Newton Gravity for spiral galaxy

[David Levy]

Newton's law of universal gravitation http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

"In Spiral galaxies the orbiting of stars around their centers seems to strongly disobey to Newton's law of universal gravitation."

I assume that spiral galaxy fully obey to Newton low of universal gravitation.

The answer had already been given by Newton:

"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 (star for example) attracts every other point mass (another star) by a force which is called gravity force.

Therefore, the equivalent gravity force vector which attracts any star should be the sum of the total gravity force vectors from all the stars in the system.

Based on Newton, the gravity force is:

"The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them"

In the solar system for example, we can use Newton gravity formula to calculate the gravity force which any planet, moon or sun attract the Earth.

We should find that the total gravity force vectors of all the other planets are neglected with regards to the gravity force vector from the sun. (As the sun includes over than 99% of the total mass in the whole solar system)

Hence, the Gravity force on Earth is based on the gravity force with the sun.

However, Spiral galaxy isn't Solar system!

In Spiral galaxy the total mass of the whole stars in the galaxy is much higher than the mass of the supper massive black hole.

Based on Wiki, the milky way black hole (For example) is 4.1 million M☉, while the total mass of the Milky way galaxy is: 0.81.5×1012 M☉. This is a ration of about 4 to 1,000,000.

Therefore, the mass of the Milky way black hole is almost neglected with regards to the total mass in the galaxy (about 400 billion stars). This is the opposite scenario from a solar system. Therefore, we need to make the correct calculation in order to verify the real gravity results in spiral galaxy.

 

Based on Newton, the total gravity force on any star in spiral galaxy is the sum of all the gravity force vectors which attracts this star in the galaxy. That's mean that we need to verify the gravity contribution of the black hole with the gravity contribution of the stars in the spiral arm (and even consider the contribution of all other stars in the galaxy). The outcome of this calculation will be called the equivalent gravity force vector. For each star in the galaxy we need to calculate its own equivalent gravity force vector. This gravity force has a direct impact on the rotation energy of that star.

In order to understand how to calculate the equivalent gravity force, let's use a very simple system – Sun and Earth (without any other planets or moons)

We know very well how to calculate the gravity force which attracts the Earth to the Sun. Let's call it as follow:

Gravity force – F, Radius – R, Sun mass – M.

In the following examples we will set the following:

- Split the sun into two equally stars. The mass of each star will be 1/2 x M.

- One star will be set at the same place as the Sun. This star will be called Hoster.

- The Earth will revolve around the hoster. The radius is – R.

-The second star will be moved on the vector of the radius to different distances from the Earth.

- So, while the Earth revolve around the hoster, the second star will keep its position on the on the same vector line of the radius R (but at different distances), so those two stars should be in one line with the Earth.

- We will calculate the gravity force vector that each star attracts the Earth. The sum of those forces should is the equivalent gravity force on Earth.

- This equivalent gravity force can be represented by Virtual hoster. Based on the radius R, we will calculate the Equivalent virtual hoster mass. The Idea is that instead of using two stars system, we will find the Equivalent mass which can replace those two stars.

1. Let's keep the second star at the same place as the Hoster. In this case, each star attracts the Earth at a gravity force of 1/2 x F. The equivalent gravity force vector on Earth is as follow:

Equivalent gravity force vector = 1/2 x F (From the Hoster) + 1/2 x F (From the second star) = F.

As stated, this equivalent gravity force vector could be represented by a virtual hoster. It's equivalent mass will be M (as the sum). This is quite clear, as those two stars together, have the same mass as the Sun.

2 . Let's move the second star further away from the hoster, so its distance will be doubled than the current distance between the Earth to Hoster. In this case, the gravity contribution of second star will be decreased by four. (Please remember that - The gravity force is inversely proportional to the square of the distance between any star and Earth). Therefore, its contribution to the gravity on earth will be 1/8 x F.

Hence, the new equivalent gravity force vector on earth is:

Equivalent gravity force vector = 1/2 x F (From the first star) + 1/8 x F (From the second star) = 5/8 x F.

Therefore, the equivalent Virtual hoster mass is 5/8 x M.

Hence, by using a virtual hoster at a mass of 5/8 x M we can get the same gravity force on earth as the two stars system.

3. Let's position the second star at the opposite direction from the hoster with regards to Earth. Hence, the Earth will be placed in between the hoster and the second star. The distance to each direction will be the same. In this case, the gravity force from each side of the earth will be half of F, but at the opposite polarity. Therefore, the sum of those gravity force vectors will be zero. Hence, the equivalent virtual hoster mass is zero.

4. Let's position the second star at the opposite direction from the hoster with radius of 2 x R (doubled distance of R). The new equivalent gravity force on Earth is:

Equivalent gravity force = 1/2 x F (From the first star) - 1/8 x F (From the second star) = 3/8 x F.

Therefore, the Equivalent virtual hoster mass is 3/8 x M.

5. Let's position the second star at the opposite direction from the hoster with radius of 1/2 x R (Half distance of R). The new equivalent gravity force on Earth is:

Equivalent gravity force = 1/2 x F (From the first star) - 2 x F (From the second star) = -1.5 x F.

This time, we have got a negative gravity force. That's mean that the Earth should disconnect from the hoster.

 

6. Let's split the second star to several stars with different mass at each one. Let's position those divided stars in the same vector line as R but at different random locations of both sides of earth.

The Equivalent force vector should be the sum of all gravity force vectors that those stars and the Hoster attract the Earth.

However, the stars in one side should contribute gravity force in one direction while the stars in the side of the Earth should contribute a gravity force in the opposite direction. Therefore, in order to get the equivalent force vector, we need to sum all vectors in one direction and subtract it from the sum of all vectors in the other direction.

Based on the equivalent gravity force, we can calculate the equivalent virtual hoster mass.

This equivalent virtual hoster mass can replace all of this system.

Now, let's go back to the Spiral galaxy. In the Milky Way there are 400 Billion stars in about seven spiral arms. Therefore, in each arm there are about 70 Billion stars (assuming that all stars are located in the arms and the arms are equally). Our sun is located in one of those arms.

To make it simpler, let's assume that all the stars in that arm are located in a long straight line and ignore the gravity influence of the other arms.

In order to calculate the equivalent gravity force that attracts the sun, we need to verify the following:

-The gravity force vector that the black hole attracts the sun.

-The gravity force vectors of all the stars in the arm which are located between the sun and the black

-The gravity force vectors of all the stars in the arm which are located between the sun and the outwards side of the arm. Please be aware that all those vectors contribute a negative gravity force.

The equivalent gravity vector is the sum of all the gravity vectors. However, it's quite difficult to calculate 70 billion gravity vectors. There is a simple way.

The rotation energy force of the sun is excellent indication for the equivalent gravity force vector that attracts the sun in the galaxy. (There must be full balance between those two forces)

The rotation energy is a direct outcome of the Sum mass and its velocity.

The radius of the sun is about 26,000 light year, Therefore, it's quite easy to calculate its virtual hoster equivalent mass.

The Sun revolves around this virtual hoster and complete one cycle in 240 million years.

Actually, any star in the galaxy revolves around its own virtual hoster. Each virtual hoster has a unique Equivalent mass which fits to that specific star.

Now, let's move the sun inwards and outwards in the spiral arm inorder to verify how the equivalent gravity force (and as an outcome – the virtual hoster equivalent mass) should be.

So, if the sun is located at the most outwards side of the arm, all the gravity vectors forces will be in the same direction. Therefore, it should have the greatest gravity force. As we will move the sun inwards in the arm, the stars at the outwards side (with relevant to the sun location) will contribute a negative gravity force. Hence, the equivalent gravity force vector will be decreased as the sun moves inwards in the spiral arm.

Therefore, the maximal gravity force (Maximal virtual hoster' equivalent mass) will be achieved by placing the sun at the most outwards side of the arm. The minimal gravity force (Minimal virtual hoster' equivalent mass) will be achieved by placing it at the most inwards side of the arm.

That is the opposite scenario from a typical solar system.

This proves Newton gravity low for spiral galaxy. It is also explains why the star velocity increases as its location is further from the center of the spiral galaxy.

Edited January 25, 2015 by David Levy

[David Levy]

a

Edited February 14, 2015 by David Levy

[Strange]

It is also explains why the star velocity increases as its location is further from the center of the spiral galaxy.

True. But this doesn't tell us by how much the velocity should increase with distance.

Edited February 14, 2015 by Strange

[David Levy]

True. But this doesn't tell us by how much the velocity should increase with distance.

Thanks!

I will explain it.

Spiral Galaxy – Orbital speed of star (in spiral arm) vs. its distance from the center

By wiki:

http://en.wikipedia.org/wiki/Galaxy_rotation_curve

"If disc galaxies have mass distributions similar to the observed distributions of stars and gas then, the orbital speed would always decline at increasing distances in the same way as do other systems with most of their mass in the centre, such as the Solar System or the moons of Jupiter".

This is incorrect!

As I have already proved spiral galaxy isn't solar system. In a solar system, most of the mass is in the center, while in spiral galaxy there are significant portion of mass outside the center. In order to evaluate the orbital speed vs. distance, let's assume that all the stars in spiral arms keep their position and do not drift them inwards or outwards. Therefore, it is like a rigid spiral disc. In this case, it is expected that all points of mass should complete one cycle at the same time.

The circumference of a circle for is

 

Where: C is circumference, r is Radius and π is a dimensionless constant approximately equal to 3.14159.

Let's use the following example:

For first star - R1 = 10,000 ly

C = 2 x 3.14 x 10,000 = 62,800 ly

For second star – R2 = 20,000 ly

C = 2 x 3.14 x 20,000 = 125,600 ly

Both stars complete one cycle in the same time (rigid spiral disc). The circumference of second star is double than the first one, therefore, its orbital speed must be doubled with regards to the first one.

 

Hence, the orbital speed must increase at increasing distance, assuming that the stars keep their position in the spiral arms.

 

 

 

Rotation Curve problem in spiral galaxy

By wiki:

http://en.wikipedia.org/wiki/Galaxy_rotation_curve

"galaxy rotation problem is the discrepancy between observed galaxy rotation curves and the theoretical prediction, assuming a centrally dominated mass associated with the observed luminous material. When mass profiles of galaxies are calculated from the luminosity profiles and mass-to-light ratios in the stellar disks, then they do not match with the masses derived from the observed rotation curves and the law of gravity."

"The rotation curve of a disc galaxy (also called a velocity curve) is a plot of the magnitude of the orbital velocities (i.e., the speeds) of visible stars or gas in that galaxy versus their radial distance from that galaxy's centre, typically rendered graphically as a plot."

Let's see the following diagram:

http://en.wikipedia.org/wiki/File:M33_rotation_curve_HI.gif

R1 = 10,000 ly the observation velocity is 90 km/s.

R2 = 20,000 ly the observation velocity is 105 km/s.

Based on our calculations, in a rigid spiral disc the orbital velocity must increase at increasing distance. At a double distance, the orbital speed must be doubled.

So, if R1 = 10,000 Ly and observation orbital velocity is 90 km/s, than for R2 = 20,000 Ly the expected speed must be 90 x 2 = 180 km/s.

However, this isn't the case. We need to explain why the speed is only 105 km/s while based on our expectation from rigid spiral disc it should be 180 km/s.

The answer is quite simple - Spiral galaxy disc isn't rigid disc.

We must give some freedom to the stars to drift inwards or outwards in the arm.

If we move a star outwards in the arm it has two vectors.

One vector is vertically to the center – let's call it outwards vector

The other vector is horizontally to the center – let's call it backwards vector.

Let's make the mathematical calculation:

The backwards vector should decrease the speed from 180 km/s (expected speed) to 105 km/s (observed speed).

So we need to decrease the speed by 75 km/s.

This is achievable by decreasing the circumference of a circle which the star will have to go in that "same period of time –T.

The ratio between 75 to 180 is 0.41666.

So if we will decrease the distance in the circumference of a circle from 125,600 ly by:

125,600 x 0.41666 = 52,332 ly.

We will achieve the 105 km/s as observed.

Hence, in the same period of time – T, the star should move in total:

125,600 – 52,332 = 73,268 ly (instead of 125,600 ly).

So, the star will start at R=10,000 Ly, set a complete movement of 73,268 Ly and get to the point where R = 20,000 at the same time -T. By doing this movement the star speed at R=20,000 will be exactly 105 km/s as observed.

Technically, we can make a calculation for a star which is drifting inwards.

In this case, the star will drift inwards from 20,000 Ly to 10,000 Ly in the same time - T.

If we move a star inwards in the arm it has two vectors.

One vector is vertically to the center – let's call it inwards vector

The other vector is horizontally to the center – let's call it forwards vector.

The star will have to increase the circumference distance inorder to meet the observation speed.

However, the mathematical calculation proves that spiral galaxy isn't a rigid disc. The stars must drift outwards or inwards in the arm while the whole arm rotates around the center of the galaxy.

Edited February 14, 2015 by David Levy

[Strange]

This is incorrect!

 

You have no calculations that support that claim.

 

As I have already proved spiral galaxy isn't solar system. In a solar system, most of the mass is in the center, while in spiral galaxy there are significant portion of mass outside the center.

 

That doesn't require proof. It is blindingly obvious.

 

In order to evaluate the orbital speed vs. distance, let's assume that all the stars in spiral arms keep their position and do not drift them inwards or outwards.

 

This is an incorrect assumption. Stars move in and out of the spiral arms.

 

Therefore, it is like a rigid spiral disc. In this case, it is expected that all points of mass should complete one cycle at the same time.

 

Except we know (from observing the movements and velocities of stars) that it isn't like a rigid disk.

 

Hence, the orbital speed must increase at increasing distance, assuming that the stars keep their position in the spiral arms.

 

Which they don't.

 

However, the mathematical calculation proves that spiral galaxy isn't a rigid disc. The stars must drift outwards or inwards in the arm while the whole arm rotates around the center of the galaxy.

 

As your initial assumptions are false, your conclusion is also false.

[Janus]

Thanks!

I will explain it.

Spiral Galaxy – Orbital speed of star (in spiral arm) vs. its distance from the center

By wiki:

http://en.wikipedia.org/wiki/Galaxy_rotation_curve

"If disc galaxies have mass distributions similar to the observed distributions of stars and gas then, the orbital speed would always decline at increasing distances in the same way as do other systems with most of their mass in the centre, such as the Solar System or the moons of Jupiter".

This is incorrect!

As I have already proved spiral galaxy isn't solar system. In a solar system, most of the mass is in the center, while in spiral galaxy there are significant portion of mass outside the center.

No a galaxy is not a solar system, but neither is it a uniformly dense disk or sphere. It actually falls somewhere between a solar system and a uniformly dense disk. It has a central bulge of higher than average density which contains a significant amount of the galaxy's mass and then a gradually thinning disk. This means that the orbital speeds as you move outward from the center should behave somewhere between how they should behave in the solar system and a uniform disk.

The point is, that when you take the observed mass distribution of the galaxy into account, the orbital speeds should still fall off with distance, not as much as they do in the Solar system, but there should still be some decrease with distance. This is what the passage you quoted means: If you take into account how the mass we see in a galaxy is distributed, and then use Newton's laws of gravity to calculate the expected orbital speed according to that mass distribution, you still get orbital velocities that should decrease as you move from the center. Nobody is ignoring the fact that all the mass is not in the center, it is just that even accounting for the mass i the disk, it isn't enough cause the orbital speeds we see.

In order to evaluate the orbital speed vs. distance, let's assume that all the stars in spiral arms keep their position and do not drift them inwards or outwards. Therefore, it is like a rigid spiral disc. In this case, it is expected that all points of mass should complete one cycle at the same time.

 

The star's don't stay in the spiral arms. Their orbits carry them in and out of them. (spiral arms are somewhat like "traffic jams", where the stars and gasses crowd closer together while passing through them and spread back out again when clear of them.)

 

The measured orbital speeds we get from the star in the disk stay pretty constant in terms of km/sec as you move outward. And since the furthest from the center stars have to travel a longer distance to complete an orbit, the galaxy as a whole doesn't rotate as if it were in one piece. The only way you can get situation where the orbital speed and orbital distance has a one to one relationship using Newton's laws of gravity is inside of a spherical volume of uniform density. Spiral galaxies are not uniformly dense spheres.

Edited February 14, 2015 by Janus

[David Levy]

 

This is an incorrect assumption. Stars move in and out of the spiral arms.

How do you know that?

Do we have any real proof?

Out of the 400 billion stars in the Milky Way galaxy – would you kindly show me just one that moves in and out?

What about the sun? Does it also move in and out?

 

Thanks Janus

Do appreciate your reply.

No a galaxy is not a solar system, but neither is it a uniformly dense disk or sphere. It actually falls somewhere between a solar system and a uniformly dense disk. It has a central bulge of higher than average density which contains a significant amount of the galaxy's mass and then a gradually thinning disk. This means that the orbital speeds as you move outward from the center should behave somewhere between how they should behave in the solar system and a uniform disk..

Somehow the science does not consider the most critical issue - the spiral arm.

Why the science does not even think about the contribution of the spiral arm to the velocity and the gravity?

The star's don't stay in the spiral arms. Their orbits carry them in and out of them. (spiral arms are somewhat like "traffic jams", where the stars and gasses crowd closer together while passing through them and spread back out again when clear of them.)

Please explain.

Why do you claim that the stars don't stay in the spiral arm? Do you mean that there are stars outside the spiral arms? How do we know that?

How do we also know that the orbits carry them in and out?

If there is traffic jams, than the outcome should be severe collisions.

We know it in our motor highway.

If stars are moving in and out in the arm, than some of them should collide with each other. Do we see significant stars collisions in spiral arms?

The only way you can get situation where the orbital speed and orbital distance has a one to one relationship using Newton's laws of gravity is inside of a spherical volume of uniform density. Spiral galaxies are not uniformly dense spheres.

So, the science has no idea for this problem. I have offered a valid solution for this issue. Edited February 15, 2015 by David Levy

[Mordred]

Entire galaxies can pass through each other or merge without a star collision. There is a huge amount of volume between stars for other stars to pass through.

Edited February 15, 2015 by Mordred

[Janus]

How do you know that?

Do we have any real proof?

Out of the 400 billion stars in the Milky Way galaxy – would you kindly show me just one that moves in and out?

What about the sun? Does it also move in and out?

 

Thanks Janus

Do appreciate your reply.

Somehow the science does not consider the most critical issue - the spiral arm.

Why the science does not even think about the contribution of the spiral arm to the velocity and the gravity?

Please explain.

No one, except you have ever claimed that science ignores the spiral arms. It is just that the spiral arms do not have the type of influence that you seem to believe they do.

Why do you claim that the stars don't stay in the spiral arm? Do you mean that there are stars outside the spiral arms? How do we know that?

We know how fast the stars travel in their orbits around the galaxy because we can measure these speeds. From this information, we know that once you get out away from the central bulge, the orbital velocities remain fairly constant. Since the size of the orbit increases as you move outward, stars further from the center take longer to complete an orbit than stars closer in.

 

If these stars remained in a a single spiral arm during all their orbits, this means that the inner part of the spiral arm must be going around the galaxy in less time than the outer part of the arm. Eventually it would catch up to and pass the outer part of the arm. This "winding up" of the spiral arm would eventually tear it apart so that you would no longer any spiral arm structure. Since we know the orbital speeds of the stars, we can work out how long spiral arms should exist before they disappear. In terms of the age of the universe, this is not very long. As a result, as we look at at the universe we should see very few spiral galaxies, as the odds of that we are looking at a galaxy during the short period of time in which it displays spiral arms would be small.

 

However, when we look out at the universe, we see lots of spiral galaxies. This indicates that the spiral structure is fairly long-lived. But, as noted above, this cannot be if the stars remain in the same spiral arm throughout its orbit around the galaxy. The only conclusion left is that the orbital motion of the stars and the rotation speed of spiral arms are independent from each other and stars pass through the spiral arms as they orbit the galaxy.

 

Yes, there are stars outside of the spiral arms. We know they are there because we can detect them. In fact, the number of stars outside of the the spiral arms are not that much less than those inside the spiral arms. It is just that the spiral arms have a larger percentage of bright stars. The reason for this is the higher density in the spiral arms. The higher density promotes the birth of stars. Of these stars a certain percentage will be on the high end of the mass scale. The more massive a star, the brighter it is, but also the shorter its life span. It takes a few million years for a star born in the spiral arm to drift out. This is longer than the lifetime of the bright massive stars so they burn out before they get out of the spiral arms. The "gaps" between the arms are populated by the stars that lived long enough to leave the arm. The "gaps" only look dark in comparison to the spiral arms because the spiral arms have all the bright stars. (this is much like how Sun spots look black against the surface of the Sun, but are actually very bright in an absolute sense. Again, we know that the "gaps" contain stars because we can measure their output with instrumentation and at different wavelengths

How do we also know that the orbits carry them in and out?

Explained above. Given the measured orbital velocities of the stars, if they stayed in the Spiral arms at all times, the spiral arms would dissipate in a time period that does not match observation.

If there is traffic jams, than the outcome should be severe collisions.

We know it in our motor highway.

If stars are moving in and out in the arm, than some of them should collide with each other. Do we see significant stars collisions in spiral arms?

The "traffic jam" is not that severe. The actual difference in average star distance in the spiral arms is not that much greater than it is in between them. The average distance between stars in the spiral arms is measured in light years. It is just slightly larger in the areas between them. The brightness of the spiral arms is due to the greater percentage of very bright stars and not due to that great a difference in the actual amount of material present. The one thing you do see because of the slightly more crowded conditions is that it pushes more gas clouds past the trigger point where they collapse and form stars, giving rise to the bright stars that make the spiral arms so noticeable

So, the science has no idea for this problem. I have offered a valid solution for this issue.

No, science has a perfectly good grasp on this situation. On the other other hand your "solution" is contradicted by actual observation. Your solution requires the orbital speeds of stars in spiral galaxies to increase directly with distance form the center of the galaxy when real life measurement of these orbital speeds show no such thing. This makes it a non-starter from the get go.

[David Levy]

Thanks again

Let me start from the end of your reply.

On the other hand your "solution" is contradicted by actual observation..

No, my solution perfectly fits the actual observation!

Your solution requires the orbital speeds of stars in spiral galaxies to increase directly with distance form the center of the galaxy

No. Sorry that I was not clear in my message.

I have just used an example of a rigid spiral galaxy in order to prove that a real spiral galaxy isn't rigid!

And yes, based on this example I have found that theoretically – the orbital speeds of stars in rigid spiral galaxies should increase directly with distance form the center of the galaxy.

However, this is not the case in the real life.

..when real life measurement of these orbital speeds show no such thing.

Yes, that is fully correct. Therefore, Spiral galaxy isn't a rigid one. The meaning is that stars must move inwards or outwards.

I have also made a mathematical calculation in order to find the real distance that stars must move in spiral arm in order to meet the real life measurements.

Hence, my solution proves that:

-Spiral galaxy isn't rigid

-Stars must move outwards or inwards in order to meet the measurements.

Therefore, my solution perfectly fits the real life measurements!

We know how fast the stars travel in their orbits around the galaxy because we can measure these speeds. From this information, we know that once you get out away from the central bulge, the orbital velocities remain fairly constant. Since the size of the orbit increases as you move outward, stars further from the center take longer to complete an orbit than stars closer in.

 

If these stars remained in a a single spiral arm during all their orbits, this means that the inner part of the spiral arm must be going around the galaxy in less time than the outer part of the arm. Eventually it would catch up to and pass the outer part of the arm. This "winding up" of the spiral arm would eventually tear it apart so that you would no longer any spiral arm structure..

.

Yes, I fully agree with this explanation.

It is 100% correct.

The winding up problem is a key element in my explanation.

However, at this stage I didn't explain why it is so important.

I will explain it later on.

Yes, there are stars outside of the spiral arms. We know they are there because we can detect them. In fact, the number of stars outside of the the spiral arms are not that much less than those inside the spiral arms. It is just that the spiral arms have a larger percentage of bright stars.

Please explain how can we detect them.

Would you kindly direct me to an article which claims that there are stars outside the spiral arms?

If there are stars outside the spiral arm, than are they also in spiral shape?

If so, how many spiral arms there are in spiral galaxy?

If those outside stars are not in spiral shape - in what shape they are?

How do we know that?

It is just that the spiral arms have a larger percentage of bright stars. The reason for this is the higher density in the spiral arms.

Why there is higher density in spiral arm? What kind of force could create this kind of density? Is it based on Newton gravity low or some dark power?

 

It takes a few million years for a star born in the spiral arm to drift out..

It is estimated that the age of the sun is a few billion years - So, why it is still in a spiral arm?

How could it be that out of the 400 billions stars in the galaxy, we can't find even one bright star between the arms?

Edited February 15, 2015 by David Levy

[Mordred]

What Janus is referring to is called Density wave theory, coincidentally this also works for Saturn's rings.

 

http://en.m.wikipedia.org/wiki/Density_wave_theory

 

here is a detailed paper on it.

 

http://adsabs.harvard.edu/abs/1969ApJ...155..721Lthe pdf is on this page.

Edited February 16, 2015 by Mordred

[David Levy]

 

What Janus is referring to is called Density wave theory, coincidentally this also works for Saturn's rings.

 

http://en.m.wikipedia.org/wiki/Density_wave_theory

 

here is a detailed paper on it.

 

http://adsabs.harvard.edu/abs/1969ApJ...155..721Lthe pdf is on this page.

Thanks

I will read it carefully.

However, I have few questions with regards to the Saturn rings:

Spiral arm: Do we see any form of spiral arms in Saturn rings?

Velocity vs. distance from the center: What is the reference between the velocity of an object and its distance from Saturn. Is it behaved like a normal solar system, or like a spiral arm?

Would you also direct me to an article which claims that there are stars outside the spiral arms in spiral galaxy?

Edited February 16, 2015 by David Levy

[Mordred]

Would you also direct me to an article which claims that there are stars outside the spiral arms in spiral galaxy?

I just supplied you that article, try googling the term Density wave theory... you will see all the links describing how young stars form in which locations due to the different metallicity in each region. Before drifting out. The reason why the spiral arms are more luminous is due to the types of stars that form in the spiral arms.

 

 

You have to examine the plasma density within and outside the spiral arms. The arms simply have a higher density than the regions between the arms. Look at not just the different densities, but what elements occupy those regions. The spiral arms is a plasma cloud....

That plasma cloud behaves as a fluid.

For that matter so does galaxy rotation curves. Your calculations are only considering the mass of stars. What about all the other mass density contributors? Where is your analysis of the plasma? What about dark matter? All these contribute to mass.

 

This is why rotation curves use the virial power laws and Not strictly Newtonian force. You need to apply the hydrodynamics of a gas to galaxy rotation curves.

 

 

Funny story about rotation curves and dark matter. For years MOND tried to use Newtonian laws to rotation curves, they had to modify Newtonian for different mass distributions. They were sort of successful in better predicting rotation curves of certain types of galaxies. When LCDM had a tough time. However in order to explain spiral galaxies MOND had to use..... you guessed it dark matter.

 

Kind of defeated the purpose of MOND.

 

MOND= modified Newtonian dynamics.

 

Here is a few links on virial theorem

 

http://www.astro.cornell.edu/academics/courses/astro201/vt.htm

 

http://arxiv.org/abs/1212.2980

 

here is a simple breakdown of rotation curve see formulas page ,30

http://www.google.ca/url?sa=t&source=web&cd=2&ved=0CB0QFjAB&url=http%3A%2F%2Fwww.astro.caltech.edu%2F~george%2Fay20%2FAy20-Lec16x.pdf&rct=j&q=galaxy%20rotation%20curve%20of%20the%20milky%20way%20pdf&ei=ZJXhVLO7B8yoogSsq4GgDw&usg=AFQjCNF7bjGbX9QdzY-6n9_GMQwN-cVAag&sig2=Cc5IFvhx1kiBXKhIoD1ZwQ

Edited February 16, 2015 by Mordred

[David Levy]

Thanks

 

I just supplied you that article, try googling the term Density wave theory..

Newton gives full explanation for spiral galaxy including the winding up problem. Saturn rings are not spiral system. Therefore, the Density wave theory is not applicable for spiral galaxy. It actually gives a solution for just one aspect of spiral galaxy – winding up problem. But it confuses us with all the other aspects. This none relevant theory can't replace Newton low!

You have to examine the plasma density within and outside the spiral arms. The arms simply have a higher density than the regions between the arms. Look at not just the different densities, but what elements occupy those regions. The spiral arms is a plasma cloud....

That plasma cloud behaves as a fluid.

Why are we so sure that there is some kind of matter between the arms?

What kind of real proof the science has to support this assumption? (Please don't use a theory as a proof).

Based on Newton low, there shouldn't be any matter between the arms.

This is the key element for spiral galaxy! I will explain it later on.

Edited February 18, 2015 by David Levy

[Mordred]

The proof is every particle has an influence regardless of how small. The ideal gas laws is an average of those influences. This is true regardless of scale.

 

It's too bad to refuse to accept this. You would be amazed at the complexity and correlations to the fluidic dynamics if you did.

 

Let me ask you one question. "What is the difference between a gas and a solid ?"

 

The proof is in the distribution of hydrogen,helium etc Duh look at it.

 

You need the plasma to form stars

Do you honestly believe your smarter than every scientist that fought dark matter over close to 50 years of research?.

 

Come on give me a break learn why the current models exist.

I've been on forums since the late 80's prior to WMAP. I can attest to how hard DM was fought. No this isn't my only forum. I've seen far more stronger arguments on galaxy rotation curves than you have presented.

As to your question as to how can we assume their is some kind of matter between the arms? The answer is every point in space has a thermodynamic relation. There is always an energy/mass relationship.

Posters of alternative models never look at the history of the current model development. They refuse to accept contractions to their ideas. They refuse to learn, the history and struggles to accept a models.

 

Too many times I have seen " I have all the answers" with no proof.......

PS you've had counter points presented in this thread by a variety of accredited physics degrees. You chose to argue with them. I won't say who right of privacy and all that

PS I fully expect them and accept any corrections they have to offer. Even after 30 years of self study, and over 40+ textbooks I still learn.

[Strange]

Therefore, the Density wave theory is not applicable for spiral galaxy.

 

From the article you didn't read:

 

Beginning in the late 1970s, Peter Goldreich, Frank Shu, and others applied density wave theory to the rings of Saturn.^[5][6][7] Saturn's rings (particularly the A Ring) contain a great many spiral density waves

https://en.wikipedia.org/wiki/Density_wave_theory#Application_to_Saturn.27s_rings

 

Why are we so sure that there is some kind of matter between the arms?

 

Because we can see it.

[David Levy]

 

Because we can see it.

As you claim that we can see matter between spiral arms – please prove it.

So far I couldn't find an article which confirms this theory by real evidence.

[Janus]

As you claim that we can see matter between spiral arms – please prove it.

So far I couldn't find an article which confirms this theory by real evidence.

Then you are just not looking hard enough or just not seeing what you don't what to see.

 

http://www.spitzer.caltech.edu/images/5212-sig12-008-Hot-and-Cold-in-the-M100-Galaxy-

 

An image of a typical spiral galaxy in infrared.

 

From the article:

 

Infrared light with wavelengths of 3.6 and 4.5 microns is shown as blue/cyan, showing primarily the glow from starlight. 8 micron light is rendered in green, and 24 micron emission is red, tracing the cooler and warmer components of dust, respectively.

You will note that the blue/cyan in the image that indicates star light is fairly evenly distributed throughout the galaxy revealing a robust stellar population even "in between" the spiral arms.

[David Levy]

You will note that the blue/cyan in the image that indicates star light is fairly evenly distributed throughout the galaxy revealing a robust stellar population even "in between" the spiral arms.

Thanks

However, please be aware for the following:

" The blue dots covering the entire image are stars that lie between us and M100."

Therefore, I assume that by eliminating those blue dots, we should get a normal view of spiral galaxy.

In any case, I couldn't find in this article a support for the idea that the stars are fairly evenly distributed throughout the galaxy.

[Mordred]

http://www.atlasoftheuniverse.com/milkyway.html

 

"It should be emphasized that there are almost as many stars between the spiral arms as in the spiral arms. The reason why the arms of spiral galaxies are so prominant is that the brightest stars are found in the spiral arms. Spiral arms are the major regions of star formation in spiral galaxies and this is where most of the major nebulae are found"

[Strange]

In any case, I couldn't find in this article a support for the idea that the stars are fairly evenly distributed throughout the galaxy.

 

The arms appear brighter because there are more young stars (hence more massive, bright stars). These massive, bright stars also die out quickly, which would leave just the darker background stellar distribution behind the waves, hence making the waves visible.

While stars, therefore, do not remain forever in the position that we now see them in, they also do not follow the arms. The arms simply appear to pass through the stars as the stars travel in their orbits.

http://en.wikipedia.org/wiki/Spiral_galaxy

 

As you are challenging accepted science, I think it is up to you to provide evidence that there are no stars in the arms and that they rotate as a solid disk.

[Mordred]

I have to agree with Strange on this, considering the amount of material we provided in particular on density wave, virial theory and rotation curve links.

 

Further detail is look at the distribution of pop1 vs pop2 stars

[imatfaal]

!

Moderator Note

 

me too - and I am going to move this topic to Speculations (not sure why I didnt do this earlier. )

 

David - please take a read of the guidelines and rules of the Speculations Forum

 

[David Levy]

 

http://en.wikipedia.org/wiki/Spiral_galaxy

 

As you are challenging accepted science, I think it is up to you to provide evidence that there are no stars in the arms and that they rotate as a solid disk.

Sorry, this isn't an evidence for any existence of mass between the arms.

This is only a theory for spiral galaxy shape which is already well known.

You have claimed that we can see those stars.

Never the less, so far we couldn't find any real evidence which confirms it.

In the M100 galaxy, there is no any proof for this assumption.

Based on Newton low there shouldn't be any mass between the arms.

The explanation is quite simple -

Let's assume that one star had been drifted out from the spiral arm and try to figure what should be the outcome:

We all know that the gravity force is : " inversely proportional to the square of the distance between them"

Therefore, the nearby stars contribute significant portion of the equivalent gravity force which attracts this star in the galaxy. If a miserable star is drifted out from the spiral arm, (not inwards or outwards in the arm – but just out of the arm) its equivalent gravity force should be decreased.

Hence, there will be no balance between its rotation energy to its new decreased equivalent gravity force. Therefore, it will be kicked out from the arm and eventually from the galaxy.

Hence, Theoretically we might see some stars between the arms. But those stars are in transient mode. They had been kicked out from the galaxy.

This proves that there are no stars in between the spiral arms!

With regards to our solar system -It's better for us to keep our position in the spiral arm.

Edited February 19, 2015 by David Levy

[Mordred]

It's too bad you don't listen or read any articles that contradicts your feelings. We've shown you articles and references that show otherwise.

 

As Strange mentioned we can see stars outside the spiral arms in the halo regions. Janus and I also supplied links. You obviously never read or understood the article on density wave I posted.

 

That's your loss not ours. You would think my reference to MOND and its failures would have clued you in.

 

 

As Strange mentioned you have demanded papers from us, that we supplied. It's your turn. Find us a peer reviewed paper showing your correct.

By the way even if there is no stars,

 

There is still mass, all forms of particles generate mass......

Stars DO NOT form without a sufficient amount of gas clouds. They form in nebulae. The Milky way wasn't always a spiral galaxy. There was star formation prior to being a spiral galaxy. Originally it was one big gas cloud with uniform dispersion. This is why older stars reside in the arms pop 2 stars. Pop 1 stars require a different combination of elements that are formed from pop 2 stars. Pop 2 stars are in the halo. Pop 1 stars are primarily in the spiral arms

Stars can and do drift into and out of the spiral arms.

 

Newtons laws does not prove otherwise. Your attempts to make it do so is by ignoring any conflicting mass. Such as dust and dark matter

 

 

"A near-spherical halo of stars, including many in globular clusters"

 

Spherical not flat like the spiral arms.

 

http://en.m.wikipedia.org/wiki/Spiral_galaxy

 

Like I stated you choose to ignore the evidence against your ideas. This is similar to statements posted in the numerous links we provided.

 

The majority of those links state

 

"Young stars form in the spiral rings "

 

 

In one of the articles I posted it has the CORRECT rotation curve formula for the milky way.

http://www.google.ca/url?sa=t&source=web&cd=2&ved=0CB0QFjAB&url=http%3A%2F%2Fwww.astro.caltech.edu%2F~george%2Fay20%2FAy20-Lec16x.pdf&rct=j&q=galaxy%20rotation%20curve%20of%20the%20milky%20way%20pdf&ei=ZJXhVLO7B8yoogSsq4GgDw&usg=AFQjCNF7bjGbX9QdzY-6n9_GMQwN-cVAag&sig2=Cc5IFvhx1kiBXKhIoD1ZwQ