into the SMBH, although they monitor the SMBH almost constantly with several X-ray detectors.
3. There is a magnetic field around the accretion disc. This magnetic field sets the Molecular jet steam above and below the accretion disc. We have already found that the jet stream velocity is almost 0.8 c while it moves up to 27,000 LY above and below the disc.
I hope that at least you agree with those observations.

If so, let's assume that there is a particle outside the magnetic field which is bagging to be eaten by the SMBH.
So, first it must cross the magnetic field.
Can you please explain how any particle can cross that magnetic field? Why it won't be boosted by the magnetic field and join all the other particles in that molecular jet stream? What is the chance that it can do it?  Is it one to 1 Million or 1 to 1 Billion?
Let's assume that somehow one particle was very lucky and could cross the magnetic field.
So, now this particle is there trying to cross the last gap to the accretion disc, while it is facing all of that 99% stream of matter that are ejected outwards.
If those 99% are ejected outwards, don't you agree that there must be some sort of force that drives them outwards?
I assume that you call this force - tidal.
Why that tidal don't push outwards also this particle?
How could it be that the tidal ejects those 99% of the matter in the accretion disc, while it lets that particle to get inwards to the accretion disc?
Do you think that there is a policeman that knows where the particle is coming from?
How could it be that out of 100 particles in the accretion disc, more than 99 are ejected outwards, while the way for our particle will be clear to fall into the accretion disc?
Let's assume that against all odds it was so lucky and finely arrived to the accretion disc.
However, after all the difficulties in its way to the accretion disc, now it has a chance of less than 1% to fulfill its goal and fall finely into the SMBH. Actually it has a chance of more than 99% to be ejected outwards.
Did we try to find how many times that poor particle must come back again to the accretion disc in order to be eaten by the mighty SMBH? Do you agree on at least 99 times?
I wonder how you still believe that a particle which had been ejected at least 99 times from the accretion disc will come back again with great hope that this is the last time.

[Kryptid]

1. More than 99% of the matter in the accretion disc is ejected outwards. So out of 100 particles - more than 99 are ejected outwards.

That appears to be the case with Sagittarius A*, yes.

2. So far our scientists couldn't find even one particle that is falling into the accretion disc from outside, or from the accretion disc into the SMBH, although they monitor the SMBH almost constantly with several X-ray detectors.

I'm not so sure that is true. Indirect evidence of matter accreting onto the disk or falling into the hole would come from detected X-ray emissions.

3. There is a magnetic field around the accretion disc. This magnetic field sets the Molecular jet steam above and below the accretion disc. We have already found that the jet stream velocity is almost 0.8 c while it moves up to 27,000 LY above and below the disc.

I don't remember if you ever answered this question, but where did you get the 0.8c figure from?

Can you please explain how any particle can cross that magnetic field? Why it won't be boosted by the magnetic field and join all the other particles in that molecular jet stream? What is the chance that it can do it?  Is it one to 1 Million or 1 to 1 Billion?

That strongly depends on the trajectory of the particle. Remember, charged particles move in spirals around magnetic field lines. If the particles are already moving in an upward or downward direction, then they could spiral up the field lines and into the jets. If they hit the field lines at a right angle, however, then the particles would be stuck moving in circles without significant movement up or down. This spiraling also causes the charged particles to release electromagnetic radiation, which makes them lose energy and slow down. Eventually, they should slow down enough to fall into the hole because the magnetic force on a charged particle becomes weaker as the particle slows down.

Alternatively, the particle that is stuck moving in circles could be pushed in by other particles hitting it from behind.

So, now this particle is there trying to cross the last gap to the accretion disc, while it is facing all of that 99% stream of matter that are ejected outwards.
If those 99% are ejected outwards, don't you agree that there must be some sort of force that drives them outwards?
I assume that you call this force - tidal.
Why that tidal don't push outwards also this particle?

The magnetic field is what does this, not tidal forces. As I said before, the trajectory of the particle is what determines whether the magnetic field can move it towards the poles or not.

Do you think that there is a policeman that knows where the particle is coming from?

Was that a joke?

How could it be that out of 100 particles in the accretion disc, more than 99 are ejected outwards, while the way for our particle will be clear to fall into the accretion disc?

Not all particles are moving at the same velocity or in the same direction. You wouldn't expect all of them to react to the magnetic field in exactly the same way.

Let's assume that against all odds it was so lucky and finely arrived to the accretion disc.

I presumed we were talking about a particle that was already in the accretion disk. It's not hard for particles to add themselves to the outer regions of an accretion disk because the outer regions are cool and slow-moving. That would make any magnetic fields present there weak. At least, that is the case with sufficiently large disks.

However, after all the difficulties in its way to the accretion disc, now it has a chance of less than 1% to fulfill its goal and fall finely into the SMBH. Actually it has a chance of more than 99% to be ejected outwards.
Did we try to find how many times that poor particle must come back again to the accretion disc in order to be eaten by the mighty SMBH? Do you agree on at least 99 times?

Who said anything about the particle coming back into the accretion disk? How could it do so if it was already being ejected by the jet? I think you are also underestimating the sheer number of particles that exist in clouds of gas. Even a single gram of hydrogen plasma contains 5.975 x 10²³ electrons and 5.975 x 10²³ protons. Even if 99.999% of those particles were ejected, that's still 1.195 x 10¹⁹ particles that fall in. There are lots and lots of grams in an accretion disk.

I also came to another realization. If you propose that the magnetic field is too strong to allow charged particles from the accretion disk to fall into the black hole, then it must also prevent movement in the opposite direction. That is, any charged particles generated around the black hole must not be able to get out into the accretion disk. So your idea that the accretion disk is actually an excretion disk is unworkable if you consider the magnetic field to be impassable.

[Halc]

Thanks Halc
It is clear to me that you don't agree with (almost) any idea that I offer.

You do seem to be going out of your way to misrepresent or ignore anything you read here or on sites, so yes, what do you expect from us?

Let me understand if you agree with the observation about the Milky Way
1. More than 99% of the matter in the accretion disc is ejected outwards.
 So out of 100 particles - more than 99 are ejected outwards.

One site said something like that, yes.  It doesn't say how the figure is measured.  Sgr-A masses less than 0.000005 the mass of the galaxy, so obviously less than 1% (0.01) has fallen in.  How is the 99% figure computed?  If 99% of the accretion disk is ejected outward, why is there still an accretion disk sitting there?  Just saying the article wasn't very specific about how it got those numbers.

2. So far our scientists couldn't find even one particle that is falling into the accretion disc from outside, or from the accretion disc into the SMBH, although they monitor the SMBH almost constantly with several X-ray detectors.

This is blatantly wrong. This was stated on no site. In fact, the opposite was recently observed (see posts 597 and 651) , totally countering the 99% thing you state in point 1.  Yes, when the disk is not fed, almost nothing falls further in, just like in the solar system where far less than 1% of the matter orbiting the sun is seen to fall in.
The article you reference in point 3 talked about 10,000 solar masses falling in, which is quite a bit more than 'not even one particle'.

3. There is a magnetic field around the accretion disc. This magnetic field sets the Molecular jet steam above and below the accretion disc. We have already found that the jet stream velocity is almost 0.8 c while it moves up to 27,000 LY above and below the disc.

This is a misrepresentation of what the article says.

If so, let's assume that there is a particle outside the magnetic field which is bagging to be eaten by the SMBH.
So, first it must cross the magnetic field.
Can you please explain how any particle can cross that magnetic field? Why it won't be boosted by the magnetic field and join all the other particles in that molecular jet stream? What is the chance that it can do it?  Is it one to 1 Million or 1 to 1 Billion?

I have posted little opinions on how the magnetic fields work. All I know is that if the energy is boosted, it needs to be taken from somewhere else.  If it falls in slow and is ejected at 0.8c, then it took energy away from the system, and that loss of energy must be reflected in either a weaker disk energy or lower potential energy (the mass of the disk drops towards/into the black hole). Conservation of mass will not allow otherwise.
Only charged particles will be affected by a magnetic field, no? That excludes anything that the Hawking radiation from the black hole itself produces.

Let's assume that somehow one particle was very lucky and could cross the magnetic field.
So, now this particle is there trying to cross the last gap to the accretion disc, while it is facing all of that 99% stream of matter that are ejected outwards.

If it is coming from anywhere except the plane of the disk or the poles, it meets none of this.  The main reason it it ejected is perhaps because its aim was poor and it misses the black hole. That's the same reason almost all comets (at least 99% of them in fact) are ejected by the sun after they fall in.  Most of them simply are not aimed close enough to the sun, just like Earth isn't.

If those 99% are ejected outwards, don't you agree that there must be some sort of force that drives them outwards?

Inertia isn't a force.  If there's a force involved, there needs to be energy expended by something to apply that force.  If that comes from the magnetic field, then the field is weakened by this expenditure.  That's why we don't get free electricity just by spinning magnets.  The spinning slows down if they are used to exert a force on anything.  If the field was generated by an orbit of particles around something, that energy loss would cause them to drop further towards the thing orbited.  Reduction of kinetic energy like that is how they take the space shuttle out of orbit.

I assume that you call this force - tidal.

No, of course not.  Tidal forces are not involved in the ejection of material. Tidal forces act to separate connected objects. Such forces are greatest where the gravitational gradient is greatest, and an SMBH has a low gravitational gradient.

Why that tidal don't push outwards also this particle?
How could it be that the tidal ejects those 99% of the matter in the accretion disc, while it lets that particle to get inwards to the accretion disc?

No such thing happens. Tides play no role. I do recall you saying that magnetic force was involved. The articles I read say the physics is not fully known, which is why I'm not claiming to know the correct answers.

Let's assume that against all odds it was so lucky and finely arrived to the accretion disc.
However, after all the difficulties in its way to the accretion disc, now it has a chance of less than 1% to fulfill its goal and fall finely into the SMBH. Actually it has a chance of more than 99% to be ejected outwards.
Did we try to find how many times that poor particle must come back again to the accretion disc in order to be eaten by the mighty SMBH? Do you agree on at least 99 times?

I don't know what those numbers represented, or how accurate they are, especially when negligible material is falling in most of the time.  I agree that it is long odds that a random particle on a random trajectory will find itself anywhere near our black hole.  It's a dang insignificant target.  Earth is in danger of falling in, but the odds are still incredibly small.  If it were to be deflected straight in, nowhere near 99% of it would be ejected, but it also wouldn't all go in no matter how well aimed.

[Kryptid]

I don't know what those numbers represented, or how accurate they are, especially when negligible material is falling in most of the time. 

Although that 99%+ figure was obtained using observational data, the irony of it is that assumptions and models were used in order to get to that number (which Dave Lev seems to be leery of). Here are some quotes from the paper:

The flat density profile of the flow (i.e., s ∼ 1) suggests the presence of an outflow that nearly balances the inflow (26). As a result, <∼ 1% of the matter initially captured by the SMBH reaches the innermost region around Sgr A*, limiting the accretion power to <∼ 10³⁹ erg s⁻¹.

And about that "flat density profile":

The RIAF model constructed above gives an excellent fit to the spectrum, both globally (χ²/n.d.f. = 187/218) and in terms of matching individual lines (Fig. 2b). The best-fit model gives γ = 1.9(1.4, 2.4). If θ ∼ 1 as typically assumed [e.g., (14,17)], this suggests a very flat density profile of the flow (i.e., s ∼ 1), indicating an outflow mass-loss rate that nearly balances the inflow.

So it's not like they directly measured the flow. They had to model it based on spectral data.

[Halc]

The RIAF model constructed above gives an excellent fit to the spectrum, both globally (χ²/n.d.f. = 187/218) and in terms of matching individual lines (Fig. 2b). The best-fit model gives γ = 1.9(1.4, 2.4). If θ ∼ 1 as typically assumed [e.g., (14,17)], this suggests a very flat density profile of the flow (i.e., s ∼ 1), indicating an outflow mass-loss rate that nearly balances the inflow.

So it's not like they directly measured the flow. They had to model it based on spectral data.

Fair enough.  I wonder what the model would say about the percentage when significant material falls in like it did last May?  Did they capture spectral data from that event and match it to this model to get an idea of the percentage of that significant influx that ended up in the outflow?

[Dave Lev (OP)]

The RIAF model constructed above gives an excellent fit to the spectrum, both globally (χ2/n.d.f. = 187/218) and in terms of matching individual lines (Fig. 2b). The best-fit model gives γ = 1.9(1.4, 2.4). If θ ∼ 1 as typically assumed [e.g., (14,17)], this suggests a very flat density profile of the flow (i.e., s ∼ 1), indicating an outflow mass-loss rate that nearly balances the inflow.

So it's not like they directly measured the flow. They had to model it based on spectral data.

Thanks Krypid
So you agree that they don't measured directly the inflow.
This article is actually based on assumptions:
"If θ ∼ 1 as typically assumed"
"assuming that the magnetic field is near equipartition,"
"With this upper limit, assuming that ri ∼ 102 rs we can infer s > 0.6"
"where PBondi is assumed to have a 10% efficiency of the Bondi mass accretion rate"
"The dashed circle around Sgr A* marks its Bondi capture radius (assumed to be 400)"
"..assuming collisional ionization equilibrium"
"..because the measurement then depends sensitively on the assumed thermal plasma model"
" The best-fit model gives γ = 1.9(1.4, 2.4). If θ ∼ 1 as typically assumed.

If we will assume that (θ ∼ 1) is incorrect, does it mean that our assumption is correct or incorrect?
Sorry - is it about science, or assumption game?
I can't understand how those scientists from arxiv do whatever it takes to show that somehow there must be in falling matter.
Are they looking to find the real understanding about our galaxy or they just wish to fit the current mainstream to the observations against all odes.
So, would you agree to say: "They had to model it based on spectral assumption"?

Let me offer the following articles from the same arxiv:
https://arxiv.org/pdf/1104.5443.pdf
"In regions close to the black hole, the AGN outflows are revealed through blueshifted absorption lines in X–ray emission (Pounds et al. 2003a, b; King 2010a). Tombesi et al. (2010a, b) show that they are present in more than 35 percent of a sample of over 50 local AGN, and deduce that their solid angles are large (certainly > 0.6 × 2π, and probably greater). The observed absorption columns imply that in many cases the outflows are quite recent (few years), suggesting that outflows are an almost ubiquitous feature of central black hole activity (King 2010b)"

Therefore, they clearly say that there is an outflow from the SMBH –" the AGN outflows are revealed through blueshifted".
In any case, so far I couldn't find any real X-Ray observation for any in falling matter into the SMBH or from outside into the accretion disc.
We only get clear observations for OUTFLOWS.

In the following article (from arxiv):
https://arxiv.org/ftp/arxiv/papers/1501/1501.07664.pdf
"Wind from the black-hole accretion disk driving a molecular outflow in an active galaxy"
"Recent observations of large-scale molecular outflows3,4,5,6,7,8 in ultraluminous infrared galaxies (ULIRGs) have provided the evidence to support these studies, as they directly trace the gas out of which stars form.
So, now they claim that the outflow wind from the BH (or SMBH) drives the star formation activity.
This fits perfectly with Theory D.

[Halc]

So you agree that they don't measured directly the inflow.

To measure it directly, you'd have to be there and be able to capture everything going in and measure the mass before letting it go on, and even then it can be argued to be not direct.  So any measurement has levels of indirection.  I don't directly measure the ball hitting me in the head, I only measure the experience of pain, and possibly the experience of an image of a ball coming, and the rest is a conclusion arrived at from this more direct evidence.

If we will assume that (θ ∼ 1) is incorrect, does it mean that our assumption is correct or incorrect?

Depends if the actual θ is near 1.  If a significantly different value fits the other date more closely, the assumption of being near 1 might not be warranted.  Personally, I didn't even look to see what θ was.

Sorry - is it about science, or assumption game?

I assume the sun will rise tomorrow, despite my lack of proof.  It's still a scientific assumption.  This is effectively what they're doing, and something you don't do:  State your assumptions.  The conclusions are then only as strong as the assumptions.

Let me offer the following articles from the same arxiv:
"In regions close to the black hole, the AGN outflows are revealed through blueshifted absorption lines in X–ray emission (Pounds et al. 2003a, b; King 2010a). Tombesi et al. (2010a, b) show that they are present in more than 35 percent of a sample of over 50 local AGN, and deduce that their solid angles are large (certainly > 0.6 × 2π, and probably greater). The observed absorption columns imply that in many cases the outflows are quite recent (few years), suggesting that outflows are an almost ubiquitous feature of central black hole activity (King 2010b)"

Therefore, they clearly say that there is an outflow from the SMBH –" the AGN outflows are revealed through blueshifted".

They clearly say an outflow is seen in about a third the cases, and that some are quite young, implying that it isn't continuous.  Just trying to read what you quoted there.

In any case, so far I couldn't find any real X-Ray observation for any in falling matter into the SMBH or from outside into the accretion disc.

Don't think matter outside the accretion disk is likely to radiate X-rays.

[Dave Lev (OP)]

To measure it directly, you'd have to be there and be able to capture everything going in and measure the mass before letting it go on, and even then it can be argued to be not direct.  So any measurement has levels of indirection.

How could it be that we see so clearly the outflow from the SMBH or from the accretion disc?
Do you agree that we see the outflow directly by X-ray and by the blueshift: " the AGN outflows are revealed through blueshifted".
So, do you agree that we don't need to be there in order to see the outflow?
However, when it comes to the Inflow - Why we need modeling?
Why we can't see directly the inflow as we clearly see the outflow?
If the total quantity of outflow is almost identical to the Inflow than why we can't see both of them in the same observation tools?

Don't think matter outside the accretion disk is likely to radiate X-rays.

It seems to me that as the matter radiates when it is ejected outwards, it should also radiates when it is falling inwards.

[Halc]

To measure it directly, you'd have to be there and be able to capture everything going in and measure the mass before letting it go on, and even then it can be argued to be not direct.  So any measurement has levels of indirection.

How could it be that we see so clearly the outflow from the SMBH or from the accretion disc?

They don't see it.  They gather data and infer it from the data.  It's a 'pathetic weak' outflow, but so is the inflow, so it might still be 99% of it.

Do you agree that we see the outflow directly by X-ray and by the blueshift: " the AGN outflows are revealed through blueshifted".

The X-rays are just light.  What those rays come from must be deduced.  The jet doesn't point straight at us, so there is no direct measurement of it.

Why we can't see directly the inflow as we clearly see the outflow?

Neither are seen directly.  The measurements taken are consistent with the conclusions drawn.

If the total quantity of outflow is almost identical to the Inflow than why we can't see both of them in the same observation tools?

They are seen with the same tools, else they'd not be able to posit a percentage like that.