David Levy Posted January 24, 2015 Posted January 24, 2015 (edited) http://en.wikipedia....sive_black_hole "From the motion of star S2, the object's mass can be estimated as 4.1 million M☉,[15][16] " However, the mass of the whole milky way galaxy is : 0.8–1.5×1012 M☉. This is astonishing ratio of four/one million... (On 4 Kg of mass in the black hole, there are 1,000,000 Kg (1,000 Tons) of mass in the spiral arms. Actually, the black hole is like an engine which rotates the whole spiral galaxy while spiral arms are similar to the biggest propeller in the universe. It is similar to an engine of 4 Kg which rotatse a propeller of 1,000,000 kg (or 1,000 tons)? In this kind of ratio, the spiral arms should be disconnected from the galaxy! Is it real? How could it be? Edited January 24, 2015 by David Levy
imatfaal Posted January 24, 2015 Posted January 24, 2015 Remember that for any star you observe orbiting the galactic centre that (to an approximation) all the mass of all the stars inside the observed star's orbit acts as if it were at the centre - and you can ignore the mass outside the observed star's orbit. So your observed star is not just gravitationally attracted to the central SMBH - it is attracted to all the stars inside its orbit as well. In fact if we look at the simple stellar masses and how quickly stars orbit the centre we find that it is not as you have suggested that there is too little mass - we find that there is more mass than we can see. The outside stars orbit quicker than we would expect if we count the stars inside and work out their mass; this is, very simplistically, where the idea of dark matter comes from 2
David Levy Posted January 24, 2015 Author Posted January 24, 2015 (edited) Thanks Remember that for any star you observe orbiting the galactic centre that (to an approximation) all the mass of all the stars inside the observed star's orbit acts as if it were at the centreSo, do you mean that S2 see (feel) more gravity than the black hole by itself (assuming that there are stars inside the rotation cycle of S2)? ..and you can ignore the mass outside the observed star's orbit.Why the mass outside can be ignored? If the inside mass effects the gravity on S2, why the mass outside has no effect? What about the mass ratio between the spiral arms to the Back hole? How can we justify a ratio of 4/one million? Edited January 24, 2015 by David Levy
imatfaal Posted January 24, 2015 Posted January 24, 2015 It is an approximation - the exact form is that in symmetric spherical cluster of objects that are attracted with an inverse square law a test object will feel a force to the centre equal to that which would be felt if all the mass closer to the centre than the test object was actually located at the centre and that the net effect of all the objects further from the centre than the test object is zero. This is Gauss' Law of Gravity or Newtons Shell Theorem. Now a spiral galaxy is neither spherical nor homogeneous and only partially symmetric - but I expect a rough approximation will apply
David Levy Posted January 24, 2015 Author Posted January 24, 2015 (edited) This is Gauss' Law of Gravity or Newtons Shell Theorem.Based on Newton's law of universal gravitation http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation "Every point mass attracts every single other point mass by a force pointing along the line intersecting both points." In other words, every point mass attracts every other point mass by a gravity force. Therefore, all the mass in the spiral arm should also have an effect on S2. Do you agree? It is an approximation - the exact form is that in symmetric spherical cluster of objects that are attracted with an inverse square law a test object will feel a force to the centre equal to that which would be felt if all the mass closer to the centre than the test object was actually located at the centre and that the net effect of all the objects further from the centre than the test object is zero.O.K. If spiral galaxy was a perfect symmetrical system, and the Black hole will be located just in the center, than we could claim that the net power on the center is zero. However, a normal spiral galaxy is not a perfect system. Even the black hole is not located just at the center. Therefore, the net force isn't zero. Please also be aware that S2 rotates between the center and the inwards side of the spiral arms. Therefore, as it moves close to the arms, the net force should be much more than zero. Edited January 24, 2015 by David Levy
Strange Posted January 24, 2015 Posted January 24, 2015 Actually, the black hole is like an engine which rotates the whole spiral galaxy while spiral arms are similar to the biggest propeller in the universe. Actually, it isn't. As you say, the mass of the black hole, while being large, is insignificant compared to the rest of the stars in the galaxy. So, do you mean that S2 see (feel) more gravity than the black hole by itself (assuming that there are stars inside the rotation cycle of S2)? It would. But the mass of the black hole is so large (at this distance) that any other stars that are there will not have a significant effect. Why the mass outside can be ignored? If the inside mass effects the gravity on S2, why the mass outside has no effect? Because it (roughly) balances out: all the mass outside the orbit ends up pulling equally in all directions so has no net effect. http://en.wikipedia.org/wiki/Shell_theorem In this case, as imatfaal says, it is only an approximation. You would need to do a very detailed calculation and/or simulation to work out the overall effect. But it is almost certainly very small - especially for the stars very close to the black hole (because of the inverse square law). What about the mass ratio between the spiral arms to the Back hole? How can we justify a ratio of 4/one million? Why does it need justifying?
David Levy Posted January 25, 2015 Author Posted January 25, 2015 (edited) Regards to the elliptical center orbits http://en.wikipedia.org/wiki/File:Galactic_centre_orbits.svg How come that all of those stars have an elliptical orbits? S14 for example, is located very close to the center and its elliptical shape is the sharpest one. How could it be? This is the opposite scenario from any solar system. In the solar system we can find elliptical orbits only with stars that are far away from the sun. Those stars have low gravity force due to their distance. Therefore, it is quite logical that Pluto (for example) has an elliptical orbit. However, in the spiral galaxy, those stars which are located quite close to the center have elliptical orbits, while a star which is located 26,000 lY (sun – for example) has a symmetrical cycle orbit. It shows that the gravity force in the spiral arm is much greater than the gravity force in the center. This is amazing phenomena. Edited January 25, 2015 by David Levy
Strange Posted January 25, 2015 Posted January 25, 2015 How come that all of those stars have an elliptical orbits? All orbits are elliptical. S14 for example, is located very close to the center and its elliptical shape is the sharpest one. How could it be? There is no connection between distance and eccentricity. For example: http://en.wikipedia.org/wiki/Elliptic_orbit#Solar_System It shows that the gravity force in the spiral arm is much greater than the gravity force in the center. This is amazing phenomena. It shows nothing of the sort.
David Levy Posted January 25, 2015 Author Posted January 25, 2015 (edited) All orbits are elliptical.Please see the following orbit diagram: http://en.wikipedia.org/wiki/File:Oort_cloud_Sedna_orbit.svg It shows that most of the planets in the solar system have quite symmetrical cycle. (O.K. – it is not fully symmetrical – but it looks symmetrical.) However, Sedna has clearly an elliptical orbit which looks similar to S14. So the one of the further planets in the solar system, orbits in a similar way as the closest one in the spiral galaxy. We have to understand the source for that. Edited January 25, 2015 by David Levy
MigL Posted January 25, 2015 Posted January 25, 2015 Hey David Levy, welcome back. Haven't heard from you for over a year. I would have thought you'd take that time to learn Newton's law of gravity and Kepler's orbital laws.
David Levy Posted January 25, 2015 Author Posted January 25, 2015 Hey David Levy, welcome back. Haven't heard from you for over a year. I would have thought you'd take that time to learn Newton's law of gravity and Kepler's orbital laws. Thanks Yes, sure. Newton might have an answer to this issue.
Strange Posted January 25, 2015 Posted January 25, 2015 However, Sedna has clearly an elliptical orbit which looks similar to S14. So the one of the further planets in the solar system, orbits in a similar way as the closest one in the spiral galaxy. We have to understand the source for that. But there is clearly no connection between distance and eccentricity.
David Levy Posted January 26, 2015 Author Posted January 26, 2015 (edited) Let's look again on the following diagram: http://en.wikipedia.org/wiki/File:Galactic_centre_orbits.svg Based on S2 it is estimated that the total mass of the black hole is as follow: "From the motion of star S2, the object's mass can be estimated as 4.1 million M☉,[15][16] or about 8.2×1036 kg" In order for the science to prove this mass, they need to verify the black hole mass also based on the other stars. So, if based on S1 (for example) the mass of the black hole isn't exactly the same, than it proves that this isn't the real total mass of the black hole. Please be aware that as those stars are quite close together, it is expected that their equivalent host mass might be quite similar, but not the same. So we have to set an accurate calculation! Edited January 26, 2015 by David Levy
Mordred Posted January 26, 2015 Posted January 26, 2015 Galaxy rotation curves is a complex subject. First you have to examine the distribution of visible matter as well as hydrogen and intergalactic plasma. This using keplers laws still won't get the right answer. This is when you need to look at the distribution of dark matter. This article is very basic. However gives the general idea. http://en.m.wikipedia.org/wiki/Galaxy_rotation_curve more detail is here http://cds.cern.ch/record/1023900/files/0703430.pdf
David Levy Posted January 26, 2015 Author Posted January 26, 2015 Galaxy rotation curves is a complex subject. First you have to examine the distribution of visible matter as well as hydrogen and intergalactic plasma. This using keplers laws still won't get the right answer. This is when you need to look at the distribution of dark matter. This article is very basic. However gives the general idea. http://en.m.wikipedia.org/wiki/Galaxy_rotation_curve more detail is here http://cds.cern.ch/record/1023900/files/0703430.pdf Based on Newton, the Galaxy rotation curves is quite simple! I will explain it later on.
imatfaal Posted January 26, 2015 Posted January 26, 2015 Let's look again on the following diagram: http://en.wikipedia.org/wiki/File:Galactic_centre_orbits.svg Based on S2 it is estimated that the total mass of the black hole is as follow: "From the motion of star S2, the object's mass can be estimated as 4.1 million M☉,[15][16] or about 8.2×1036 kg" In order for the science to prove this mass, they need to verify the black hole mass also based on the other stars. So, if based on S1 (for example) the mass of the black hole isn't exactly the same, than it proves that this isn't the real total mass of the black hole. Please be aware that as those stars are quite close together, it is expected that their equivalent host mass might be quite similar, but not the same. So we have to set an accurate calculation! "In order for the science to prove this mass, they need to verify the black hole mass also based on the other stars." Did you bother to look up the papers? The German group analysed the motion of around 6000 stars We measured the proper motions of more than 6000 stars within ~1.0 pc of the supermassive black hole Sgr A*. http://arxiv.org/abs/0902.3892
Airbrush Posted January 26, 2015 Posted January 26, 2015 (edited) Supermassive black holes have a mass of about 1% of the mass of the galaxy, or less, I'm not sure. Does anyone remember the ratio? The mass of the galaxy depends upon the mass of the SBH. During its' quasar phase it pushes gas and dust away which terminates the growth of the galaxy. You would think that SBH must form very early in the universe because you need the critical amount of matter close enough together. So how long after the big bang did the first SBH form? How long after the big bang until the first matter congealed from energy? Edited January 26, 2015 by Airbrush
David Levy Posted January 26, 2015 Author Posted January 26, 2015 (edited) Did you bother to look up the papers? The German group analysed the motion of around 6000 stars http://arxiv.org/abs/0902.3892 Thanks It is stated: "We measured the proper motions of more than 6000 stars within ~1.0 pc of the supermassive black hole Sgr A*". "They imply a best-fit black hole mass of 3.6 (+0.2/-0.4) x 10^6 solar masses." However, based on S2 it is: "From the motion of star S2, the object's mass can be estimated as 4.1 million M☉,[15][16]" So, there is a difference in the B.H. Mass (as I have expected). Can you please explain the following: "The point mass of Sagittarius A* is not sufficient to explain the velocity data. In addition to the black hole, the models require the presence of an extended mass of 0.5-1.5x10^6 solar masses in the central parsec." Why the science need extended mass in the center? Please be aware that this extended mass is 41% of the B.H. mass (up to 1.5 out of 3.6). Why this extended mass isn't part of the black hole itself? Where exactly this extended mass is located? Is it possible that the science get different black hole mass for different stars? Therefore, this extended mass is needed to balance between the different results If so, this is great news! Edited January 26, 2015 by David Levy
Strange Posted January 26, 2015 Posted January 26, 2015 It is stated: "We measured the proper motions of more than 6000 stars within ~1.0 pc of the supermassive black hole Sgr A*". "They imply a best-fit black hole mass of 3.6 (+0.2/-0.4) x 10^6 solar masses." However, based on S2 it is: "From the motion of star S2, the object's mass can be estimated as 4.1 million M☉,[15][16]" If you analyse more stars, you will get a more accurate estimate. Although, when you look at the error bounds, there isn't a very significant difference between these values. Can you please explain the following:"The point mass of Sagittarius A* is not sufficient to explain the velocity data. In addition to the black hole, the models require the presence of an extended mass of 0.5-1.5x10^6 solar masses in the central parsec." Why the science need extended mass in the center? Please be aware that this extended mass is 41% of the B.H. mass (up to 1.5 out of 3.6). Why this extended mass isn't part of the black hole itself? Where exactly this extended mass is located? It is a very complex and detailed paper so I have only skimmed it. However, the velocities of stars some distance from the black hole are determined by the mass of the black hole plus the surrounding stars. It is the mass of the surrounding stars that is referred to as the "extended mass". Is it possible that the science get different black hole mass for different stars?Therefore, this extended mass is needed to balance between the different results No, that is not what it means.
David Levy Posted January 27, 2015 Author Posted January 27, 2015 (edited) No, that is not what it means.What does it mean? "The point mass of Sagittarius A* is not sufficient to explain the velocity data." Please see the following: Orbital speed http://en.wikipedia.org/wiki/Orbital_speed "The orbital speed of a body, generally a planet, a natural satellite, an artificial satellite, or a multiple star, is the speed at which it orbits around the barycenter of a system, usually around a more massive body. Mean orbital speed." It should be quite easy to extract the mass of the black hole. In this formula there is no room for: "extended mass". What is the source for this requirement? Unless, there were different B.H. mass results for different stars. Is it possibale to get the real velocity data and further explanation about thier calculations? If you analyse more stars, you will get a more accurate estimate..Why? They have analyzed 6000 stars. Is it not enough? Don't you think that it should be accurate enough? Although, when you look at the error bounds, there isn't a very significant difference between these values.Really? Let's look again: "They imply a best-fit black hole mass of 3.6 (+0.2/-0.4) x 10^6 solar masses." So, the B.H. should be 3.2 min to 3.8 max. It isn't in the range of 4.1.(based on S2). However, (+0.2/-0.4, which is 0.6 out of 3.6 - 16.6%) by itself indicates that they have got different results from different stars. Please remember that those stars are located " within ~1.0 pc of the supermassive black hole Sgr A" Edited January 27, 2015 by David Levy
Strange Posted January 27, 2015 Posted January 27, 2015 (edited) What does it mean? "The point mass of Sagittarius A* is not sufficient to explain the velocity data." Correct. The other stars in the cluster need to be taken into account. (Note that this work was about the star cluster, not just the black hole.) Orbital speedhttp://en.wikipedia.org/wiki/Orbital_speed "The orbital speed of a body, generally a planet, a natural satellite, an artificial satellite, or a multiple star, is the speed at which it orbits around the barycenter of a system, usually around a more massive body. Mean orbital speed." It should be quite easy to extract the mass of the black hole. In this formula there is no room for: "extended mass". What is the source for this requirement? It says usually around a more massive body. In this case, it is a cluster os stars orbiting their centre of mass. Near the black hole, the mass of the black hole is the only thing that matters. Further away the mass of other stars becomes significant. Is it possibale to get the real velocity data and further explanation about thier calculations? Have you not read the paper? There is an eye-watering amount of detail. Why? They have analyzed 6000 stars. Is it not enough?Don't you think that it should be accurate enough? They analysed 6,000 stars and got a better result than just looking at S2. Really? Let's look again:"They imply a best-fit black hole mass of 3.6 (+0.2/-0.4) x 10^6 solar masses." So, the B.H. should be 3.2 min to 3.8 max. It isn't in the range of 4.1.(based on S2). If you look at the source for the 4.1 figure, it is +/- 0.6, therefore it overlaps with the smaller figure. This is as expected. However, (+0.2/-0.4, which is 0.6 out of 3.6 - 16.6%) by itself indicates that they have got different results from different stars. No, it just indicates the accuracy of the calculation. Obviously, if you look at individual stars, you will get different results because of the errors in estimating the speeds, distances and masses of each star. That is why looking at a large number gives you a better result: the errors tend to average out. Edited January 27, 2015 by Strange 2
David Levy Posted January 28, 2015 Author Posted January 28, 2015 (edited) Thanks. Well, based on Newton: " "Every point mass attracts every single other point mass by a force pointing along the line intersecting both points" Therefore, every point mass (star for example) attracts every other point mass (another star) by a force which is called gravity force. So, if we try to calculate the gravity force of S2 for example, we need to take in account all the gravity force vectors which attract this star. (Inwards to the center plus outwards from the center). Hence, it should be the sum of the black hole gravity force on S2, plus The gravity force vectors of all the stars (or clusters) between S2 and the Black hole, minus The gravity force vectors of all the stars (or clusters) from S2 and outwards The outcome of all those gravity force vectors will be the equivalent gravity force which attracts that star. This equivalent gravity force should be in full balance with the proper motion of that star. I have explained this issue in the following thread: http://www.scienceforums.net/topic/87496-newton-gravity-for-spiral-galaxy/ The equivalent gravity force can be represented by a virtual hoster mass. So, the scientists found the proper motion of S2 (and also for 6,000 stars). However, as stated, this proper motion is balanced with the equivalent gravity force. Therefore, the scientists actually found the virtual hoster mass for S2 (or the virtual hoster of the 6000 stars). Never the less, this is not the real Black hole mass. So, the 4.1 Million Solar mass isn't the real mass of the Sagittarius A*, it is only the virtual hoster mass for S2 which the scientists found in their modeling. If we move from the core of the Sagittarius A* outwards we will find different gravity forces. Therefore, it is expected that for any radius vector there will be a specific virtual hoster mass. We could actually detect it if our calculations were accurate. Unfortunately, the error factor of the calculation is quite high. In the 6000 stars it is 16.5% (+0.2/- 0.4 out of 3.6), while for S2 it is 30% (+/- 0.6 out of 4.1). Therefore, it is overlap due to wide error factor. Edited January 28, 2015 by David Levy -1
Strange Posted January 28, 2015 Posted January 28, 2015 Thanks. Well, based on Newton: " "Every point mass attracts every single other point mass by a force pointing along the line intersecting both points" Therefore, every point mass (star for example) attracts every other point mass (another star) by a force which is called gravity force. So, if we try to calculate the gravity force of S2 for example, we need to take in account all the gravity force vectors which attract this star. (Inwards to the center plus outwards from the center). All the "outward" forces cancel (especially this close to the center, where the surrounding mass can more accurately be modelled as speherical ditribution). So the stars outside the orbit can be ignored (do the clauclations to check this if you want). Therefore, the scientists actually found the virtual hoster mass for S2 (or the virtual hoster of the 6000 stars). Never the less, this is not the real Black hole mass. So, the 4.1 Million Solar mass isn't the real mass of the Sagittarius A*, it is only the virtual hoster mass for S2 which the scientists found in their modeling. As the stars outside the orbit can be ignored, the stars closest to the black hole (e.g. 2S) can be used to more accurately determine the black hole mass. As the paper points out, within a certain distance, the only signifcant contribution to the orbital speed is the black hole mass. Once you get further out the masses of the other stars becomes significant and so it is possible to estimate the mass of the stars in the cluster (which seems to be what they were more interested in). I don't know what a "virtual hoster mass" is. Unfortunately, the error factor of the calculation is quite high. Given the extreme difficulty of these observations, I think those error bounds are pretty extraordinary.
Mordred Posted January 28, 2015 Posted January 28, 2015 There is one other consideration you also need to consider the mass distribution of the intergalactic plasma. Which is significant.
David Levy Posted January 28, 2015 Author Posted January 28, 2015 (edited) All the "outward" forces cancel (especially this close to the center, where the surrounding mass can more accurately be modelled as speherical ditribution). So the stars outside the orbit can be ignored (do the clauclations to check this if you want). Sorry, this is a severe mistake! Let me use your following answer: However, the velocities of stars some distance from the black hole are determined by the mass of the black hole plus the surrounding stars. It is the mass of the surrounding stars that is referred to as the "extended mass".So, the extended mass is surrounding stars. Those stars do not just hang there in one spot. They must be in full rotate motion around the Black hole. (Please remember that the black hole attracts any mass, even if it is called - "extended mass". Let's assume that there are no spiral arms and that it is a perfect system: So those stars set a perfect ring around the black hole. Actually, let assume that there are two perfect rings. Let's position S2 (for example) in a rotate motion between those two rings. If I understand you correctly, you believe that the inwards ring will attracts S2 while the outwards ring will not. Please explain why. Edited January 28, 2015 by David Levy
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