[Malamute Lover]

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

Thanks
Before 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?

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. Decades of data on the position of S2 is available and was used to produce the estimate of 4.31 million solar masses.  I linked to this in my previous post but here it is again.
https://en.wikipedia.org/wiki/S2_(star)#Orbit

[Dave Lev (OP)]

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.

Sorry
Gravity is gravity.
Do you claim that there is a friction in gravity?
Based on Newton formula we can extract the real mass of the SMBH.

Therefore:

The formula is as follow:
V^2 = M G / R
M = V^2 R / G

Hence,
If  V = 0.3c = 0.3 * 300,000 Km /s = 90,000Km /s = 90,000,000 m/s = 90 *10^6 m/s
R = 88 BKm = 88 * 10^9 Km = 88 * 10^12 m

M = (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 Kg

M(SMBH) = 6.13*10^38 / 1.9891*10^30  = 3.08 *10^8 Sm = 0.3 B Sun mas
Therefore, 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?

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.

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.

[Bored chemist]

Do you claim that there is a friction in gravity?

No, he didn't.
Did you try reading what he wrote?

Based on Newton formula we can extract the real mass of the SMBH.

As long as the velocities involved are not close to the speed of light.
Unfortunately,

If  V = 0.3c

then you obviously can't use Newtonian physics.

So why did you try?
And on a related note

Quote from: Bored chemist on 25/07/2020 11:17:26
No, 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?

[Malamute Lover]

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.

Sorry
Gravity is gravity.
Do you claim that there is a friction in gravity?

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.

Based on Newton formula we can extract the real mass of the SMBH.

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.

Therefore:

The formula is as follow:
V^2 = M G / R
M = V^2 R / G

Hence,
If  V = 0.3c = 0.3 * 300,000 Km /s = 90,000Km /s = 90,000,000 m/s = 90 *10^6 m/s
R = 88 BKm = 88 * 10^9 Km = 88 * 10^12 m

M = (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 Kg

M(SMBH) = 6.13*10^38 / 1.9891*10^30  = 3.08 *10^8 Sm = 0.3 B Sun mas
Therefore, based on this calculation the mass of the SMBH is 0.3 B Sun mass.

You are using the extent of the cold gas disc as the part that is moving at 0.3 c. Totally wrong. Also, your value for G is wrong.

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?

The 0.3 c figure for orbital speed estimate has a lot of intervening variables as I discussed above. But I have no idea what you are trying to convey here.

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.

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.

So you think there is another 4.3 billion solar mass object very near to Sgr A* that S2 is really orbiting around that has somehow escaped notice.

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?

[Dave Lev (OP)]

your value for G is wrong.

Thanks
You are correct.
https://en.wikipedia.org/wiki/Gravitational_constant
"The measured value of the constant is known with some certainty to four significant digits. In SI units its value is approximately 6.674×10−11 m3⋅kg−1⋅s−2."
I have used 1.67 instead of 6.674.
So, the mass should be lower by about 4 times.
SMBH mass (by newton) = 75M solar mass.

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.

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)?

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.

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 c?
If 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?

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?

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???

[Bored chemist]

S13 sets almost a pure orbital cycle.

All orbits are orbital.
Do you mean that you think it is circular?
It's plainly elliptical and as far as I can tell Sgr5 is at the focus.

Why do you refuse to use this vital information???

Well, because the disk is subject to other forces - like friction.
Also it's more difficult to define the exact location, and we don't know the velocity.
It's not a matter of refusing to use the information.
The information does not exist.

Quote from: Bored chemist on Yesterday at 10:12:53
Quote from: Bored chemist on 27/07/2020 13:53:30
Quote from: Bored chemist on 25/07/2020 11:17:26
No, 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?

[Malamute Lover]

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.

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)?

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.

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.

Actually, the plasma temp at the accretion disc is estimated at the range of 10^9K.

My source says 10 million = 10^7 K
https://www.space.com/milky-way-monster-black-hole-cool-disk.html
What is your source?

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 c
If 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?[ ?

The high temperature due to friction in the innermost past of the accretion disc might result in local motions that could bias spectrographic readings and slightly modify the speed estimate. The temperature is not going to change the actual orbital speed. Please learn some physics.

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?

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???

Keplerian orbits are ellipses. The center of gravitation lies at a focal point of the ellipse. An ellipse can be rotated on its long axis at less than a right angle and still be recognized as an ellipse. The focii do not move. A 2D projection of a 3D reality does not change that. Please learn some math.

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.

[Dave Lev (OP)]

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.

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.gif
We 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).
https://danspace77.files.wordpress.com/2018/02/18-028d.png
It 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 SMBH?

Now 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.html
https://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.

[Bored chemist]

Orbits.png (107.69 kB . 1280x916 - viewed 2819 times)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?

[Halc]

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?

Yes, in the case of each of those orbits, the BH is on the major diameter, but that major diameter is not obvious when you're looking at a 2D projection of a 3D orbit. You're used to pretty 2D diagrams of the solar system.  The orbit of Venus is pretty circular, but from our perspective, it just moves back and forth, sometimes hitting (eclipsing) the sun.

So I look at that diagram, and the most circular orbit is definitely S1, not because of its shape from our perspective, but because the BH is most centered in that path.  The more off to the side it is, the more eccentric the orbit.

[Bored chemist]

Are the orbits planar?
Because, if they are, I still see a problem.

[Malamute Lover]

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.

What do you mean by relativistic effects? How that effect can impact the orbital path?

One of the early confirmations of General Relativity was explaining the odd orbit of Mercury. Because the Sun creates a 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.

http://www.sci-news.com/astronomy/s2-star-general-relativity-08339.html
https://www.sciencealert.com/a-star-dancing-around-a-supermassive-black-hole-is-another-win-for-relativity
https://www.space.com/41291-relativity-revealed-milky-way-core.html
https://www.mpg.de/14692117/detection-of-schwarzschild-precession-in-the-orbit-of-star-s2
https://www.aanda.org/articles/aa/abs/2017/12/aa31148-17/aa31148-17.html
https://www.aanda.org/articles/aa/full_html/2020/04/aa37813-20/aa37813-20.html

There is a lot more but I got tired of copy/paste. If you really wanted to know about S2 and relativity you could have done a search on ‘s2 star relativity’.

Please look at the following diagram:
https://www.lcas-astronomy.org/articles/images/orbit.gif
We 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=click
It 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 SMBH

Now 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.

Here is the picture presented earlier.



If you count graph boxes, you will see that the orbit of S13 is an ellipse. The orbit is Keplerian, not Copernican.

Here is a fun video.

You have a link to a png that is in garbage form. I extracted the actual url
danspace77.files.wordpress.com/2018/02/18-028d.png
and it does not work. danspace77 is there in wordpress, but the picture is not.

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.html
https://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.

The 100,000 solar mass black hole is 200 light years away. That’s 10^15 kilometers. It is utterly irrelevant to this discussion. If there were another black hole in the neighborhood powerful enough to influence the orbits of the stars being discussed, those orbits would be wildly chaotic, not just because of the two source of serious gravitation but because being so close to the black hole at the galactic center it would be in orbit around that black hole. There is no other SMBH in the neighborhood.

As heavily documented at the beginning of this post, the orbit of S2 is not Copernican, nor is it exactly Keplerian due to relativistic effects.

[Halc]

Are the orbits planar?
Because, if they are, I still see a problem.

They're all (more or less) planar.

I can look at any ellipse from just the right perspective and it will appear to be circular. S13 is a good approximation of that.
I can look at any circle from a non-perpendicular perspective and it will appear elliptical. Dave can't envision that. You can't either?  Here's a hula hoop:

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.

S8 and S14 are really as eccentric as they appear.

"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"

What lovely writing by indianexpress. Why put the word 'toxic' in there. Toxic to what exactly? It there to discourage tourists from trying to breathe in the gasses near this black hole, like most gas clouds elsewhere might not be as toxic as this one?

[Bored chemist]

I think I was running into the same problem as Dave Lev, and I think I have got it sorted out
The BH should be on the major axis but it doesn't look like it.

It's not a matter of making an ellipse look like a circle- if you consider the ellipse as a conic section, it's obvious that, from the point of the cone, it looks like a circle.
Nor is it a matter of making a circle look like an ellipse- there are old poems about that.


art.png (179.71 kB . 504x441 - viewed 2948 times)

The issue is that the real major axis may not be where it "looks" like it is.

If you draw an ellipse and its major axis, then look at it along a line parallel to that axis and nearly parallel to the plane of the paper, you see the line at right angles to major axis of the projection.

[Dave Lev (OP)]

Thank you all

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.

Yes, your explanation is very clear.
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?

Based on Kepler law:
https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#/media/File:Kepler_laws_diagram.svg
"The two shaded sectors A1 and A2 have the same surface area and the time for planet 1 to cover segment A1 is equal to the time to cover segment A2."
Let's look at the S2 orbital cycle:
https://astro.swarthmore.edu/ir/ir_results.html
It seems to me that the surface aria that is locked between the SMBH to the arch of 2001.50 to 2002.25 (31 weeks) is significantly bigger than the surface aria that is locked between the SMBH to the arch of 2002.25 to 2002.58 (33 Weeks).
Could it be that it is due to general relativity as we see with the mercury motion:

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.

https://courses.lumenlearning.com/astronomy/chapter/tests-of-general-relativity/
Due to Einstein general relativity the perihelion point moves from one orbital cycle to the next one.
Therefore, if I understand it correctly, the real impact will take place on the next orbital cycle and the change is quite minor.
So how can we explain that significant difference in the aria surface for a similar given time.
In the same token:
The surface aria that is locked between the SMBH to the arch of 1992.23 to 1994.32 (two years +9 months) is bigger than the surface aria that is locked between the SMBH to the arch of  1994.32 to 1996.43 (two years +9 months).
I assume that in this case we can't blame that it is due to general relativity.
So, how can we explain those problems?
Could it be that S2 diagram itself is incorrect?
Could it be that S2 is under the impact of other gravity forces (and not only under the SMBH gravity)?
Please be aware that in the whole galaxy there are about 400 Billions stars. So why nearby gas clouds, millions nearby stars, Bar, Ring, spiral arms... of the galaxy couldn't set some gravity force on S2 orbital cycle?

What about dark matter?
Our scientists claim that it must be there at quite high density (as we go closer to the center). So why it has no impact on S2 orbital cycle? (Especially, If the real orbital cycle in 3D is much more than just 10 Light days).
Why don't use the dark matter for any S stars?
If I understand it correctly, the dark matter increases it density as we get closer to the SMBH, sets different densities based on the size of the spiral galaxy and has no negative impact on S stars or any other star or gas clouds in the center of the galaxy.
How could it be that we use the dark matter only when we need it?


Would you kindly also answer the following problem with S2?

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.

So, do you agree that this is one more confirmation that the orbital cycle of S2 is not just effected by the SMBH gravity?

[Bored chemist]

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?

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.
On the other hand...

Could it be that it is due to general relativity as we see with the mercury motion:

It's because we are not looking at it from a viewpoint perpendicular to teh plane of the orbit.

We can see clearly that S2 is not always detected at the expected orbital path. It moves around it. In and out.

We don't see that at all.

What do you think the crosses through the plotted positions are there for?
https://datavizcatalogue.com/methods/error_bars.html

Wouldn't it be better if you  understood the diagram before you tried to say it was wrong?

So, do you agree that this is one more confirmation that the orbital cycle of S2 is not just effected by the SMBH gravity?

No, it is evidence that you do not understand the subject.

[Bored chemist]

You keep forgetting to answer this

Quote from: Bored chemist on 28/07/2020 16:30:30
Quote from: Bored chemist on Yesterday at 10:12:53
Quote from: Bored chemist on 27/07/2020 13:53:30
Quote from: Bored chemist on 25/07/2020 11:17:26
No, 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?

[Dave Lev (OP)]

Quote from: Dave Lev on 01/08/2020 05:28:11
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?

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.

Are you sure about it?
Let's use Pluto as an example.
Assuming that we can only see the following orbital of Pluto
https://en.wikipedia.org/wiki/Pluto#/media/File:Plutoorbit1.5sideview.gif
Can we extract the mass of the Sun from this orbital cycle?
Assuming that we don't see the Sun.
Can we extract the correct location of the sun from this orbital shape and calculate the real Sun mass?

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.

Well, S2 is clearly not good enough.

"It's estimated up to 85 percent of all stars could be in binary pairs, or even triple or quadruple systems; and over 50 percent of all Sun-like stars are in binary pairs."
https://www.sciencealert.com/we-may-have-found-our-sun-s-long-lost-identical-twin-star
Even the sun has a twin:
"It's thought that somewhere out there, the Sun has a twin - born not just in the same stellar nursery, but an almost identical twin, a binary companion made of the same star-stuff. And astronomers think they might have just found it.
Located roughly 184 light-years away, it's called HD 186302, and it's almost certainly at least a long-lost sibling of our home star.
So, why are we so sure that none of the S stars have no companion star or even companion BH?

So, S2 is not good enough for the estimation of the SMBH mass due to the following:
1. Do we know the real orbital cycle of S2 in 3D?
2. Do we know how many other companion stars it might have?
3. Do we know the impact of the millions of nearby stars or even the billions in the galaxy?
4. Do we know the impact of the Dark matter on S2

Therefore, it is quite clear to me that it should be much more accurate to extract the real mass of the SMBH from the accretion disc due to the following arguments:

1. The accretion disc clearly orbits in almost a pure circular orbital cycle around the SMBH.
2. It is located very nearby the SMBH and therefore it isn't affected by dark matter or any other real matter as stars or gas clouds.

I fully agree with you that it is affected by the magnetic field or heat that could add some friction.
However that friction can't slow down the plasma accretion orbital velocity.
If that was the case, than all the plasma in the accretion disc had to fall immediately into the SMBH
As the plasma still orbits around the SMBH, it proves that the plasma orbits at the "magic velocity" that is needed to keep it in the orbital loop..

Based on Newton formula I have found that the total mass should be 75 M sun mass.
However, if you consider that Newton formula is not applicable to that ultra high velocity of 0.3c, than please use Einstein formula.
Please estimate the orbital velocity at the correct radius, add the impact of the magnetic field/heat/friction and set the calculation for the SMBH mass based on this accretion disc valid data..

[Bored chemist]

You keep forgetting to answer this

Quote from: Bored chemist on 28/07/2020 16:30:30
Quote from: Bored chemist on Yesterday at 10:12:53
Quote from: Bored chemist on 27/07/2020 13:53:30
Quote from: Bored chemist on 25/07/2020 11:17:26
No, 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?

[Dave Lev (OP)]

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.

Do you claim that there is no need Negative mass particle, Dark matter, dark energy and many other none realistic ideas as density wave to justify the BBT and the activity of the spiral galaxies in the Universe?
Let's focus on dark matter:
https://www.businessinsider.com/less-dark-matter-in-milky-way-2014-10
The team found the dark matter in our galaxy weighs 800 billion times the mass of the Sun, half of previous estimates.
"Rubin found that galaxies rotated nothing like our own Solar System. The outer stars did not rotate slower than the inner stars, but just as fast. There had to be dark matter on the outskirts of every galaxy.

Now, astronomer Prajwal Kafle, from The University of Western Australia, and his colleagues have once again observed the speed of stars on the outskirts of our own galaxy, the Milky Way. But he did so in much greater detail than previous estimates.
From a star's speed, it's relatively simple to calculate any interior mass. The simple equation below shows that the interior mass (M) is equal to the distance the star is from the galactic center (R) times its velocity (V) squared, all divided by the gravitational constant (G):

M = R * V^2 / G"

So, our scientists have no basic clue how spiral galaxy really works.
They hope that somehow the dark matter will help to fulfill the requested Newton Formula,
They also hope that stars are migrating from outside just to be eaten by the SMBH.
As they can't explain the orbital velocity of the stars in the spiral arm, they have called the dark matter for some help.

So dear scientists
You have a fatal error in your understanding.
Stars do no migrate from outside inwards. This is a pure imagination.
Stars in spiral arms can't migrate inwards. Never ever!!!
Actually, nothing can migrate into the Milky Way while it crosses the open space at almost 600 Km/s.
Any star/ globular cluster or galaxy (that is not connected to the MW) must be shifted away and clear the path for our mighty galaxy.
All the stars and dwarf galaxies that we see around the MW galaxy are direct products of the galaxy itself.
Therefore, NOTHING from outside could come closer to the galactic disc plane.

Let's look at the SPIRAL arm.
The local gravity of the spiral arm holdes it in place.
However, all the stars in the arms MUST be drifted outwards over time.
That outwards drift keeps the constant (or almost constant) orbital velocity of the stars at almost any radius - as long as they are at the galactic disc.
At the ring (the starting point of the spiral arm - 3KPC) the thickness of the arm is about 3,000Ly.
While at the outer side of the arm (10-15KPC) the thickness of the arm is 400Ly.
So, that by itself proves that our Sun doesn't orbit around the center of the galaxy as our scientists wish.
If that was the case, than as you move outwards, the chance to move above and below the orbital disc is higher.
Please look at the orbital path of Pluto
https://en.wikipedia.org/wiki/Pluto#/media/File:Animation_of_Pluto_orbit.gif

Therefore, the following formula for the total requested mass at the center is clearly none realistic:
M = R * V^2 / G

I can tell you by 100% that you won't find even one star that try to penetrate into the Milky Way or even back to the galactic disc. Never ever!

Our scientists also hope that the MW galaxy increases its mass due to a collision with other massive galaxy.
That is also pure imagination.
When Andromeda will come closer to the MW, the Bigger must be the winner.
So, the Milky Way would probably clear the way the mighty Andromeda.
If not, the collision would set a severe impact at both galaxies.
Andromeda would survive, but the Milky Way would probably lose some portions of the arms if not all the arms.
Those broken arms would be ejected into the open space including the MW SMBH.
Andromeda won't take even one star or moon from the Milky way.
So, they would never ever integrate with each other. Eventually, each one of those galaxies will continue to cross the space but with less stars on board.
The only way for galaxies to gain mass is by creating new mass and new stars.
S stars (including S2) are close to the SMBH as they all are new born stars.
S2 had been created from the molecular that had been created and ejected from the accretion disc.
Once upon a time, when the solar system were very young, it was there orbiting around the center as S2 does.
One day, our sun will be ejected from the Orion arm to the open space.
That would be a very tough time for us.

In any case, based on the current BBT theory, without the unproved dark matter you can't explain the orbital rotation velocity of the stars at the spiral arms.
While based on theory D there is no need for dark matter or density wave hypothetical idea.
Local Newton gravity at the spiral arms is good enough for that job.