Milky way galaxy mass vs. black hole mass

[David Levy]

 

You need to sum all the relevant forces to show that.

That is quite easy;

Let's make a brief calculation for a star which is located in the same arm as the sun, but further away from the center. Let's call is S100. It will have same mass as the Sun.

In this case, the total equivalent gravity force vectors from the center and from (m) are quite similar.

Therefore, the real difference will be based on (n)

[math]\vec{f_n} =\vec{f_x}+\vec{f_y} [/math].

While –

(x) - Representing all the stars in the inwards side of the arm (with relative to the star location in the arm).

(y) - Representing all the stars in the outwards side of the arm (with relative to the star location in the arm).

However, those vectors have opposite polarities.

Therefore, (y) is a negative force vector with regards to (x).

So, if S100 is located further away from the center;

There will be fewer stars which are located further away from its location to the end of the spiral arm (with relative to the sun).

Therefore, the contribution of the (y) vectors will be lower than the one for the Sun.

In the same token, the (x) vector will be stronger, as there will be more stars which are located in the inwards side from its location to the center of the spiral arm (with relative to the sun).

Hence, the equivalent gravity force vector on S100 will be stronger than the gravity force on the Sun.

Edited February 7, 2015 by David Levy

[Strange]

 

 

That is quite easy;

Let's make a brief calculation for a star which is located in the same arm as the sun, but further away from the center. Let's call is S100. It will have same mass as the Sun.

In this case, the total equivalent gravity force vectors from the center and from (m) are quite similar.

Therefore, the real difference will be based on (n)

[math]\vec{f_n} =\vec{f_x}+\vec{f_y} [/math].

While –

(x) - Representing all the stars in the inwards side of the arm (with relative to the star location in the arm).

(y) - Representing all the stars in the outwards side of the arm (with relative to the star location in the arm).

However, those vectors have opposite polarities.

Therefore, (y) is a negative force vector with regards to (x).

So, if S100 is located further away from the center;

There will be fewer stars which are located further away from its location to the end of the spiral arm (with relative to the sun).

Therefore, the contribution of the (y) vectors will be lower than the one for the Sun.

In the same token, the (x) vector will be stronger, as there will be more stars which are located in the inwards side from its location to the center of the spiral arm (with relative to the sun).

Hence, the equivalent gravity force vector on S100 will be stronger than the gravity force on the Sun.

 

You need to sum all the relevant forces to show that.

[David Levy]

 

You need to sum all the relevant forces to show that.

As I have stated, we just wasting our time.

Thanks.

[Strange]

 

As I have stated, we just wasting our time.

Thanks.

 

If you are not prepared to do the necessary work then yes, you are wasting our time.

[David Levy]

 

The calculation should be quite simple –

The equivalent gravity force vector on any star in spiral arm should be as follow:

F(equivalent) = F (center) + F(Arm)+ F(all other)

While -

F(center) is the equivalent gravity force of all the mass in the center including the black hole

F(arm) is the equivalent gravity force of all the stars in that arm.

Please be aware that if a star is located somewhere in the arm, than all stars which are located inwards from its location should contribute a positive gravity force, while the stars in the outwards side should contribute negative gravity force.

Therefore, the equivalent force of F(arm) is as follow:

F(arm) = F(Arm Inwards) - F(Arm outwards)

Please also be aware that due to a distance, only the nearby stars will have a significant effect on the gravity force.

F(all other) is the equivalent gravity force of all the stars in the other arms. This force is also quite neglected due to a distance.

Hence:

F (equivalent) = F(center) + F(Arm Inwards) - F(Arm outwards)+ F(all other)

So, let's set a brief calculation for the equivalent gravity force vector:

If star A is located at the most outwards side of the spiral arm, (let's assume that it is located at about 45,000 light years from the center) than its equivalent gravity force vector is as follow:

Fa (equivalent) = Fa(center) + Fa(Arm Inwards) – Fa(Arm outwards)+ Fa(all other)

However, as there are no stars outwards from its location, therefore:

Fa(Arm outwards) = 0

If star B is located 1,000 light years inwards from star A in the arm (44,000 light years from the center), than its equivalent gravity force vector is as follow:

Fb (equivalent) = Fb(black hole) + Fb(other mass in the center) + Fb(Arm Inwards) – Fb(Arm outwards)+ Fb(all other)

In this case Fb(Arm outwards) contributes a real negative value.

So let's calculate:

The contribution of gravity force by center on each star is almost the same due to this high distance.

Therefore we can assume that:

Fa(center) = Fb(Center)

In the same token, the contribution of F(all other) on each star should be similar, therefore, we can also assume that:

Fa(all other) = Fb(all other).

Hence,

Fa(equivalent) is greater than Fb(equivalent).

This brief calculation proves that the gravity force increases as the star is located further away from the center.

[Strange]

 

This brief calculation proves that the gravity force increases as the star is located further away from the center.

 

I don't think anyone will argue with this. (Despite your rather hand-waving derivation.)

 

However, because you haven't done any calculations, you can't say anything about how much the gravitational force increases with distance. So your conclusion is correct but not very useful.

[imatfaal]

David this is assertion upon baseless assertion. Unfortunately you need to either simulate with a very powerful computer or calculate with a vastly simplified scenario. You cannot just assume - from a quick reading you assume that Fb(Inward)-Fb(outward)<Fa(inward); as an example explain why this must (MUST) be the case and isn't just the result of one of a myriad of possible scenarios.

 

Centripetal force varies with the inverse of r - simple gravitational attraction varies with the inverse square of r

[Strange]

 

This brief calculation proves that the gravity force increases as the star is located further away from the center.

I don't think anyone will argue with this. (Despite your rather hand-waving derivation.)

 

Reading this again, it isn't clear that this is true. The total mass contributing to the inward gravitational force will increase as you go further out but this is offset by the increasing distance. So without some calculations from actual data, it isn't clear whether the force increases, decreases or stays the same.

Centripetal force varies with the inverse of r - simple gravitational attraction varies with the inverse square of r

 

And the mass of stars inside the orbit increases (roughly) as r cubed at the center of the galaxy and then moves closer to r squared as you move through the disk.

[David Levy]

 

David this is assertion upon baseless assertion. Unfortunately you need to either simulate with a very powerful computer or calculate with a vastly simplified scenario. You cannot just assume - from a quick reading you assume that Fb(Inward)-Fb(outward)<Fa(inward); as an example explain why this must (MUST) be the case and isn't just the result of one of a myriad of possible scenarios.

 

Centripetal force varies with the inverse of r - simple gravitational attraction varies with the inverse square of r

Thanks

Well, there are about 400 billion stars in a Milky Way galaxy. We need to map them all and verify the correct gravity force. This is quite difficult for me as I don't have the full map, the mass of each star, it's velocity, its direction and the distance to each other star.

I assume that NASA could do it better than me.

However, as usual, we can set a simple model which can give us an indication for the gravity force. I have already done it in page 27.

In this model I have proved that a star in the inwards side of the arm has less gravity force than the one which is located outwards from its position.

Theoretically, I can use more complex module with few hundreds of stars. But the results should be the same.

[David Levy]

As I have already stated, the ratio between the Black hole core mass and the whole mass in the Milky Way is 4 to one million. Therefore, it is a ratio of 1 to 250,000.

This is similar to the ratio between the Earth and Sun which is 1 to 330,000.

So, is it possible that the Sun will rotate around the Earth?

Even if we consider the bulge as symmetric spherical cluster of objects, it still has significantly less mass that the whole mass in the arms.

Please be aware that the dark matter was needed due to the missing mass at the center of the galaxy.

I just would like to highlight that there might be more mass in the B.H than our expectation.

Therefore, this could minimize our need for dark matter.

Edited March 13, 2015 by David Levy

[Strange]

Please be aware that the dark matter was needed due to the missing mass at the center of the galaxy.

 

Missing mass throughout the whole galaxy (and beyond the visible part).

 

 

I just would like to highlight that there might be more mass in the B.H than our expectation.

Therefore, this could minimize our need for dark matter.

 

For that to be the case, the black hole would need to have about the same mass as all the stars in the disk. In that case it would be rather obvious.

[imatfaal]

As I have already stated, the ratio between the Black hole core mass and the whole mass in the Milky Way is 4 to one million. Therefore, it is a ratio of 1 to 250,000.

This is similar to the ratio between the Earth and Sun which is 1 to 330,000.

So, is it possible that the Sun will rotate around the Earth?

Even if we consider the bulge as symmetric spherical cluster of objects, it still has significantly less mass that the whole mass in the arms.

Please be aware that the dark matter was needed due to the missing mass at the center of the galaxy.

I just would like to highlight that there might be more mass in the B.H than our expectation.

Therefore, this could minimize our need for dark matter.

 

In dealing with interactions you need to be cautious when you assume that you can treat an aggregation of individual masses as a single mass. From a decent distance you can treat the mass of the milky way as a point mass at the centre of mass. When you are dealing with an object that can receive different forces from different parts of the disc you must calculate individually. In this particular case you can say very little about the movement of the galactic centre when considering a only single point mass for the rest of the galaxy

[David Levy]

 

When you are dealing with an object that can receive different forces from different parts of the disc you must calculate individually.

 

 

Let's verify how different forces affect different parts of the spiral galaxy.

So in spiral galaxy there are three main parts (or sections): Bulge, spiral arms, Halo.

Rotation curve is an indication for the available force at each section.

In each section there is different rotation velocity curve as follow:

Bulge –The rotation curve meets the science expatiation – Therefore, no need for dark mass.

Spiral arms - The rotation curve is high above the science expatiation – Therefore, dark mass is needed.

Halo - The rotation curve meets the science expatiation – Therefore, no need for dark mass.

Let's verify the stellar orbits at each section. Please see pg 16:

http://www.ifa.hawai...10/MilkyWay.pdf

.

Disk stars (yellow) – all move in the same direction on roughly circular orbits.

Stars in the bulge (red) - and Halo (green) move in fairly random orbits.

Hence, in the Bulge and Halo, all stars move in fairly random orbits, and the rotation curve meets science expectation.

However, in spiral arms, all stars move in the same direction on roughly circular orbits, and the rotation curve is high above science expectation.

Therefore, we see 100% correlation between the rotation curve and the steller orbits at each section.

Don't you think that it could be an indication that the same force which is responsible for the high curve velocity at the spiral arm section has also a direct effect on the shape of the arm?

Now, let's look again on the dark mass.

It is only needed for the spiral arm section. However, if we add it, it might have negative effect on the Bulge and halo.

I don't think that there is a way to tell the dark matter to set an influence only in one section of the galaxy. So we need to figure out what kind of force could have significant effect on spiral arm section without causing any problems at bulge and halo.

Remember the doctor – If there is a correlation between those phenomenons, than we should consider one explanation for all we see.

Edited March 14, 2015 by David Levy

[Mordred]

I posted a detailed articles in your other thread

http://www.google.ca/url?sa=t&source=web&cd=5&ved=0CCwQFjAE&url=https%3A%2F%2Fwikis.uit.tufts.edu%2Fconfluence%2Fdownload%2Fattachments%2F9440479%2Fchemouni_bach_GE_dec07.pdf%3Fversion%3D1&rct=j&q=singular%20isothermal%20sphere%20profile%20of%20spiral%20arms&ei=xNQDVaDFLsfwoATsxoCACQ&usg=AFQjCNGm931PDgYo5WOdtIksZVMLKqwVSQ&sig2=fa628v9sgDG0sloyufaLVg&bvm=bv.88198703,d.eXY

 

"Constructing basic galactic models"

 

This should show you the complexity of the various factors. Keep in mind the title specified Basic

[pavelcherepan]

 

 

Halo - The rotation curve meets the science expatiation – Therefore, no need for dark mass.

 

I thought it was just the unusually high velocities of stars in globular clusters in the Halo that was one of the main evidences for Dark Matter.

[Strange]

 

I thought it was just the unusually high velocities of stars in globular clusters in the Halo that was one of the main evidences for Dark Matter.

 

It is the velocities of galaxies in clusters, the velocities of stars in the disk, the velocity of gas in the halo, microlensing, the velocities of stars in elliptical galaxies, the CMB, the large scale structure of the universe, the proportion of deuterium to hydrogen and probably other things I can't think of right now.

 

Not only do all these things require dark matter, but they all require the the same amount of dark matter with the same properties.

Edited March 14, 2015 by Strange