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Cougar
2017-Nov-05, 02:26 PM
This paper (https://arxiv.org/abs/1711.00612) addresses the possibility of an intermediate black hole neighbor of the supermassive black hole at the center of our galaxy....

glappkaeft
2017-Nov-05, 05:36 PM
Thanks! That was a very interesting paper. ALMA keeps delivering impressive data.

Dave Lee
2017-Dec-01, 06:59 AM
Few weeks ago, I have seen a discussion in this site about a 10 ^ 4 Sun Mass object which orbits the ‏Sagittarius A.
Would you kindly help me to find it again?

slang
2017-Dec-01, 02:24 PM
https://forum.cosmoquest.org/showthread.php?167047-Another-central-Milky-Way-massive-black-hole

Dave Lee
2017-Dec-01, 07:02 PM
Thanks

https://arxiv.org/pdf/1711.00612.pdf

In the article it is stated:

"The enclosed mass is estimated to be 10 ^ 4 M in the case of the Keplerian orbit with high eccentricity around IRS13E3."

Hence, it is estimated that IRS13E3 has 10 ^ 4 Sun mass.
However, IRS13E3 is just one object in the IRS13E complex.

It is stated:

"The presence of an IMBH in the IRS13E complex is strongly supported by a large enclosed mass of IRS13E3."

"The observed velocity width is ∆V ∼ 600 km s−1, and the enclosed mass of the object is estimated to be M ∼ 4 − 7 × 10 ^4 M, comparable to the mass of an IMBH, for e = 0.96 − 0.98 and a ∼ 1.4 × 1017 cm"

"Even in the case that the compact ionized gas is not belong to the Keplerian orbit, the compactness and large velocity width would indicate ∼ 10 ^ 4 M in the IRS13E complex".

So, the first question is:
What is the full estimated IRS13E complex mass?

It is also stated:


"If there is an IMBH orbiting around Sgr A∗, the position of Sgr A∗ must be affected by it.
The position of Sgr A∗ on the celestial sphere has been monitored using VLBA for a long time over 15 years (e.g. Reid&Brunthaler 2004). The positional shift along the Galactic plane is as large as -6.4 mas year−1. However, one can not distinguish between the perturbation by the massive object and the proper motion by the Galactic rotation."

"On the other hand, the positional shift crossing the Galactic plane is found to be as small as -0.2 mas year−1. This indicates that the upper limit mass of the second black hole is M .10 ^4 M in the area of r ∼ 10 ^3 −10 ^5 AU from Sgr A∗"

Hence, if I understand it correctly, the orbital radius of the IMBH (in the IRS13E complex), from Sgr A∗ is 10 ^3 to 10 ^5 AU.

As 1 Astronomical Unit = 1.58128451 × 10 ^-5 light year, the radius is:

1.58128451 × 10 ^-2 to 1.58128451 light Year (or about 1.6 10 ^-2 to 1.6 Light year)

In the same token, it is also stated that this orbital cycle has no impact on the position of Sgr A∗.

Therefore, as we know that there is an object with estimated mass (or a full complex) that orbit the Sgr A* with a verified radius:
Can we try to estimate the expected mass of Sgr A* from that orbital cycle?

Shaula
2017-Dec-01, 07:20 PM
So, the first question is:
How could it be that the whole IRS13E complex mass is estimated to be as IRS13E3 (which is just one object in that complex)?
Thanks to the wonders of rounding. The largest stars are only at most 200-300 Ms. Or 0.02-0.03 x 10^4. In a similar manner the solar system weighs about 2x10^30kg. The Sun weighs about 2x10^30kg. This is not a contradiction - it is simply that to one or two significant figures the mass of the solar system is the mass of the Sun. It looks like the IRS13E complex is similar.

Dave Lee
2017-Dec-01, 07:44 PM
Thanks

Can we try to estimate the expected mass of Sgr A* from that orbital cycle?
Why do we only based our current estimation for Sgr A* mass on S2?

If I recall correctly, S2 size is significantly smaller than IRS13E3 (by almost 10 ^ 4) and its maximal orbital distance from Sgr A* is longer (estimated to be 17 LY).

It is simmilar as we try to estimate the Sun Mass based on one of the asteroids instead of using one of the planets.

Shaula
2017-Dec-01, 09:47 PM
S2 has been observed for longer than one complete orbit, is a single compact object and is better characterised. IRS13E3 has greater uncertainty on the mass, is part of a tidally disrupted diffuse structure and is comprised of a number of interacting bodies.

It's not about the mass of a system, it is about how well you can capture orbital and body parameters. S2 is considerably easier to work with. I'm sure you can get something out of IRS13E3, but at the moment I think the error bars on this have to be pretty large. Larger than for S2.

Dave Lee
2017-Dec-01, 11:41 PM
S2 is considerably easier to work with. I'm sure you can get something out of IRS13E3, but at the moment I think the error bars on this have to be pretty large. Larger than for S2.

Why do you think that the error bar of IRS13E3 would be larger than S2.

Based on the example which I have used in order to evaluate the mass of the Sun: Could it be that an orbital cycle of a planet near the sun will have larger error bar than a far end asteroid in a kepler belt?

Let's look at the following criteria:

Do you agree that inorder to get better estimation of the host size in Keplerian orbit, we would prefer the following:
Cycle shape and distance - Less ellipse shape and closer distance is better.
Mass - Higher mass of the orbital object is better.
As:
The ellipse shape of S2 is 17 LY to almost Zero, while the ellipse shape of IRS13E3 is only 1.6 Ly to 0.016 Ly.
The mass of IRS13E3 is almost 10 ^ 4 the size of S2.

So don't you think that IRS13E3 could give better estimation about the Sun mass?
Why we don't even try to set any sort of estimation based on IRS13E3?

actually, as it is stated that the orbital path of IRS13E3 has no effect of the Sgr A*, it's quite clear that the ratio between the two must be very high.
The earth orbital cycle has no effect on the sun as 1,300,000 Earths should fit inside the Sun (about a ratio of 1: 10 ^6).

However, the IRS13E3 mass is evaluated at a range of 10 ^ 4 Sun mass while Sgr A* is estimated at a range of 10 ^ 6 Sun mass
So it is about 1 :100 ratio.
Don't you think that it is significantly low than expected?
Don't you think that based on a similar Sun/Earth ratio, the estimated mass of Sgr A* could be as high as 10 ^ 10 Sun mass?

01101001
2017-Dec-02, 01:12 AM
What's the antonym of nostalgia, the feeling I have for this contextual matter: The galactic center (https://forum.cosmoquest.org/showthread.php?161160-The-galactic-center)

What if everything in the world were a misunderstanding, what if laughter were really tears?
--Søren Kierkegaard

Shaula
2017-Dec-02, 08:13 AM
Why do you think that the error bar of IRS13E3 would be larger than S2.
I explained why in my post. In fact that was basically what the post was about:

S2 has been observed for longer than one complete orbit, is a single compact object and is better characterised. IRS13E3 has greater uncertainty on the mass, is part of a tidally disrupted diffuse structure and is comprised of a number of interacting bodies.


Do you agree that inorder to get better estimation of the host size in Keplerian orbit, we would prefer the following:
Cycle shape and distance - Less ellipse shape and closer distance is better.
Disagree. It is completely dependent on what is present closer in to the central object. As I said, what is important is the precision with which you can determine the orbit and how well you can characterise it (so a combination of how stable it is and how closely the components match the simplifying assumptions you make in looking at a Keplerian orbit)


Mass - Higher mass of the orbital object is better.
Disagree. Since once of the simplifying assumptions you make in solving the two body problem is that the mass of one component is small compared to the other this is just wrong. As I said in my post what matters is how well you can determine the mass of the system.


So don't you think that IRS13E3 could give better estimation about the Sun mass?
Nope.


Why we don't even try to set any sort of estimation based on IRS13E3?

S2 has been observed for longer than one complete orbit, is a single compact object and is better characterised. IRS13E3 has greater uncertainty on the mass, is part of a tidally disrupted diffuse structure and is comprised of a number of interacting bodies.


However, the IRS13E3 mass is evaluated at a range of 10 ^ 4 Sun mass while Sgr A* is estimated at a range of 10 ^ 6 Sun mass
So it is about 1 :100 ratio.
Don't you think that it is significantly low than expected?
No, why would I? Can you explain your logic on this? I don't see what the ratio of masses of the Sun and the Earth have to do with this. Why not pick, say, the ratio of masses of alpha Centauri to Proxima, or any one of a number of other systems. Proxima orbits alpha Centauri AB at about the same distance and the mass ratio is about 1:20. Not that any of these ratios are actually relevant as far as I can tell.


Don't you think that based on a similar Sun/Earth ratio, the estimated mass of Sgr A* could be as high as 10 ^ 10 Sun mass?
No, because that would be inconsistent with observations and for no reason I can see.

Ara Pacis
2017-Dec-02, 05:57 PM
"This is getting out of hand! Now, there are two of them!"

Dave Lee
2017-Dec-02, 06:25 PM
No, why would I? Can you explain your logic on this? I don't see what the ratio of masses of the Sun and the Earth have to do with this. Why not pick, say, the ratio of masses of alpha Centauri to Proxima, or any one of a number of other systems. Proxima orbits alpha Centauri AB at about the same distance and the mass ratio is about 1:20. Not that any of these ratios are actually relevant as far as I can tell.
Thanks for the question.
In orbital System, the ratio between the two objects is very critical on the rotation cycle effect of each other.
If the two objects have the same mass, they should orbit around a center of mass which is located exactly at the middle of their distance.
Therefore, we should see them both moving in cycles around each other.
With regards to your example about Proxima which orbits alpha Centauri AB:
As their mass ratio is 1:20, it is quite clear that both stars orbits a virtual point of center of mass (which is located closer to the bigger object).
That is the same scenario with the Earth - Moon orbit (even if their ratio of mass is 1:100).
Both of them orbits a virtual point that represents their center of mass (this point is located very close to earth - but we still can verify the movement of earth due to the Moon cycle).
Hence, in a ratio of 1:100 the small object has a noticeable impact on the big object.
This is very important and critical.
Therefore, with regards to - IRS13E3 orbit around Sgr A*:
As they have the same mass ratio as Earth/moon, (1:100) it is expected to see the impact of the orbit cycle on Sgr A* (as we see the impact on the earth due to the Moon orbital cycle).
However, it is stated clearly in the article, that IRS13E3 orbit around Sgr A* has absolutely no impact on the location of Sgr A*.
Therefore, why we shouldn't assume that the mass ratio should be significantly higher than 1:100?

There is also one more important issue -
The IRS13E3 is just one object in the IRS13E complex.
In the article it is stated that the main members of the IRS13E complex are physically bound.
"This fact indicates that the main members of the complex are physically bound although the complex should be disrupted by the strong tidal force of Sgr A∗"
Let me use the earth/Moon example.
If around the moon we will set a complex of moons, could it be that this complex would be physically bound?
What is needed to set that kind of bound?

Shaula
2017-Dec-02, 10:01 PM
As they have the same mass ratio as Earth/moon, (1:100) it is expected to see the impact of the orbit cycle on Sgr A* (as we see the impact on the earth due to the Moon orbital cycle). However, it is stated clearly in the article, that IRS13E3 orbit around Sgr A* has absolutely no impact on the location of Sgr A*.
Therefore, why we shouldn't assume that the mass ratio should be significantly higher than 1:100?
First up - where in the article does it clearly state this? It says that they detected no effect, and the precision of the available measurements from the VLBA puts an upper bound on the system mass. This is very different. So to answer your question - the reason we shouldn't assume a significantly higher mass ratio is because there are no measurements that support a higher mass ratio and several that support the stated mass ratio.

If you disagree with their calculated limits then please feel free to present your own analysis of this, and by analysis I don't mean a wall of text. As the paper clearly says the position of SgrA* is expected to be effected by the IRS13E complex. And the fact that we don't see any effect puts its mass to be of the order of 10^4 solar masses or less. Which is consistent with their other results.

Dave Lee
2017-Dec-03, 06:38 AM
If you disagree with their calculated limits then please feel free to present your own analysis of this, and by analysis I don't mean a wall of text. As the paper clearly says the position of SgrA* is expected to be effected by the IRS13E complex. And the fact that we don't see any effect puts its mass to be of the order of 10^4 solar masses or less. Which is consistent with their other results.

I'm not sure that I fully understand their explanation.

As it is stated:

"If there is an IMBH orbiting around Sgr A∗, the position of Sgr A∗ must be affected by it. The position of Sgr A∗ on the celestial sphere has been monitored using VLBA for a long time over 15 years (e.g. Reid&Brunthaler 2004). The positional shift along the Galactic plane is as large as -6.4 mas year−1. However, one can not distinguish between the perturbation by the massive object and the proper motion by the Galactic rotation"

What shall we understand from that?
My understanding is that it is expected to see the IMBH orbiting effect on the position of Sgr A*, but "one one can not distinguish between the perturbation by the massive object and the proper motion by the Galactic rotation".
In other words - they don't see any Sgr A* position effect due to the IMBH orbiting.
What is your understanding from that?

However, they continue with the following message:

"On the other hand, the positional shift crossing the Galactic plane is found to be as small as -0.2 mas year−1. This indicates that the upper limit mass of the second black hole is M .104 Min the area of r ∼ 103 −105 AU from Sgr A∗ (Reid&Brunthaler 2004). Therefore, there is no conflict between our derived enclosed mass and the upper limit mass of the second black hole inferred from the VLBA observations."

So it seems to me that they want to solve the problem by blaming the VLBA calculated limits observation.
Could it be that there is no problem with the VLBA observation?
Could it be that we just need to think what is the outcome if what we see is correct?
Why it is so important to them to keep the Sgr A∗ mass estimation as had been discovered by S2 orbit cycle?

If I recall correctly, Sgr A* and S2 had been observed in the same spot.
Therefore, inorder to solve the conflict even with the S2 they have used the benefit of the observation error bar limits.

Please remember that error bar works in all directions.
So, as they take the error bar to the maximal value in one direction to solve their problem, we can ask them what is the outcome if we will take the error bar at the same value to the other direction.
With S2 it is clear that it will force Sgr A* to be outside the orbital cycle of that star.

Why they need to use the error limits of the observation to square the cycle?
Why they don't believe in what they see?
Just think what could be the impact if they will accept the observation as is.

However, if you think that we have to blame the calculated error bar limits of the VLBA, than this is OK with me.

Shaula
2017-Dec-03, 07:50 AM
"If there is an IMBH orbiting around Sgr A∗, the position of Sgr A∗ must be affected by it. The position of Sgr A∗ on the celestial sphere has been monitored using VLBA for a long time over 15 years (e.g. Reid&Brunthaler 2004). The positional shift along the Galactic plane is as large as -6.4 mas year−1. However, one can not distinguish between the perturbation by the massive object and the proper motion by the Galactic rotation"

What shall we understand from that?
My understanding is that it is expected to see the IMBH orbiting effect on the position of Sgr A*, but "one one can not distinguish between the perturbation by the massive object and the proper motion by the Galactic rotation".
In other words - they don't see any Sgr A* position effect due to the IMBH orbiting.
What is your understanding from that?
So the paper doesn't 'clearly say' anything resembling what you said it did. Please try to be more accurate when making these kinds of claims, I reread the paper three or four times trying to find what I had missed.

My understanding from that is exactly what the paper says. That SgrA* has a number of components of motion that add up to its measured displacement and that they cannot disentangle a unique component from IRS13E from it. I'd add that if they were able to observe the system for many cycles and perhaps with other wavelengths where the sources are more compact it might be more possible to do that via harmonic analysis.


"On the other hand, the positional shift crossing the Galactic plane is found to be as small as -0.2 mas year−1. This indicates that the upper limit mass of the second black hole is M .104 Min the area of r ∼ 103 −105 AU from Sgr A∗ (Reid&Brunthaler 2004). Therefore, there is no conflict between our derived enclosed mass and the upper limit mass of the second black hole inferred from the VLBA observations."

So it seems to me that they want to solve the problem by blaming the VLBA calculated limits observation.
Could it be that there is no problem with the VLBA observation?
Could it be that we just need to think what is the outcome if what we see is correct?
However, if you think that we have to blame the calculated limits of the VLBA, than this is OK with me.
They are not blaming anyone and there is no problem. They are saying any obvious perturbations are below the detection limits of the VBLA data or cannot be distinguished from other motions and thus deriving an upper mass limit on the IRS13E system. They are not saying there is a problem with their hypothesis they need to find a way around, they are not using measurement uncertainties to 'fix' anything.

You are rather bizarrely twisting the narrative here. This is a scientific investigation of the properties of the systems near the galactic core. No one is doing anything so negative as placing blame, claiming there are problems with other pieces of work/measurements or defending a dogmatic hypothesis. It is a simple study using the best available measurements and taking into account other pieces of work on the components of the systems. Nor am I using measurement uncertainties to avoid dealing with an 'issue' I don't want to think about.

I don't know why you are fixating on this system - not only is the evidence for what you are suggesting not there but even if it were it would be outweighed by the evidence from other systems.
https://arxiv.org/pdf/0810.4674.pdf - 28 stars used to derive the mass of SgrA*
https://arxiv.org/pdf/0902.3892.pdf - 6000 stars used to constrain the mass of SgrA*

Strange
2017-Dec-03, 05:23 PM
However, they continue with the following message:

"On the other hand, the positional shift crossing the Galactic plane is found to be as small as -0.2 mas year−1. This indicates that the upper limit mass of the second black hole is M .104 Min the area of r ∼ 103 −105 AU from Sgr A∗ (Reid&Brunthaler 2004). Therefore, there is no conflict between our derived enclosed mass and the upper limit mass of the second black hole inferred from the VLBA observations."

So it seems to me that they want to solve the problem by blaming the VLBA calculated limits observation.

[In case a different approach will clarify what Shaula has said]

My understanding of this is that if you assume that all of the observed motion is caused by the IMBH then you can calculate a mass for the IMBH. However, as some (or all) of the motion may be due to other causes, this calculated mass must be an upper limit.


Why it is so important to them to keep the Sgr A∗ mass estimation as had been discovered by S2 orbit cycle?


Because that is the most accurate estimate we current have. And confirmed by observing other stars orbiting Sgr A*.

Reality Check
2017-Dec-04, 02:18 AM
Why do we only based our current estimation for Sgr A* mass on S2?
As mentioned by Shaula - we do not base our current estimation of the mass of Sgr A* (https://en.wikipedia.org/wiki/Sagittarius_A*#Central_black_hole) just on the orbit of S2. We used the orbits of many individual stars, S0-2 and the motions of thousands of stars.
MONITORING STELLAR ORBITS AROUND THE MASSIVE BLACK HOLE IN THE GALACTIC CENTER (http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075/meta) (28 stars)
Measuring Distance and Properties of the Milky Way's Central Supermassive Black Hole with Stellar Orbits (http://iopscience.iop.org/article/10.1086/592738/meta) (looks like just SO-2 but other stars are discussed)
The nuclear star cluster of the Milky Way: proper motions and mass (https://www.aanda.org/articles/aa/abs/2009/28/aa10922-08/aa10922-08.html)(more than 6000 stars)

"If there is an IMBH orbiting around Sgr A∗," then we have the same situation as a planet orbiting a star - the limit to the mass we can detect is the detection limit of our instruments. If we cannot detect a planet of mass M tugging on its star then M is that limit. Likewise if we cannot detect an IMBH of mass M tugging on Sgr A∗ then M is that limit. The VLBA observations give a mass to the suggested IMBH below M. Thus not detecting perturbations of Sgr A∗ by an IMBH is not an argument against the IMBH.

Dave Lee
2017-Dec-04, 04:02 PM
I have a brilent Idea.

Why don't we base our verification for Srg A* mass on its accretion disc?

http://www.scholarpedia.org/article/Accretion_discs

"The accretion disc is surrounded by moving gas clouds and encircled by a large torus of gas and dust. Very fast (almost speed of light) jets emerge from many AGNs."

The Gas is perfectly OK, as it orbits at almost the speed of light.
Based on Newton low:

http://slideplayer.com/slide/5105616/

The orbital velocity for orbital motion is:
V ^ 2 = G * M / R
V (For accretion disc) = Speed of Light.
G = 6.67 * 10 ^ (-11)
M = the total mass of the Srg A*
R = the radius of Srg A* accretion disk.

As you can see, we do not need the mass of the gas in order to calculate Srg A* total mass.
Unfortunately, I don't know the radius of Srg A* accretion disk. Would you kindly help me with that.

Do you agree that based on a very simple data we can easly extract the real value of Srg A* mass?

Strange
2017-Dec-04, 05:11 PM
Why don't we base our verification for Srg A* mass on its accretion disc?

Does it have one?

Shaula
2017-Dec-04, 08:36 PM
I have a brilent Idea.

Why don't we base our verification for Srg A* mass on its accretion disc?

...

Do you agree that based on a very simple data we can easly extract the real value of Srg A* mass?
It is not a new idea. If the system was isolated and simple then I would say it was a technique that was worth a shot. However the whole core area is a complex system of supernovae remnants, inflows, outflows, stars and other stuff along our line of sight. There is a lot of scattering so the ability to focus the image for IR and radio wavelengths is limited by this fundamental constraint. It is also a long way away, and to get an accurate result you need a fine resolution over the disc.

Sgr A* is measured as about 40 microarcseconds in size. But of course it may be smaller than that in a given frequency band. VLBA resolutions range from 0.1mas to 21mas depending on frequency.

So ultimately, while this is something worth looking at (as are all dynamics of the SgrA* system) we have other, more accurate and simpler, ways to do this. I'm still not sure why you think the other methods are not convincing.

Edit to add: https://arxiv.org/abs/1204.4454 has some modelling of the accretion disk in. You can see some of the complexity there - and that is leaving out the effects of magnetic fields.

Dave Lee
2017-Dec-05, 04:31 AM
It is not a new idea.
Thanks
At least I'm not the first one which consider this idea.


If the system was isolated and simple then I would say it was a technique that was worth a shot. However the whole core area is a complex system of supernovae remnants, inflows, outflows, stars and other stuff along our line of sight. There is a lot of scattering so the ability to focus the image for IR and radio wavelengths is limited by this fundamental constraint. It is also a long way away, and to get an accurate result you need a fine resolution over the disc.

Sgr A* is measured as about 40 microarcseconds in size. But of course it may be smaller than that in a given frequency band. VLBA resolutions range from 0.1mas to 21mas depending on frequency.

So ultimately, while this is something worth looking at (as are all dynamics of the SgrA* system) we have other, more accurate and simpler, ways to do this. I'm still not sure why you think the other methods are not convincing.

Edit to add: https://arxiv.org/abs/1204.4454 has some modelling of the accretion disk in. You can see some of the complexity there - and that is leaving out the effects of magnetic fields.

That is clear.
However, which system is more isolated and simple?
Is it the accretion disk or S2 orbit cycle?
The S2 has to cross a maximal distance of almost 17 LY from Srg A* (While its minimal distance is almost Zero).
What kind of a complex aria it has to cross?
Could it be that it is a very complex system of supernovae remnants, inflows, outflows, stars, clouds and other stuff?
It is quite clear to me that the aria of the accretion disc is much more stable.
Even if its difficult for us to monitor it, we can get much accurate results.
If we have thought about it, why we didn't even try to set a brief calculation.
It is so simple.

We only need to find the inwards radius of the accretion disk.
If we don't know that value (As our view isn't face on), lets look at nearby galaxies and try to estimate this value.
I do recall that for one galaxy it was stated that the inwards radius of the accretion was about 3 AU.

For this value, the SMBH mass should be as follow:

M = V ^ 2 * R / G

V (speed of light) = 299 792 458 meter / second = (about) 300,000,000 meter / second = 3 * 10 ^ 8 m/sec

G = 6.67 * 10 ^ (-11)

AU = 1 Astronomical Unit = 149 597 871 kilometers = (about) 150,000,000,000 m = 15 * 10 ^ 10 m.

R = 3 AU = 3 * 15 * 10 ^ 10 m.

M= (3 * 10 ^ 8) ^ 2 * 3 * 15 * 10 ^ 10 / 6.67 * 10 ^ (-11)

M= 9 * 10 ^16 * 45 * 10 ^10 * 1.5 * 10 ^ 10

M = 607.5 * 10 ^ 36 (Is it in kg?)

If So,

Sun Mass = 1.99 * 10 ^ 30 kilograms. = (about) 2 *10 ^ 30 kilograms

SMBH Mass = (about) 300 * 10 ^ 6 Sun Mass = 3 * 10 ^ 8

Do you see any error in the calculation?

If so, it is about 100 bigger than our expectation based on S2 calculation (assuming that R = 3 AU and the outcome is in Kg)

Shaula
2017-Dec-05, 06:34 AM
However, which system is more isolated and simple?
Is it the accretion disk or S2 orbit cycle?
S2. You don't have to consider the magnetohydrodynamics of an accretion disk. Your Newton only calculation is a huge simplification of a complex problem with a lot more approximations than the S2 method (and, as I linked to the 28 and 6000 star methods which all give the same result).


The S2 has to cross a maximal distance of almost 17 LY from Srg A* (While its minimal distance is almost Zero).
What kind of a complex aria it has to cross?
Could it be that it is a very complex system of supernovae remnants, inflows, outflows, stars, clouds and other stuff?
Yup. A complex area of that will have a significant effect on observations of extended sources and on gas dynamics. Much less effect on a star barrelling through it. S2 and the other nearby stars are in a sweet spot - they are heavy enough to simplify to a two body problem and far enough out that GR can mostly be ignored.


It is quite clear to me that the aria of the accretion disc is much more stable.
Even if its difficult for us to monitor it, we can get much accurate results.
If we have thought about it, why we didn't even try to set a brief calculation.
It is so simple.
Well, sorry, but you are wrong. The accretion disk is actually very complex, as the paper I linked shows. First off it is tilted. There are GR effects. There are effects due to the magnetic field. Your model is simple, but that is because you are ignoring the complexity deliberately.



Do you see any error in the calculation?

If so, it is about 100 bigger than our expectation based on S2 calculation (assuming that R = 3 AU and the outcome is in Kg)
At least one error and several omissions. Apart from all the things I mentioned you ignoring your 3ly radius isn't even for our galaxy. Seems pretty clear you are determined to 'prove' that the SMBH is much heavier than we think. Take a moment and look at what you are saying. You are saying that a back of envelope calculation that ignores almost all of the relevant physics and is based on a number you vaguely remember might apply to some galaxy somewhere is better than the results from 28 analyses of Keplerian orbits (including at least one fully characterised one) and a study of the motions of 6000 stars around the core.

Edit: Also - the speed of rotation of the accretion disk is NOT c, and certainly not c at its outer edge. I just reread your original post. Your whole method is wrong from the first step.

Dave Lee
2017-Dec-05, 04:42 PM
S2. You don't have to consider the magnetohydrodynamics of an accretion disk. Your Newton only calculation is a huge simplification of a complex problem with a lot more approximations than the S2 method (and, as I linked to the 28 and 6000 star methods which all give the same result).
Yup. A complex area of that will have a significant effect on observations of extended sources and on gas dynamics. Much less effect on a star barrelling through it. S2 and the other nearby stars are in a sweet spot - they are heavy enough to simplify to a two body problem and far enough out that GR can mostly be ignored.

Well, sorry, but you are wrong. The accretion disk is actually very complex, as the paper I linked shows. First off it is tilted. There are GR effects. There are effects due to the magnetic field. Your model is simple, but that is because you are ignoring the complexity deliberately.

Also - the speed of rotation of the accretion disk is NOT c, and certainly not c at its outer edge.



O.K

We have the results from 28 analyses of Keplerian orbits (including at least one fully characterised one) and a study of the motions of 6000 stars around the core.
So let's agree that we have full knowledge about the estimated SMBH mass in the Milky way.


However, The Milky way is just one galaxy in the whole Universe.
What about the other galaxies?
How can we estimate the SMBH mass of Andromeda, Triangulum, Virgo and 400 Billion other galaxies?

Can we detect their S stars?
Don't you think that at least for some of them we can get quite good information about their accretion disk?
If so, why can't we use this information to get even basic estimation on their SMBH mass?
Why can't we use Newton gravity formula to estimate their SMBH mass?

Let me ask the following:

1. Newton Formula -
What is wrong with the following formula
M = V ^ 2 * R / G
Do you think that it is invalid?

2. magnetohydrodynamics in Accretion disk

How could it be that any kind of magnetohydrodynamics will have any sort of affect on gravity?
Why Newton didn't take it into consideration?
Why do we have to consider the magnetohydrodynamics when we try to calculate the gravity?
Do we have a formula which represents the impact of magnetohydrodynamics on gravity and Vice versa?

3. Radius of accretion disk.

Why can't we use the estimated radius in the formula?

4. Speed of plasma in the Accretion disk
In most of the article it is stated clearly that the speed of the plasma/gas in the accretion disk is almost the seed of light.
So why can't we just assume that it is 95% or even 90% of the speed of light?
even if we will assume that it is just 50%, than we just need to divide the outcome by 2 ^ 2 = 4.
Therefore, instead of the following outcome based on speed of light:
M (at speed of light) = 607.5 * 10 ^ 36 Kg
It should be the following for 50% the speed of light:
M (at 50% the speed of light) = (607.5 * 10 ^ 36)/4 = 151 * 10 ^ 36.
No big difference.
At least we have some estimation.

5. GR effects.
How GR effects the formula?

6. complexity deliberately
What do you mean by that?

I still don't understand why are you so opposed to this Idea.

Strange
2017-Dec-05, 05:26 PM
How could it be that any kind of magnetohydrodynamics will have any sort of affect on gravity?
Why Newton didn't take it into consideration?
Why do we have to consider the magnetohydrodynamics when we try to calculate the gravity?

Because there will be other forces acting on the accreting material because of electric charge, magnetic fields, fluid dynamics, and so on. If you ignore these and pretend that the only force is gravity, you will get the wrong results.


Do we have a formula which represents the impact of magnetohydrodynamics on gravity and Vice versa?

There is no single, simple formula. Plasma physics is complex and there are many variables to take into account. (Magnetohydronamics doesn't affect gravity, it affects the material in the accretion disk.)

"The set of equations that describe MHD are a combination of the Navier–Stokes equations (https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations) of fluid dynamics (https://en.wikipedia.org/wiki/Fluid_dynamics) and Maxwell’s equations (https://en.wikipedia.org/wiki/Maxwell%E2%80%99s_equations) of electro*magnetism (https://en.wikipedia.org/wiki/Electromagnetism). These differential equations (https://en.wikipedia.org/wiki/Differential_equation) must be solved simultaneously (https://en.wikipedia.org/wiki/Simultaneous_equation), either analytically or numerically (https://en.wikipedia.org/wiki/Numerical_analysis)."
https://en.wikipedia.org/wiki/Magnetohydrodynamics

As far as I know, the only way of modelling the behaviour of the accretion disk accurately is by massive simulations.


3. Radius of accretion disk.

Why can't we use the estimated radius in the formula?

Which estimated radius? Do you have a radius for the accretion disk around Sgr A*?

Shaula
2017-Dec-05, 07:13 PM
We have the results from 28 analyses of Keplerian orbits (including at least one fully characterised one) and a study of the motions of 6000 stars around the core.
So let's agree that we have full knowledge about the estimated SMBH mass in the Milky way.
I did not say that, I'd be a fool to. I said we have better ways than you have suggested to estimate the mass of SgrA* and they seem to be in agreement. But happy to leave SgrA* there.


How can we estimate the SMBH mass of Andromeda, Triangulum, Virgo and 400 Billion other galaxies?
It is harder - generally people rely on scaling estimates for more distant galaxies or orbital characteristics of stars that are further out. Both methods are less certain.


Can we detect their S stars?
Generally no. I think some objects have been resolved in Andromeda but it is not often we can do this.


Don't you think that at least for some of them we can get quite good information about their accretion disk?
No. It is generally smaller in angular size than the orbit of the innermost stars. So if you can't see them, you can't see it.


If so, why can't we use this information to get even basic estimation on their SMBH mass?
Why can't we use Newton gravity formula to estimate their SMBH mass?
See above. And we do use Newton. On stars where possible. Or on velocity dispersions of core stars if we cannot get good measurements of orbits.


1. Newton Formula -
What is wrong with the following formula
M = V ^ 2 * R / G
Do you think that it is invalid?
Why would I think Newtonian mechanics were globally invalid? I've never said anything close to that. I have said you are applying them in situations considerably more complex that they are good approximations for though. I also said that your estimates were wildly off because you claimed that the accretion disk rotated at c, which is wrong.


2. magnetohydrodynamics in Accretion disk

How could it be that any kind of magnetohydrodynamics will have any sort of affect on gravity?
Why Newton didn't take it into consideration?
Why do we have to consider the magnetohydrodynamics when we try to calculate the gravity?
Do we have a formula which represents the impact of magnetohydrodynamics on gravity and Vice versa?
As Strange says, it doesn't affect gravity. It affects the motion of matter affected by gravity. An obvious example - how do you think jets are formed from SMBH if gravity is all you have to consider?
Newton didn't take it into account because he was looking at simple two rigid body problems, in general. Not complex scenarios involving plasma. We have to take MHD into account because it exerts a strong effect when you are dealing with gases, magnetic fields and high charged particle fluxes. And as Strange also says ... it is not trivial to do in the slightest.


3. Radius of accretion disk.

Why can't we use the estimated radius in the formula?
Because you can't even say where you got the number you did from. You said "I do recall that for one galaxy it was stated that the inwards radius of the accretion was about 3 AU." - no source, no context. You may as well say it was 10 AU, 5 AU, 1 AU. Claiming that a number you think you remember that applied to some galaxy somewhere, maybe, is a valid input that allows you to refute other studies is just bizarre.


4. Speed of plasma in the Accretion disk
In most of the article it is stated clearly that the speed of the plasma/gas in the accretion disk is almost the seed of light.
So why can't we just assume that it is 95% or even 90% of the speed of light?
even if we will assume that it is just 50%, than we just need to divide the outcome by 2 ^ 2 = 4.
Therefore, instead of the following outcome based on speed of light:
M (at speed of light) = 607.5 * 10 ^ 36 Kg
It should be the following for 50% the speed of light:
M (at 50% the speed of light) = (607.5 * 10 ^ 36)/4 = 151 * 10 ^ 36.
No big difference.
At least we have some estimation.
In most articles? Can you provide some references to this apparently voluminous literature? Call me crazy but I thought it varied with the mass of the black hole, the radiative efficient of energy loss and so on. I also thought that it varied with radius. So - evidence please. In this thread you have repeatedly not given sources and already been very wrong about what a paper is saying. So, please, provide evidence that there is gas moving near c around SgrA*.

And why do you think we need estimates based almost entirely on guesses, approximations and assumptions when we have better methods available?


5. GR effects.
How GR effects the formula?
Complexly as you get very close to the SMBH. For a full answer please complete the equivalent of about 4 years of undergraduate studies. Seriously, it is not a trivial topic. There are some reviews of accretion disk physics out there on arxiv - https://arxiv.org/pdf/1104.5499.pdf looks OK at a quick glance (I'd have to look more and revisit some old notes to be sure)


6. complexity deliberately
What do you mean by that?
What I mean by that is that you have been pointed to papers, you have had the issues explained, you have had errors and other stuff pointed out to you over and over by at least a couple of people. And yet you have ignored all of that an insisted that a superficial model based on guesswork and over simplification is as good as a peer reviewed study using actual observations. You are deliberately ignoring the complexity of the topic. Either that or you are not actually reading what people are saying or the links they have given.


I still don't understand why are you so opposed to this Idea.
I am not opposed to accretion disk measurements being used. I am opposed to your analysis, which I think is a simplistic model loaded with guesswork, and I am very opposed to your assertions that it deserves equal weight to carefully performed studies based on observation.

Dave Lee
2017-Dec-05, 08:21 PM
Thanks

Dave Lee
2017-Dec-06, 03:09 PM
Newton Formula:




Why would I think Newtonian mechanics were globally invalid? I've never said anything close to that. I have said you are applying them in situations considerably more complex that they are good approximations for though. I also said that your estimates were wildly off because you claimed that the accretion disk rotated at c, which is wrong.

Please advice if I understand you correctly:

1. The Newton formula is perfectly OK to extract Central mass from orbital cycle based on real value verification for R (radius) and V (Velocity).
2. However, as the accretion disK is very complex, we can't verify the correct values for R and V.

If that is correct, Let's try to understand how the complexity effects the formula.

So, lets start by

magnetohydrodynamics in Accretion disk


As Strange says, it doesn't affect gravity. It affects the motion of matter affected by gravity. An obvious example - how do you think jets are formed from SMBH if gravity is all you have to consider?

This is quite complicate. I need further explanation.
Lets assume that the accertion disc is not complex. There is no magnetohydrodynamics or any other effect.
In this case it is quite clear that the Velocity of the matter is a direct outcome of the total mass of the SMBH and R.
Now, by adding the magnetohydrodynamics in Accretion disk, what might be the effect on the Velocity?
Is it going to change the speed? If so, increase or decrease?

However, if it change the speed of matter, than by definition it effects the gravity force.

Therefore, we need to change the gravity formula so it represents the change of the velocity due to the magnetohydrodynamics in Accretion disk.

Could it be that this is a real violation of the gravity formula?
So please, can we assume that the magnetohydrodynamics can't have any effect on the speed of matter in the Accretion disk?

With regards to the velocity of matter in the accretion disc:
Please see tread 19:

http://www.scholarpedia.org/article/Accretion_discs

"The accretion disc is surrounded by moving gas clouds and encircled by a large torus of gas and dust. Very fast (almost speed of light) jets emerge from many AGNs."

So, can we assume that this almost speed of light ONLY depends on gravity force?

If no, what might be the impact of magnetohydrodynamics in Accretion disk on this velocity?
Is it going to increase or decrease?

Swift
2017-Dec-06, 04:00 PM
Dave Lee

There is a fuzzy grey line between asking and arguing against mainstream science. You have been flirting with that line for most of this thread. It is the opinion of the moderators that you are are now getting into advocating non-mainstream ideas (ATM) outside of the ATM forum.

There is a difference between starting with the available data and asking where it leads, or starting with the conclusion you wish to reach, and trying to bend the data to fit that conclusion. You seem to be doing the later.

Normally I would just close this thread before you get yourself infracted, but you are not the OP, and there are others interested in this topic, so we'll leave the thread open for now. But I suggest you be a little more careful in your arguing, or you will be infracted. Or go start an ATM thread and present your arguments there.

Dave Lee
2017-Dec-06, 04:08 PM
Thanks

Your message is clear.
In any case, I do not try to argue, I just try to get better understanding.
Sorry if I ask too many or too difficult questions.
I do appreciate the great support.

Strange
2017-Dec-06, 04:56 PM
Now, by adding the magnetohydrodynamics in Accretion disk, what might be the effect on the Velocity?
Is it going to change the speed? If so, increase or decrease?

Both. And reverse it. And turn it round. It is a bit like asking if a turbulent wind will increase or decrease the rate at which a leaf falls to the ground.


However, if it change the speed of matter, than by definition it effects the gravity force.

It doesn't. It is just another complex set of forces that affect the motion of the material in the accretion disk. Gravity still works the same way. You wouldn't say that your chair affects the force of gravity, would you.


So please, can we assume that the magnetohydrodynamics can't have any effect on the speed of matter in the Accretion disk?

That is like asking can ignore the winds in a hurricane, and just assume the rain will fall straight down.


"The accretion disc is surrounded by moving gas clouds and encircled by a large torus of gas and dust. Very fast (almost speed of light) jets emerge from many AGNs."

So, can we assume that this almost speed of light ONLY depends on gravity force?

If no, what might be the impact of magnetohydrodynamics in Accretion disk on this velocity?

That is describing jets moving away from the black hole at nearly the speed of light. So:

1. It doesn't say the accretion disk is moving at the speed of light. (Although it might be, in places. I don't know.)

2. As the jets are going outward, that cannot be due to gravity so we can't ignore MHD - it is what is driving the jets. (In complex ways that are not fully understood.)

slang
2017-Dec-06, 05:43 PM
http://www.scholarpedia.org/article/Accretion_discs

"The accretion disc is surrounded by moving gas clouds and encircled by a large torus of gas and dust. Very fast (almost speed of light) jets emerge from many AGNs."

This says nothing about the accretion disc. It says something about jets, not about discs. They are two different things.

(ETA: as Strange also said)

Swift
2017-Dec-06, 06:46 PM
Thanks

Your message is clear.
In any case, I do not try to argue, I just try to get better understanding.
Sorry if I ask too many or too difficult questions.
I do appreciate the great support.
Just to clarify further.... it is not that you ask too many or too difficult questions. It is that you are framing your questions with assumptions that are clearly against the mainstream, or that you are "asking"
for people to prove or disprove things you are proposing that are clearly not mainstream. For example:

<snip>
However, if it change the speed of matter, than by definition it effects the gravity force.

Therefore, we need to change the gravity formula so it represents the change of the velocity due to the magnetohydrodynamics in Accretion disk.

Could it be that this is a real violation of the gravity formula?
So please, can we assume that the magnetohydrodynamics can't have any effect on the speed of matter in the Accretion disk?

Yes, two of those sentences end in question marks, but you are not really asking about mainstream science. You are proposing non-mainstream ideas (like MHD does not effect the speed of matter in the disk)
and asking for consent. That is essentially advocating an ATM idea.

Shaula
2017-Dec-06, 09:20 PM
Newton Formula:

Please advice if I understand you correctly:

1. The Newton formula is perfectly OK to extract Central mass from orbital cycle based on real value verification for R (radius) and V (Velocity).
2. However, as the accretion disK is very complex, we can't verify the correct values for R and V.
That is correct. Just like you cannot use Newton's laws of motion alone if there are strong magnetic or electric effects.


If that is correct, Let's try to understand how the complexity effects the formula.
So, lets start by
magnetohydrodynamics in Accretion disk
This is quite complicate. I need further explanation.
Magnetohydrodynamics is a post graduate study subject, if you are going to do it in detail. It is sufficiently complex that you need good grasp of fluid mechanics, plasma physics and electromagnetic theory to get started. Even without reading ahead I know what is coming so I will say this now: You are not going to get a simple, one line answer to this question. To get to the right answer you need computer modelling, large amounts of it. That is why I linked to a paper about the modelling of accretion disks in post 21.


Lets assume that the accertion disc is not complex. There is no magnetohydrodynamics or any other effect.
In this case it is quite clear that the Velocity of the matter is a direct outcome of the total mass of the SMBH and R.
Now, by adding the magnetohydrodynamics in Accretion disk, what might be the effect on the Velocity?
Is it going to change the speed? If so, increase or decrease?
Would you also like to assume the Earth is flat or that the sky is a painted dome? Your assumption is unrealistic and serves no purpose in illuminating this topic. As I said the effects of MHD are complex. Some of the gas will speed up, some will slow down. Some will get caught up in flares.


However, if it change the speed of matter, than by definition it effects the gravity force.
Therefore, we need to change the gravity formula so it represents the change of the velocity due to the magnetohydrodynamics in Accretion disk.
No, this is ... this isn't even logic! The effects due to gravity remain the effects due to gravity. Adding MHD modifies the result but has no effect at all on the theoretical framework underlying gravity. You may as well say that because I can throw a ball in the air we need to change the laws of gravity.


Could it be that this is a real violation of the gravity formula?
No, it is taking into account more than one effect. I honestly don't see how you can conclude that because there are things other than gravity involved gravity has to change.


So please, can we assume that the magnetohydrodynamics can't have any effect on the speed of matter in the Accretion disk?
This is also broken logic. And trivially proven wrong. If gravity was all there was we wouldn't see flares, jets or emission due to inspiral.


With regards to the velocity of matter in the accretion disc:
Please see tread 19:

http://www.scholarpedia.org/article/Accretion_discs

"The accretion disc is surrounded by moving gas clouds and encircled by a large torus of gas and dust. Very fast (almost speed of light) jets emerge from many AGNs."
As others have pointed out - this is about jet. It is completely irrelevant to the speed of the infalling matter. As the jets are accelerated by a mix of MHD and GR effects.


So, can we assume that this almost speed of light ONLY depends on gravity force?

If no, what might be the impact of magnetohydrodynamics in Accretion disk on this velocity?
Is it going to increase or decrease?
I've pointed out the issues with your arguments above. These questions are either of the "Summarise all of the effects of quantum physics in one sentence" type or based on chains of reasoning that don't work.

Dave Lee
2017-Dec-07, 04:33 PM
Thanks for the productive answers.

With regards to the velocity of matter in the accretion disc:
http://www.scholarpedia.org/article/Accretion_discs
"The accretion disc is surrounded by moving gas clouds and encircled by a large torus of gas and dust. Very fast (almost speed of light) jets emerge from many AGNs."



As others have pointed out - this is about jet. It is completely irrelevant to the speed of the infalling matter. As the jets are accelerated by a mix of MHD and GR effects.


O.K.
If I understand it correctly:
Jets emerge from many AGNs at almost speed of light, and it is accelerated by a mix of MHD and GR effects.
In other words, you don't see any connection between the orbital velocity of the palsma in the accretion disc to this jet.

So, how can we verify the orbital velocity of the plasma?

Let me offer an example:
Lets assume that there is a spinning wheel.
We have no technology to measure its spinning velocity.
But we have the technology to monitor an outflow jet from that wheel.
So, how can we find the spinning velocity of that wheel?

I have an Idea.
If we drop water on the spinning wheel, we can get an outflow jet of water from the wheel.
however, it is expected that this water jet must be fully tangent with the direction of the spinning wheel.
Now, with that knowledge, let's look again on the jet which emerge from many AGNs.

We can't drop water on the spinning accretion disc, but we can monitor a jet of dust which outflow from the accertion disc.
So now we need to verify if that outflow jet is it fully tangent with the direction of the spinning accretion disc, or is it verical to the disc?
If it is vertical, than it is clear that it can't give any real information about the spinning accretion velocity.
However, if it is tangent with the spinning accretion than we use it.

I couldn't find an answer for the direction of that jet.
However, I have found the following article:

the Accretion disk
https://arxiv.org/pdf/1701.04627.pdf

On the other hand, our simulations cannot explain some observables. For example, (Calvet et al. 1993) have estimated an outflow rate of 10−5M yr−1 for FU Ori which has an accretion rate of 2 × 10−4M yr−1 . In our simulations, the outflow rate from R=0.5 to 5 is only 0.4% of the accretion rate, while observations suggest that the outflow rate is 5% of the accretion rate. One solution is that the outflow originates from a wide disk range (from the stellar surface to 1 AU). Another solution is that the disk is threaded by a stronger net vertical magnetic fields. Strong magnetic fields have been observed in FU Orionis system (Donati et al. 2005).

It is stated clearly that our scientists expected to see a very low outflow jet of about 0.4% from the accretion disc, but they have discovered that it is as high as 5%.
It is also stated that the disk is threaded by a stronger net vertical magnetic fields.
So, does it mean that the outfow Jet is also moving vertically to the disc or does it mean that a vertical magnetic fields force the jet to move tangent with the spinning accretion disc?

Shaula
2017-Dec-07, 05:39 PM
O.K.
If I understand it correctly:
Jets emerge from many AGNs at almost speed of light, and it is accelerated by a mix of MHD and GR effects.
In other words, you don't see any connection between the orbital velocity of the palsma in the accretion disc to this jet.

So, how can we verify the orbital velocity of the plasma?
If it is resolved, spectroscopically. Modelling also helps. Tracking the motion of flaring parts of the disk. There are methods, but they are complicated by the highly dynamic and complex nature of the area around the core. Lots of inflowing gas, twisted magnetic field lines, dust etc etc.


If it is vertical, than it is clear that it can't give any real information about the spinning accretion velocity.
It is perpendicular to the plane of the accretion disk, or nearly so. So it doesn't act like a wheel throwing out water at all.
See: https://en.wikipedia.org/wiki/Astrophysical_jet#Relativistic_jets
https://en.wikipedia.org/wiki/Accretion_disk#Accretion_disk_physics

Dave Lee
2017-Dec-07, 07:20 PM
If it is resolved, spectroscopically. Modelling also helps. Tracking the motion of flaring parts of the disk. There are methods, but they are complicated by the highly dynamic and complex nature of the area around the core. Lots of inflowing gas, twisted magnetic field lines, dust etc etc.

Thanks
It is perpendicular to the plane of the accretion disk, or nearly so. So it doesn't act like a wheel throwing out water at all.
See: https://en.wikipedia.org/wiki/Astrophysical_jet#Relativistic_jets
https://en.wikipedia.org/wiki/Accretion_disk#Accretion_disk_physics

I have just noticed that the article is about - active galaxies, radio galaxies or quasars.

However, we discuss about accretion disk in spiral galaxies (Milky Way, andromeda and so on).

Hence, do we see any kind of outflow jet from the accretion disk in spiral galaxies?
If so, is it also prependicular to the plane of the accretion disk?

Dave Lee
2017-Dec-07, 07:55 PM
In the following article about SMBH in spiral galaxy it is stated:
https://en.wikipedia.org/wiki/Supermassive_black_hole

Some of the best evidence for the presence of black holes is provided by the Doppler effect whereby light from nearby orbiting matter is red-shifted when receding and blue-shifted when advancing. For matter very close to a black hole the orbital speed must be comparable with the speed of light, so receding matter will appear very faint compared with advancing matter, which means that systems with intrinsically symmetric discs and rings will acquire a highly asymmetric visual appearance.

So can we understand that the orbital speed of matter close to the black hole is almost (or comparable) to the speed of light?
If so, what do the mean by "very close".
Please be aware that they do not speak about the accretion disc (no accretion disc - no MHD).
Therefore, if we can find the radius of that matter orbital cycle and we know that it is already comparable to speed of light (and hopefully, there is no MHD) than why can't we use Newton formula to extract SMBH mass?

Reality Check
2017-Dec-07, 08:54 PM
If so, what do the mean by "very close".
The Wikipedia authors mean really, really close to the speed of light :D. If you want to find the actual speeds than that would be in the scientific literature.
Even in a "spherical cow" approximation that ignores the existence of the rest of the accretion disc, using any Newton formula is wrong. We have a relativistic mass orbiting a supermassive black hole. Relativistic velocity + possible strong gravitational field = Newton is invalid.

Shaula
2017-Dec-07, 09:21 PM
Hence, do we see any kind of outflow jet from the accretion disk in spiral galaxies?
If so, is it also prependicular to the plane of the accretion disk?
Jets can occur with just about any black hole. They are always close to perpendicular. Possible SgrA* jet: http://chandra.harvard.edu/photo/2013/sgra/

The article I linked to covered this question (my bold):

Relativistic jets are beams of ionised matter accelerated close to the speed of light. Most have been observationally associated with central black holes of some active galaxies, radio galaxies or quasars, and also by galactic stellar black holes, neutron stars or pulsars.

Shaula
2017-Dec-07, 09:37 PM
Therefore, if we can find the radius of that matter orbital cycle and we know that it is already comparable to speed of light (and hopefully, there is no MHD) than why can't we use Newton formula to extract SMBH mass?
If there is no accretion disk what matter are you talking about observing?

Plus - the matter in an accretion disk is falling in. It is not in a stable orbit. Depending on the density of the matter and other effects the speed of the gas as it crosses the event horizon can vary. It is not c, as you claimed most of the available material said. It is high.

As I have said a fair few times now. Newton's laws are too simple for this situation. You need GR and MHD. I'm not going to engage any more with you on this topic. Simply because it is clear that you are not listening, you are not reading most of the more advanced material being posted and you are determinedly ignoring the majority of what is said in order to insist on your highly simplistic models. And, frankly, I am bored of rehashing the same argument with you over and over.

Summary: We have a good handle on the mass of SgrA* from a couple of different methods. Direct observations of the central object are not yet accurate enough to be a better method of doing it. And simplified models filled with guesses are not better than detailed observational evidence. They are certainly not going to convince anyone that we have the mass wrong by orders of magnitude.