Supermassive black hole

[David Levy]

Please look at the following orbits of 6 stars around suppermassive black hole candidate Sagittarius A* at the Milky Way's centre:

https://en.wikipedia.org/wiki/Sagittarius_A*#/media/File:Galactic_centre_orbits.svg

Please focus on S13.

Its path is very round and circular.

Therefore, based on Newton and Kepler, I would expect to see the B.H just at the center of that circular path.

 

So how could it be that the Supper massive B.H. isn't located at the center?

Edited May 1, 2016 by David Levy

[Strange]

I think the biggest factor is that you are judging it to be a circle "by eye". You need to use some actual data. (Better than nothing, print out the picture and measure the ellipse and calculate the expected position of the focus. See how close it is to the black hole on the diagram.)

 

The eccentricity is about 0.395 which is about the same as S1. On that basis, the position of the black hole as one of the foci seems quite plausible.

[swansont]

!

Moderator Note

If this is asking a question it shouldn't be in speculations. If you have a speculation to present, then present it. These threads where we're waiting for the other shoe to drop is getting annoying.

 

Also, super ≠ supper

[Janus]

Please look at the following orbits of 6 stars around suppermassive black hole candidate Sagittarius A* at the Milky Way's centre:[/size]

[/size]

https://en.wikipedia.org/wiki/Sagittarius_A*#/media/File:Galactic_centre_orbits.svg[/size]

[/size]

Please focus on S13.[/size]

Its path is very round and circular.[/size]

Therefore, based on Newton and Kepler, I would expect to see the B.H just at the center of that circular path.[/size]

 

So how could it be that the Supper massive B.H. isn't located at the center?[/size]

The listed ecentricity is ~0.395 This works out to minor to major axis ratio of ~0.919, which fits the figure shown for S13. Calculations also gives a periapis distance to major axis distance ratio of ~0.3 for an ellipse of that eccentricity. which again matches the figure. Maybe you would have been better off looking up how to determine what the math says the orbit would look like rather than going by what you "felt" the orbit should look like before posting. Especially since, that up to now, your intuition on these types of subjects has had a pretty dismal track record.

[David Levy]

Also, super ≠ supper[/modtip]

Would you kindly fix it?

I have looked again on that diagram.

The size of S13 Circle in that diagram is as follow:

Horizontally -21 Unites. Vertically - 20 Units.

That is almost full round Circle.

Therefore, it is expected to see the B.H. almost at the center.

However, the B.H is located as follow:

Vertically - 6 unites from the top and 15 unites from the bottom. - That represents a severe shift from the center

Horizontally - 10 Unites from the left and 8 Unites from the right. - That represents a minor shift from the center.

I think that all of us should agree that especially the vertical shift is not excepted (based on Newton and kepler).

So, assuming that S13 diagram is fully correct, let me ask the following questions:

How do we know the exact location of the B.H? Why did we place it at its current location in the diagram? Based on what evidences?

Edited May 1, 2016 by David Levy

[Strange]

How do we know the exact location of the B.H? Why did we place it at its current location in the diagram? Based on what evidences?

 

From the data.

Note the source of the image you are using: https://commons.wikimedia.org/wiki/User:Cmglee

 

So it was created by an amateur illustrator with an unknown degree of accuracy from an unknown source of data. You have downloaded the file and (I assume) printed it on a printer of unknown calibration and accuracy and then measured it using an unknown method with an unknown ruler of unknown accuracy.

 

Not surprisingly, your conclusions are meaningless.

 

The data you need is readily available. For example: http://www.astrophysicsspectator.com/tables/MilkyWayCentralStars.html

[David Levy]

 

From the data.

 

Well, let me try to guess.

The B.H had been placed at its current location due to evidences from S2 star path.

As it is stated:

https://en.wikipedia.org/wiki/Sagittarius_A*

"Observations of the star S2 in orbit around Sagittarius A* have been used to show the presence of, and produce data about, the Milky Way's central supermassive black hole, and have led to the conclusion that Sagittarius A* is the site of that black hole.^[8]"

So, if I understand it correctly, based on S2 path, it had been decided to place the B.H. at its current location (and off course to evaluate its total mass).

That is a severe mistake. The science didn't find the location of the real B.H.

They have actually found the location/mass of the equivalent hoster for S2.

So, for each star, based on its location, there could be a different equivalent hoster.

Therefore, it was stated: " Observations of the star S2…".

Why not S13 or S1… or S12?

 

I would expect that for each star path, the B.H. mass might be different.

Is it correct?

Edited May 1, 2016 by David Levy

[Strange]

No. If you read the referenced papers, instead of a summary on Wikipedia, you would see that all of stars were used to estimate the position of the black hole.

 

For example: "Our main results are: all stellar orbits are fit extremely well by a single-point-mass potential to within the astrometric uncertainties"

http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075

 

The significance of S2 is that it was the first star around the black hole to have its complete orbit observed.

 

Please try and do a little bit of research instead making things up.

[David Levy]

Well, I'm quite sure that there is an error:

Actually, based on the path shape we can estimate the gravity force.

If the path is circular - than the gravity force is high. (That is similar path for all the planets in the solar system except of Pluto and Sedna)

If the Path is ellipse - than the gravity force is weak. (That is similar path for Pluto and especially Sedna which is located at the farthest distance from the Sun).

 

So, I'm quite sure that based on the S13 Path (as it is circular), the equivalent mass hoster of that star should be significantly higher than the one for S2 (as it is ellipse).

No. If you read the referenced papers, instead of a summary on Wikipedia, you would see that all of stars were used to estimate the position of the black hole.

For example: "Our main results are: all stellar orbits are fit extremely well by a single-point-mass potential to within the astrometric uncertainties"

http://iopscience.iop.org/article/10.1088/0004-637X/692/2/1075

The significance of S2 is that it was the first star around the black hole to have its complete orbit observed.

Please try and do a little bit of research instead making things up.

 

 

That article is quite old - Published 2009 February 23

It is stated clearly that S2 is the only star which has completed full cycle (revolution?):

"The combination of a long-time baseline and the excellent astrometric accuracy of adaptive optics data allows us to determine orbits of 28 stars, including the star S2, which has completed a full revolution since our monitoring began."

 

Therefore, it's quite clear that this star was the main source for this estimation.

However, now after more than 7 years we might have more information about the other stars.

Some of them might also complete the cycle.

So, I would like to know if we get different results based on any star which has completed its cycle by now (especially - S13).

 

Actually, it is quite clear for me that S14 has the weakest gravity force, while S13 has the highest gravity force.

I would mostly appreciate to get the calculated equivalent hoster mass for those two stars.

Edited May 1, 2016 by David Levy

[pzkpfw]

What do you mean by "hoster"?

 

It seems like you think that different things in a system can be gravitationally bound to different things.

[Strange]

Well, I'm quite sure that there is an error:

 

Please show the details of your calculations that lead to this conclusion. Please provide references to all data sources used and the error bounds on your results.

 

If you fail to do that, I will suggest the moderators close this thread and ban you as a time-wasting troll.

 

 

Actually, based on the path shape we can estimate the gravity force.

If the path is circular - than the gravity force is high. (That is simmilar path for all the planets in the solar system except of Pluto and Sedna)

If the Path is ellipse - than the gravity force is weak

 

Utter nonsense.

 

That article is quite old - Published 2009 February 23

It is stated clearly that S2 is the only star which has completed full cycle (revolution?)

 

Yes, it is quite old. Since then at least one other star has been observed to do a complete orbit. And more detailed estimates of the orbits have been made.

 

Therefore, it's quite clear that this star was the main source for this estimation.

 

No it isn't. Please stop making up nonsense.

 

However, now after more than 7 years we might have more information about the other stars.

 

We do, indeed.

 

So, I would like to know if we get different results based on any star which has completed its cycle by now (especially - S13).

 

So go and find out. Stop wasting our time with your ludicrous and ignorant guesswork.

Edited May 1, 2016 by Strange

[Janus]

Would you kindly fix it?

 

I have looked again on that diagram.[/size]

The size of S13 Circle in that diagram is as follow:[/size]

Horizontally -21 Unites. Vertically - 20 Units.[/size]

That is almost full round Circle.[/size]

Therefore, it is expected to see the B.H. almost at the center.[/size]

[/size]

However, the B.H is located as follow:[/size]

Vertically - 6 unites from the top and 15 unites from the bottom. - That represents a severe shift from the center[/size]

Horizontally - 10 Unites from the left and 8 Unites from the right. - That represents a minor shift from the center.[/size]

I think that all of us should agree that especially the vertical shift is not excepted (based on Newton and kepler).[/size]

[/size]

So, assuming that S13 diagram is fully correct, [/size]let me ask the following questions:[/size]

How do we know the exact location of the B.H? [/size]Why did we place it at its current location in the diagram? [/size]Based on what evidences?[/size]

Even if we were to take your measurements as accurate , this still gives a eccentricity of 0.305( for an eccentricty of 0.395, the listed eccentricity the minor radius would be 19.3 units which is well within the likely margin of error of your measurements), which would put the focus at a point some 7.3 units from one end of the ellipse and 13.7 units from the other end. ( a bit less than twice as far from one end than from the other and not nearly at the center, as you propose)

 

The placement of the foci of ellipses just don't behave like your intuition seems to be telling you that they do. Consider the orbit of Mars with an eccentricity of just 0.09. if we scale it down to have a major axis of 21 units, then its minor to major axis ratio will be a mere 0.996 and its minor axis would be 20.9 units. Even harder to visually distinguish from a circle. The focus of the orbit would be 9.55 units from one end and 11.5 from the other, a small, but still measurable offset from the center.

If you look at this image of Mars orbit compared to the Earth's, you will see how even such a small eccentricity can produce a visibly noticeable offset of the orbit

So it is of no surprise that an eccentricity of 0.395 or even 0.305 would produce even a more noticeable offset. So, no matter what you may think, the image of the orbit of S13 does have the position of the focus properly located.

[swansont]

 

Well, I'm quite sure that there is an error:

Actually, based on the path shape we can estimate the gravity force.

If the path is circular - than the gravity force is high. (That is similar path for all the planets in the solar system except of Pluto and Sedna)

If the Path is ellipse - than the gravity force is weak. (That is similar path for Pluto and especially Sedna which is located at the farthest distance from the Sun).

 

 

 

!

Moderator Note

You will need to present your alternative model of gravity and evidence that supports it, in keeping with the rules of speculations.

[Janus]

Well, I'm quite sure that there is an error:[/size] [/size]

Actually, based on the path shape we can estimate the gravity force.[/size]

If the path is circular - than the gravity force is high. (That is [/size]similar path for all the planets in the solar system except of Pluto and Sedna)[/size]

If the Path is ellipse - than the gravity force is weak. (That is [/size]similar path for Pluto and especially Sedna which is located at the farthest distance from the Sun).[/size]

How convenient for your argument that you left Mercury off of your list, the planet closest to the Sun and in the strongest gravity but with an eccentricity of 0.21, which is over twice that of Mars which, as I have already pointed out, has a noticeably offset orbit. Edited May 2, 2016 by Janus

[David Levy]

Even if we were to take your measurements as accurate , this still gives a eccentricity of 0.305( for an eccentricty of 0.395, the listed eccentricity the minor radius would be 19.3 units which is well within the likely margin of error of your measurements), which would put the focus at a point some 7.3 units from one end of the ellipse and 13.7 units from the other end. ( a bit less than twice as far from one end than from the other and not nearly at the center, as you propose)

 

The placement of the foci of ellipses just don't behave like your intuition seems to be telling you that they do. Consider the orbit of Mars with an eccentricity of just 0.09. if we scale it down to have a major axis of 21 units, then its minor to major axis ratio will be a mere 0.996 and its minor axis would be 20.9 units. Even harder to visually distinguish from a circle. The focus of the orbit would be 9.55 units from one end and 11.5 from the other, a small, but still measurable offset from the center.

If you look at this image of Mars orbit compared to the Earth's, you will see how even such a small eccentricity can produce a visibly noticeable offset of the orbit

 

So it is of no surprise that an eccentricity of 0.395 or even 0.305 would produce even a more noticeable offset. So, no matter what you may think, the image of the orbit of S13 does have the position of the focus properly located.

 

 

Thanks for the Answer.

Sorry, but Mars can't be used as an example for S13 path due to the following:

Let's try to analyze the S13 Path (Based on Units from the diagram):

 

Circular Path drift -

The ratio between the Vertical and the horizontal path is:

21 / 20 = 1.05 = 105% (a drift of 5%)

So, we can claim the path is a pure circular by a drift/error of only 5%.

 

However, with regards to the location of the B.H.:

 

Radius drift -

The longest distance (Radius) from S13 to the B.H.is 15 Units.

The shortest radius from S13 to the B.H. is 6 Units.

Therefore, the ratio between the Longest to the shortest radius is:

15 / 6 = 2.5 = 250%. (a drift of 150%)

 

So, how a 5% drift in the circular path could be supported by a drift of 150% in the radius?

 

Can we prove this phenomenon by Newton or Kepler?

Can we introduce even one planet path (as an example) which has similar drifts?

 

Actually, S2 path has high similarity to Sedna path. Therefore, we can easily claim that the location of the B.H with reference to S2 path is FULLY correct.

 

However, that isn't the case with S13.

 

Is it possible to get the full data about the path of S13? I would like to analyze it by myself.

Edited May 2, 2016 by David Levy

[Strange]

 

Is it possible to get the full data about the path of S13? I would like to analyze it by myself.

 

See post 6:

 

The data you need is readily available. For example: http://www.astrophysicsspectator.com/tables/MilkyWayCentralStars.html

 

And, of course, it is on the Wikipedia page you initially referenced. Please do not post anything else until you have calculated the position of the black hole for each star.

Edited May 2, 2016 by Strange

[swansont]

 

Is it possible to get the full data about the path of S13? I would like to analyze it by myself.

 

 

!

Moderator Note

You've been provided a link to its orbital characteristics, which is more than enough to decide if the illustrator made a mistake in rendering the orbits. The rest of this thread is unsupported nonsense.

 

Posting nonsense isn't against the rules, but the unsupported part is. You were asked for evidence for your claims, and you didn't provide any.