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Dave Lee
2015-Sep-26, 06:42 AM
Please see the following presentation at pg 15
http://www.ifa.hawaii.edu/~barnes/ast110/MilkyWay.pdf
There is an image of the galactic rotation random vectors in the nearby Solar Neighborhood.
It is stated:
"Most stars near the Sun have random velocities of a few tens of km/sec. These stars orbit the galactic center at ~230 km/sec."
Let' assume that few tens of km/sec is about 25 to 50.
Hence, each star in the Solar Neighborhood, has a random velocity of about 10% to 20% with related to the orbit velocity.
In this image, we see that those random velocity vectors are pointing to all directions. Up, down, left and right.
Therefore, if those stars will maintain their momentum and their velocity vectors, they will be ejected from the spiral arm in about one million year. Each one will be ejected at a random direction from the arm.
For example, the Sun is moving with some degree to the galactic plane.
http://www.americanscientist.org/libraries/documents/2005620101154_306.pdf
"The sun’s path is inclined about 25 degrees to the plane of the galaxy and is headed toward a region in the constellation of Hercules near its border with Lyra."
By keeping this momentum and without an external force, the Sun might disconnect from the disc plane. However, it is also stated:
"The sun oscillates through the plane of the galaxy with an amplitude of about 230 lightyears, crossing the plane every 33 million years."
So, how could it be???

Shaula
2015-Sep-26, 07:10 AM
The random velocities are relative to the Sun (the Sun appears to have no random velocity in that picture).

The other clue as to what is going on is that phrase about the Sun oscillating. Any in or out of plane motion is damped by the mass of the disk. So these random motions will change, bring any star moving too far out of plane back into plane. They won't maintain their vectors and be ejected.

Edit: A good way to think about these relative motions is to think about the relative motions of the planets - they can appear to be moving in all sorts of directions relative to the Earth as they (and we) orbit the Sun.

Jeff Root
2015-Sep-26, 08:42 AM
Shaula,

It isn't clear whether you caught that he said "ejected from
the spiral arm", not "ejected from the galaxy". I read it that
way for a second myself.


Animation shows the Sun's last 250-million-year orbit around
the center of the galaxy (GC), and other stars converging toward
the Sun, ending with all of them near the Sun today. Credit: ESO

Said to be 14,000 stars plotted.
ESO is the European Southern Observatory.

Found it on two websites. The animation is larger and looks
better on the first; the second site, in Japan, has more info:

http://thehandstand.org/archive/may2005/index.htm
http://blog.goo.ne.jp/trova/e/b16c682fe40311c523fe48282d3402cf

-- Jeff, in Minneapolis
.

Shaula
2015-Sep-26, 10:12 AM
Shaula,
It isn't clear whether you caught that he said "ejected from the spiral arm", not "ejected from the galaxy". I read it that way for a second myself.
Oh, good point. I read it as ejected from the galaxy. So I guess I would add that the spiral pattern is just a star formation burst - it is not a fixed structure. Stars can 'leave' it at any time - the shock waves or density waves just carry on sweeping around regenerating the leading edge of the spiral structure.

Dave Lee
2015-Sep-26, 11:58 AM
The random velocities are relative to the Sun (the Sun appears to have no random velocity in that picture).

The other clue as to what is going on is that phrase about the Sun oscillating. Any in or out of plane motion is damped by the mass of the disk. So these random motions will change, bring any star moving too far out of plane back into plane. They won't maintain their vectors and be ejected.

Edit: A good way to think about these relative motions is to think about the relative motions of the planets - they can appear to be moving in all sorts of directions relative to the Earth as they (and we) orbit the Sun.

Thanks
With regards to your example: "The Earth orbits the Sun".
Gravity force – There is a clear gravity force between the Sun and the Earth which holds them together.
So, do you think that a gravity force holds the Sun in the galactic plane?
If so, what is the main source for this gravity?
In other words, the Sun must rotate around something. So where is the hosting star or the Center of Mass which holds the sun in its orbit? (Is it the black hole, Nearby stars, dark mass..)
However, in a basic gravity force the rotation is quite smooth. There is a simple disc rotation, without any oscillation.
I'm not sure that I fully understand your answer with regards for the Sun oscillation phenomenon.
Please elaborate.

Cougar
2015-Sep-26, 12:37 PM
So, do you think that a gravity force holds the Sun in the galactic plane?
If so, what is the main source for this gravity?

Yes, it's just the gravity of the stars in the disk that pull the sun back down when it wanders too high above the disk, and pull it back up when it wanders too low, all while orbiting the center of the galaxy.


In other words, the Sun must rotate around something. So where is the hosting star or the Center of Mass which holds the sun in its orbit? (Is it the black hole, Nearby stars, dark mass..)...

The sun's 250-million-year orbit around the center of the galaxy is determined by the mass of all the stars, gas, and dust interior to that orbit, including the central supermassive black hole (which is actually just a small percentage of that overall mass). Plus, apparently, a bunch of unseen dark matter. As mentioned, as it orbits, it wanders -- oscillates -- above and below the plane of the disk.

cjameshuff
2015-Sep-26, 12:45 PM
Thanks
With regards to your example: "The Earth orbits the Sun".
Gravity force – There is a clear gravity force between the Sun and the Earth which holds them together.
So, do you think that a gravity force holds the Sun in the galactic plane?
If so, what is the main source for this gravity?
In other words, the Sun must rotate around something. So where is the hosting star or the Center of Mass which holds the sun in its orbit? (Is it the black hole, Nearby stars, dark mass..)
However, in a basic gravity force the rotation is quite smooth. There is a simple disc rotation, without any oscillation.
I'm not sure that I fully understand your answer with regards for the Sun oscillation phenomenon.
Please elaborate.

Your assumption that the sun must rotate around something is incorrect. Galaxies can not be treated as point masses when working with the trajectories of stars within them. Galaxies like the Milky Way include a disk of stars, gas, and dust, a central bulge with a higher density of all of these, and a diffuse halo of dark matter, and you must sum up the contributions of all of these. When near the disk, a large component of the overall gravitational acceleration is toward the plane of the disk, which causes the up/down oscillation with a period much shorter than the orbital period.

Shaula
2015-Sep-26, 01:46 PM
I cannot add much more to Cougar's and cjameshuff's explanations - save to say the picture of gravitational forces reducing to an object simply orbiting something is only really accurate for the case that the mass distribution is circularly symmetric in the orbital plane. When you have a case like the galaxy, where this is not the case the picture is far more complicated and the simple two body approximation cannot be used.

Dave Lee
2015-Sep-26, 06:46 PM
Yes, it's just the gravity of the stars in the disk that pull the sun back down when it wanders too high above the disk, and pull it back up when it wanders too low, all while orbiting the center of the galaxy.

Thanks
So it is gravity force of the stars in the disk.
However, it is known that the gravity force is reduced dramatically as we move away.
Therefore, I would expect that if the Sun is wander too high from the disc, than the gravity force from the stars in the disc should be reduced by definition. Hence, there will be less and less gravity power (from the stars at the disc) as the Sun moves higher and higher from the disc. So, if the sun is too high, there won't be enough gravity power to pull it back.
Do you agree?
Based on my understanding, in real gravity force there is no oscillation. It should be a smooth movement. For example:
Earth and moon – Smooth movement in a disc shape.
Sun and Earth – Smooth movement in a disc shape.
Do you have any example of oscillation in gravity force?

01101001
2015-Sep-26, 07:08 PM
Look instead, in the solar sytem, for gravity-bound systems with massive numbers of members.

http://saturn.jpl.nasa.gov/news/newsreleases/newsrelease20101101/

Cassini Sees Saturn Rings Oscillate Like Mini-Galaxy

Scientists believe they finally understand why one of the most dynamic regions in Saturn's rings has such an irregular and varying shape, thanks to images captured by NASA's Cassini spacecraft. And the answer, published online today in the Astronomical Journal, is this: The rings are behaving like a miniature version of our own Milky Way galaxy.

DaveC426913
2015-Sep-26, 09:26 PM
So it is gravity force of the stars in the disk.
However, it is known that the gravity force is reduced dramatically as we move away.
Therefore, I would expect that if the Sun is wander too high from the disc, than the gravity force from the stars in the disc should be reduced by definition. Hence, there will be less and less gravity power (from the stars at the disc) as the Sun moves higher and higher from the disc. So, if the sun is too high, there won't be enough gravity power to pull it back.
Do you agree?

No. Look at a simple system just for a moment. As Pluto moves away from the Sun, it too feels a drop in pull from the Sun. Do we see Pluto drifting off into space under the lesser gravity? No, it has simply reached its aphelion. Pluto's orbital velocity is lower than the Sun's escape velocity.



So, back to the galaxy: the sun is moving out of the plane of the galaxy, true, but all the while it is slowing wrt to its distance from the plane. It will eventually lose all its motion "upward" and will have reached apocenter, then it will start coming back. It will oscillate, back and forth in a sinusoidal motion, between upper and lower limits.

Or, simply put, the galactic disc's escape velocity is higher than the Sun's velocity. It will always be pulled back.

http://www.centauri-dreams.org/wp-content/uploads/2007/07/galaxy_radiation.jpg

pzkpfw
2015-Sep-26, 09:55 PM
... Therefore, I would expect that if the Sun is wander too high from the disc, than the gravity force from the stars in the disc should be reduced by definition. Hence, there will be less and less gravity power (from the stars at the disc) as the Sun moves higher and higher from the disc. So, if the sun is too high, there won't be enough gravity power to pull it back. ...

If you throw a ball up, it is moving away from the Earth, so - yes - there will be less and less gravity as it travels upwards. But does the ball fall back to Earth? Yes. Unless thrown with sufficient force. As the ball travels upwards, it is slowing down. It takes a lot of initial speed to counteract that.

Edit: so with your issue with the movements of stars relative to the galactic plane, it's not just about them moving "up" or "down" - but how fast.

https://en.wikipedia.org/wiki/Escape_velocity


... Based on my understanding, in real gravity force there is no oscillation. It should be a smooth movement. For example:
Earth and moon – Smooth movement in a disc shape.
Sun and Earth – Smooth movement in a disc shape.
Do you have any example of oscillation in gravity force?

Orbits are elliptical - which you could see as "oscillation" in the distance between the orbiting thing and the thing it's orbiting. A perfectly circular orbit would be something very rare. The up-and-down motion of the sun relative to the galactic plane is sort of similar to that.

https://en.wikipedia.org/wiki/Elliptic_orbit

Hornblower
2015-Sep-26, 10:58 PM
Let me point out that no gravitation toward the plane is needed to make a star in an inclined orbit bob up and down. If the gravitation of the disk is vanishingly slight, the star will do so with one complete cycle in a circuit around the core in its orbit. The gravitational action of the disk just makes it bob up and down more quickly, thus doing several cycles during one orbital trek.

Jeff Root
2015-Sep-26, 11:09 PM
Somewhat relevant and quite interesting is that the gravity
from a large plane -- like the disk of a galaxy -- does NOT
decrease as you move away from it. As long as you don't
get so far from it that the edge of the disk is no longer far
off in the distance. If the plane were infinite (an infinitely
large galactic disk), then the gravity would never diminish
nomatter how far you got from it. Such an infinite galaxy
would also have infinite mass, which helps to understand
how the gravity could behave in such a strange way.

The stars in typical spiral galaxies are distributed quite
evenly. The density of stars, gas, dust, and dark matter in
the arms isn't much different from the density between the
arms. But the arms are where the clouds of gas and dust
have collapsed and produced many new stars, which are
much brighter than older main-sequence stars. Those bright
stars light up the nearbly dust clouds. The arms are the
pattern of waves of star production around the galaxy.

The disk has thickness. Although it thins out gradually so
that saying where the "surfaces" of the two sides are located
is entirely arbitrary, the center plane of the disk is generally
quite well defined. Any object on either side of the plane is
pulled toward the plane. But the net pull close to the plane is
extremely weak. The pull is strongest near the fuzzy surface,
where almost all matter in the disk is pulling in one direction,
toward the central plane. Anywhere inside the disk, inside
the fuzzy surface, some matter is pulling the object away
from the central plane, reducing the net force toward it.

Because the matter in a spiral galaxy is spread out so evenly
in a broad disk, rather than concentrated all in one object like
the Sun at the center of the Solar System, gravitational force
toward the central plane is significant. So objects orbiting the
galaxy in the disk bob "up and down" as well as in and out.

Similarly with the net force toward the center of the galaxy.
All the matter in the entire galaxy pulls on everything in the
galaxy, but if the galaxy is radially symmetrical, the net force
toward the center can be closely approximated by considering
just the pulls from matter inside the orbit. Pulls from matter
outside the orbit cancel each other out.

-- Jeff, in Minneapolis
.

cjameshuff
2015-Sep-27, 02:29 AM
Thanks
So it is gravity force of the stars in the disk.
However, it is known that the gravity force is reduced dramatically as we move away.
Therefore, I would expect that if the Sun is wander too high from the disc, than the gravity force from the stars in the disc should be reduced by definition. Hence, there will be less and less gravity power (from the stars at the disc) as the Sun moves higher and higher from the disc. So, if the sun is too high, there won't be enough gravity power to pull it back.
Do you agree?
Based on my understanding, in real gravity force there is no oscillation. It should be a smooth movement. For example:
Earth and moon – Smooth movement in a disc shape.
Sun and Earth – Smooth movement in a disc shape.
Do you have any example of oscillation in gravity force?

Those are two-body systems. Why would you expect behavior similar to a many-body system such as a galaxy?
Also, circular or elliptical motion is oscillation, every bit as much as a pendulum or the solar system swinging back and forth through the galactic plane.

Gravity is an attractive force that follows the inverse square law, nothing about it makes circular motion the necessary result...in fact, for systems of 3 bodies or more, exact circular motion is only possible for a few special cases. In very large systems such as spiral galaxies, the motions can average out to an overall rotating disk, but on the level of individual stars the motions can be very different, just as the motion of individual molecules of a gas are different from the overall motion of the gas.

Dave Lee
2015-Sep-27, 02:29 AM
No. Look at a simple system just for a moment. As Pluto moves away from the Sun, it too feels a drop in pull from the Sun. Do we see Pluto drifting off into space under the lesser gravity? No, it has simply reached its aphelion. Pluto's orbital velocity is lower than the Sun's escape velocity.



So, back to the galaxy: the sun is moving out of the plane of the galaxy, true, but all the while it is slowing wrt to its distance from the plane. It will eventually lose all its motion "upward" and will have reached apocenter, then it will start coming back. It will oscillate, back and forth in a sinusoidal motion, between upper and lower limits.

Or, simply put, the galactic disc's escape velocity is higher than the Sun's velocity. It will always be pulled back.

http://www.centauri-dreams.org/wp-content/uploads/2007/07/galaxy_radiation.jpg


Wow
It's a great image.

http://www.centauri-dreams.org/wp-content/uploads/2007/07/galaxy_radiation.jpg

I think that I have an idea how to generate this kind of movement with two simple rotation disc systems.
So, let's take the example of Sun, Earth and moon system.
As I have stated,
Earth and moon – Smooth movement in a disc shape.
Sun and Earth – Smooth movement in a disc shape
However, let's assume that those two disc shapes are vertically to each other.
Hence, if we stay at the sun, looking at the moon, (while the Earth is invisible) than we should see the same sinusoidal image.
The moon in this case, should cross the Sun/Earth disc plane in a clear sinusoidal movement and complete one cycle in one month (or 28 days).
Therefore, if the Sun rotates around some invisible object in a pure disc shape, while that invisible object rotate around the galactic plane in a pure disc shape, than the outcome is a clear sinusoidal movement.
Do you agree?

DaveC426913
2015-Sep-27, 02:46 AM
The moon in this case, should cross the Sun/Earth disc plane in a clear sinusoidal movement and complete one cycle in one month (or 28 days).
Therefore, if the Sun rotates around some invisible object in a pure disc shape, while that invisible object rotate around the galactic plane in a pure disc shape, than the outcome is a clear sinusoidal movement.
Do you agree?

Not quite. The Moon's path around the sun will form a cycloid:
http://jwilson.coe.uga.edu/EMAT6680Fa05/Brown/Assign%2010/image7.gif

The sun's motion through the galaxy will form a proper sinusoid:
http://4.bp.blogspot.com/-I--i1MvrtiU/T-7XriX59iI/AAAAAAAABEk/_0yllNcQKjA/s320/Sine%2Bwave%2Bgenerator%2Boutput-745946.png

OK, if you normalized the Sun's motion (say, by following along with its planar motion through the galaxy), you would see an ellipse, sure but it would be an ellipse with an eccentricity of - effectively - infinity. (minor axis of zero)

DaveC426913
2015-Sep-27, 02:51 AM
Somewhat relevant and quite interesting is that the gravity
from a large plane -- like the disk of a galaxy -- does NOT
decrease as you move away from it..

Of course! I'd forgotten that.

Dave Lee
2015-Sep-27, 03:00 AM
Not quite. The Moon's path around the sun will form a cycloid:
http://jwilson.coe.uga.edu/EMAT6680Fa05/Brown/Assign%2010/image7.gif



Thanks
Fully agree.



The sun's motion through the galaxy will form a proper sinusoid:
http://4.bp.blogspot.com/-I--i1MvrtiU/T-7XriX59iI/AAAAAAAABEk/_0yllNcQKjA/s320/Sine%2Bwave%2Bgenerator%2Boutput-745946.png

I'm not sure about it.
Please don't forget that we stay near the Sun.
So, if we could see the Sun movement from the center of the galaxy, I assume that we should get that cycloid movement.
However, as we stay near the Sun, I assume that it's quite difficult to see the cycloid movement.
Do you have an illustration for what kind of image we might see from the Sun in this case?

Hornblower
2015-Sep-27, 03:19 AM
Thanks
Fully agree.


I'm not sure about it.
Please don't forget that we stay near the Sun.
So, if we could see the Sun movement from the center of the galaxy, I assume that we should get that cycloid movement.
However, as we stay near the Sun, I assume that it's quite difficult to see the cycloid movement.
Do you have an illustration for what kind of image we might see from the Sun in this case?

What sort of mechanical action would make you expect that cycloid rather than a sinusoidal pattern?

Dave Lee
2015-Sep-27, 10:35 AM
What sort of mechanical action would make you expect that cycloid rather than a sinusoidal pattern?

Let me ask the following:
Why the science estimates that it is a sinusoidal pattern instead of cycloid.
As It is stated: "The sun oscillates through the plane of the galaxy with an amplitude of about 230 lightyears, crossing the plane every 33 million years."
Therefore, in order to verify just one cycle, 33 million years is needed.
Hence, do you agree that in order to get a clear visibility, we must send a probe to the center of the galaxy and monitor the Sun pattern for about 33 Million years?
How can we estimate the correct pattern by monitoring less than 0.00001% of the wave and from wrong location?
Therefore, do you agree that currently we can't get to solid conclusion if the Sun pattern is sinusoidal or cycloid?

Dave Lee
2015-Sep-27, 11:04 AM
Not quite. The Moon's path around the sun will form a cycloid:
http://jwilson.coe.uga.edu/EMAT6680Fa05/Brown/Assign%2010/image7.gif

Sorry, but after reconsidering this issue I estimate that there is no chance for cycloid pattern.
In order to get a cycloid pattern, the velocity of the moon around the Earth, should be higher than the velocity of the Earth around the Sun. Therefore, we get a negative movement.
It is clear the Earth rotates at much higher velocity than the moon.
In the same token, the average velocity of the Sun around the galaxy is much faster than the expected velocity of the 230 Light years amplitude (in 33 million years).
Therefore, the whole discussion about the cycloid is none relevant.
We should get a sinusoidal pattern in both - (but not a pure one). In one side of the wave it might be narrow, while in the other side it might be wide.
Do you agree?

cjameshuff
2015-Sep-27, 11:34 AM
Let me ask the following:
Why the science estimates that it is a sinusoidal pattern instead of cycloid.
As It is stated: "The sun oscillates through the plane of the galaxy with an amplitude of about 230 lightyears, crossing the plane every 33 million years."
Therefore, in order to verify just one cycle, 33 million years is needed.
Hence, do you agree that in order to get a clear visibility, we must send a probe to the center of the galaxy and monitor the Sun pattern for about 33 Million years?
How can we estimate the correct pattern by monitoring less than 0.00001% of the wave and from wrong location?
Therefore, do you agree that currently we can't get to solid conclusion if the Sun pattern is sinusoidal or cycloid?

You are trying to go back to epicycles, in a setting where they are even less applicable than the one where they already failed. The sun is simply not traveling in a circular path hundreds of light years across around some point in the galaxy that is orbiting the center of the galaxy. Gravitational motions are not a composition of circular motions. If you want to understand gravity, you need to give up on this approach, as it has absolutely nothing to do with real world physics.

The vertical oscillation is roughly sinusoidal...nothing about gravity would cause forwards and backwards acceleration with respect to the overall galactic rotation as the sun moves through the disk. It is not exactly sinusoidal, it depends on the mass distribution as a function of distance from the plane of the disk, and how far a given star goes from that plane.

Hornblower
2015-Sep-27, 01:15 PM
Some general remarks: If a moon is orbiting a planet rapidly enough, it can trace an epicycloid path relative to the Sun similar to the first one in post 19. Io's motion around Jupiter is such a case, as its velocity relative to Jupiter is approximately as fast as Jupiter's velocity relative to the Sun. The Moon's motion relative to Earth is far too slow by comparison, so its heliocentric motion relative to Earth's orbit is more nearly like that in the second example, but with the reference path being a big circle rather than a straight line. The Sun's motion around the center of the galaxy is more nearly like the second example.

We should get a sinusoidal pattern in both - (but not a pure one). In one side of the wave it might be narrow, while in the other side it might be wide.
Do you agree? Yes. You appear to be getting an improved understanding of the situation.

Let me add that we have not actually observed the carousel-horse-like bobbing of the Sun's motion, because it is too slow to see more than a tiny fraction of a cycle in recorded history. That motion is predicted by numerical integration of the Sun's motion, based on the galaxy's gravitational signature which has been inferred by other means.

cjameshuff
2015-Sep-27, 06:43 PM
Let me add that we have not actually observed the carousel-horse-like bobbing of the Sun's motion, because it is too slow to see more than a tiny fraction of a cycle in recorded history. That motion is predicted by numerical integration of the Sun's motion, based on the galaxy's gravitational signature which has been inferred by other means.

And we can test the overall behavior of those numerical models against the behavior of galaxies and other many-body systems such as Saturn's rings or the planets and moons of the solar system. The oscillation through the galactic plane is just the expected result of gravity behaving as we see it behave. We can also look at the distribution of doppler shifts of the stars in face-on spiral galaxies to get an overall measurement of the amount of motion perpendicular to the plane.

Jeff Root
2015-Sep-27, 07:28 PM
I agree with everything Hornblower just said.



You are trying to go back to epicycles,
I don't think that's what he's trying to do, but it isn't
at all clear.

The orbital motion of a star around the Galaxy and its
up-and-down, roughly sinusoidal bobbing motion from
one side of the galactic plane to the other can be looked
at as two separate motions. Both are caused by the net
gravitational forces from ALL the masses in the Galaxy,
but there is no reason not to consider them seperately.

The orbital motion is caused by the constant net force in
the direction of the center of the Galaxy, and the bobbing
motion is caused by the varying net force toward the
central plane of the Galaxy. Both motions are calculated,
based primarily on the observed velocities (speed and
direction of motion) of thousands of stars relative to us
and relative to very distant reference objects which are
considered to be fixed in the transverse direction.

As has been said, orbits in the Galaxy are not "Keplerian"
ellipses, but they do follow the same "rules" which cause
keplerian orbits: They speed up as they fall deeper into
a gravitational well and slow down as they rise out of it.
The shape of the Galaxy's gravitational well is such that
both a very roughly elliptical orbital motion around the
center of the Galaxy and a very roughly sinusoidal side-
to-side motion around the central plane of the Galaxy are
the predominant motions.

Dave Lee,

You are correct that the Moon's path around the Sun is
not a cycloid. Here is a diagram I made of a portion of
the paths of the Moon and the Earth-Moon barycenter
around the Sun, to scale:

http://www.freemars.org/jeff2/MoonOrb1.png

The Sun is of course at the point where the orange lines
meet.

-- Jeff, in Minneapolis

Dave Lee
2015-Sep-27, 07:44 PM
Some general remarks: If a moon is orbiting a planet rapidly enough, it can trace an epicycloid path relative to the Sun similar to the first one in post 19. Io's motion around Jupiter is such a case, as its velocity relative to Jupiter is approximately as fast as Jupiter's velocity relative to the Sun. The Moon's motion relative to Earth is far too slow by comparison, so its heliocentric motion relative to Earth's orbit is more nearly like that in the second example, but with the reference path being a big circle rather than a straight line. The Sun's motion around the center of the galaxy is more nearly like the second example.
Yes. You appear to be getting an improved understanding of the situation.


Thanks
It is stated:
http://www.americanscientist.org/lib...101154_306.pdf
"The sun’s path is inclined about 25 degrees to the plane of the galaxy and is headed toward a region in the constellation of Hercules near its border with Lyra."
Do you estimate that there is similarity between the expected Sun oscillation through the galactic plane and the oscillation pattern of the following examples:
First example: Let's assume that there is a 25 degrees shift between the Sun/Earth rotation disc to the Earth/moon rotation disc. Hence, by looking at the moon from the Sun shall we get the requested oscillation pattern?
Second example: Let's assume that the Sun orbits an invisible object which rotates around the center of the galaxy – and with a shift of 25 degrees between their rotation discs. Hence, by looking at the Sun from the center of the galaxy shall we also get the requested oscillation pattern?

Dave Lee
2015-Sep-27, 07:54 PM
Dave Lee,

You are correct that the Moon's path around the Sun is
not a cycloid. Here is a diagram I made of a portion of
the paths of the Moon and the Earth-Moon barycenter
around the Sun, to scale:

http://www.freemars.org/jeff2/MoonOrb1.png

The Sun is of course at the point where the orange lines
meet.

-- Jeff, in Minneapolis
Wow
That's great!
Thanks

cjameshuff
2015-Sep-27, 11:12 PM
I don't think that's what he's trying to do, but it isn't
at all clear.

It seems pretty clear that that's exactly what he's doing.



The orbital motion of a star around the Galaxy and its
up-and-down, roughly sinusoidal bobbing motion from
one side of the galactic plane to the other can be looked
at as two separate motions. Both are caused by the net
gravitational forces from ALL the masses in the Galaxy,
but there is no reason not to consider them seperately.

You certainly can. You could omit the orbit entirely and consider a the trajectory of a test mass through a universe containing an infinite, uniform plane of mass, like an infinite, non-rotating galactic disk. It'll bob up and down, just like the sun does in the disk of the Milky Way.

What it absolutely will not do is start a circular orbit around some point in that plane, which is how Dave Lee is trying to treat the problem. There is no physical force that would cause such a motion. Gravitation is not a series of circular motions, it is an inverse-square attractive force. You could indeed mathematically represent any curve as an infinite series of such motions, but it would tell you nothing about the physical cause, nor would it give you any way to predict future motions.

DaveC426913
2015-Sep-27, 11:30 PM
Sorry, but after reconsidering this issue I estimate that there is no chance for cycloid pattern.
In order to get a cycloid pattern, the velocity of the moon around the Earth, should be higher than the velocity of the Earth around the Sun. Therefore, we get a negative movement.

Perceptive. Though you still have a cycloid. Cycloids come in many shapes:
http://www.math10.com/en/geometry/analytic-geometry/geometry5/11-10,11.jpg



It is clear the Earth rotates at much higher velocity than the moon.
Therefore, the whole discussion about the cycloid is none relevant.

The point here, is that it is not a sine wave.

Remember how we got here. You were suggesting the the sun's motion to and fro through the plane of the galaxy was as if it were orbiting a fixed body. This is not so. If it were orbiting a fix body - an elliptical path - you would see the shape like that of the Moon around the Earth around the Sun. (I was simplifying, to make my point.)



We should get a sinusoidal pattern in both - (but not a pure one). In one side of the wave it might be narrow, while in the other side it might be wide
Good. You have arrived at the correct interpretation.
It is still called a cycloid. See diagram above.

Jeff Root
2015-Sep-28, 12:36 AM
I agree that the Moon's orbit is a cycloid, or very close to
being a cycloid. I was over-reacting to that first diagram
when I said it isn't.

-- Jeff, in Minneapolis

Dave Lee
2015-Sep-28, 05:35 AM
There is no physical force that would cause such a motion. Gravitation is not a series of circular motions, it is an inverse-square attractive force.
Sorry. I'm not sure about this statement.
I assume that gravitation could be a series of circular motions.


You could indeed mathematically represent any curve as an infinite series of such motions, but it would tell you nothing about the physical cause, nor would it give you any way to predict future motions.

Let me use the again the solar system example:
https://en.wikipedia.org/wiki/Solar_mass
It is stated:
"The Sun already holds 99.86% of the Solar System's total mass,"
"One solar mass, M☉, can be converted to related units:
• 27,068,510 ML (Lunar mass)
• 332,946 M⊕ (Earth mass)
• 1047.56 MJ (Jupiter mass)"

The sun mass with related to the Moon (Lunar)mass is - 2,7068,510 (almost infinite).
So, why the Moon prefers to orbit the Earth instead of orbiting the Sun?
This example by itself proves that gravity could be series of circular motions. Currently we know that the moon orbits the Earth, which orbits the Sun, which orbits the galactic center.
So why there is no room for: Moon orbits the Earth, which orbits the Sun, which orbits invisible object, which orbits the galactic center.
Technically, we could even add several invisible objects which orbit each other in a chain.
So, do you agree that Gravity shouldn't have any limitation in series of circular motions?
Actually, you have already gave the answer:
" Gravitation …is an inverse-square attractive force."
Therefore, local small mass (with low gravity force) could have higher impact than far end huge mass.
Do you agree that this could be the reason why the moon prefers to orbit the Earth instead of the Sun? Or why the Earth prefers to orbit the Sun instead of the center of the galaxy.
Therefore, could it be that in the solar neighborhood there is enough local mass (or gravity) to support a local sun rotation?

Dave Lee
2015-Sep-28, 07:43 AM
With regards to the location of that invisible object;
It is stated:
"The sun oscillates through the plane of the galaxy with an amplitude of about 230 lightyears, crossing the plane every 33 million years."
So, the max amplitude could be an indication for the orbital diameter.
Therefore, the radius between that invisible object to the Sun should be 115 Light years.
If the amplitude is symmetrical with regards to the galactic plane, (above and below the plane), than this object should be located directly at the galactic plane.
So do you agree that technically, we can find its exact location in the galaxy and the Sun orbital velocity (One cycle every 33 x 2 million years)?

Jeff Root
2015-Sep-28, 08:18 AM
Oh dear.

-- Jeff, in Minneapolis

Grashtel
2015-Sep-28, 08:36 AM
With regards to the location of that invisible object;
It is stated:
"The sun oscillates through the plane of the galaxy with an amplitude of about 230 lightyears, crossing the plane every 33 million years."
So, the max amplitude could be an indication for the orbital diameter.
Therefore, the radius between that invisible object to the Sun should be 115 Light years.
If the amplitude is symmetrical with regards to the galactic plane, (above and below the plane), than this object should be located directly at the galactic plane.
So do you agree that technically, we can find its exact location in the galaxy and the Sun orbital velocity (One cycle every 33 x 2 million years)?
The problem is that if there was such an object it would need to be very massive, easily large enough to produce noticeable gravitational effects on other stars, particularly the ones closer to it than the Sun. The fact that we don't see such gravitational effects is very strong evidence that the sinusoidal motion of the Sun through the galactic disk is due to the overall gravitational effect of all the other stars and other material making up the disk rather than one single massive invisible object

Dave Lee
2015-Sep-28, 11:16 AM
The problem is that if there was such an object it would need to be very massive, easily large enough to produce noticeable gravitational effects on other stars, particularly the ones closer to it than the Sun. The fact that we don't see such gravitational effects is very strong evidence that the sinusoidal motion of the Sun through the galactic disk is due to the overall gravitational effect of all the other stars and other material making up the disk rather than one single massive invisible object

Thanks.
Excellent question!
I Have thought about three options as follow:
1. Very massive Object
Let's assume that it should be at least 100 times the size of the sun mass. (Some sort of black hole). In this case, you are fully correct – it should produce noticeable gravitational effects on other stars. Hence, it is expected that all stars in our neighborhood should also orbit this object. So, all of those stars should behave as a planets for this object, and rotate at some sort of order around it. This isn't the case. We know that each star in the solar neighborhood has its own random velocity and direction. Therefore this isn't a feasible solution.
2. Binary star–
https://en.wikipedia.org/wiki/Binary_star
"A binary star is a star system consisting of two stars orbiting around their common center of mass". In this case, the twin sun brother could be in the sun size, and it won't produce noticeable gravitational effects on other stars. However, it can't support the expected sinusoidal amplitude of the sun above and below the galactic plane. Therefore, it should also considered as none feasible solution.

3. CoM (Center of Mass)
https://en.wikipedia.org/wiki/Center_of_mass
" In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero or the point where if a force is applied causes it to move in direction of force without rotation. The distribution of mass is balanced around the center of mass and the average of the weighted position coordinates of the distributed mass defines its coordinates. Calculations in mechanics are often simplified when formulated with respect to the center of mass."
" In the case of a distribution of separate bodies, such as the planets of the Solar System, the center of mass may not correspond to the position of any individual member of the system."
" The center of mass is a useful reference point for calculations in mechanics that involve masses distributed in space, such as the linear and angular momentum of planetary bodies and rigid body dynamics. In orbital mechanics, the equations of motion of planets are formulated as point masses located at the centers of mass."
So, the idea is that each star in the Solar neighborhood has its own CoM. However, the CoM mass and location for each star is a total product of any individual mass star and location in the neighborhood (and also outside the nearby aria). Therefore, it meets the following main criteria: A. It is massive - Therefore is can support the requested Sun amplitude. B. As each star has its own CoM, Each star can rotate at random velocity and direction around its own CoM. C. As the majority of stars are located at the galactic plane, it is quite clear that also the CoM of each star is located at the galactic plane.
I assume that it isn't easy task for us to calculate the exact CoM for each star. However, the nature can do it automatically without any help from our computers.
Hence, do you agree that CoM could be a feasible solution?

cjameshuff
2015-Sep-28, 12:19 PM
The problem is that if there was such an object it would need to be very massive, easily large enough to produce noticeable gravitational effects on other stars, particularly the ones closer to it than the Sun. The fact that we don't see such gravitational effects is very strong evidence that the sinusoidal motion of the Sun through the galactic disk is due to the overall gravitational effect of all the other stars and other material making up the disk rather than one single massive invisible object

Specifically, it'd need to be a black hole of around 350000 solar masses, if you were to assume Newtonian gravitation. But if you were to assume Newtonian gravitation, the disk of the galaxy would have to not exist, because its gravitation would be sufficient to cause the bobbing motion, and you are trying to cause it with a point mass instead. You also need a separate black hole for nearly every star in the now-nonexistent disk of the galaxy, and for each to mysteriously not affect other stars that are much nearer. And it still wouldn't more than vaguely resemble the motions predicted from the actual application of Newtonian gravitation to the system of masses that are actually there. It's safe to say that this idea has nothing to do with reality...it's such a mess of contradictions and misunderstandings that it's not even wrong.

cjameshuff
2015-Sep-28, 12:28 PM
I assume that it isn't easy task for us to calculate the exact CoM for each star. However, the nature can do it automatically without any help from our computers.
Hence, do you agree that CoM could be a feasible solution?

The center of mass of the galaxy does not jump around the galaxy for each star, only a spherically symmetric body can be approximated as a point mass, and a spiral galaxy is not spherically symmetric, so no, it is not a feasible solution.

mkline55
2015-Sep-28, 12:54 PM
Does the bobbing motion of the Sun above and below the galactic plane also cause the Sun to oscillate in and out from the galactic center? Sorry to jump in late, but I just caught up with this thread, and I didn't see any mention of whether this effect occurs. Also, if there is an in-out oscillation, at what points in the bobbing cycle would it be nearest to and farthest from the galactic center? I.e. would the Sun be nearest the galactic center when it is most distant from the galactic plane, when it is centered in the galactic plane, or somewhere in between? I think I could create an analog model using a flat plane, but the added complexity of having a gaseous type of plane . . .

DaveC426913
2015-Sep-28, 01:05 PM
Does the bobbing motion of the Sun above and below the galactic plane also cause the Sun to oscillate in and out from the galactic center?No.

DaveC426913
2015-Sep-28, 01:11 PM
Thanks.
Excellent question!

He did not actually ask a question. He pointed out the flaw in your idea.


I Have thought about three options as follow:
1. Very massive Object
2. Binary star–
3. CoM (Center of Mass)

4. Sun is behaving exactly as it expected to do in the gravitational pull of a disclike galaxy.


Dave Lee, you are inventing solutions for fictional problems. This is the Astronomy Q&A section, not the ATM section.

mkline55
2015-Sep-28, 01:15 PM
No.

That's surprising. So, aside from the elliptic nature of its orbit around the galactic center, the Sun has varying velocities as it bobs above and below the plane and still maintains the same distance from the center.

DaveC426913
2015-Sep-28, 01:20 PM
That's surprising. So, aside from the elliptic nature of its orbit around the galactic center, the Sun has varying velocities as it bobs above and below the plane and still maintains the same distance from the center.
Why would it bob in and out though (I mean, beyond its elliptical orbit)?
There's a simple mechanism for its elliptical planar orbit, and there's a simple mechanism for its vertical bobbing. What mechanism would cause and in-out bobbing?

mkline55
2015-Sep-28, 01:24 PM
Why would it bob in and out though (I mean, beyond its elliptical orbit)?
There's a simple mechanism for its elliptical planar orbit, and there's a simple mechanism for its vertical bobbing. What mechanism would cause and in-out bobbing?

I assumed it would accelerate toward the plane. I expected that to increase its orbital velocity relative to the center of the galaxy as well, and as a result, effect it's distance from the center.

Hornblower
2015-Sep-28, 02:30 PM
I assumed it would accelerate toward the plane. I expected that to increase its orbital velocity relative to the center of the galaxy as well, and as a result, effect it's distance from the center.
No, the gravitational acceleration toward the disk plane is perpendicular to that toward the core of the galaxy.

mkline55
2015-Sep-28, 02:58 PM
No, the gravitational acceleration toward the disk plane is perpendicular to that toward the core of the galaxy.

Understood. The orbital velocity is also perpendicular to the core. I am surprised that adding an acceleration perpendicular to the plane would not also increase the total velocity or distance relative to the center. Simple vector math would indicate that the total velocity and/or distance relative to the center should be varying, What cancels out the vectors? As the Sun moves above from the plane, does it also decelerate in its orbital direction to maintain a constant distance and velocity, as if it were moving along the surface of an imaginary sphere?

Hornblower
2015-Sep-28, 04:12 PM
Understood. The orbital velocity is also perpendicular to the core. I am surprised that adding an acceleration perpendicular to the plane would not also increase the total velocity or distance relative to the center. Simple vector math would indicate that the total velocity and/or distance relative to the center should be varying, What cancels out the vectors? As the Sun moves above from the plane, does it also decelerate in its orbital direction to maintain a constant distance and velocity, as if it were moving along the surface of an imaginary sphere?
In a rigorous, high precision calculation, yes indeed there will be some disturbance of the forward speed around the orbit because the vertical gravity component will not always be exactly perpendicular to the Sun's forward velocity vector. However, the angle of inclination of the Sun's path as it crosses the plane is small, so the magnitude of the horizontal component will be a very small fraction of the magnitude of the vertical component. My previous remarks were for rough estimates in a model in which the actual mass distribution in the galaxy is uncertain. In a face-on view of the Sun's orbit I would expect some small periodic departures from a reference circle or ellipse with a period equal to that of the up and down bobbing, but with a much smaller magnitude.

Clear as mud? When I post in a hurry in a forum like this, I acknowledge that my technical writing probably is not as clear as it should be. It would be so much easier to have us at a desk where I could make sketches on the fly as needed to illustrate the vector combinations. A picture can easily be worth a thousand words or more.

Hornblower
2015-Sep-28, 04:23 PM
In a rigorous, high precision calculation, yes indeed there will be some disturbance of the forward speed around the orbit because the vertical gravity component will not always be exactly perpendicular to the Sun's forward velocity vector. However, the angle of inclination of the Sun's path as it crosses the plane is small, so the magnitude of the horizontal component will be a very small fraction of the magnitude of the vertical component. My previous remarks were for rough estimates in a model in which the actual mass distribution in the galaxy is uncertain. In a face-on view of the Sun's orbit I would expect some small periodic departures from a reference circle or ellipse with a period equal to that of the up and down bobbing, but with a much smaller magnitude.

Clear as mud? When I post in a hurry in a forum like this, I acknowledge that my technical writing probably is not as clear as it should be. It would be so much easier to have us at a desk where I could make sketches on the fly as needed to illustrate the vector combinations. A picture can easily be worth a thousand words or more.

Addendum: My bold. I saw a reference earlier in this thread which described the Sun's path as inclined some 25 degrees from the disk plane. It is my understanding that this is reckoned in a rotating frame of reference that is matching the averaged motions of the stars in our vicinity. The path as reckoned in an inertial frame of reference should be inclined by a much smaller angle.

ngc3314
2015-Sep-28, 07:12 PM
Does the bobbing motion of the Sun above and below the galactic plane also cause the Sun to oscillate in and out from the galactic center?

In principle such oscillations happen. A distributed mass (like a galaxy) produces orbits which may often be well described as epicyclic about a so-called guiding center (a fictitious point in a circular mean orbit). For a central point mass, the frequencies of these cyclic motions are the same radially and tangentially, giving the familiar elliptical closed path, while for a distributed mass they differ. So you end up with radial behavior which is like a rosette, while the "vertical" bobbing can have a different period so the motion even within such a rosette petal does not repeat if the mass distribution is itself flattened toward a disk plane. (For the really interested, the standard textbook description of this is in Binney and Tremaine's Galactic Dynamics). In our neighborhood, the amplitude of "vertical" velocities depends on stellar age, because such processes as gravitational scattering with other stars or giant clouds of interstellar gas can pump energy into stars over time.

cjameshuff
2015-Sep-28, 09:32 PM
If you are concerned with the varying distance from the center of the galaxy as the solar system bobs through the disk, you are working at a level of detail where it's probably not useful to keep adding ever-higher-order effects together. If there is negligible exchange of momentum with other matter, the sum of kinetic and potential energy will remain constant through the trajectory. Kinetic energy will be somewhat lower at the peaks, potential energy will increase with distance from the disk or from the core, and the non-uniformity of real galaxies means you're probably going to have to just build an overall mass distribution model and crunch numbers on that to get anything like realistic numbers at that level of detail.



Addendum: My bold. I saw a reference earlier in this thread which described the Sun's path as inclined some 25 degrees from the disk plane. It is my understanding that this is reckoned in a rotating frame of reference that is matching the averaged motions of the stars in our vicinity. The path as reckoned in an inertial frame of reference should be inclined by a much smaller angle.

It goes through 2.7 cycles in a galactic year with an amplitude of 100-150 light years as it travels along a circumference of about 170000 light years, so yes, the actual angle in the rest frame of the galaxy is much less than 25 degrees.

Jeff Root
2015-Sep-28, 10:29 PM
It goes through 2.7 cycles in a galactic year ...
So the number of cycles around the orbit is much less
than shown in the diagram in post #11.

-- Jeff, in Minneapolis

cjameshuff
2015-Sep-28, 10:47 PM
So the number of cycles around the orbit is much less
than shown in the diagram in post #11.

And lower in amplitude, yes. That's nothing more than a highly exaggerated representational diagram.

mkline55
2015-Sep-29, 12:18 PM
If you are concerned with the varying distance from the center of the galaxy as the solar system bobs through the disk, you are working at a level of detail where it's probably not useful to keep adding ever-higher-order effects together. If there is negligible exchange of momentum with other matter, the sum of kinetic and potential energy will remain constant through the trajectory. Kinetic energy will be somewhat lower at the peaks, potential energy will increase with distance from the disk or from the core, and the non-uniformity of real galaxies means you're probably going to have to just build an overall mass distribution model and crunch numbers on that to get anything like realistic numbers at that level of detail.




It goes through 2.7 cycles in a galactic year with an amplitude of 100-150 light years as it travels along a circumference of about 170000 light years, so yes, the actual angle in the rest frame of the galaxy is much less than 25 degrees.

Thanks. That describes the effect I was expecting. However, this is the Q&A section, and if the mainstream answer is, "you're probably going to have to just build an overall mass distribution model and crunch numbers", does that mean the model has not been created by someone already?

DaveC426913
2015-Sep-29, 01:02 PM
And lower in amplitude, yes. That's nothing more than a highly exaggerated representational diagram.
Lordy, yes. That was a quick Google to make a visual point. No idea of the pic's provenance.

Hornblower
2015-Sep-29, 01:35 PM
Thanks. That describes the effect I was expecting. However, this is the Q&A section, and if the mainstream answer is, "you're probably going to have to just build an overall mass distribution model and crunch numbers", does that mean the model has not been created by someone already?

My bold. It does not mean any such thing. Models have been works in progress over the past century, as astronomers have observed and analyzed the distribution and velocities of stars and interstellar gas in the galaxy.

Let me stress that our astronomers have not seen a star move around the center of the galaxy, let alone see a carousel-horse-like bobbing component in such motion. The elapsed time of human history is too short for that. The only motion-related data we have to go on are the radial velocity vectors of the stars at the present time, along with some transverse proper motions of nearby stars. We have inferred the mass distribution of the galaxy by other means, and by means of gravitational calculations based on that distribution we have predicted how the stars will move over tens and hundreds of millions of years.

mkline55
2015-Sep-29, 01:40 PM
My bold. It does not mean any such thing. Models have been works in progress over the past century, as astronomers have observed and analyzed the distribution and velocities of stars and interstellar gas in the galaxy.

Let me stress that our astronomers have not seen a star move around the center of the galaxy, let alone see a carousel-horse-like bobbing component in such motion. The elapsed time of human history is too short for that. The only motion-related data we have to go on are the radial velocity vectors of the stars at the present time, along with some transverse proper motions of nearby stars. We have inferred the mass distribution of the galaxy by other means, and by means of gravitational calculations based on that distribution we have predicted how the stars will move over tens and hundreds of millions of years.

Are any of these models accessible to the public for download?

Hornblower
2015-Sep-29, 01:51 PM
Here is a complete paper that popped up when I Googled "galaxy mass distribution". It was at the top of the list. It looks like pretty heavy reading. There are many others that I have not looked at. As always it is caveat lector. The ones that are short and easy to read may or may not be accurate.
http://www.ioa.s.u-tokyo.ac.jp/~sofue/htdocs/2013psss/sofue2013psss.pdf

mkline55
2015-Sep-29, 03:58 PM
Here is a complete paper that popped up when I Googled "galaxy mass distribution". It was at the top of the list. It looks like pretty heavy reading. There are many others that I have not looked at. As always it is caveat lector. The ones that are short and easy to read may or may not be accurate.
http://www.ioa.s.u-tokyo.ac.jp/~sofue/htdocs/2013psss/sofue2013psss.pdf

Nice paper. I appreciate the link. It has a lot about galaxy rotation curves, mass distribution, measurement techniques, various theories, etc., but having read through the entire paper front to back, I found no mention of the motion I asked about, i.e. variations in orbital velocity for individual stars as a result of the bobbing motion above and below the plane of the galaxy. I am not asking for anyone to perform a search for such a model/paper, as I can do that, but if anyone already knows of a good mainstream paper describing that particular aspect of motion, I'd like to know.

kzb
2015-Sep-29, 04:47 PM
Bear in mind that the circular velocity at our galactic radius is 239 km/s according to a recent estimate. The peculiar motions of the sun and the stars around us are only in the region of 10 km/s, very small in comparison.

In fact, the peculiar motion of the sun seems to be a major confounding factor in establishing just what is the average circular velocity. Apparently the solar neighbourhood is going slightly faster than average for our radius, but on top of that, the sun is going slightly faster than the neighbourhood.

On the motion perpendicular to the disk plane, it makes a difference to the expected motion depending on whether you consider the mass is concentrated in the disk plane (i.e most mass is visible matter) or if you think most of the mass is dark matter distributed as a spheroid. There's been quite a controversy over this in the literature. See this link for an example:

On local dark matter density
C. Moni Bidin et al (Nov 2014)

http://arxiv.org/abs/1411.2625

mkline55
2015-Sep-29, 05:29 PM
I found this paper (http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1998MNRAS.298..387D&data_type=PDF_H IGH&whole_paper=YES&type=PRINTER&filetype=.pdf) which does take into account for the (here's where the right name for this motion would be useful) motion. "It's complicated" is almost an understatement. Add to that the earlier notes from others about the limited observational data and the likely small contribution to overall motion, and I'm thinking of taking DaveC's first response of "No" as the easiest way out.

cosmonut
2015-Sep-29, 07:17 PM
It's fairly easy to draw up a simple model of the sun's motion around the milky way using two component motions. Everything inside the sun's orbit can be regarded as a point mass at the center of the galaxy and everything outside the sun's orbit can be regarded as a planar mass through which the sun oscillates. It's simple Newtonian physics. Keep in mind that the sun's sinusoidal motion around the galaxy, its period and its amplitude, has never been observed. It's only been predicted through modeling and simulations.

Jeff Root
2015-Sep-30, 02:42 AM
Even without knowing the mass of the galaxy or density of
matter in the disk, an average amplitude of the "up-down"
motion can be inferred from the distribution of stars across
the thickness of the disk. By comparing the velocities, the
amplitudes of specific stars could be found without having
to calculate forces.

-- Jeff, in Minneapolis

Cougar
2015-Sep-30, 12:23 PM
...and everything outside the sun's orbit can be regarded as a planar mass through which the sun oscillates. It's simple Newtonian physics.

I believe this is contrary to simple Newtonian physics. Perhaps you could explain.

mkline55
2015-Sep-30, 01:18 PM
I think you can find an interesting description of the Newtonian physics here (http://arxiv.org/ftp/arxiv/papers/0903/0903.1962.pdf). I am not certain how well accepted this paper was, and I'm certainly not the person to check the math.

DaveC426913
2015-Sep-30, 01:19 PM
I believe this is contrary to simple Newtonian physics. Perhaps you could explain.
What do you mean? If you construct a planar mass, you can demonstrate how Newtonian gravity will operate in its vicinity. It operates perpendicular to the plane.


I think you can find an interesting description of the Newtonian physics here (http://arxiv.org/ftp/arxiv/papers/0903/0903.1962.pdf). I am not certain how well accepted this paper was, and I'm certainly not the person to check the math.
Let's be clear though - there's nothing exotic about the Newtonian gravity in the vicinity of a planar mass.

mkline55
2015-Sep-30, 01:39 PM
Let's be clear though - there's nothing exotic about the Newtonian gravity in the vicinity of a planar mass.

Right. It's just that you cannot treat a disc the same as a sphere.

Hornblower
2015-Sep-30, 01:42 PM
Right. It's just that you cannot treat a disc the same as a sphere.
Experts doing the math correctly do not do any such thing. Either way is an exercise in calculus.

Hornblower
2015-Sep-30, 05:30 PM
I found this paper (http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1998MNRAS.298..387D&data_type=PDF_HIGH&whole_paper=YES&type=PRINTER&filetype=.pdf) which does take into account for the (here's where the right name for this motion would be useful) motion. "It's complicated" is almost an understatement. Add to that the earlier notes from others about the limited observational data and the likely small contribution to overall motion, and I'm thinking of taking DaveC's first response of "No" as the easiest way out.
If there was any mention of a periodic departure from uniform forward motion around the ellipse in an idealized, uniform bulge and disk combination, I did not recognize it. Can you point us to whatever you are talking about? They appeared to be interested in perturbations from mass concentrations such as spiral arms and a central bar. I would expect these effects in a real galaxy to swamp the small variation you are concerned with in an idealized smooth model, and to be a whole lot more consequential and interesting.

mkline55
2015-Sep-30, 05:56 PM
If there was any mention of a periodic departure from uniform forward motion around the ellipse in an idealized, uniform bulge and disk combination, I did not recognize it. Can you point us to whatever you are talking about? They appeared to be interested in perturbations from mass concentrations such as spiral arms and a central bar. I would expect these effects in a real galaxy to swamp the small variation you are concerned with in an idealized smooth model, and to be a whole lot more consequential and interesting.

You would have to add that particular function to what is already in the paper I linked. The point was that galactic gravitational-rotational motion is usually treated as equal to (exactly equal to) motion around a point at the center of the galaxy. The paper shows that a disk-shaped galaxy rotation as a sphere, and cannot be treated as such. It's precisely because anything outside the plane of the galactic disk is subject to different gravitational pull than just "treat the mass of the entire galaxy as if it were a point around which everything orbits."

ngc3314
2015-Sep-30, 06:01 PM
The point was that galactic gravitational-rotational motion is usually treated as equal to (exactly equal to) motion around a point at the center of the galaxy.


I have to call for references here. There are textbooks loaded with detailed treatments of nonspherical galaxy potentials, and a massive research literature both analytic and numerical on stellar orbit in realistic galaxy gravitational fields. Where are galaxies usually treated as point masses, since the earliest numerical work circa 1972 had to use that approximation?

mkline55
2015-Sep-30, 06:13 PM
Which of those textbooks explicitly describe all 3-dimensions of motions of stars including multi-dimensional velocity perturbations related to the bobbing motions through the disk?

mkline55
2015-Sep-30, 06:39 PM
I have to call for references here. There are textbooks loaded with detailed treatments of nonspherical galaxy potentials, and a massive research literature both analytic and numerical on stellar orbit in realistic galaxy gravitational fields. Where are galaxies usually treated as point masses, since the earliest numerical work circa 1972 had to use that approximation?

I may be mistaken, but I thought I recalled being told/taught that gravity for any object could be treated the same as an equivalent point mass located at the center of gravity. Perhaps I misunderstood.

antoniseb
2015-Sep-30, 06:46 PM
I may be mistaken, but I thought I recalled being told/taught that gravity for any object could be treated the same as an equivalent point mass located at the center of gravity. Perhaps I misunderstood.
You misunderstood. If that was said to you it was an oversimplification. As a thought experiment, imagine two spheres orbiting each other. some object very close to them will not experience uniform attraction regardless of placement given the same distance to the center of mass. A test mass close to one and distant from the other will be drawn to the close one. Now tie those two objects together with a thread. They are now one object, with the same non-uniform gravity.

Shaula
2015-Sep-30, 06:53 PM
I may be mistaken, but I thought I recalled being told/taught that gravity for any object could be treated the same as an equivalent point mass located at the center of gravity. Perhaps I misunderstood.
Only really true of spherically symmetric (or effectively so) objects.

Jeff Root
2015-Sep-30, 07:31 PM
The gravity of a nonspherical object can be treated as a point
source if the test object is sufficiently far away. In this case
the test object is *inside* the nonspherical gravitating object!

-- Jeff, in Minneapolis

chornedsnorkack
2015-Sep-30, 08:11 PM
How big is the collective velocity of solar neighbourhood relative to circular orbit? And what is its direction?

ngc3314
2015-Sep-30, 08:40 PM
Which of those textbooks explicitly describe all 3-dimensions of motions of stars including multi-dimensional velocity perturbations related to the bobbing motions through the disk?

One that comes to mind is Binney and Tremaine, Galactic Dynamics.

Amber Robot
2015-Sep-30, 09:21 PM
Keep in mind that the sun's sinusoidal motion around the galaxy, its period and its amplitude, has never been observed. It's only been predicted through modeling and simulations.

We're in the process of observing it.

cjameshuff
2015-Sep-30, 09:59 PM
We're in the process of observing it.

Pretty sure the acceleration isn't something we can reasonably expect to observe directly...tens of km/s over millions of years. We can at best observe it indirectly in a statistical sense by looking at the distribution of velocities in face-on galaxies.

However, it's a prediction based on models of gravity that are tested in great detail against observation. It's not in the slightest bit controversial or poorly understood, it's a very straightforward result of Newtonian gravity with a non-spherical mass distribution, and for it not to happen would require something very strange to be going on.

DaveC426913
2015-Oct-01, 02:17 AM
Right. It's just that you cannot treat a disc the same as a sphere.
You don't need to. Newtonian physics applies quite nicely to any mass (that isn't super-massive).


The gravity of a nonspherical object can be treated as a point
source if the test object is sufficiently far away. In this case
the test object is *inside* the nonspherical gravitating object!


You guys are oversimplifying Newtonian physics wrt gravity.

Any object of any shape can be considered as a sum of its component parts (such as atoms), each contributing their own tiny pull in their own direction.

Newtonian gravitation as it applies to bodies with spatial extent:
https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation#Bodies_wit h_spatial_extent

True, there is one special case, where an object distant enough can be considered a point - which is what you guys are saying, but that is just one tiny facet.

In the case of a galactic plane, you can satisfy yourself what the gravity will be at any point in or around the plane by summing up the pulls of individual stars. And there's a generalized formula for it if you don't want to do the calcs.

Dave Lee
2015-Oct-01, 02:37 AM
Does the bobbing motion of the Sun above and below the galactic plane also cause the Sun to oscillate in and out from the galactic center? Sorry to jump in late, but I just caught up with this thread, and I didn't see any mention of whether this effect occurs. Also, if there is an in-out oscillation, at what points in the bobbing cycle would it be nearest to and farthest from the galactic center? I.e. would the Sun be nearest the galactic center when it is most distant from the galactic plane, when it is centered in the galactic plane, or somewhere in between? I think I could create an analog model using a flat plane, but the added complexity of having a gaseous type of plane . . .

Let's set a brief verification:
The available info is as follow:
Sun’s path inclined to the plane of the galaxy - about 25 degrees
Amplitude – 230 Ly
Calculation:
Vertical Amplitude vector = 230 x 0.423 = 97 Ly
Horizontal Amplitude vector = 230 x 0.906 = 208 Ly
Hence,
The bobbing motion amplitude of the Sun (above and below the galactic plane) = 97 Ly
In-out oscillation Amplitude, (nearest to and farthest from the galactic center) = 208 Ly
This is a significant motion.
Therefore, I can understand the following question:

I assumed it would accelerate toward the plane. I expected that to increase its orbital velocity relative to the center of the galaxy as well, and as a result, effect it's distance from the center.

So far the only real example which had been used to explain this phenomenon was – Saturn ring:

Look instead, in the solar sytem, for gravity-bound systems with massive numbers of members.

http://saturn.jpl.nasa.gov/news/newsreleases/newsrelease20101101/
It is stated:
"Cassini Sees Saturn Rings Oscillate Like Mini-Galaxy"
"In this image, Cassini's narrow angle camera captured a 1,200-kilometer-long (750-mile-long) section arcing along the outer edge of the B ring. Here, vertical structures tower as high as 2.5 kilometers (1.6 miles) above the plane of the rings -- a significant deviation from the vertical thickness of the main A, B and C rings, which is generally only about 10 meters (about 30 feet)." "
However, this outer edge of the B ring represents less than 0.001% of the whole ring aria. It is just one picture and it's not fully clear what going inside of this outer edge of the B ring. As the ring is mainly based on broken rocks and stones which might collide with each other, than it could create some abnormal phenomenon.
So, how can we compare those collided broken rocks to The Sun and actually to all the stars in the galaxy? How can we compare max amplitude of 2.5 km to 97 Ly. Please also be aware that there is no vertical Amplitude in this picture – So there is no explanation for the Sun' In-out oscillation Amplitude of 208 Ly.
Even so, we see no oscillation at 99.99% of the whole Saturn ring. So technically, we could use this kind of example to explain some abnormal activity at few stars in the whole galaxy.
But we know that all the stars in the arm disc have some kind of oscillation. Therefore – this example is just irrelevant to our discussion.
Do you have better example?

Reality Check
2015-Oct-01, 03:35 AM
...So, how can we compare those collided broken rocks to The Sun and actually to all the stars in the galaxy? How can we compare max amplitude of 2.5 km to 97 Ly.
No problem there, Dave Lee.
We have a big system that has scales of light years.
We have a similar smaller system with the same physics that has scales of kilometers.
The ice particles (not "broken rocks") in Saturn's rings (https://en.wikipedia.org/wiki/Rings_of_Saturn) in Cassini Sees Saturn Rings Oscillate Like Mini-Galaxy (http://saturn.jpl.nasa.gov/news/newsreleases/newsrelease20101101/) oscillate as we expect for stars in the Milky Way , e.g. the Sun. The main difference is the scale of the system - light years versus kilometers.

The B Ring (https://en.wikipedia.org/wiki/Rings_of_Saturn#B_Ring) is more than 0.01% of the rings and is the first place that the oscillations have been measured (in 2010). They may be observed elsewhere in the future.

There is no claim that the dynamics of the rings of Saturn are exactly the dynamics of stars in the Milky Way. For example there are also oscillations of the B Ring caused by perturbations from the moon Mimas. Nice video here: Oscillations at B Ring Edge (http://saturn.jpl.nasa.gov/video/videodetails/?videoID=218)

Dave Lee
2015-Oct-01, 05:07 AM
No problem there, Dave Lee.
We have a big system that has scales of light years.
We have a similar smaller system with the same physics that has scales of kilometers.
The ice particles (not "broken rocks") in Saturn's rings (https://en.wikipedia.org/wiki/Rings_of_Saturn) in Cassini Sees Saturn Rings Oscillate Like Mini-Galaxy (http://saturn.jpl.nasa.gov/news/newsreleases/newsrelease20101101/) oscillate as we expect for stars in the Milky Way , e.g. the Sun. The main difference is the scale of the system - light years versus kilometers.

The B Ring (https://en.wikipedia.org/wiki/Rings_of_Saturn#B_Ring) is more than 0.01% of the rings and is the first place that the oscillations have been measured (in 2010). They may be observed elsewhere in the future.

There is no claim that the dynamics of the rings of Saturn are exactly the dynamics of stars in the Milky Way. For example there are also oscillations of the B Ring caused by perturbations from the moon Mimas. Nice video here: Oscillations at B Ring Edge (http://saturn.jpl.nasa.gov/video/videodetails/?videoID=218)

Yes, the B ring is more than 0.01%. However, It is stated clearly that this activity take place at the edge of B ring. Please look again. This edge should be less than 0.001% than the total A, B,C ring aria.
So, all of this discussion is about the edge of ring B which is less than 0.001%.
However, with regards to the Oscillation:
In the description of the image it is not stated that there is a proof for up-down and in-out oscillation. The scientists didn't set any real measurement for those "ice particles". They also didn't claim that there is a solid proof for any oscillation in any of the Ice particles. Actually, there is no way to verify any sort of oscillation just by looking at the image. How can we distinguish between the ice particles? How can we see any oscillation in this image?
So, the scientists do not claim that they see any real oscillation. They actually discuss about a rotating wave pattern – which is the reality!
However, this rotating wave pattern had been suddenly transformed to Oscillation (or oscillations):
"Now, analysis of thousands of Cassini images of the B ring taken over a four-year period has revealed the source of most of the complexity: at least three additional, independently rotating wave patterns, or oscillations, that distort the B ring's edge".
And from this moment – instead of using "rotating wave patter", they start to call it oscillation.
Sorry – I do not see any proof for real oscillation in those "Ice particles" (Up-down and especially in-out).
I don't think that 0.001% of ring edge which we see in an image can give us any real knowledge about the Milky way.
I still think that there is some gap between 2.5Km and 97 Ly (which is 917,671,000,000,000 Km.)
In any case, those "Ice particles" are located in a ring which is obeyed to Newton gravity law. So if we compare the sun to colided Ice particle, we should also compare ring to ring. Hence, do we claim that the Sun rotation ring should also obey to Newton law?
However, beside those collided ice particles – do we have any better example?

cjameshuff
2015-Oct-01, 11:41 AM
Let's set a brief verification:
The available info is as follow:
Sun’s path inclined to the plane of the galaxy - about 25 degrees
Amplitude – 230 Ly
Calculation:
Vertical Amplitude vector = 230 x 0.423 = 97 Ly
Horizontal Amplitude vector = 230 x 0.906 = 208 Ly

No. If the amplitude of the oscillation is 230 light years, the vertical component of the oscillation is 230 light years. The oscillation is vertical, not at 25 degrees to the plane.



So far the only real example which had been used to explain this phenomenon was – Saturn ring:

No. There is every observation of objects moving as predicted by Newtonian gravity, and our continued success at sending spaceprobes where we want them to go, even when it requires extremely non-circular orbits.



So, how can we compare those collided broken rocks to The Sun and actually to all the stars in the galaxy?

Gravity doesn't behave differently for stars.

Again, you are arguing against extremely basic and well-established physics, and inventing problems that simply don't exist in reality. I really have no idea why you are so fixated on this...there is absolutely no physical basis for your attempt at modeling motions as a composition of circular motions, and it was a technique that was abandoned centuries ago because it failed even in a system vastly simpler than the galaxy. There is nothing in the slightest bit questionable about the vertical oscillation of stars in a disk shaped galaxy. There is no conflict with a rule requiring circular motion because there is no rule requiring circular motion.

Cougar
2015-Oct-01, 11:58 AM
What do you mean? If you construct a planar mass, you can demonstrate how Newtonian gravity will operate in its vicinity. It operates perpendicular to the plane.

The full quote I was questioning is this:


Everything inside the sun's orbit can be regarded as a point mass at the center of the galaxy and everything outside the sun's orbit can be regarded as a planar mass through which the sun oscillates.

Well, first of all, "everything outside the sun's orbit" is not matter "through which the sun oscillates" simply because it's outside the sun's orbit -- the sun doesn't get there. I'm not questioning the sun's minor vertical oscillation during its galactic orbit.

Second, anytime I hear about mass exterior to a body's orbit and its potential gravitational effect on that orbit, I am reminded of Newton's Principia wherein he proved geometrically, in 1687!, that such exterior mass results in no gravitational effect in the case of a spherically symmetric scenario. OK, the Milky Way is not spherically symmetric; however, as a simplified model on the large scale, I should think it could be considered as a flattened sphere that has 2-dimensional symmetry (instead of 3-D). Exactly what that dimensional collapse does to Newton's proof, I'm not sure. But I note that the situation is quite similar with respect to mass exterior to an orbit, i.e., the exterior mass on the near side of an orbiting body will have a certain effect due to its close proximity, but there's much more exterior mass on the far side of the body's orbit. These are counteracting effects, and the resulting gravity is largely cancelled, as in the spherical symmetric case.

cjameshuff
2015-Oct-01, 12:23 PM
The full quote I was questioning is this:



Well, first of all, "everything outside the sun's orbit" is not matter "through which the sun oscillates" simply because it's outside the sun's orbit -- the sun doesn't get there. I'm not questioning the sun's minor vertical oscillation during its galactic orbit.

Second, anytime I hear about mass exterior to a body's orbit and its potential gravitational effect on that orbit, I am reminded of Newton's Principia wherein he proved geometrically, in 1687!, that such exterior mass results in no gravitational effect in the case of a spherically symmetric scenario. OK, the Milky Way is not spherically symmetric; however, as a simplified model on the large scale, I should think it could be considered as a flattened sphere that has 2-dimensional symmetry (instead of 3-D). Exactly what that dimensional collapse does to Newton's proof, I'm not sure. But I note that the situation is quite similar with respect to mass exterior to an orbit, i.e., the exterior mass on the near side of an orbiting body will have a certain effect due to its close proximity, but there's much more exterior mass on the far side of the body's orbit. These are counteracting effects, and the resulting gravity is largely cancelled, as in the spherical symmetric case.

The exterior mass in a disk does not fully cancel. Remember why the Ringworld is unstable?

Jeff Root
2015-Oct-01, 12:54 PM
Cougar,

I agree with Cjameshuff: The gravitational forces from
masses exterior to the orbit do not entirely cancel if they are
not spherically symmetric. In this case cosmonut is talking
only about the forces pulling the Sun toward the central plane
of the galaxy. He is saying that the Sun oscillates through
that plane. The fact that the matter is entirely outside the
Sun's orbit (the way he has divided it up for convenience)
is pretty much irrelevant. I don't think his way of dividing
it up will give a very accurate result, but for describing the
overall effects, it seems not too bad. He's describing the
part of the galaxy interior to the Sun's orbit as being like a
point source of gravity at the galaxy's center, causing the
Sun's circular motion, and the part of the galaxy exterior
to the Sun's orbit as being like a planar source of gravity,
causing the Sun's bobbing motion. Crude, but effective.

-- Jeff, in Minneapolis

Dave Lee
2015-Oct-01, 01:10 PM
No. If the amplitude of the oscillation is 230 light years, the vertical component of the oscillation is 230 light years. The oscillation is vertical, not at 25 degrees to the plane.
It is stated:
http://www.americanscientist.org/lib...101154_306.pdf
"The sun’s path is inclined about 25 degrees to the plane of the galaxy and is headed toward a region in the constellation of Hercules near its border with Lyra."
"The sun oscillates through the plane of the galaxy with an amplitude of about 230 lightyears, crossing the plane every 33 million years."
As you can see the Sun' path is inclined about 25 degrees to the plane of the galaxy.
So, what does it mean?
What about the horizontal vector?

DaveC426913
2015-Oct-01, 01:48 PM
It is stated:
http://www.americanscientist.org/lib...101154_306.pdf
"The sun’s path is inclined about 25 degrees to the plane of the galaxy and is headed toward a region in the constellation of Hercules near its border with Lyra."
"The sun oscillates through the plane of the galaxy with an amplitude of about 230 lightyears, crossing the plane every 33 million years."
As you can see the Sun' path is inclined about 25 degrees to the plane of the galaxy.
So, what does it mean?
What about the horizontal vector?

The definition of a wave's amplitude is its vertical magnitude, described independently of any horizontal component ( i.e. wavelength).

The sun's oscillation has a vertical component of 230ly, or 115ly above and below the galactic plane.

Dave Lee
2015-Oct-01, 02:15 PM
The definition of a wave's amplitude is its vertical magnitude, described independently of any horizontal component ( i.e. wavelength).

The sun's oscillation has a vertical component of 230ly, or 115ly above and below the galactic plane.

Wow.
In this case,
The Horizontal magnitude is:
230 x 2.144 = 493 LY
While the total Sun magnitude is: 544 Ly
That is amazing.
Do you agree with this calculation?

mkline55
2015-Oct-01, 03:16 PM
Wow.
In this case,
The Horizontal magnitude is:
230 x 2.144 = 493 LY
While the total Sun magnitude is: 544 Ly
That is amazing.
Do you agree with this calculation?

I don't if you are using the 25 degree angle. The angle would should over time.

DaveC426913
2015-Oct-01, 03:23 PM
Wow.
In this case,
The Horizontal magnitude is:
230 x 2.144 = 493 LY
While the total Sun magnitude is: 544 Ly
That is amazing.
Do you agree with this calculation?
No.
The magnitude of the oscillation is 230ly. Where do you get 544 from?
And you don't have a "horizontal magnitude", you have a wavelength or period of oscillation. I'm not sure where the 2.144 comes from.

Dave Lee
2015-Oct-01, 03:45 PM
No.
The magnitude of the oscillation is 230ly. Where do you get 554 from?
And you don't have a "horizontal magnitude", you have a period of oscillation. I'm not sure where the 2.144 comes from.

O.K.
If the Sun path was incline at 90 degrees to the plane of the galaxy, than the Vertical amplitude is by definition - the total amplitude.
In this case, the horizontal amplitude is zero!
However, it is stated:
"The sun’s path is inclined about 25 degrees to the plane of the galaxy and is headed toward a region in the constellation of Hercules near its border with Lyra."
Therefore, there must be vertical amplitude and horizontal Amplitude.
As we have the Vertical amplitude and we have the alpha
Than
The Total amplitue = Vertical amplitude / (Sin-alpha)
The Horizontal amplitude = Total amplitude x (cos-alpha) or Vertical amplitude x (Cotan-alpha)
This is basic.
Do you agree with that explanation?

Amber Robot
2015-Oct-01, 04:00 PM
Pretty sure the acceleration isn't something we can reasonably expect to observe directly...tens of km/s over millions of years.

I didn't say we were done observing it. I'll let your descendents know to keep taking measurements and they can let their descendents know....

Hornblower
2015-Oct-01, 05:11 PM
O.K.
If the Sun path was incline at 90 degrees to the plane of the galaxy, than the Vertical amplitude is by definition - the total amplitude.
In this case, the horizontal amplitude is zero!
However, it is stated:
"The sun’s path is inclined about 25 degrees to the plane of the galaxy and is headed toward a region in the constellation of Hercules near its border with Lyra."
Therefore, there must be vertical amplitude and horizontal Amplitude.
As we have the Vertical amplitude and we have the alpha
Than
The Total amplitue = Vertical amplitude / (Sin-alpha)
The Horizontal amplitude = Total amplitude x (cos-alpha) or Vertical amplitude x (Cotan-alpha)
This is basic.
Do you agree with that explanation?

No, no, a thousand times no. As explained earlier, this motion attributed to the Sun is kinematically reckoned in a moving frame of reference in which the velocity vectors of the nearby stars average out to zero. This is calculated from the astrometrically observed transverse proper motions and the spectroscopically observed radial velocities. From these observations we have no means of distinguishing the paths of these stars from straight lines. By themselves they tell us nothing about how the stars will move over millions of years in response to the galaxy's gravity field. To relate these observations to galactic gravity we need to transform the reckoning of the motions to an inertial frame of reference in which the galactic center is stationary, and ascertain the galaxy's mass distribution by other means. Only then can we predict how a given star will move over a very long period.

It appears to me that you have nowhere near enough understanding of this topic for further dialogue in a forum like this to be of much value. There is no disgrace in that. We all have to start somewhere, and for you and to some extent for me it is at mathematically elementary levels far below that of the techniques used analytically in the papers linked in this thread. What is important is that we trust the experts while we strive to improve our understanding by striving to wrap our heads around simplified models in which a rough idea of the gravitational effects on a star's motion can be visualized. I and some other well-informed posters have done just that.

Jeff Root
2015-Oct-01, 05:57 PM
It is stated:
http://www.americanscientist.org/lib...101154_306.pdf
"The sun’s path is inclined about 25 degrees to the
plane of the galaxy and is headed toward a region in
the constellation of Hercules near its border with Lyra."

"The sun oscillates through the plane of the galaxy
with an amplitude of about 230 lightyears, crossing
the plane every 33 million years."

As you can see the Sun' path is inclined about 25
degrees to the plane of the galaxy.
So, what does it mean?
What about the horizontal vector?
Your link to the article wasn't made correctly, and I'm
not going to try to search for it myself.

"The Sun's path" is an approximately circular orbit
modified by a sine wave. The Sun is basically orbiting
around the center of the galaxy, and slowly bobbing
up and down as it orbits. Currently the Sun is headed
in the direction of the constellation Hercules.

As it moves along the "path", the direction it is headed
changes. As it nears the peak of the sine wave, the
angle decreases. When it reaches the peak, it will be
headed parallel to the plane of the galaxy. The angle
will be zero degrees. Then the direction along the path
will gradually change so that the Sun is headed at an
increasingly steep angle into the plane of the galaxy in
the opposite direction. The maximum angle would be
something greater than 25 degrees, as the Sun passes
through the central plane of the galaxy.

The "horizontal vector" is, as always, that the Sun orbits
the center of the galaxy in about 230 million years, at
a speed of about 828,000 km/hr. The bobbing through
the galaxy's plane doesn't change that. The Sun's speed
along its "path" will increase and decrease slightly as it
bobs -- fastest when passing through the central plane
and slowest when it is farthest from the central plane.
But the overall "horizontal vector" isn't changed by that.

-- Jeff, in Minneapolis

mkline55
2015-Oct-01, 07:19 PM
The Sun's speed
along its "path" will increase and decrease slightly as it
bobs -- fastest when passing through the central plane
and slowest when it is farthest from the central plane.
But the overall "horizontal vector" isn't changed by that.

-- Jeff, in Minneapolis

If by "horizontal vector" you mean the distance from the galactic center that a non-bobbing star might have, then what counteracts the speed change to hold it at the same distance? By my limited understanding, the Sun would have the greatest velocity perpendicular to the plane of the disk at the time it is passing through the disk. Simply by vectors, the net velocity relative to the galactic center at that time is greater than that of another star which is not bobbing. That net velocity alone should be enough to push the Sun outward from the center, or else something must be preventing it. Similarly, at its greatest distance from the disk, the Sun has no velocity relative to the disk, and its net velocity would be at its lowest relative to the galactic center, resulting in a closer position. What prevents this from happening?

chornedsnorkack
2015-Oct-01, 07:39 PM
No, no, a thousand times no. As explained earlier, this motion attributed to the Sun is kinematically reckoned in a moving frame of reference in which the velocity vectors of the nearby stars average out to zero. This is calculated from the astrometrically observed transverse proper motions and the spectroscopically observed radial velocities.

If Sun´s speed were zero with respect to that average, does it need to mean lack of oscillations?
I see a reason why it should not.

Reality Check
2015-Oct-01, 08:36 PM
However, with regards to the Oscillation:
In the description of the image it is not stated that there is a proof for up-down and in-out oscillation.

Look at the video - the observation is of "in-out oscillations". The proof that the oscillations exist is observing the edge of the B Ring oscillating! Any claim that the oscillations do not exist is a fantasy, Dave Lee.
Read about the rings of Saturn - they are actually made of ice particles.
The scientists set a real measurement for the oscillations of the ice particles.
We can see oscillation in the video by opening our eyes and seeing the edge of the B Ring oscillating: http://saturn.jpl.nasa.gov/video/videodetails/?videoID=218

Yes - there is a gap between 2.5 km and 97 light years. That gap is 97 light years - 2.5 km :p! But this is comparing two different things
* The measurement of the edge of the B Ring oscillating in and out along the plane of Saturn's rings.
* An assertion about the amplitude of the Sun's oscillation up and down through the galaxy plane.

Cassini Sees Saturn Rings Oscillate Like Mini-Galaxy (http://saturn.jpl.nasa.gov/news/newsreleases/newsrelease20101101/) is giving knowledge about the dynamics of all gravitationally bound disk systems

"We have found what we hoped we'd find when we set out on this journey with Cassini nearly 13 years ago: visibility into the mechanisms that have sculpted not only Saturn's rings, but celestial disks of a far grander scale, from solar systems, like our own, all the way to the giant spiral galaxies," said Carolyn Porco, co-author on the new paper and Cassini imaging team lead, based at the Space Science Institute, Boulder, Colo.

The ice particles are located in a ring which obeys Newton's law of gravitation like all masses do.
Spiral galaxies have stars that are located in a "ring" (the outer part of the galaxy disk) which obeys Newton's law of gravitation like all masses do.
The edge of the B Ring spontaneously oscillate in and out.
We can expect the edge of some galaxy disks to also spontaneously oscillate in and out. The conditions seem to be that the edge of the disk is sharp and density of the disk high. Thus the "giant spiral galaxies" above - they are massive with sharp edges.

cjameshuff
2015-Oct-01, 09:29 PM
If by "horizontal vector" you mean the distance from the galactic center that a non-bobbing star might have, then what counteracts the speed change to hold it at the same distance? By my limited understanding, the Sun would have the greatest velocity perpendicular to the plane of the disk at the time it is passing through the disk. Simply by vectors, the net velocity relative to the galactic center at that time is greater than that of another star which is not bobbing. That net velocity alone should be enough to push the Sun outward from the center, or else something must be preventing it. Similarly, at its greatest distance from the disk, the Sun has no velocity relative to the disk, and its net velocity would be at its lowest relative to the galactic center, resulting in a closer position. What prevents this from happening?

There is no outward force to counter, greater velocity just means more momentum in the direction of motion. The sun accelerates as it falls toward the plane of the galaxy, deeper into the gravitational potential well of the galaxy. It will reach its highest velocity when it is as deep in the potential well as it gets, then start to rise out again on the other side, reaching its lowest velocity when it is furthest out...exactly as one would expect.

Hornblower
2015-Oct-01, 10:29 PM
If by "horizontal vector" you mean the distance from the galactic center that a non-bobbing star might have, then what counteracts the speed change to hold it at the same distance? By my limited understanding, the Sun would have the greatest velocity perpendicular to the plane of the disk at the time it is passing through the disk. Simply by vectors, the net velocity relative to the galactic center at that time is greater than that of another star which is not bobbing. That net velocity alone should be enough to push the Sun outward from the center, or else something must be preventing it. Similarly, at its greatest distance from the disk, the Sun has no velocity relative to the disk, and its net velocity would be at its lowest relative to the galactic center, resulting in a closer position. What prevents this from happening?

Here is my take after having plenty of time to think about it. Forget what I said in my earliest post on this detail, as I was thinking of rough approximations and taking this detail as negligible for small inclinations.

Suppose the Sun is orbiting exactly in the plane of an idealized uniform disk, and that the orbit is perfectly circular. Its speed will be constant and the orbit will remain circular.

Now let us deflect it into a new orbit that is inclined to the disk plane while keeping the speed unchanged, and for the sake of argument make the disk's mass vanishingly small compared to the spherical core and halo. The Sun will now stay in that new orbital plane and do one complete "bobbing up and down and back" cycle through the disk plane during one circuit of the orbit.

Now let us make the disk significantly massive as in a real galaxy. As the Sun rises above the crossing point (ascending node) the disk's gravity will attract it downward and make it fall back to the plane sooner than it would have if unperturbed. This is like the precession of an inclined satellite orbit around Earth because of the equatorial bulge, but proportionately faster because of the greater mass of the disk. This gives us the carousel-horse-like bobbing motion that is so grossly exaggerated in some popular media illustrations.

Now for the variation mkline55 is mentioning. When the Sun is above the plane and still rising, or below the plane and still falling, the disk's gravity will have a small retrograde component against the Sun's forward motion, slowing it down a bit. This should cause it to curve a bit more toward the galaxy's center, like an orbiting spacecraft when its retro rocket is fired. After the Sun levels off and starts approaching the plane, this perturbation becomes prograde, increasing the speed and making it veer out slightly. This should make a small wavy component superimposed on the reference circle, as viewed from a face-on position. For small inclination angles I would expect this effect to be very small and thus masked by the larger perturbations from the mass concentrations we would find in a real galaxy. Thus I would not expect research papers or textbooks to make a big deal about mentioning it.

Jeff Root
2015-Oct-02, 12:54 AM
One of the many differences between Saturn's rings and the
galactic disk is that Saturn's rings contain so little mass that
the gravitational force they apply on a test particle is minute.
It is Saturn's equatorial bulge that keeps the rings in plane.
The mass of the bulge is vastly greater than the mass of the
rings.

Stars in the galaxy are scattered much farther apart than
particles in Saturn's rings, relative to their size, but each star
contains so much mass that their individual and combined
gravitational effects are far greater.

-- Jeff, in Minneapolis

Jeff Root
2015-Oct-02, 01:02 AM
When the Sun is above the plane and still rising, or
below the plane and still falling, ...
Confusing! What direction is someone in Australia
moving when they are rising? Up or down? :p

Edit to add: Otherwise an outstandingly excellent post!

-- Jeff, in Minneapolis

Hornblower
2015-Oct-02, 01:46 AM
When the Sun is above the plane and still rising, or below the plane and still falling,...In hindsight I see that I should have said that as, "When the Sun has just crossed the plane and is still separating from it..." to avoid giving a misleading impression of gravitational action.

Dave Lee
2015-Oct-02, 02:58 AM
No, no, a thousand times no. As explained earlier, this motion attributed to the Sun is kinematically reckoned in a moving frame of reference in which the velocity vectors of the nearby stars average out to zero. This is calculated from the astrometrically observed transverse proper motions and the spectroscopically observed radial velocities. From these observations we have no means of distinguishing the paths of these stars from straight lines. By themselves they tell us nothing about how the stars will move over millions of years in response to the galaxy's gravity field.

No need for thousand no.
One image is enough.
Please see pg 56 Figure 5:
http://www.americanscientist.org/libraries/documents/2005620101154_306.pdf
Figure 5. Nearby interstellar medium
In this figure we can see the direction of Sun's movement (Yellow) with reference to the galactic center (blue).
What do you understand from this figure?
Don't you think that the Sun is currently moving closer to the center of the galaxy?
Therefore, do you agree that there must be Horizontal amplitude?

Jeff Root
2015-Oct-02, 03:33 AM
Figure 5. Nearby interstellar medium
In this figure we can see the direction of Sun's movement
(Yellow) with reference to the galactic center (blue).
What do you understand from this figure?
Don't you think that the Sun is currently moving closer
to the center of the galaxy?
Therefore, do you agree that there must be Horizontal
amplitude?
The last sentence of the diagram's caption is "The
motions are shown with respect to nearby stars."
Namely Alpha Centauri, Altair, Procyon and Sirius.
The motion relative to the center of the galaxy could
be entirely different from that.

-- Jeff, in Minneapolis

Dave Lee
2015-Oct-02, 06:01 AM
The last sentence of the diagram's caption is "The
motions are shown with respect to nearby stars."
Namely Alpha Centauri, Altair, Procyon and Sirius.
The motion relative to the center of the galaxy could
be entirely different from that.

-- Jeff, in Minneapolis

I thought that I shouldn't answer this respond.
However, let me tell you a story about the smallest Elephant.
One day my friend told me that he saw the smallest Elephant in the whole word.
It was in the size of a mouse and even looked like a mouse.
So I have asked him why he considered it as an Elephant.
He claimed that he had seen it an Elephant cage in a Zoo. He was sure by 100% that it was written that this is an Elephant.
So, sometimes we have to believe in what we see…

Swift
2015-Oct-02, 12:52 PM
I thought that I shouldn't answer this respond.
However, let me tell you a story about the smallest Elephant.
One day my friend told me that he saw the smallest Elephant in the whole word.
It was in the size of a mouse and even looked like a mouse.
So I have asked him why he considered it as an Elephant.
He claimed that he had seen it an Elephant cage in a Zoo. He was sure by 100% that it was written that this is an Elephant.
So, sometimes we have to believe in what we see…
Dave Lee,

Q&A is for asking questions about astronomy and related topics and for getting the mainstream answers to them. One is allowed to question those answers, so as to get clarification of something you don't understand, or to ask questions on related topics. Those questions are not for questioning mainstream science; if you wish to do that, take it to Against the Mainstream (ATM). You seem to be rapidly approaching that boundary. Be careful you don't cross it.

And if your post contains anything along the lines of "I probably shouldn't post this, but..." you should generally follow your own advice and not post it.

Dave Lee
2015-Oct-02, 10:06 PM
Sorry

So let me ask again if based on Figure 5 pg 56 you agree that the Sun is moving ahead to the center of the galaxy?

http://www.americanscientist.org/lib...101154_306.pdf

If you agree, please advice how can we explain this kind of movement.

cjameshuff
2015-Oct-02, 11:07 PM
Sorry

So let me ask again if based on Figure 5 pg 56 you agree that the Sun is moving ahead to the center of the galaxy?

http://www.americanscientist.org/lib...101154_306.pdf

If you agree, please advice how can we explain this kind of movement.

You mean http://www.americanscientist.org/libraries/documents/2005620101154_306.pdf...you linked the URL incorrectly. And the figure shows the direction of the sun's movement with respect to the nearby stars: Alpha Centauri, Altair, Procyon, and Sirius. It says nothing about the motion of the sun with respect to the center of the galaxy, since those stars themselves are moving with respect to it.

The motion with respect to the nearby stars could be in any direction, no explanation is needed...stars in a galaxy do not move in rigid formation, they freely move past each other as they follow independent paths through the galaxy. However, that figure doesn't even show the sun moving toward the center of the galaxy with respect to the nearby stars, those arrows clearly point in different directions.

Hornblower
2015-Oct-03, 12:16 AM
Sorry

So let me ask again if based on Figure 5 pg 56 you agree that the Sun is moving ahead to the center of the galaxy?

http://www.americanscientist.org/lib...101154_306.pdf

If you agree, please advice how can we explain this kind of movement.

Dave Lee, none of the illustrations to which you have linked have enough information to determine the orbital elements of the Sun's orbit around the center of the galaxy. The various nearby stars were formed from different primordial nebulae at different times, and primordial turbulence in the system has given these stars different initial velocities that set them on different orbits. In general these orbits are not perfect circles but are eccentric ellipses whose eccentricities, line-of-apsides orientations and inclinations to the disk plane are all over the place. My educated guess is that the Sun's orbit is eccentric while the average of the velocity vectors of the nearby stars is somewhere near that of a circular orbit, but I would need more information on the galaxy's mass distribution to know for sure. Experts have done extensive observations of the galaxy in the infrared and radio frequency parts of the spectrum, which can penetrate the dust that hides most of it in visible light, and by painstaking statistical analysis have estimated the mass distribution.

Dave Lee
2015-Oct-03, 04:55 AM
It says nothing about the motion of the sun with respect to the center of the galaxy, since those stars themselves are moving with respect to it.


Please, look again at that Figure and ignore all the nearby stars.
Please start by looking at the blue arrow (To galactic center).
Do you agree that this arrow represents the direction and distance from the Sun to the center of the galaxy?
Now, let's look at the Yellow arrow (Sun movement).
It looks to me that the angle between those two arrows is about 45 degrees. Do you agree?
Please look at head of the Yellow arrow. Don't you think that it is closer to the galactic center than the current Sun location?
So, do you agree that if the Sun will move in this direction (Yellow arrow), it will get closer and closer to the Center of the galaxy?

chornedsnorkack
2015-Oct-03, 06:15 AM
Dave Lee, none of the illustrations to which you have linked have enough information to determine the orbital elements of the Sun's orbit around the center of the galaxy. The various nearby stars were formed from different primordial nebulae at different times, and primordial turbulence in the system has given these stars different initial velocities that set them on different orbits. In general these orbits are not perfect circles but are eccentric ellipses whose eccentricities, line-of-apsides orientations and inclinations to the disk plane are all over the place. My educated guess is that the Sun's orbit is eccentric while the average of the velocity vectors of the nearby stars is somewhere near that of a circular orbit, but I would need more information on the galaxy's mass distribution to know for sure.

My guess is not.
If you take the average velocity of Earth-crossing asteroids and comets, would you get the speed of a circular orbit?

Jeff Root
2015-Oct-03, 07:52 AM
If you take the average velocity of Earth-crossing
asteroids and comets, would you get the speed of
a circular orbit?
Yes. The average would be the speed of a circular
orbit close to the average radius of the asteroids'
elliptical orbits.

-- Jeff, in Minneapolis

Jeff Root
2015-Oct-03, 08:52 AM
Please, look again at that Figure and ignore all the
nearby stars.
And ignore the caption which says "The motions are
shown with respect to nearby stars."



Please start by looking at the blue arrow (To galactic center).
Do you agree that this arrow represents the direction and
distance from the Sun to the center of the galaxy?
No. It represents the direction, but not distance.



Now, let's look at the Yellow arrow (Sun movement).
It looks to me that the angle between those two arrows
is about 45 degrees. Do you agree?
I get an angle closer to 55 degrees than 45, but 45
isn't very far off.



Please look at head of the Yellow arrow. Don't you
think that it is closer to the galactic center than the
current Sun location?
Unfortunately, there are no clues within the illustration
to determine whether the arrow is pointing up out of the
plane of the galaxy or down into the plane of the galaxy.
That affects the angle of the line in a 2-dimensional
image. Also, the plane of the diagram may be tipped
relative to the plane of the galaxy, similar to how it is
tipped in figures 1 and 2. Maybe you can find somewhere
in the text that actually says whether the Sun is currently
moving toward or away from the center of the galaxy.



So, do you agree that if the Sun will move in this direction
(Yellow arrow), it will get closer and closer to the Center
of the galaxy?
No. I would need more information than that diagram.

-- Jeff, in Minneapolis

cjameshuff
2015-Oct-03, 12:23 PM
Please, look again at that Figure and ignore all the nearby stars.
Please start by looking at the blue arrow (To galactic center).
Do you agree that this arrow represents the direction and distance from the Sun to the center of the galaxy?
Now, let's look at the Yellow arrow (Sun movement).
It looks to me that the angle between those two arrows is about 45 degrees. Do you agree?
Please look at head of the Yellow arrow. Don't you think that it is closer to the galactic center than the current Sun location?
So, do you agree that if the Sun will move in this direction (Yellow arrow), it will get closer and closer to the Center of the galaxy?

Again, no. You can not ignore the nearby stars when the diagram is a depiction of the sun's motion with respect to those stars, it could point in any direction depending on how the sun and those stars are moving past each other. The arrow pointing at the center of the galaxy indicates its direction and nothing else. The diagram simply does not show the motion of the sun with respect to the center of the galaxy.

Dave Lee
2015-Oct-03, 01:56 PM
Stop!
I have just found the most important image of the Sun movement.
http://www.biocab.org/Motions_of_the_Solar_System.jpg
In this image we can see clearly that the sun rotates around the orbital motion axis.
I didn't draw this image. However, I was expecting to see exactly this kind of orbital motion.
So, the sun isn't just moving up and down from the galactic center. It also moves in and out to the center of the galaxy.
If I recall correctly, you had rejected the idea of the Sun orbital motion. So, based on this image, do you finely agree that the Sun is moving in orbital motion around the main galactic orbital motion axis???

Hornblower
2015-Oct-03, 03:56 PM
Stop!
I have just found the most important image of the Sun movement.
http://www.biocab.org/Motions_of_the_Solar_System.jpg
In this image we can see clearly that the sun rotates around the orbital motion axis.
I didn't draw this image. However, I was expecting to see exactly this kind of orbital motion.
So, the sun isn't just moving up and down from the galactic center. It also moves in and out to the center of the galaxy.
If I recall correctly, you had rejected the idea of the Sun orbital motion. So, based on this image, do you finely agree that the Sun is moving in orbital motion around the main galactic orbital motion axis???

First things first: I never rejected motion of the Sun toward or away from the galactic center, nor did anyone else in this thread as far as I can tell. All we are saying is that your previous references do not tell us one way or the other.

This latest link looks flawed in more ways than one. I don't know what that helix is supposed to mean. The solar system's overall galactic motion is nothing of the sort. The sketch also has the vertical component backward. The Sun's current motion relative to the nearby stars is toward galactic north. The problems with this sketch are enough for me to reject it as not being a reliable source.

Let's get back to the americanscientist.org piece, where the author specifically says that her work does not address the Sun's motion relative to the galactic center, but only relative to the nearby stars and the local interstellar medium. This motion is directed toward galactic north, that is, the direction toward the galactic celestial pole in Coma Berenices, and is angled toward the center of the galaxy. This is consistent with any one of the following possibilities:

1. Nearby stars averaging fixed distance from center while Sun is moving closer to it.

2. Sun at fixed distance from center (circular orbit) while nearby stars are drifting away from center.

3. Some in-between combination thereof.

4. Everything drifting away from center, with the Sun being slower than the average of the nearby stars.

My hunch is that #1 is closest to the truth, but once again you have not given us enough information to make the call.

cjameshuff
2015-Oct-03, 04:22 PM
Stop!
I have just found the most important image of the Sun movement.
http://www.biocab.org/Motions_of_the_Solar_System.jpg
In this image we can see clearly that the sun rotates around the orbital motion axis.
I didn't draw this image. However, I was expecting to see exactly this kind of orbital motion.
So, the sun isn't just moving up and down from the galactic center. It also moves in and out to the center of the galaxy.
If I recall correctly, you had rejected the idea of the Sun orbital motion. So, based on this image, do you finely agree that the Sun is moving in orbital motion around the main galactic orbital motion axis???

Again, no.

The helical motion in that diagram is evidently intended to be caused by the gravitation of the planets orbiting the sun, and...something else vaguely stated that is for some reason 90 degrees out of phase with the former. The description is inaccurate in several ways...for starters, the galactic plane is inclined by about 60 degrees to the plane of the solar system, not 90 degrees, and the changes in the sun's motion with the orbital motions of the planets (primarily Jupiter) are far smaller than stated. In any case, apart from being incorrect in multiple details, when taken in context (http://www.biocab.org/coplanarity_solar_system_and_galaxy.html), this is yet another diagram that doesn't actually depict what you say it does.

Even if you eventually find a diagram that shows exactly what you think is happening, what you think is happening is wrong, and the same will go for that diagram. You're desperately looking for a cherry to pick, but finding one won't validate your approach.

DaveC426913
2015-Oct-03, 04:53 PM
Stop!
I have just found the most important image of the Sun movement.
http://www.biocab.org/Motions_of_the_Solar_System.jpg


This website is highly suspect.
It is called Biology Cabinet, yet seems to contain little more than random links to physicsy stuff.

chornedsnorkack
2015-Oct-03, 08:08 PM
Yes. The average would be the speed of a circular
orbit close to the average radius of the asteroids'
elliptical orbits.

-- Jeff, in Minneapolis

Um, what?
Let´s split it into groups.
I´d divide them into three:

Asteroids on Earth-crossing orbits with large radial velocities on inbound or outbound legs of ellipse;
asteroids with low radial velocity near perihelion
asteroids with low radial velocity near aphelion

Is there any generalization you can give about the tangential speeds of the first group?

Dave Lee
2015-Oct-03, 08:41 PM
First things first: I never rejected motion of the Sun toward or away from the galactic center, nor did anyone else in this thread as far as I can tell.
Thanks
At Last, an answer with some supportive elements. Thanks again. I do appreciate even this minor support.


Again, no.

The helical motion in that diagram is evidently intended to be caused by the gravitation of the planets orbiting the sun
.
Why always No? Don't you agree with something?
With regards to Sun' helical motion;
Do you agree that the Sun is moving in helical motion? Yes or no.
Do you agree that the sun is moving toward or away from the galactic center? Yes or no?

cjameshuff
2015-Oct-03, 10:02 PM
Why always No? Don't you agree with something?

If you continue to persistently cling to an idea that is just completely wrong...then no.



With regards to Sun' helical motion;
Do you agree that the Sun is moving in helical motion? Yes or no.
Do you agree that the sun is moving toward or away from the galactic center? Yes or no?

Both of those are inaccurate, incomplete, and misleading descriptions, so no.

The sun follows a complex motion around the barycenter of the solar system, swinging around by a bit more than its radius as the planets orbit around it. This motion is within the plane of the solar system as a whole, ignoring the relatively tiny effects of the variations in inclination between planets, and is pretty much meaningless when discussing the path of the sun through the galaxy or relative to its neighbors. On a galactic scale, the sun follows a roughly elliptical orbit that oscillates vertically through the galactic plane about 2.7 times per orbit. With respect to the galaxy as a whole, it always moves at a near-perpendicular to the center of the galaxy. With respect to whatever stars happen to be nearby at a given time, it can move in any direction, depending on how those stars themselves are moving.

What it certainly does not do is orbit some point within the disk, and that will be the answer no matter how many times or how many ways you ask.

Jeff Root
2015-Oct-03, 10:33 PM
Yes. The average would be the speed of a circular
orbit close to the average radius of the asteroids'
elliptical orbits.
Um, what?
Let´s split it into groups.
I´d divide them into three:

Asteroids on Earth-crossing orbits with large radial
velocities on inbound or outbound legs of ellipse;
asteroids with low radial velocity near perihelion
asteroids with low radial velocity near aphelion

Is there any generalization you can give about the
tangential speeds of the first group?
I don't know why you are analyzing the speeds into
radial and tangential components, but I'll play along.

The tangential speeds of asteroids on Earth-crossing
orbits with large radial velocities inbound and outbound
will be much higher than the speed of Earth when the
asteroids are near perihelion, and much lower than the
speed of Earth throughout the rest of their orbits.

Regarding the other two groups:

Radial velocity at perihelion and aphelion is always zero,
so I don't know what you mean by "low radial velocity".
Maybe that was just a typo?

-- Jeff, in Minneapolis

Jeff Root
2015-Oct-03, 10:49 PM
I don't know what that helix is supposed to mean.
I don't, either.



The solar system's overall galactic motion is nothing of the
sort. The sketch also has the vertical component backward.
The Sun's current motion relative to the nearby stars is
toward galactic north.
The diagram has a label that says "Solar System" which
points to an object which looks like the Sun (sorry George!)
That object is NOT on the helix, but at the center of the
helix. I don't see any way that you could connect either
the Sun or the Solar System to a unique point on the helix
as representing the current position, so I don't get your
comment that the "vertical component" is "backward".

Edit:

Oh! I see now. The label on the south end of the galactic
north-south arrow. That's supposed to be some kind of
general, overall speed along that axis, I think, not the
speed at the current time, in the current direction.

-- Jeff, in Minneapolis

DaveC426913
2015-Oct-04, 02:50 AM
Thanks
At Last, an answer with some supportive elements. Thanks again. I do appreciate even this minor support.

The only reason one would need "support" is if one is trying to assert something and is meeting resistance.
This is the Questions and Answers forum, not the Assertions forum.

Dave Lee
2015-Oct-04, 11:19 AM
.Both of those are inaccurate, incomplete, and misleading descriptions, so no.

On a galactic scale, the sun follows a roughly elliptical orbit that oscillates vertically through the galactic plane about 2.7 times per orbit. With respect to the galaxy as a whole, it always moves at a near-perpendicular to the center of the galaxy. With respect to whatever stars happen to be nearby at a given time, it can move in any direction, depending on how those stars themselves are moving.

.
So, you disagree that the Sun is moving toward or away from the galactic center.
You claim that it follows a roughly elliptical orbit that oscillates vertically through the galactic plane. Hence, there is no Horizontal oscillation.
O.K.
Let assume that this is correct. Let's assume that the Sun has no horizontal oscillation.

However, please see the image of the galactic rotation random vectors in the nearby Solar Neighborhood, which I have used in the first pg of this thread.
In the following presentation at pg 15
http://www.ifa.hawaii.edu/~barnes/ast110/MilkyWay.pdf
It is stated:
"Most stars near the Sun have random velocities of a few tens of km/sec. These stars orbit the galactic center at ~230 km/sec."
So the nearby stars are moving in Random velocities to all directions. Therefore, by definition, some of them are moving horizontally towards the galactic center and some of them are moving away from the galactic center. Actually, in statistics, if you have enough stars which are moving at random velocities to each other, than it is expected that the average horizontal movement should be similar to the average vertical movement. Therefore, if we consider that this movement is represented by oscillation amplitude, than the average vertical amplitude should be similar to the average horizontal amplitude. (This is based on my formal knowledge in mathematics and statistics).
Hence, those random velocities prove that the average vertical oscillation amplitude of the nearby stars should be similar to their average horizontal oscillation amplitude.
So, how could it be that the only exception is the Sun? How could it be that the Sun is the only star in the system which has no horizontal oscillation?
If you still disagree, please try to back up your statment by real evidence.

cjameshuff
2015-Oct-04, 12:53 PM
So, you disagree that the Sun is moving toward or away from the galactic center.
You claim that it follows a roughly elliptical orbit that oscillates vertically through the galactic plane. Hence, there is no Horizontal oscillation.

What I've said is plainly visible in this thread, and it is not what you claim I said. There is a small variation that comes with the vertical oscillation, due to the galaxy's non-uniform gravity field. It simply doesn't even slightly resemble the motion you require.



However, please see the image of the galactic rotation random vectors in the nearby Solar Neighborhood, which I have used in the first pg of this thread.
In the following presentation at pg 15
http://www.ifa.hawaii.edu/~barnes/ast110/MilkyWay.pdf
It is stated:
"Most stars near the Sun have random velocities of a few tens of km/sec. These stars orbit the galactic center at ~230 km/sec."
So the nearby stars are moving in Random velocities to all directions. Therefore, by definition, some of them are moving horizontally towards the galactic center and some of them are moving away from the galactic center. Actually, in statistics, if you have enough stars which are moving at random velocities to each other, then it is expected that the average horizontal movement should be similar to the average vertical movement. Therefore, if we consider that this movement is represented by oscillation amplitude, than the average vertical amplitude should be similar to the average horizontal amplitude. (This is based on my formal knowledge in mathematics and statistics).

It is also just as wrong as your idea that gravitational motions have to be "smooth rotation". There is nothing that requires the distribution of motions to be isotropic. Obvious counterexample: the solar system.



Hence, those random velocities prove that the average vertical oscillation amplitude of the nearby stars should be similar to their average horizontal oscillation amplitude.
So, how could it be that the only exception is the Sun? How could it be that the Sun is the only star in the system which has no horizontal oscillation?
If you still disagree, please try to back up your statment by real evidence.

No. The burden of proof is on you, as you are the one making the exceptional claim that stars follow motions radically different from those predicted by Newtonian gravity. Since it has been several centuries since Newtonian gravity proved to be far more useful at describing gravitational motions than the approach you are using, you are going to hear "no" a lot.

Swift
2015-Oct-04, 02:23 PM
Dave Lee,

Q&A is for asking questions about astronomy and related topics and for getting the mainstream answers to them. One is allowed to question those answers, so as to get clarification of something you don't understand, or to ask questions on related topics. Those questions are not for questioning mainstream science; if you wish to do that, take it to Against the Mainstream (ATM). You seem to be rapidly approaching that boundary. Be careful you don't cross it.


If you still disagree, please try to back up your statment by real evidence.
Dave Lee,

I warned you, but you ignored my warning. You continued to advocate a non-mainstream idea in Q&A and continued arguing with the multiple mainstream answers you were given. This will earn you an infraction and the closing of this thread. If you wish to discuss this topic ever again, you will start a thread in ATM about it. And your next infraction will earn you a suspension.