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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.

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.

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

Space-time:https://en.wikipedia.org/wiki/SpacetimeSpace-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)^2That 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.

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.

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)^2So, does it mean that this formula is correct?What Mr. Pythagoras would say about that?

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,

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 = 0I 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?

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.

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.

Any idea about Mr. Einstein approach to this formula?

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, what comes first?

However, due to the Imaginary space, there is a situation where (Δs)^2 = (Δx)^2 - (Δt)^2 = 0I 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?

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

QuoteTherefore, we should get the following:(Δs)^2 = (Δx)^2 + (Δt)^2That 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-dimensionsBe 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.

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

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

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

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

Quote from: Halc on 30/08/2019 14:42:53QuoteTherefore, we should get the following:(Δs)^2 = (Δx)^2 + (Δt)^2That 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-dimensionsBe 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 HalcSo, 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)^2How 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)^2We 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.

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

I really want to understand this calculation.

QuoteI'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

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

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.

Quote from: Halc on 31/08/2019 06:07:12Why 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,

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).

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).

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, 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:

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):

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 outwardsSo 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?

Quote from: Halc on 31/08/2019 13:16:551) 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.

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

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.

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?

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?

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?

Thanks HalcIt 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 Way1. More than 99% of the matter in the accretion disc is ejected outwards. So out of 100 particles - more than 99 are ejected outwards.

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?

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.

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 <∼ 10^{39} erg s^{−1}.

The RIAF model constructed above gives an excellent fit to the spectrum, both globally (χ^{2}/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.

QuoteThe RIAF model constructed above gives an excellent fit to the spectrum, both globally (χ^{2}/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.

QuoteThe 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.

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

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?

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".

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.

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.

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

Quote from: Halc on 06/09/2019 17:37:52To 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".

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?