What is the speed of rotation of stars close to the center of a galaxy? Is the acceleration close to the center inversely proportional to the square of the distance from the center? Approximately how is the mass distributed across the radius?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

The small figure on the right on this page answers all those questions, I think. It shows the rotation speed close to the center of a typical spiral is on or near what one would expect - based on what mass we can detect using the entire electromagnetic spectrum.... Then the observations, uh, diverge from that expectation. The rotation curve traces how the mass is distributed across the radius, assuming that mass acts in an inverse-square fashion.

__________________
Everyone is entitled to his own opinion, but not his own facts.

Apparently spiral galaxies do not wind up tightly as would be expected if the speed of the outer stars was only twice the speed of the stars 5% of the way from the center. If the mass was mostly near the center we would expect the inner stars to be at least twice as fast, so there is a discrepency of at least 4 times, which it is explained as possible due to dark matter. I don't think so Tim. Perhaps the spiral galaxies started spiraling 137,000 years ago instead of 13,700,000,000 years ago. I don't know, but it does not compute. Neil

Thanks, guys. Does the density fall approximately with 1/r?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Well, I have been attempting to work out the integrations for the gravity of a disk, but still haven't made too much progress that way, so I finally broke down and wrote a computer program to find the gravitational force and potential energy for a cylinderical mass of uniform density which is 100000 light years across and 12000 light years in height, by going through it point by point. At the very edge of the disk, the force is about twice that of a spherical mass with the same mass at the same distance. The energy is about 1.2 times as great.

Here is the interesting part. A long while back in another thread, while working out the gravity of a wire and then trying to use that to equate it with a short, wide wire that might then imitate the disk of a galaxy, Richard suggested I try a density that varies with 1/r, so I tried that as well in the program. The force then came out to exactly the same as that for a sphere at the same distance and so did the energy. That seems very natural somehow.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

The rotation speed for a density that varies with 1/r doesn't seem too natural, however. It apears to climb, approximately with 1/sqrt(r), so gets larger toward the center. That would be because the density at the center is 1/r, so approaches infinity there. A graph of the density would be similar to a hyperbola, and climb to infinity at the center, so I will just top it off there.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

From the diagram on Wiki Cougar linked to, and from other links I have seen, it looks to me like the constant speed, at least for the Milky Way, I believe, begins at about one-third of the radius, or about 5 kiloparsecs. Closer to the center, it looks like a roughly steady drop to zero, so half the speed should be observed at about half the distance to the point where it becomes constant, or at about one-sixth of the radius instead of 5%, shouldn't it?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

This is very strange. I noticed that if I use a density that falls with 1/r, 1/r^2, 1/r^3, even 1/r^1.1 or 1/r^1.01, the force and energy at the edge of the disk are still the same as that of a sphere with the same total mass at the same distance. The only way it doesn't appear to work this way is with 1/r^0, for uniform density. Apparently, as long as the density climbs toward infinity at the center, we will get the same force and energy as that of a sphere. I'll play around with it some more to be sure.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Oh, okay. I see what is happening now. For any 1/r^n where n>0, the density at the center climbs toward infinity and overwhelms the mass of the rest of the disk, so it basically just acts like a point mass at the center. I will have to top it off.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Okay, I added a particular parameter to top off the density in the center of the disk, and guess what? It reproduces the shape of the rotation curve as seen in that Wiki link. At the edge of the disk to some point close to the center, the rotation speed is almost constant. Then it begins to drop off in proportion to the distance from the center, in just the same way as the link shows. The value of the parameter apparently determines where the drop off begins, but it is also sensitive. I will keep working with it to try to determine what the value would be for our galaxy. So my question now is, should this be moved to ATM or astronomy when I am done, or should it just stay here?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Actually, it's because only those stars within any given star's orbit contributes to it's orbital velocity. Thus, the innermost star is only affected by our galaxy's supermassive black hole, wile the outermost star is being affected by all other stars in our galaxy.

__________________
I am Mugs, of the Alien clan of Usa, Nordamerica, a Terran, of Sol.

Perception isn't reality. It's merely an abstraction thereof, and quite often not a very good one at that.

"Staying young requires the unceasing cultivation of the ability to unlearn old falsehoods." - Heinlein

"Freedom begins when you tell Ms. Grundy to go fly a kite." - Heinlein

I wish that were true. It would simplify things tremendously. Unfortunately, it's not true for the gravitation of a disk, but applies to points within a sphere only, where the mass outside of the orbit cancels itself out naturally due to the geometry. I think it is interesting that a point that is located anywhere within the central cavity of a hollowed out sphere will experience no gravity whatsoever, anywhere from the center of the sphere to the boundary of the cavity. However, the stars outside of a given star's orbit in a disk will also make a difference with the force and orbital speed. That's part of what makes it so complicated.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Is there a simple formula for the circular speed of a particle in a thin,
completely uniform disk?

-- Jeff, in Minneapolis

__________________
http://www.FreeMars.org/jeff/

"The other planets?
Well, they just happen to be there, but the point of rockets is to explore them!"
-- Kai Yeves

Well, yes and no. If we integrate for the acceleration of gravity of a disk plane on a particle, we have Int G (M/A) dx dy (R-x) / d^3, where M/A is the mass per area which is uniform, R is the distance of the particle from the center, d is the distance from the particle to each point in the disk, and (x,y) are the coordinates of a point using the center of the disk as the origin. From this, we get

Int G (M/A) dx dy (R-x) / sqrt[(R-x)^2 +y^2]^3

= Int G (M/A) dx (R-x) * [2 * sqrt(r^2 - x^2) / (R-x)^2 / sqrt[(R-x)^2 + (r^2 - x^2)] ] for y = -sqrt(r^2 - x^2) to sqrt(r^2 - x^2)

but the last integration for dx does lead to complicated elliptic integrals, three different kinds, in fact, so let's avoid that by finding a solution for a specific R, such as R=r, for acceleration of gravity acting on a particle right at the edge of the disk. Then we get

Int 2 G (M/A) dx sqrt(r^2 - x^2) / (r-x) / sqrt[(r-x)^2 - r^2 -x^2]

= Int 2 G (M/A) dx sqrt(r+x) sqrt(r-x) / (r-x) / sqrt[2r * (r-x)]

= Int 2 G (M/A) dx sqrt(r+x) / (r-x) / sqrt(2r)

= 2 G (M/A) / sqrt(2r) * [ 2 sqrt(2r) atanh[sqrt(r+x) / sqrt(2r)] - 2 sqrt(r+x)]

from x = -r to r, so

= 4 G (M/A) [ atanh(1) - 1]

Here, since x = -r to r, it reduces to just zero for the whole thing for -r, but for +r, the part in atanh becomes sqrt(r+x)/sqrt(2r) = 1, and atanh(q) = (1/2)[ln(1+q) - ln(1-q)], so if q=1, then - ln(1-q) = -ln(0) = - (-infinity) = infinity. We have nothing else that might cancel out the infinity here, so that is the result for a disk plane. The gravity is infinite at the edge of the disk plane.

Now, I am thinking that finite masses cannot result in infinite force, so the only thing I see that might cause this is due to considering any and all point masses that lie directly in front of the particle, even within zero distance, like singularities. I am thinking that a better method would be to allow some distance between the particle and the closest possible point mass. For the gravity of something like a planet or the sun, this could possibly be as small as the distance between atoms, so wouldn't make much difference in the integration for those, at least that of spheres. With galaxies, however, especially being disks, I'm thinking an average or minimum possible distance between stars must be accounted for, at the very least being that of the diameter of a typical star.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Here is a link I found a little while ago that summarizes some methods. It looks like I have already tried most of the ones mentioned. Is this pretty much all that has been explored of the subject so far?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Incidently, I have also tried the integration for a galaxy with some thickness of 2h, where h is the height from the plane. That would be

Int G D dx dy dz (R-x) /d^3, where D is the density

= Int G D dx dy dz / sqrt[(R-x)^2 +y^2 + z^2]^3

= Int 2 G D h dx dy (R-x) / [(R-x)^2 + y^2] / sqrt[(R-x)^2 + y^2 + h^2]

= Int 4 G D dx atan[h sqrt(r^2-x^2) / (R-x) / sqrt[(R-x)^2 + r^2 -x^2 + h^2]

but I have not been able to get past that last integration. It also lead to a long set of elliptic integrals. If I make R=r, and use an algorithm for the square root in the denominator to the fourth order, as well as expanding the atan itself to the fourth order using the Taylor series, then I can barely manage to bypass the elliptics. But what pops up in their place is the adding and subtracting of infinities anyway, which I'm sure reduces down, but I cannot be sure in what manner.

I've just realized something, though. Instead of that program I'm using to find for the gravity of a galaxy point by point, I can just use that last integration linearly for x only, since the integrations for y and z have already been found. That way, instead of running it for a long time to find, say, 100 points along the x axis and 100 along the y axis, as well as 12 along the z axis (which I'm thinking I should probably figure for only 1000 light lears for that for stars instead of 12000 which includes gas and dust), which would find for pi * 100^2 * 12 = 377, 991 points in all, I can just run it linearly using that last integration, allowing me to run the program about 378 times faster, or 378 times more precisely. In fact, I'm wondering now if I might be able to deduce the integration along the x and y axes originally, leaving just that little bit along the z axis for the height where I could run the program even a hundred times faster and/or more precise than that, even. I'll try it.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

The result of that is very interesting. I did manage to reduce the integration down to just that along the z axis without any complication, but there are complex numbers in the result, all in direct association with z. I noticed that if I make d = sqrt[(R-x)^2 + y^2 - z^2] instead of d = sqrt[(R-x)^2 + y^2 + z^2] as it should be the latter, then the integration reduces to practically the same thing, but without any complex numbers. I wonder what that means. I'll explore it further.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Oh, wow! I couldn't get a direct solution for the integration of atanh incorporating complex numbers, but by using the Taylor series for atanh, so far I have been able to find solutions to the first and second order where the complex numbers just fall out each time. The first order doesn't involve elliptics or anything complicated and I'm thinking (hoping) the other orders might not either. I may well be on my way to a solution for a uniformly dense disk after all. The result would still be in the form of an infinite series, though, unless I can find the way the pattern for it runs and use another function for that as well. I'll keep working on it.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Oops. Looks like I missed a square root. That figures. Disregard the last couple of posts.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."