Post by: CliffordK on 27/04/2012 13:38:44
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2F8%2F89%2F236084main_MilkyWay-full-annotated.jpg%2F480px-236084main_MilkyWay-full-annotated.jpg&hash=688107264b2b68add1e565b00bc9fd71)
One of the problems with Spiral Galaxies (including the Milky Way) is describing the movement of the stars in the spiral arms, and the “Winding Problem”.
If one considers our Solar System.
Mercury orbits the sun in 88 days.
Venus in 225 days.
Earth in 365 days
Mars in 687 days
Jupiter in 12 years.
Saturn in 30 years.
Uranus in 84 years.
Neptune in 165 years.
A spiral pattern of planets would not be stable with the outer planets orbiting the Sun much slower than the inner planets.
This problem is known as the Winding Problem (http://en.wikipedia.org/wiki/Density_wave_theory) with respect to the Galaxy.
The solution appears to be the gravity experienced by each body in the galaxy must be calculated not by the center of mass of the galaxy, but rather the gravity between the body and every additional body in the galaxy. In the case of our Sun, there are some stars both more central, and more peripheral to our Sun in the Milky Way. So, the gravity vector is both towards the center, and away from the center of the Milky Way.
At this point, one can assume the tangential gravity vectors cancel out.
So, I created a spreadsheet with a simplified galaxy consisting of 330 stars, approximately evenly distributed around a central star.
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For 10 exemplar stars arranged along the X axis at distances 1 through 10 from the central star, I calculated the distances to every other star in my Galaxy.
The exemplars are on the X axis so y1=0.
Initially fixing all the masses the same (1), I was then able to calculate the gravity between every pair of stars:
G = KG
KG being a constant, normalized later.
The X (inward) component of that gravity was:
Gx=
, which was then summed over all 331 stars to give me my total gravity GT.Centripital Force, F =
V=KV
With the mass (m) of the exemplar stars being initially fixed = 1, the constant KV was again normalized later.
I then calculated the orbital period as P=Ko
.Again normalizing the constant Ko, and the circumference being 2π r being swallowed up in the constant.
All values were normalized to the minimum (or maximum) as appropriate being normalized to 1, independent of units.
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What I found for my stylized galaxy was that the inward component of the gravity was lowest at the middle of the galaxy, and increased towards the outer edge of the galaxy, not too unlike theories of zero gravity in the middle of the Earth.
Likewise the orbital velocity of the stars increased the further out one went.
The calculated orbital period was most unique, however. With a fixed mass or density of the stars in the stylized galaxy, the orbital period was shortest in the middle as expected, but also decreased towards the outer edge of the galaxy.
Thus, the difference in gravity, decreasing towards the middle of the galaxy slowed the inner stars with respect to the outer stars, to give a similar orbital period for all the stars beyond a certain point.
Obviously the slight relative decrease in orbital period near the outer edge of the galaxy would still be an issue for a spiral galaxy with the outer stars speeding too fast to maintain the spiral.
So, what I did was diddled with the stellar mass, which is equivalent to the stellar density (number of stars) at different distances from the center of my Galaxy.
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What I came up with was by slightly decreasing the stellar density towards the outer part of the spiral, I arrived at a constant orbital period beyond a certain point. And thus, one could create a stable spiral galaxy. Noting, of course, that the spiral arms in the Milky Way don't extend all the way to the middle.
And, no exotic dark mater/dark energy was necessary. Nor are there any "density waves".
Post by: CliffordK on 29/04/2012 15:48:49
Let me add the updated, merged force of gravity VECTOR equation:
is the gravitational force vector experienced by the xth star in the galaxy (the exemplar stars used in the calculations).n is the number of stars in the galaxy.
rxi is the distance between the xth and ith stars in the galaxy.
is the unit coordinate vector between the xth and ith stars sx & si in the galaxy.G is the gravitational constant (6.67384 × 10-11m3/kgs2)
mx & mi are the masses of the xth and ith stars in the galaxy.
OUCH!!!
It looks like I messed up on my normalizing the vector above in the first draft.
I had divided by |X|+|Y|, but instead, I should have divided by the distance (r).
It bumped the curves around a little bit, but the end result is just about the same.
Oh, and here is the Wikipedia page on the Galaxy Rotation Curve.
http://en.wikipedia.org/wiki/Galaxy_rotation_curve
Post by: MikeS on 30/04/2012 09:40:32
The problem is only apparent when considering velocities without applying a time dilation correction factor.
When we view a galaxy face on we are seeing an object that that has a very large radius, differences in gravitational potential and differences in time dilation factors. It exists in many different time frames simultaneously. Due to the gravitational gradient and time dilation gradient from center to periphery, time at the periphery passes faster than at the center. Speed = distance/time. As time contracts at the periphery so speed of rotation at the periphery decreases in relation to the speed of rotation at the center.
edit
With a fixed rotating body like a wheel for example all of the components are fixed relative to each other. To rotate together a point on the periphery has to travel at a higher velocity to maintain its relative position. A spiral galaxy is not a fixed body but is only held together by gravity. As it is not fixed we would expect the spiral arms to wind up over time. Also we would expect the galaxy to be thrown apart by centripetal forces due to the velocity at the periphery. This is not what we observe. What we do observe is the whole galaxy rotating essentially as one body. One revolution at the center being the same as one revolution at the periphery. The center of mass of the galaxy is at the center. Therefore, time is more dilated at the center becoming less so further from the center. There is a time dilation gradient from the periphery across the radius to the center. Time is contracted (passes faster) at the periphery in comparison to the center. A faster passage of time means that the velocity of the periphery is slower than would be expected. (The faster you travel in the TIME aspect of space-time the slower you travel in the SPACE aspect of space-time.) There is less tendency for the arms to wind up. A lower velocity also equals less centripetal force. end edit
What we observe is the time line of points across the radius of the galaxy smeared across space-time. The speed of rotation of a galaxy is meaningless in any reference frame other than our own local frame. With the periphery rotating much slower than expected there is no tendency for the spiral arms to 'wind up'. Likewise with centripetal force, the velocity of the periphery being smaller than expected, so to is the centripetal force.
Dark matter and dark energy are therefore surplus to requirements when viewed as part of the bigger picture.
Post by: Guthers on 30/04/2012 12:13:26
Post by: yor_on on 01/05/2012 18:30:40
What strikes me immediately though is "What I found for my stylized galaxy was that the inward component of the gravity was lowest at the middle of the galaxy, and increased towards the outer edge of the galaxy, not too unlike theories of zero gravity in the middle of the Earth."
Wouldn't that imply, if applied as a model of our universe, that we now have a center defined?
==
That is, if I assume that this can be applied on all galaxies together, as the main idea is that it must be decided from all mass in the universe :)
That is assuming the universe to be isotropic and homogeneous. The same everywhere. then treat it like a fractal :) This is also assuming that gravity is a 'frame of reference' constantly measurable for us in the universe. Although? Ouccch :) need to think about that one..
Post by: CliffordK on 01/05/2012 19:49:17
http://arxiv.org/pdf/astro-ph/0507619.pdf
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It turns out that there is about a hundred fold difference between the density of the Milky Way in the middle, and towards the outer spiral arms.
Attempting to match the the density data, I recalculated the rotation curves and got a decreasing velocity curve, and a significantly increasing orbital period. So much for my idea.
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This, however, only deals with average density, and doesn't take into account the spiral arms. I'll have to re-think how to build my spreadsheet to take into account the spiral arms which may actually create a force vector that deviates from the center of the galaxy for many of the stars.
Some of the smaller spiral galaxies such as M33 & NGC300 may not have such an extreme difference in density, and thus may have a more stable rotation.
Perhaps there is a limit in the size of galaxy that can support a spiral structure. Andromeda is believed to be somewhat larger than the Milky Way, and seems to have more of a ring structure than a spiral structure, or ring + spiral.
Now if I could only figure out how to derive high resolution density data from digital photographs.
Post by: MikeS on 02/05/2012 07:11:04
If the Milky Way has a density gradient of 100 fold (thanks Clifford) from center to periphery then presumably it has a 100 fold time dilation gradient. If time at the periphery is passing 100 times faster than at the center, the velocity at the periphery is 100 times slower than would be expected.
Fcentripetal=m v2/r
So the centripetal force will decrease by the square of the decrease in velocity.
Post by: Guthers on 02/05/2012 13:48:53
Continuation of reply # 2 of this post.That doesn't follow. Time dilation might be 100 times greater, but that wouldn't mean time itself is running 100 times faster.
If the Milky Way has a density gradient of 100 fold (thanks Clifford) from center to periphery then presumably it has a 100 fold time dilation gradient. If time at the periphery is passing 100 times faster than at the center,.
Post by: MikeS on 03/05/2012 09:00:09
Continuation of reply # 2 of this post.That doesn't follow. Time dilation might be 100 times greater, but that wouldn't mean time itself is running 100 times faster.
If the Milky Way has a density gradient of 100 fold (thanks Clifford) from center to periphery then presumably it has a 100 fold time dilation gradient. If time at the periphery is passing 100 times faster than at the center,.
Well I did say if.
quote Guthers
"Time dilation might be 100 times greater, but that wouldn't mean time itself is running 100 times faster."
I didn't say that. What I did say was "If the Milky Way has a density gradient of 100 fold from center to periphery then presumably it has a 100 fold time dilation gradient." Meaning that time contraction is 100 fold, therefore time is running 100 times faster at the periphery as compared to the center. (From the reference frame of a distant observer.)
To answer you question I need to understand it better.
What is it that you are questioning or have I just explained it?
That time contraction does not lead to time passing faster (from the reference frame of a distant observer).
or
That time contraction may be 100 times greater but time would not be running 100 times faster. Are you questioning the numeric relationship between time contraction and time passing faster.
Post by: Guthers on 03/05/2012 21:50:19
That time contraction does not lead to time passing faster (from the reference frame of a distant observer).The latter. If time contraction at the periphery is say 0.01% (and I have no idea whether that is the right figure, probably nowhere near) and at the centre is 100 times this because the density is also 100x, then time dilation at the centre would be 1%. Hence time at the centre would run at about 99% that of the periphery.
or
That time contraction may be 100 times greater but time would not be running 100 times faster. Are you questioning the numeric relationship between time contraction and time passing faster.
Just because time dilation is 100 times greater does not necessarily mean time runs 100 times slower. It might, but not necessarily and probably not in the case of the Milky Way, whose density is very low anyway.
Post by: Phractality on 03/05/2012 23:09:39
This is a common mistake, applying Newton's shell theorem to the universe while pretending not to assume the universe is finite. The shell theorem assumes that the outermost shell is finite and nothing exists outside of it. If the universe is homogeneous, isotropic and infinite, then every point is equivalent to every other point. Therefore, gravitational potential is the same everywhere and the gradient of gravitational potential (i.e. field strength) is zero everywhere. A big bang universe may be self-consistent, but an infinite universe can also be self-consistent.
Quite nice work Clifford :)
What strikes me immediately though is "What I found for my stylized galaxy was that the inward component of the gravity was lowest at the middle of the galaxy, and increased towards the outer edge of the galaxy, not too unlike theories of zero gravity in the middle of the Earth."
Wouldn't that imply, if applied as a model of our universe, that we now have a center defined?
==
That is, if I assume that this can be applied on all galaxies together, as the main idea is that it must be decided from all mass in the universe :)
That is assuming the universe to be isotropic and homogeneous. The same everywhere. then treat it like a fractal :) This is also assuming that gravity is a 'frame of reference' constantly measurable for us in the universe. Although? Ouccch :) need to think about that one..
Post by: MikeS on 04/05/2012 06:44:38
That time contraction does not lead to time passing faster (from the reference frame of a distant observer).The latter. If time contraction at the periphery is say 0.01% (and I have no idea whether that is the right figure, probably nowhere near) and at the centre is 100 times this because the density is also 100x, then time dilation at the centre would be 1%. Hence time at the centre would run at about 99% that of the periphery.
or
That time contraction may be 100 times greater but time would not be running 100 times faster. Are you questioning the numeric relationship between time contraction and time passing faster.
Just because time dilation is 100 times greater does not necessarily mean time runs 100 times slower. It might, but not necessarily and probably not in the case of the Milky Way, whose density is very low anyway.
Your quite right.
However, regardless of the actual figures the principle remains the same. The velocity of the periphery of a galaxy is less than it seems. Therefore the centripetal force is also less than it seems.
It is easy to assume that a very small difference in time dilation is insignificant but it is not. The difference in time dilation between the Earths surface and a few hundred miles above the surface is minute but it is enough for the surface of the Earth to accelerate at 1g.
Scaling that up to the size of a galaxy it must become very significant and should not be ignored.
Post by: CliffordK on 04/05/2012 10:06:19
Which brings up the problem with 100:1 ratio of the density within the center of the galaxy and the spiral arms.
Considering our sun's orbital velocity of about 220 km/s, with the speed of light 300,000 km/s, or the sun travelling at 1/1000 the speed of light.
The Lorentz factors at 0.001 c will be small.
I suppose I should put into my spreadsheet values to create the constant velocity profile that is observed.
http://en.wikipedia.org/wiki/Galaxy_rotation_curve
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2Fb%2Fb9%2FGalacticRotation2.svg%2F250px-GalacticRotation2.svg.png&hash=486291821a5ff66020234d1f76086c4c)
Here there was a question about whether planets could form around closely orbiting binary stars.
http://www.thenakedscientists.com/forum/index.php?topic=43779.0
Perhaps what one would find is that in the center of the galaxy there would be much fewer planets, comets, asteroids, and other non luminous debris. I.E. All that would be left would be stars. And, further out there would be more debris outside of the stars. Nothing particularly exotic.
Suggestions are of an Oort Cloud (http://en.wikipedia.org/wiki/Oort_cloud) that extends for the better part of a lightyear from our sun. But, what would be the boundary, other than stuff controlled by our sun vs the neighboring stars. Is there the equivalent of Lagrangian points between stars? What is in them?
Post by: Guthers on 04/05/2012 20:17:15
The difference in time dilation between the Earths surface and a few hundred miles above the surface is minute but it is enough for the surface of the Earth to accelerate at 1g. Scaling that up to the size of a galaxy it must become very significant and should not be ignored.Nonsense. Don't even try to explain what you think you mean by this, I will just avoid your posts in future.
Post by: MikeS on 05/05/2012 07:43:00
The difference in time dilation between the Earths surface and a few hundred miles above the surface is minute but it is enough for the surface of the Earth to accelerate at 1g. Scaling that up to the size of a galaxy it must become very significant and should not be ignored.Nonsense. Don't even try to explain what you think you mean by this, I will just avoid your posts in future.
If it's nonsense then you will have no difficulty in falsifying it.
It should be so obvious that it is nonsense that you can falsify it with very little effort or time.
Before dismissing offhand what I have written as nonsense you should be in a position to falsify it and be prepared to falsify it, are you?
What is it you believe to be nonsense?
1) That the Earths surface accelerates (upward) at 1g.
2) That acceleration is due to differences is time dilation between the Earths surface and away from (above) the Earths surface.
3) Scaling that up to the size of a galaxy it must become very significant and should not be ignored. (That is, the time dilation factor across the radius of the Galaxy)
Post by: MikeS on 06/05/2012 06:55:25
Points 1 & 2 above I don't believe are open to dispute. They are facts. (In as much as there are facts in science.)
1) "That the Earths surface accelerates (upward) at 1g."
That corresponds with experience so is a fact.
2) "That acceleration is due to differences is time dilation between the Earths surface and away from (above) the Earths surface."
This also corresponds with experience as it has been measured. An accelerometer placed on the Earths surface at sea level will measure 1g. Time dilation has been measured and verified.
3) "Scaling that up to the size of a galaxy it must become very significant and should not be ignored. (That is, the time dilation factor across the radius of the Galaxy)"
To deny this, is to deny that either there is no time dilation factor across the radius of the galaxy due to the mass gradient, or that the decrease in speed of the periphery that it creates is so minute as to be irrelevant. Either way I would like to see evidence.
Post by: yor_on on 17/05/2012 11:06:21
This is a common mistake, applying Newton's shell theorem to the universe while pretending not to assume the universe is finite. The shell theorem assumes that the outermost shell is finite and nothing exists outside of it. If the universe is homogeneous, isotropic and infinite, then every point is equivalent to every other point. Therefore, gravitational potential is the same everywhere and the gradient of gravitational potential (i.e. field strength) is zero everywhere. A big bang universe may be self-consistent, but an infinite universe can also be self-consistent.
Quite nice work Clifford :)
What strikes me immediately though is "What I found for my stylized galaxy was that the inward component of the gravity was lowest at the middle of the galaxy, and increased towards the outer edge of the galaxy, not too unlike theories of zero gravity in the middle of the Earth."
Wouldn't that imply, if applied as a model of our universe, that we now have a center defined?
==
That is, if I assume that this can be applied on all galaxies together, as the main idea is that it must be decided from all mass in the universe :)
That is assuming the universe to be isotropic and homogeneous. The same everywhere. then treat it like a fractal :) This is also assuming that gravity is a 'frame of reference' constantly measurable for us in the universe. Although? Ouccch :) need to think about that one..
Interesting point Phractality :)
Don't know if the universe is 'infinite'. It seems as it depends on what you define as 'dimensions' and how they act, or maybe 'interact'? The thing is that you can have it both ways depending on how you think of that as I find. You can define it as a 'finite' thingie, but 'infinite' to us if using more dimensions. Or you can use 'gravity' as 'distorting/bending space' for it too.
After all, what do you expect a 'space' without 'gravity' (the metric) to become?
Can it even exist for us, as being measurable in distance and 3-D?
=
Eh, by that I question a 'flat space' btw.