Well, this is astonishing. In response to Cougar's post, I changed the dimensions of the disk in the program from 100,000 light years across and 12,000 light years thick for the Milky Way to 100,000 light years across and 1,000 light years thick, which accounts only for the stars themselves and not the dust and gas which extends out further, to see what that would give for a lesser thickness. Well, the force still comes out to almost twice as great as that of a sphere and the energy 1.2 times as great. I even found for twice as many points along each axis, for eight times as many points in all, and it comes out the same. If that continues with smaller thicknesses, then the gravity of a disk plane is finite. I'm not sure what that means yet, unless the integration must somehow be performed in a much different than the ordinary way or something. It might also have something to do with the natural "clumping" that is done when considering the mass over a finite number of points in the program instead of an even spreading out of the mass throughout the disk, as would be the case when performing integrations for it. I will also recheck the program. This sort of clumping does definitely seem to be the way to go with this.

__________________
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."

Whoops. I rechecked the program and it appears I changed the numbers of points found for along the z axis, but not the actual thickness of the galaxy itself, so it was still finding for the same thickness, but just using a lesser number of points along it. Sorry. So for a thickness that is 3/25 that of the diameter of the galaxy, the force at the rim is about 1.9 times greater than that of a sphere of the same mass and the energy is 1.2 times greater. For a thickness that is 1/12 as great, or 1/100 of the diameter, the force at the rim is 2.75 times as great as that of a sphere of the same mass and the energy is 1.25 times as great. I'm still going to try working with a "clumping" variable in the integrations, though.

__________________
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."

By the way, to be fair, this is not saying that a galaxy is regarded in the same way as a sphere. If that were the case, the force would climb in direct proportion to the distance all the way to the rim, and the rotational speed would be constant, so that the entire galaxy should rotate together with the same period of rotation for all points. The rotational speed at the rim, figuring for the total mass and radius of the Milky Way galaxy would then be about twice as great as is observed, which would actually require some form of "negative" dark matter, or lesser matter to account for the lesser speed observed at the rim. Also, if regarded as just a thin disk of uniform density, then the rotational speed would be expected to climb very rapidly at the rim, so a leveling off of the rotational speed would also require some form of "negative" dark matter. As it is, though, the rotational speed is predicted to climb in the same way it is observed to do up to some peak and then drop back down, but instead it just levels off at the peak, as shown in this link. The difference between the predicted curve and that of a geometry incorporating a uniform density would be that of the predicted mass distribution for the galaxy. I guess my next step, then, is to find the mass distribution that would give a curve that matches that for the predicted curve for a galaxy and go from 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."

Cougar,

You might want to check my post #30 again. I accidently erased it , so I redid it, but I know I responded differently to your second paragraph after rereading it than I did the first time, not that I remember my response to that anyway, but I think I got the other two responses the same.

__________________
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."

It sounds to me like you still have to get the math right before the
simulations can have any value. The fundamental problem causing
infinities must be a common one -- I'm sure it has a common solution.

-- 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

The math is right, although I am still trying to find out the predicted or observed mass distribution of galaxies, yes. The infinity is for the edge of a flat disk plane of uniform density, not one with some thickness, but as far as I know, the exact solution for a thick disk is not known either, only approximated, so that is why I am just using the program unless I can manage to work it out through the integrations. But even then, the program does the same thing as the integration, just point by point, and I can vary the mass distribution at will instead of reperforming the integrations accordingly, and it's much quicker to boot.

__________________
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."

Hi grav,
I think you are on the right track in referring to the above paper. Nicholson's calculation was discussed extensively in this forum here:

Galaxy Rotation Curve Calculation.


Nicholson's solution avoided the dreaded elliptical integral solution by using a numerical interation method in a computer program he used while working on the Space Shuttle Program. Dont stop reading the thread after the first parts, as it takes a turn later, and Nicholson joins the discussion.

TomT

Thanks, Tom. Good to hear from you again. I remember you helped out with this when I first went through it. Hh and publius have since given me a two week crash course for finding integrations and I have learned a lot from them. I have been trying to use it for this, but you say it's not necessary, at least not with a computer program for it, or by using iterations, eh? I have read through some of Nicholson's papers before. Maybe this time I can follow it better. I'll read through it again and try to use any calculations he provides to see what it comes up with.

__________________
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 don't think it can be right if it produces infinities, even if only for
the unrealistic case of zero thickness. I suspect that the added
calculations for points out of the plane are partially masking the
error which is apparent at zero thickness.

If it happens at the edge, it should happen everywhere. I suspect
that in the interior, the error in one direction is masked by the error
in the opposite direction.

Can you describe how the infinities come about in your program?
Does your program actually calculate the attraction between a mass
at one point and the same mass at the same point? Or does it
calculate attraction between masses at two points that are very
close together? Or what?

-- 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

I strongly disagree, for one simple reason - you have two stars, one on the edge of the circular field, the other positioned such that half the mass is inside it's orbit, the other half outside it's orbit. That mass is finite, and so is the gravitational attraction towards the center. Nothing goes to zero. Nothing approaches infinity.

We're dealing with finite entities, here, and the model is patterned after, and close to, our own Milky way. Clumping doesn't enter into the picture, nor do the spiral arms, because it's at such a distance, and there are so many clumps and arms, that while these might cause pertubations in our orbit, those are minimal unless we get fairly close to a clump. If that happens, then, yes - we might get tossed outwards (slingshot) or decayed inwards (orbital braking).

__________________
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."

You're really missing something, here, grav...

Instead of stars, let's use steel ball bearings, 1" in diameter, arranged in a circular plane. Let's use the same numbers as in the first problem I stated, and set things in motion such that all initial orbits are circular given the initial conditions. Let's assume we're in the center of that supermassive void they discovered so the local gravitational gradient from even the closest object is zero.

Forget the math you're using for just one second and answer this question: Is gravity "just right on the 'edge' of infinity?"

I'll answer it for you: Hardly.

In fact, it'll be so low that the outermost ball bearing will have an orbital velocity measured in micrometers per second, if not hour.

If we blow this back up to the model I proposed, the proportions increase along with it, but nothing, especially the gravitational tug at the edge of the disk, get's anywhere close to infinity, any more than it does for our own Milky way.

Again, the model is patterned after our Milky Way, and for all intents and purposes, when you get beyond about 1% of the Milky Way's radius, things can be approximated by using a thin disk model.

Thus, I conclude that since the only substantial difference between my model and the Milky Way is local clumping, I think you're tripping over your calculations, somewhere...[/quote]

Thanks, Jeff Root, and Cougar, for posts 27 and 28.

__________________
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."

To answer these last three posts at once, hopefully, the integration for a disk plane does indeed come out to a result of infinite acceleration at the very edge for a point particle. A larger sphere at the edge, half in and half out should experience the same thing, at least for the point at its center, or considering the sphere a point mass at the center. In other words, if I were to perform separate integrations for every point in the sphere toward the disk, the point in the center would still blow up regardless of the rest of the points, and other points in the sphere shouldn't make much difference adding or subtracting with their finite contributions, then, I wouldn't think.

If we took a bunch of spheres and arranged them in a disk, then they would have thickness, and the acceleration toward the disk would be finite.

The computer program doesn't blow up at the edge, since it is calculating over clumped points within the disk instead of a mass spread evenly throughout the disk, as with an integration. However, as I make the thickness of the disk smaller and smaller, the gravity at the edge climbs higher and higher.

I know it doesn't seem to make much sense that the gravity at the edge of a finite mass disk should be infinite, but I have two theories about this. The first is with the thickness. With zero thickness, the particle does not lie against a surface as it would with a sphere or thick disk. Since the force varies with the square of the distance, we might need two dimensions to cancel out the infinities. That can be tested by finding for the potential energy, which only varies with the inverse of the distance, should become finite against the one dimension of the disk's edge. The other is with the clumping. I'm thinking stars, even gas and dust, are not spread evenly throughout a geometry, so the minimum possible distance between points must be taken into account.

I am wondering about the precise relationship between the resulting force and energy in all of this.

__________________
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 "integration" you're talking about. All you need is the distance from your test particle to each object in the galaxy and the approximate mass of each object. I think you could simplify it to a single central black hole 5% to 10% of the total mass, and the rest solar-sized masses. Let the computer add up the resultant accelerations....

This "infinite gravity" is obviously an error.

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

For a central black hole, we would still perform the same integration for the disk, and then just add the gravity of a point mass for the central body, which wouldn't add much considering the mass of the rest of the galaxy. In any case, here's the integration for a particle on the rim of a flat disk plane from post #14.

__________________
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."

grav,

I'm going to reply to a couple of things you just said in a separate post in a
few minutes, but right now I want to suggest a test I just thought of.

Can you have your program calculate the force on a plane shape other than
a disk? Like, the force on a point at the middle of an edge of a square? Or
on the edge of a U-shape? Like this:

The period indicates the location of the test point.

From your descriptions, I expect that your program will give a result of
infinite force toward the bottom of the U, and the force from the upper
part of the U, while calculated, will have no effect on the result. If it
does have an effect, that could tell you something about the problem.

You might want to use an L-shape instead of a U. The lopsidedness
could provide more diagnostic data.

-- 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

That should make the problem much easier to resolve than if it only
showed up when it reaches zero thickness. This way, you will get
a definite indication that you have found the problem and fixed it,
rather than applying some arbitrary restriction which makes the
infinities go away but doesn't necessarily give correct results.

I don't understand the geometry you are describing when you say
"...lie against a surface...".

I think you shouldn't be producing infinities in the first place. Cancelling
them out sounds like a hack, and a particularly dangerous one.

-- 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