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Quote from: Malamute Lover on 27/07/2020 21:35:23The accretion disc has an outside radius of about 88 billion km. The hot spot inner region has a radius of 130-220 million km. The disc cuts off at about 11 million km at the innermost stable circular orbit. Anything below that point is going to fall into the black hole. Temperature in the inner part of the accretion disc reaches 10 million K.https://arxiv.org/abs/1810.12641ThanksBefore we continue our discussion on any other subject, let's focus on the accretion disc.In the following article it is stated that the "magnetized accretion disk/torus of ∼10 light minutes in diameter".https://arxiv.org/pdf/1810.12641.pdf"Intercontinental microwave interferometry and polarized infrared(IR)/X-ray variability on 10-30 minute timescales suggest that this emission comes from highly relativistic electrons in a hot, magnetized accretion disk/torus of ∼10 light minutes in diameter, plus perhaps a jet, just outside the innermost stable circular orbit (ISCO) of the putative massive black hole "Therefore, the radius of the accretion disc is about 10/2 = 5 light minutes. Which is about 88 Billion Km as you have stated.Can we assume that the orbital velocity of the plasma at the accretion disc (R=88Bkm) is 0.3c?Based on this data, what is the estimated mass of the SMBH?
The accretion disc has an outside radius of about 88 billion km. The hot spot inner region has a radius of 130-220 million km. The disc cuts off at about 11 million km at the innermost stable circular orbit. Anything below that point is going to fall into the black hole. Temperature in the inner part of the accretion disc reaches 10 million K.https://arxiv.org/abs/1810.12641
As I said in my previous post, the 88 billion km includes the large cool gas region. The 5 light minutes = 90 million km radius is the innermost portion of the hot region. That innermost portion is densest and moving the fastest and so has “highly relativistic electrons in a hot, magnetized accretion disk/torus”.To calculate the mass of the black hole that close to the event horizon would require Schwarzschild-Kerr dynamics, which in turn would require assuming a mass for the black hole to begin with and interpolating, making the equations seriously non-linear. In addition, there are dynamics going on in the accretion disc that could affect measured velocity independent of gravity, such as friction, temperature, the magnetic field, continually infalling material etc. Not a good way to measure.
A much easier way to determine the mass of the black hole is to look at the hyperbolic orbit of the star S2, which orbits the black hole at a significant distance, enough to avoid hard to resolve non-linearities, but follows a distinctly non-Keplerian elliptical orbit.
Do you claim that there is a friction in gravity?
Based on Newton formula we can extract the real mass of the SMBH.
If V = 0.3c
Quote from: Bored chemist on 27/07/2020 13:53:30Quote from: Bored chemist on 25/07/2020 11:17:26No, it's irrelevant.Because, when you said this "Sorry, in order to support the BBT, we need a help from Negative mass particle, Dark matter, dark energy and many other none realistic ideas."It wasn't true.So, why did you tell that lie?Smoke and mirrors are not going to help you here.You need to answer the question.Why did you say something that's not true?
Quote from: Bored chemist on 25/07/2020 11:17:26No, it's irrelevant.Because, when you said this "Sorry, in order to support the BBT, we need a help from Negative mass particle, Dark matter, dark energy and many other none realistic ideas."It wasn't true.So, why did you tell that lie?Smoke and mirrors are not going to help you here.You need to answer the question.Why did you say something that's not true?
Quote from: Malamute Lover on 28/07/2020 13:29:40As I said in my previous post, the 88 billion km includes the large cool gas region. The 5 light minutes = 90 million km radius is the innermost portion of the hot region. That innermost portion is densest and moving the fastest and so has “highly relativistic electrons in a hot, magnetized accretion disk/torus”.To calculate the mass of the black hole that close to the event horizon would require Schwarzschild-Kerr dynamics, which in turn would require assuming a mass for the black hole to begin with and interpolating, making the equations seriously non-linear. In addition, there are dynamics going on in the accretion disc that could affect measured velocity independent of gravity, such as friction, temperature, the magnetic field, continually infalling material etc. Not a good way to measure.SorryGravity is gravity.Do you claim that there is a friction in gravity?
Therefore:The formula is as follow:V^2 = M G / RM = V^2 R / GHence,If V = 0.3c = 0.3 * 300,000 Km /s = 90,000Km /s = 90,000,000 m/s = 90 *10^6 m/sR = 88 BKm = 88 * 10^9 Km = 88 * 10^12 mM = (90 * 10^6) ^2 * 88 * 10^12 / 1.67 *10^11= 8100 10^12 *88*10^12 *0.86 * 10^11 = 613,008 *10^ 33 = 6.13 * 10 ^ 38 kg?The sun mass is: Sm = 1,989,100,000,000,000,000,000 billion kg = 1.9891 *10^30 KgM(SMBH) = 6.13*10^38 / 1.9891*10^30 = 3.08 *10^8 Sm = 0.3 B Sun masTherefore, based on this calculation the mass of the SMBH is 0.3 B Sun mass.
Actually, do you agree that if your following message is correct and there is a friction due to "temperature, the magnetic field, continually infalling material etc" than the real mass of the SMBH shold be much more that that?
Quote from: Malamute Lover on 28/07/2020 13:29:40A much easier way to determine the mass of the black hole is to look at the hyperbolic orbit of the star S2, which orbits the black hole at a significant distance, enough to avoid hard to resolve non-linearities, but follows a distinctly non-Keplerian elliptical orbit.Sorry S2 doesn't orbit around the SMBH.If you monitor S2 orbital path you would notice that it moves inwards and outwards in its elliptic orbital shape. Few years ago our scientists even observe the S2 and the SMBH at almost the same line. I claim that the SMBH isn't located even at the S2 orbital plane (including all the other S stars).Therefore, S2 and all the other S stars can't give real indication about the mass of the SMBH.The real mass should be extracted from the accretion disc.So, do you agree that the Minimal mass of the SMBH should be 0.3 B Sun mass.
your value for G is wrong.
As I said, you cannot use Newtonian mechanics. The disc extends right up to the lowest stable circular orbit for massive particles. This is General Relativity territory. You have to at least use Schwarzschild dynamics.
Friction changes motion. With temperatures are high as 10^7 K, primarily due to friction, you cannot take the disc as a coherently moving body. Speeds as high as 0.3 c as determined from spectrographic readings may not represent the actual orbital speed. There could be lateral motion, or heat expansion changing the perceived motion.
Here are the orbital paths of the five stars. Sgr A* is obvious at the focus of each of the ellipses, once you factor in relativistic considerations. You do know that in general orbits are ellipses not circles, right?
S13 sets almost a pure orbital cycle.
Why do you refuse to use this vital information???
Quote from: Bored chemist on Yesterday at 10:12:53Quote from: Bored chemist on 27/07/2020 13:53:30Quote from: Bored chemist on 25/07/2020 11:17:26No, it's irrelevant.Because, when you said this "Sorry, in order to support the BBT, we need a help from Negative mass particle, Dark matter, dark energy and many other none realistic ideas."It wasn't true.So, why did you tell that lie?Smoke and mirrors are not going to help you here.You need to answer the question.Why did you say something that's not true?
Quote from: Malamute Lover on 28/07/2020 19:32:30As I said, you cannot use Newtonian mechanics. The disc extends right up to the lowest stable circular orbit for massive particles. This is General Relativity territory. You have to at least use Schwarzschild dynamics.OK.Based on Newton (and assuming that there is no friction and my calculation is correct) the requested mass of the SMBH is 0.3B Sun mass.However, you claim that it is General Relativity territory.Therefore, can you please show how General Relativity can justify a mass of only 4.1 M solar mass (almost 100 times lower than Newton) for that activity (assuming that there is no friction or high temp)?
Quote from: Malamute Lover on 28/07/2020 19:32:30Friction changes motion. With temperatures are high as 10^7 K, primarily due to friction, you cannot take the disc as a coherently moving body. Speeds as high as 0.3 c as determined from spectrographic readings may not represent the actual orbital speed. There could be lateral motion, or heat expansion changing the perceived motion.Actually, the plasma temp at the accretion disc is estimated at the range of 10^9K.
Therefore, as there is so high friction at the accretion disc, don't you agree that it should reduce the velocity or the orbital plasma?Therefore, if we measure an orbital velocity of 0.3c with the impact of the friction, than what should be the velocity without a friction?Don't you agree that without a friction the orbital velocity SMBH must be quite higher?Should it be 0.5c, 0,75c or even cIf the velocity is higher, than the requested SMBH should be higher.Therefore, based on General Relativity what is the minimal SMBH mass that is needed to suport 0.3c orbital velocity at radius of 88B Km while the friction is so high?How could it be that this estimated ultra small 4.1M Sun mass could be good enough for that job?[ ?
Quote from: Malamute Lover on 28/07/2020 19:32:30Here are the orbital paths of the five stars. Sgr A* is obvious at the focus of each of the ellipses, once you factor in relativistic considerations. You do know that in general orbits are ellipses not circles, right?Please look again at the diagram that you had offered.S13 sets almost a pure orbital cycle.Based on newton the main object should be located at the center. Therefore, how could it be that the SMBH is not located exactly at the center of this cycle?Please look at S1 and S12.S1 elliptical shape is very similar to S12 (but bigger).Take them out from the diagram and try to estimate where the center should be located for each one of them?So, as they look similar, it is expected to see the center of mass at relatively similar locations in the diagram.Surprisingly, this is not the case, while the center of S1 is not located at symmetrical spot which could be justified by Kepler.In any case, this diagram is presented in 2D. I don't know what is the accuracy of the distance to each object at 3D.Somehow it seems to me that in 3D some of the orbital cycles do not even cut each other.So, there is high chance that the SMBH is not located at the orbital plane of some of those stars.Therefore, those stars can't be used as a mass reference for the SMBH.Only the orbital accretion disc must be used as a real reference as it clearly orbits around the SMBH!!!.Why do you refuse to use this vital information???
The orbits deviate slightly from Keplerian ellipses due to relativistic effects, especially S2. The deviations are exactly what they are expected to be from a 4 million mass black hole being the gravitating body.
It can't really matter- because one BH can't make a universe but I think there is something odd about that diagram.Shouldn't the BH be on the major diameter of the orbits?
Quote from: Malamute Lover on 29/07/2020 15:53:19The orbits deviate slightly from Keplerian ellipses due to relativistic effects, especially S2. The deviations are exactly what they are expected to be from a 4 million mass black hole being the gravitating body.What do you mean by relativistic effects? How that effect can impact the orbital path?
Please look at the following diagram:https://www.lcas-astronomy.org/articles/images/orbit.gifWe see in two images:1. Copernican orbit diagram - Circular orbit while the host is located exactly at the center.2. Keplerian orbit diagram - Elliptical orbit while the host is located at a symmetrical point as the distance to 1 is identical to the distance to 7. in the same token 2 to 6 and 5 to 3.Please look at S13 orbital Path (blue line).F2018%2F02%2F18-028d.png#id=-1&iurl=https%3A%2F%2Fdanspace77.files.wordpress.com%2F2018%2F02%2F18-028d.png&action=clickIt looks as a perfect circular orbit.Therefore, it fully meets the Copernican orbit. Hence, the Host point should be located just at the center.Surprisingly, the SMBH is located somewhere at the upper side. How could it be that this idea of relativistic effects could set such an impact?Don't you see that S13 doesn't orbit directly around the SMBHNow please look at S2 orbital path.As it is elliptical, and based on Keplerian orbit, the host should be located at the symmetrical point. Hence, it is expected that the distance to point 1 should be identical to point 7.Again - surprisingly, the SMBH is located at the offside to the left.So, also in this case, how the relativistic effects could set such severe impact?There is no almost correct orbital path. If S13 or S2 were orbiting around the SMBH they have to obey to the orbital law.Therefore, it is quite clear to me that S13 and S2 don't orbit directly around the SMBH.
There is another important issue.We know that that there are many BH at the center of our galaxy near the SMBH.https://www.ajc.com/news/science/000-black-holes-hiding-center-milky-way-study-suggests/4r7M8X1ECfYNwiHrS8utIN/Scientists have found evidence of 10,000 black holes in the Milky Way galaxy, surrounding its central suppermassive black hole, Sagittarius A*.Some of those black holes are quite massive:https://indianexpress.com/article/technology/science/massive-black-hole-discovered-near-heart-of-the-milky-way/"Massive black hole discovered near heart of the Milky Way""A huge black hole – about 100,000 times more massive than our Sun – has been discovered lurking in a toxic gas cloud near the heart of the Milky Way. If confirmed, the object will rank as the second largest black hole in the Milky Way after the supermassive Sagittarius A* which is located at the very centre of the galaxy."Those BH could have an impact on the orbital Path of S2 and S13.Let's look at the orbital path of S2:https://astro.swarthmore.edu/ir/ir_results.htmlhttps://www.universetoday.com/72315/astronomy-without-a-telescope-galactic-gravity-lab/We can see clearly that S2 is not always detected at the expected orbital path. It moves around it. In and out.That shows that S2 does not orbit directly around the SMBH but it orbits around some massive object as BH or even MBH while they both move together and set the final orbital path of S2.As an example, let's assume that we could shut down the light of the Earth.So, from outside, you can only see the Sun and the Moon.If you monitor the moon, you would see that it is moving in and out with regards to the expected orbital path around the Sun.This proves that S2 is not there by itself. It must orbit around a massive object as BH or MBH while they both set the orbital path.In any case, as the orbital path of S13 and S2 do not fully obey to Copernican or keperlian orbits, than it proves that they do not orbit directly around the SMBH.
Are the orbits planar?Because, if they are, I still see a problem.
"A huge black hole – about 100,000 times more massive than our Sun – has been discovered lurking in a toxic gas cloud near the heart of the Milky Way"
The 2D image traces a highly eccentric ellipse, and yet the 3D object photographed is circular. Is that so hard?I can do the same with an ellipse of any eccentricity, viewing it low along the long axis until it projects a circle. S13 is like that, clearly a significantly eccentric orbit, as evidenced by Sgr-A being so off-center.S1 is hardly circular, but is the least eccentric of the bunch, as evidence by Sgr-A appearing closest to the middle.S2's long axis runs not vertical, but more or less in the direction of 1 O-clock in that image.
One of the early confirmations of General Relativity was explaining the odd orbit of Mercury. Because the Sun creates gravity well and because Mercury is so close, there is less space in the region of Mercury’s orbit than expected from a Euclidean model.https://courses.lumenlearning.com/astronomy/chapter/tests-of-general-relativity/In similar manner the space that S2 and the other orbiting stars are in diverges noticeably from Euclidean flat.For information about S2 and General Relativity, try these.
We can see clearly that S2 is not always detected at the expected orbital path. It moves around it. In and out.That shows that S2 does not orbit directly around the SMBH but it orbits around some massive object as BH or even MBH while they both move together and set the final orbital path of S2.As an example, let's assume that we could shut down the light of the Earth.So, from outside, you can only see the Sun and the Moon.If you monitor the moon, you would see that it is moving in and out with regards to the expected orbital path around the Sun.This proves that S2 is not there by itself. It must orbit around a massive object as BH or MBH while they both set the orbital path.
However, in this case, the real orbital cycle of S2 might be much bigger than the 2 x 10 Light days that we observe in 2D. So, how can we set any realistic calculation on S2 without understanding the real size of its orbital cycle?
Could it be that it is due to general relativity as we see with the mercury motion:
We can see clearly that S2 is not always detected at the expected orbital path. It moves around it. In and out.
So, do you agree that this is one more confirmation that the orbital cycle of S2 is not just effected by the SMBH gravity?
Quote from: Bored chemist on 28/07/2020 16:30:30Quote from: Bored chemist on Yesterday at 10:12:53Quote from: Bored chemist on 27/07/2020 13:53:30Quote from: Bored chemist on 25/07/2020 11:17:26No, it's irrelevant.Because, when you said this "Sorry, in order to support the BBT, we need a help from Negative mass particle, Dark matter, dark energy and many other none realistic ideas."It wasn't true.So, why did you tell that lie?Smoke and mirrors are not going to help you here.You need to answer the question.Why did you say something that's not true?
QuoteQuote from: Dave Lev on 01/08/2020 05:28:11However, in this case, the real orbital cycle of S2 might be much bigger than the 2 x 10 Light days that we observe in 2D. So, how can we set any realistic calculation on S2 without understanding the real size of its orbital cycle?Because the important thing is the time it takes to get back to where it started, and that's not affected by the orientation of the orbital plane wrt our line of sight.
Quote from: Dave Lev on 01/08/2020 05:28:11However, in this case, the real orbital cycle of S2 might be much bigger than the 2 x 10 Light days that we observe in 2D. So, how can we set any realistic calculation on S2 without understanding the real size of its orbital cycle?
As I have pointed out before, you continue to combine the outer radius of the cold gas cloud and speed of the inner radius of the accretion disc. Your math is wrong. You are using the wrong numbers.
You keep forgetting to answer thisQuote from: Bored chemist on 29/07/2020 10:12:12Quote from: Bored chemist on 28/07/2020 16:30:30Quote from: Bored chemist on Yesterday at 10:12:53Quote from: Bored chemist on 27/07/2020 13:53:30Quote from: Bored chemist on 25/07/2020 11:17:26No, it's irrelevant.Because, when you said this "Sorry, in order to support the BBT, we need a help from Negative mass particle, Dark matter, dark energy and many other none realistic ideas."It wasn't true.So, why did you tell that lie?Smoke and mirrors are not going to help you here.You need to answer the question.Why did you say something that's not true?
Because, when you said this "Sorry, in order to support the BBT, we need a help from Negative mass particle, Dark matter, dark energy and many other none realistic ideas."It wasn't true.