That sounds interesting also. I'll work on that just as soon as I write this other program and do a few comparisons.

__________________
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, that program is taking an extra minute or two to work out since I forgot I would need another way to add up the total mass, and I know it will be in proportion to sqrt(r^2-x^2), but I'll still have some work to do on it, so I went ahead and worked out the integration for the gravity on a particle on the apex of a cone of uniform density. This time, I made the apex the origin, and the cone lies along the x axis, so it should go like this . . .

Int G D x dx dy dz / d^3, where d = sqrt(x^2 + y^2 + z^2), (x/h)^2 r^2 = y^2 + z^2 at any point along the surface at the edge of the cone, h is the height of the cone, and r is the radius of the base.

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

= Int 2 G D dx dy sqrt[(r/h)^2 x^2 - y^2] / (x^2 + y^2) / sqrt[1 + (r/h)^2]

= Int 4 G D dx / sqrt[1 + (r/h)^2] * [sqrt[x^2 + (xr/h)^2] atan[sqrt(1 + (r/h)^2)(xr/h) / sqrt[(xr/h)^2 - (xr/h)^2]] / x - atan[(xr/h) / sqrt[(xr/h)^2 - (xr/h)^2]]] for y = +/- (xr/h), where the denominator in the atans then become zero, so we get atan(infinity) for each of them, but don't worry, these infinities work out to atan(infinity) = pi/2, so we get

= Int 4 G D dx (pi / 2) / sqrt[1 + (r/h)^2] * [sqrt[x^2 + (xr/h)^2] / x -1], for x = 0 to h

= 2 pi G D h [1 - 1 / sqrt[1 + (r/h)^2]], where D = M / V and V = pi r^2 h / 3, so

= 2 pi G M h [3 / (pi r^2 h)] [1 - 1 / sqrt[1 + (r/h)^2]]

= (G M / r^2) * 6 [1 - 1 / sqrt[1 + (r/h)^2]]

It looks like if we were to set a particle on the apex of a cone with a very small height, many times smaller than the radius of the base, then the gravity would strengthen toward a limit of 6 times greater than that of a sphere with the same mass and at a distance equal to the radius of the base.

__________________
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."

Above we found a gravity that is 6 times greater than the same mass at a distance equal to the radius of the base if h<<r. But let's say h>>r, extended indefinitely. In that case, we can use the first order for the square root with a very small r/h ratio to get

6 (G M / r^2) [1 - 1 / sqrt[1 + (r/h)^2]]

= 6 (G M / r^2) [1 - 1 / (1 + (r/h)^2 / 2)]

= 6 (G M / r^2) [ 1 - (1 - (r/h)^2 / 2)]

= 6 (G M / r^2) [(r/h)^2 / 2]

= 3 G M / h^2

so if h is very large in respect to r, the limit for the gravity works toward being 3 times stronger than that of a sphere or point mass with the same mass at a distance of h from the particle.

__________________
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."

Just for good measure, here's the solution for a pyramid with base lengths of 'a' and 'b'. Yep, it's kinda fun when an actual workable solution can be obtained.

Int G D x dx dy dz / (sqrt[x^2 + y^2 + z^2])^3, where y = (x/h)(a/2) and z = (x/h)(b/2) at the edge of the pyramid.

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

= Int 4 G D x^2 dx (b/2h) atan [[(x/h)(b/2)][(x/h)(a/2)] / x / sqrt[x^2 + (x/h)^2 (a/2)^2 + (x/h)^2 (b/2)^2]] / x / [(x/h)(b/2)]

= Int 4 G D dx atan[a b / (4 h^2) / sqrt[1 + (a/2h)^2 + (b/2h)^2]]

= 4 G D h atan[a b / (4 h^2) / sqrt[1 + (a/2h)^2 + (b/2h)^2]]

where D = M / V and V = a b h / 3, so

= 12 (G M / (a b)) atan[a b / (4 h^2) / sqrt[1 + (a/2h)^2 + (b/2h)^2]]

So that is the solution for a pyramid. If we make h very large in respect to 'a' and 'b', then we can just take the first term for atan and the square root, and it reduces to

= 12 (G M / (a b)) [a b / (4 h^2)]

= 3 G M / h^2

the same limit as with the cone for very large h.

__________________
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, grav says he's using close to 400,000 points. Whether they're "stars" or little "regions" of space with a certain mass, it shouldn't matter. Apparently he is polling each point, running its characteristics through a single closed-form algorithm to tell it where to move, then on to the next point. That is indeed considerably more efficient than polling each point and running the characteristics through a simple algorithm for every other star in the galaxy, and adding up the results to make a single move.

Somebody would have to pay me to check the math grav is churning out, though. Objects orbiting on the outer rim of the galaxy should have just a little more acceleration due to gravity than an object close to the rim.

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

grav, you can not have infinate accelleration about a finite amount of mass!

You're getting lost in the trees of your math, somewhere.

Please step back and take another look at the forest.

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

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

Heinlein's: "Staying young requires the unceasing cultivation of the ability to unlearn old falsehoods." "Freedom begins when you tell Ms. Grundy to go fly a kite."

I think we're getting confused about my posed problem being "infinately thin." It's not. It's as thick as the diameters of the stars which populate it.

The only reason that spiral galaxies have any thickness to begin with is the fact that they condensed from rather large clouds. Objects bobbing above and below the orbital plane tended to collide with other objects (stars, usually). If two stars of the same mass collided, you'd have one star with twice the mass, bobbing half as high.

The reason behind the center bulge is two-fold. First, there are a lot more stars, so the interactions between they are far more numerous (then tend to get flung out of the plane because of near-misses by a star that's slightly above or below them - both stars fling one another out of the plane).

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

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

Heinlein's: "Staying young requires the unceasing cultivation of the ability to unlearn old falsehoods." "Freedom begins when you tell Ms. Grundy to go fly a kite."

Assuming a perfect point (perfectly sharp, that is) the force at the point is indeed infinate for all angles between (not inclusive) 0 and 180 degrees.

Bad analogy, though, as myself and others have been trying to explain in several posts. You can not get anywhere near infinate gravity from a finite mass.

At best, your sharpest gravitational gradient would be experienced if you condenses all matter to a quark soup, a black hole, then measured the gravity on the surface of this quark soup (inside the event horizon, but on the surface of a glob of mass x, where x is the mass of the entire Milky Way, and of a radius r, such that the Milky Way's mass was condensed into quark soup of a sphere of radius r).

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

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

Heinlein's: "Staying young requires the unceasing cultivation of the ability to unlearn old falsehoods." "Freedom begins when you tell Ms. Grundy to go fly a kite."

mugaliens,

I'm not sure how much of the thread you are actually reading. I am in the same camp as you guys that gravity cannot actually be infinite. It only becomes so when using straightforward Newtonian gravity that is in inverse proportion to the square of the distance. But as far as gaining an actual infinite gravity is concerned, there are probably many factors that constrain it, such as the limit of the speed of light, all real bodies having magnitude in all three dimensions, and others. I am only demonstrating the potential of nonlimited gravity using Newtonian physics at the edge of a disk plane to show the great difference between the gravity of a sphere and disk. Any real disk will have some thickness, however, and as long as the thickness is at least appreciable, much greater than the Swartzschild radius for the mass, say, then Newtonian physics should suffice. The gravity at the edge of a disk with some thickness will have some gravity between just slightly greater than that of a sphere with the same mass and radius to some very large but finite value, depending on the measured non-zero thickness of the disk.

Also, that second post was quoted out of context. It referred to something Jeff posed for the non-infinite gravity of a thick disk. That is where the +h^2 comes in. Only if h is zero (for the height of the disk) do infinities show up. And as far as the gravity at the apex of a cone integration goes in the third post, the result was not infinite for that either. It showed up to 6 times as much gravity as that of a sphere with the same mass at a distance equal to the radius of the base for a cone of very short height, and a limit of 3 times as great for a cone with a very large height as compared to a sphere with the same mass at the same distance as the height of the cone.

__________________
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."

Let me put in a brief word to support grav: it really is true that, if you use the Newtonian formula for gravitational force, and integrate over a flat disk, the force at the edge of the disk will be infinite.

Yes, it's not realistic, because real distributions of mass aren't infinitely thin. But it's how the math works out.

Keep up the good work, grav!

Thank you, Stupendousman. I was running out of things to say about that, and I was wondering if anybody else was going to step in. You are truly stupendous, man.

__________________
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 finally got that program up and running. It turns out that I didn't need to find the mass along each section of the x axis after all. It was already incorporated into the integrations.

This new program just takes the two thirds of the integration I have already found and then just runs the last one along the x axis only, instead of doing it for all three, which would take an exponential amount of time. Earlier in the thread, I gained results for a 12000 light year thick galaxy and 1000 light year thick galaxy which is 100000 light years in diameter and found them to have about 1.9 and 2.75 times greater gravity than that of a sphere for at the rim of the disk, respectively. But those were just using a minimum number of points due to the time involved in running the program. I reran those with the old program using more points, letting them run up to an hour each, and gained 2.01 and 3.47 for the results that time. But now that I have the new program, I can run them much, much quicker and much more precisely. Even running it for just a couple of seconds gives results that are much more precise than what I got with the old one. Anyway, below is a graph for the ratio of gravity to a sphere for a 12000 light year thick disk and 1000 light year thick disk, or a ratio of 8.33333 and 100 for the diameter to thickness ratio (r/h) since the actual mass and scale don't matter for the resulting ratio of gravity to that of a sphere with the same radius and mass. The last one is the thin plate that Jeff proposed, with a diameter to thickness (or radius to height) ratio of 2000. I also showed three sets for each using a different number of points along the x axis.

__________________
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 in answer to the original topic of this thread, by Grav and Cougar on 4/24/08 "galaxy rotation speeds and mass distribution"

The key to finding rotation speeds from galaxy mass distribution is to realize that a disk can be represented by a series of rings. For a single ring a test mass at the center of the ring has no acceleration, but as it goes out toward the ring it is attracted toward the ring more strongly , until very near the ring. Outside the ring the attraction is again strong but reversed, and falls off as the radius increases, becoming asymptotic to that caused by the ring as a point mass. For a zero-thickness "wire" ring the acceleration becomes infinite at the ring in each direction, because the ring has mass but no volume. But a simple thought experiment shows this is wrong for galaxies. Imagine a test mass passing through a "cloudy" ring with thickness in the z direction (perpendicular to the disk) and also a little in the r direction. We know the acceleration should look like that of the wire ring until the test mass gets close to the ring, then the acceleration should smoothly change signs as it passes through the ring. Right at the ring the acceleration should be the same as if the mass of the ring was concentrated at the center (this is not obvious). A proper simulation will include this characteristic.

So the gravitational field for a ring is quite different from that for a sphere. In particular a test mass inside a spherical shell is not attracted at all by the shell. The approach that says the rotation speed at a given radius of a galaxy is set by the mass inside that radius is WRONG. The mass outside that radius can have a strong effect on that inside.

Thus the forward problem (acceleration, and thus rotation speeds, from mass distribution) becomes straightforward, given the geometry (including thickness) and mass or density distribution. Just add up the acceleration effects of all the rings on a test mass at selected radii. To do this I first compute the acceleration a(i,j) of a test mass at each desired radius j caused by the ring i with unit density. I say I here because most people do not consider thickness or density. The matrix solution for the total acceleration A(j) at ring j becomes, with p being the ring density,

a(i,j)*p(i) = A(j) , forward problem

Now the reverse problem (mass distribution from speeds) becomes easy because A(j) is known from the measured speeds, a(i,j) does not change and its inverse ainv(i,j) can be used. The solution is now for density and thus the local SMD (surface mass density, mass per unit surface area).

p(i) = ainv(i,j)*A(j), reverse problem

I think the best place to see more details of this development and application to several galaxies is in my paper, arxiv-ph/0309762 v2. Probably the best place to see how it was done in error is the original source of dark matter, van Albada et al, Apj 295 305 (1985). There they assumed that since galaxy light seemed to fall off exponentially to the rim, and mass was thought to be proportional to light, that ALL galaxy disks must have an exponential mass distribution. When they used the known analytic solution for a disk with exponential loading (correct, Freeman ApJ 160 811F, 1970) they found the measured speeds were much larger. Rather than putting the needed extra mass in the disk (perhaps because they didn't know how) they added spherical shells, centered on the galaxy center, with the density adjusted to cause the combined acceleration effects to approximate those needed. Wikipedia echoes those results.

Since there was no evidence of those shells, they called them dark matter, and it has aquired special properties as the time-and-money-consuming search for it has gone on over the years: it cannot be detected by its own or reflected radiation, only by gravitational effects. Well there is no doubt about that! After all it is imaginary. It's not hard to predict it will never be found. Fritz Zwicky coined the term back in the 30's, but he apparently thought the mass he needed was just ordinary non-star matter.

The dark-matter solution using spherical shells was obviously wrong because it allows an infinite number of solutions, as the ratio of shell matter to disk matter is changed. A good match of speeds can be found using spherical shells alone! Freeman and van Albada were good mathematically (not so their peer reviewer) so that the solution might have been a hoax. If so it has been one of the most successful scientific hoaxes of all time.

Ken Nicholson, knchlsn@sbcglobal.net

Excellent info, Ken. Thanks!

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

Thank you, Ken, for joining into the discussion. I have been going over some of your work and have decided that using rings is the best way to go with this. One reason being, we could stand on common ground in the discussion, but also, if I could work out the integration for the wires, and then for a ring of wires, then I could write a new program for just the last part of the integration for a set of rings from the center to the rim where the height and mass density of each ring could then be varied at will, allowing for various shapes and mass distributions within a very quick program. Since a particle that is very near the edge of a ring with some height will being lying against a surface with two dimensions, then the gravity should always be finite, even right up to the edge, and so there will be no need to consider the spacing between wires if I can work out the integration for it accordingly.

__________________
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."

I'm not sure what you mean by a ring of wires. If you mean the rings are made up of wires perpendicular to the disk (like my rods) then you can avoid divisions by zero like I did, by passing the test mass directly beteen 2 wires. If you mean the rings are zero-thickness wires as continuous circles, you will run into an infinity (probably with elliptic integrals) and have to decide how to dodge it. Why not send me an email with your equations and I'll see if I can help.

In the meantime you can look at arxiv:0803.0556 v1 by Feng and Gallo. They have an approach using elliptic integrals, and they show how to easily compute them, and (not so easily) fix the singularities. I prefer a discrete number of rods for each ring. I've gone as high as 720/ring, but usually I only use 180 and with symmetry that's one per degree, and get great accuracy, without double precision. Also I only need about 40 to 80 rings. I don't need any fancy math methods, just stick to sines and cosines, and the new computers handle it all just fine.

No predictions are hard, the issue is if they are correct. What evidence can you cite that your prediction is correct? Dark matter has already made several predictions that have proven correct, even if it has not yet been detected. On what basis can you predict it won't be detected?
Your approach allows an nfinite number of solutions too, in thee dimensions.

But that doesn't surprise anyone, and says nothing about dark matter.

__________________
If we understood everything going on in the head of a pin... we still wouldn't know not to step on the pointy end.

People think the problem with models is that they are limited by our minds, but the greater problem is that our minds are limited by our models.