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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.
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.
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.
The boarder of the" fully time-like" space time is at the event of horizon because nothing can escape from that point and than in order to proof that nothing can escape from there we use the space-time formula.
There must be an error in this logic.
Nothing can escape from the event horizon based on its meaning and not because of space-time.
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?
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?
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 <∼ 1039 erg s−1.
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.
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.
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.
Thanks KrypidSo 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?
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)"
We assume that Sgr A* outflow either carries with it CR protons created near the black hole, or that the CR protons are accelerated on shock fronts where the outflow runs into the interstellar medium.
If instead we assume that the gamma–ray lobes were produced in this event, we must conclude that the energy–driven outflow is still proceeding, with a mean velocity hvi ≃1000 km s−1 over its lifetime.
Our model is similar to that of Crocker & Aharonian (2011) in terms of assuming the CR protons (rather than electrons) produce the observed gamma-ray emission.
They further consider a quasi–steady state model in which the CR protons are continuously injected by supernova explosions.
Within this model, then, in galaxies with large σ>∼ 150 km s−1, Eddington outflows tend to sweep the vicinity of the hole clear of gas of density (2) and prevent further accretion and growth, establishing the M−σ relation for the black hole mass (King 2003; 2005).
For the present day Milky Way and directions well out of the Galactic plane, we expect fg to be significantly less than fc, so we parametrize fg as fg = 1.6 · 10−3 f0.01, where f0.01 ∼ 1 is a dimensionless free parameter of the model.
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.
The efficiency f = 0.2 is assumed.
We model the broad absorption at E ~ 9 keV with the XSTAR31 code v. 2.2.1bn. We consider a Γ = 2 power-law continuum, consistent with the observed value (see SI), and standard Solar abundances. A turbulent velocity of 30,000 km s−1 is assumed for the fast wind model.
This equation assumes a spherical, but clumpy, geometry.
We derive a physical characterization of this outflow using a dedicated photoionization absorption model (see Methods).