As discussed in a thread on this same topic perhaps a year ago, you have now started to produce models of mass which can create rotation curves which look like those of actual galaxies. Well done! That's quite an achievement.

You are at roughly the point mainstream astronomers reached 60 years ago. There are many _possible_ mass distributions which can reproduce the observations, so it's hard to figure out which one(s) are the right one(s). This is an example of an "inverse problem", which lacks a unique solution.

Your next mission, should you choose to accept it, is to demonstrate that your mass distribution(s) is stable over long time scales -- say, ten or twenty orbits. Or, if not stable, show that the instabilities which arise are perhaps similar to some observed properties of galaxies.

Good luck!

Thanks, stupendousman. I appreciate that. I'm sure Ken N does too.
For a particular shape for a galaxy, there is only one mass distribution that will provide for the rotation curve from the center to the rim. That is the point of the iteration, to find it. I have also tried it for different thicknesses for a disk, and the total mass required does not change very much. I'm not accounting for a central supermassive black hole, though, but I could add that in easily enough, but it shouldn't make too much difference either, and spiral arms, which Ken G said shouldn't matter much also, and I hope that's true, since they would probably be very difficult to accommodate for in the program.

That is also part of what the iteration does, kind of the point of it also. It gives the density distribution required for stable orbits of the stars at a particular rotation speed at each particular position. That means that all of the stars that are already considered to be positioned there are orbitting in that way, although I am considering only circular orbits, so highly elliptical orbits may or may not be stable, I don't know. If not, then they would have probably been thrown off a long time ago during the formation of the galaxy and only the less elliptical orbits would remain, and the iteration goes by the average density distribution over all of the current positions of the stars in their orbits. The only thing the iteration does not tell, then, basically, is how the galaxy forms in the first place.

__________________
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."

Looking back over StupendousMan's post, and thinking about the replies of a couple of other posts I was wondering about near the beginning of this thread, I think there might be a bit of a misconception about what the program is actually doing. There are simulations of galaxy formation and orbittal programs and such, and they are pretty nifty, but this program is nothing like that. In fact, it is not even a simulation. It's like freeze-framing a galaxy and just finding the accelerations for particular distances with a particular density distribution, and then for what the orbittal speed would be with that acceleration at that distance, using ordinary Newtonian gravity, without actually putting the galaxy in motion or simulating the orbits of individual stars and such. If I had been able to work out the full integration for a disk and/or other disk-like shapes to find the acceleration of gravity at some distance from the center of the body, then I wouldn't have needed the computer program to begin with. It would have just come out to some simple formula, like a = GM/r^2 does for a sphere. But since I just kept getting elliptical integrals and such for the result of the integrations, which require integrations or summations for them as well, then a computer program is the best and quickest way to go with that, so it is simply performing the mathematical application to finish the integration for finding the acceleration and orbittal speed at some distance from the center of a body with a particular shape and density distribution, more like just a fancy calculator than a simulation. Of course, using a computer program to find the required density distribution for the reverse problem does make life a heck of a lot easier for that also.

__________________
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."

Nice work grav in working out all these calculations, and so rapidly. I am glad you understand the details of how to do Ken Nicholson's calculations.

I am not sure of the contention that accounting for the spiral arms shouldn't matter much. This is saying that mass variation in the circumferential direction doesn't matter. So take the case of a 2 armed spiral galaxy. If you observe the light variation in the circumferential direction, would you say there is much variation or very little? It looks like a lot of variation to me. If M/L is important in estimating the mass, then I think circumferential variation of light is important, or should at least be checked out to see.

If you are doing a numerical/iterative solution to this problem, can you work in circumferential variation using either a table of mass values, or something like a sine wave mass variation in the circumferential direction. I am really curious about this effect.

Not that you don't have enough to do already :>).
TomT

That is not correct. In general, there are many possible density distributions which might create very similar rotation curves. That was one of the main points of my previous posting. If someone tells you the distribution of mass, then you can uniquely figure out the rotation curve; that's a "forward problem". However, if someone only provides you with the rotation curve, then it is an "inverse problem" to go back and figure out a mass distribution which could account for it.

At least, that's true in the most general sense. For some particular shapes of rotation curves, it may be true that only one sort of mass distribution will work; but I don't think that's the case for typical galactic rotation curves.

No, sorry to contradict you again, but your model -- if I understand it correctly -- does not tell you anything about the stability of the orbits. That was another point in my previous post.

For example, if you make a very simple galaxy -- a point mass of 100 units, a star of mass 1 unit at distance 10, and a star of mass 1 unit at distance 20 .... will the system be stable over many orbits of the stars? Your current method, if I understand correctly, does not address this issue. The answer for this simple system is "no", since the two orbits have a commensurate period, so perturbations between the stars will grow with time.

That's the next step in your work, I think. And it requires some new techniques in your code. Fortunately, other people have done this sort of thing already, so if you search through the literature, you can get a lot of tips.

TomT wrote:

Typical spiral arms are much more prominent in light than they are in mass, as images in the infrared (and models) show clearly. Nonetheless, they can have important dynamical effects on timescales of several rotation periods. It is this sort of effect which will require grav to use some new techniques. It _may_ be possible to do some stability analysis of rings or similar simple shapes analytically, but the most general method for dealing with this is to move to N-body calculations.

To Tom T

Thanks for pointing that out Tom. Although he's got a ways to go, Grav deserves a lot of credit for fighting it through to this point. It looks like he came up with an iteration scheme for the reverse problem on his own. Also he is only number 2 to solve for density, and that requires a knowledge of thickness. Most people solve for SMD, that doesn't require thickness, and results in a total mass that is a little light.

As you know, people other than myself that have since solved the reverse problem use a matrix inversion program. I used the iteration scheme because I didn't know how to program the matrix solution, and couldn't find a canned program. In my last paper I show the matrix solution, because it so easy to follow. An iteration approach is a perfectly valid solution. I call it an organized method of trial and error.

Ken N

To Grav

Here's my results for your reverse problem. Since I don't know your exact rotation profile, the densities are probably not a match, but total mass looks about right. Be sure to correct for the units. Answers use msuns, pcs, km/sec. The inputs I used are:

rmax = 100000/2 lt years = 30660.7/2 pcs
h = 1000 lt years = 306.607 pcs
vm(i) = 6(i)/20*220 km/sec, i <= 6
= 220 km/sec , i > 6

How are you coming along on your checks for equations and coding?

Ken N

grav's problem, 20 rings, 120 rods, reverse
nr,nt = 20 24

dimensioned results

rmax, vol, mtot, SMDav, rhoav, =
1.5330E+04 2.2638E+11 1.1041E+11 1.4954E+02 4.8774E-01
SMDmax,rhomax,vkrim,vmmx,akrim =
7.1942E+02 2.3464E+00 1.7599E+02 2.2000E+02 2.1132E-12

j _____r_________ rr________ rho _______ SMD ________rm _______ m
1 3.8326E+02 4.2925E+02 2.3464E+00 7.1942E+02 7.6652E+02 1.3279E+09
2 1.1498E+03 1.1651E+03 2.3011E+00 7.0555E+02 1.5330E+03 5.2349E+09
3 1.9163E+03 1.9255E+03 2.2177E+00 6.7995E+02 2.2996E+03 1.1510E+10
4 2.6828E+03 2.6894E+03 2.0955E+00 6.4248E+02 3.0661E+03 1.9812E+10
5 3.4493E+03 3.4544E+03 1.9189E+00 5.8834E+02 3.8326E+03 2.9586E+10
6 4.2158E+03 4.2200E+03 1.6549E+00 5.0740E+02 4.5991E+03 3.9888E+10
7 4.9824E+03 4.9859E+03 1.1213E+00 3.4379E+02 5.3656E+03 4.8137E+10
8 5.7489E+03 5.7519E+03 8.7840E-01 2.6932E+02 6.1321E+03 5.5594E+10
9 6.5154E+03 6.5181E+03 7.1791E-01 2.2012E+02 6.8987E+03 6.2501E+10
10 7.2819E+03 7.2843E+03 6.0007E-01 1.8398E+02 7.6652E+03 6.8954E+10
11 8.0484E+03 8.0506E+03 5.0842E-01 1.5588E+02 8.4317E+03 7.4996E+10
12 8.8149E+03 8.8169E+03 4.3430E-01 1.3316E+02 9.1982E+03 8.0650E+10
13 9.5815E+03 9.5833E+03 3.7253E-01 1.1422E+02 9.9647E+03 8.5920E+10
14 1.0348E+04 1.0350E+04 3.1971E-01 9.8024E+01 1.0731E+04 9.0806E+10
15 1.1114E+04 1.1116E+04 2.7345E-01 8.3841E+01 1.1498E+04 9.5294E+10
16 1.1881E+04 1.1882E+04 2.3196E-01 7.1121E+01 1.2264E+04 9.9363E+10
17 1.2648E+04 1.2649E+04 1.9375E-01 5.9405E+01 1.3031E+04 1.0298E+11
18 1.3414E+04 1.3415E+04 1.5734E-01 4.8241E+01 1.3797E+04 1.0610E+11
19 1.4181E+04 1.4182E+04 1.2081E-01 3.7042E+01 1.4564E+04 1.0863E+11
20 1.4947E+04 1.4948E+04 8.0913E-02 2.4808E+01 1.5330E+04 1.1041E+11


i _____ rt __________ at ___________ v ___________ vm
1 7.665168E+02 1.834592E-12 3.666669E+01 3.666667E+01
2 1.533034E+03 3.669180E-12 7.333335E+01 7.333334E+01
3 2.299550E+03 5.503769E-12 1.100000E+02 1.100000E+02
4 3.066067E+03 7.338361E-12 1.466667E+02 1.466667E+02
5 3.832584E+03 9.172947E-12 1.833333E+02 1.833333E+02
6 4.599101E+03 1.100754E-11 2.200000E+02 2.200000E+02
7 5.365618E+03 9.435032E-12 2.200000E+02 2.200000E+02
8 6.132134E+03 8.255653E-12 2.200000E+02 2.200000E+02
9 6.898651E+03 7.338357E-12 2.200000E+02 2.200000E+02
10 7.665168E+03 6.604522E-12 2.200000E+02 2.200000E+02
11 8.431685E+03 6.004111E-12 2.200000E+02 2.200000E+02
12 9.198201E+03 5.503769E-12 2.200000E+02 2.200000E+02
13 9.964718E+03 5.080403E-12 2.200000E+02 2.200000E+02
14 1.073124E+04 4.717517E-12 2.200000E+02 2.200000E+02
15 1.149775E+04 4.403017E-12 2.200000E+02 2.200000E+02
16 1.226427E+04 4.127829E-12 2.200001E+02 2.200000E+02
17 1.303079E+04 3.885016E-12 2.200001E+02 2.200000E+02
18 1.379730E+04 3.669183E-12 2.200001E+02 2.200000E+02
19 1.456382E+04 3.476069E-12 2.200001E+02 2.200000E+02
20 1.533034E+04 3.302267E-12 2.200002E+02 2.200000E+02
21 1.609685E+04 2.643469E-12 2.016967E+02 0.000000E+00
22 1.686337E+04 2.261172E-12 1.909326E+02 0.000000E+00
23 1.762989E+04 1.987245E-12 1.830171E+02 0.000000E+00
24 1.839640E+04 1.772754E-12 1.765761E+02 0.000000E+00

dimensionless results

j ___rrd ___ rhod _ rmd ____ md ____ rd______SMDd
1 0.0280 4.8107 0.0500 0.0120 0.0250 4.8107
2 0.0760 4.7180 0.1000 0.0474 0.0750 4.7180
3 0.1256 4.5468 0.1500 0.1042 0.1250 4.5468
4 0.1754 4.2962 0.2000 0.1794 0.1750 4.2962
5 0.2253 3.9342 0.2500 0.2680 0.2250 3.9342
6 0.2753 3.3930 0.3000 0.3613 0.2750 3.3930
7 0.3252 2.2989 0.3500 0.4360 0.3250 2.2989
8 0.3752 1.8010 0.4000 0.5035 0.3750 1.8010
9 0.4252 1.4719 0.4500 0.5661 0.4250 1.4719
10 0.4752 1.2303 0.5000 0.6245 0.4750 1.2303
11 0.5251 1.0424 0.5500 0.6792 0.5250 1.0424
12 0.5751 0.8904 0.6000 0.7304 0.5750 0.8904
13 0.6251 0.7638 0.6500 0.7782 0.6250 0.7638
14 0.6751 0.6555 0.7000 0.8224 0.6750 0.6555
15 0.7251 0.5606 0.7500 0.8631 0.7250 0.5606
16 0.7751 0.4756 0.8000 0.8999 0.7750 0.4756
17 0.8251 0.3972 0.8500 0.9327 0.8250 0.3972
18 0.8751 0.3226 0.9000 0.9609 0.8750 0.3226
19 0.9251 0.2477 0.9500 0.9838 0.9250 0.2477
20 0.9751 0.1659 1.0000 1.0000 0.9750 0.1659


i ___ rtd ____atd ____ vd
1 0.0500 0.8682 0.2083
2 0.1000 1.7363 0.4167
3 0.1500 2.6045 0.6250
4 0.2000 3.4726 0.8334
5 0.2500 4.3408 1.0417
6 0.3000 5.2089 1.2501
7 0.3500 4.4648 1.2501
8 0.4000 3.9067 1.2501
9 0.4500 3.4726 1.2501
10 0.5000 3.1254 1.2501
11 0.5500 2.8412 1.2501
12 0.6000 2.6045 1.2501
13 0.6500 2.4041 1.2501
14 0.7000 2.2324 1.2501
15 0.7500 2.0836 1.2501
16 0.8000 1.9534 1.2501
17 0.8500 1.8385 1.2501
18 0.9000 1.7363 1.2501
19 0.9500 1.6449 1.2501
20 1.0000 1.5627 1.2501
21 1.0500 1.2509 1.1461
22 1.1000 1.0700 1.0849
23 1.1500 0.9404 1.0399
24 1.2000 0.8389 1.0033

Oh, yes. In the case of a 2 armed spiral galaxy, I'm sure there is a significant difference along the circumferential direction. What I was referring to was more along the lines of a galaxy with many spiral arms, spun around and around, so that it becomes very close to that of a uniform disk. The difference for that should become insignificant whether a particular star is located in or near a spiral arm, or in the space in between, as far as the gravity of the galaxy as a whole upon the star is concerned.

I have been trying to think of a simple way to find for the gravity of a two armed galaxy where each arm might go, say, half way around the circumference, with some drop in density distribution along each arm. So far I haven't been able to come up with a way to do it in a simple enough way with a reasonable running time for the program. I'll keep working on it though, although I can't promise anything anytime soon, but I am also curious about that.

__________________
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."

StupendousMan,

I'm sorry if I'm not being clear enough about what the program is doing. I'll try again. If two stars approach too closely in the galaxy, where the gravity between them begins to overcome the overall gravity of the galaxy itself, then they might have a significant effect on each other's orbits, like what you were saying in your post, that's true. They might slightly alter each other's orbits, or collide, or start orbitting each other, or one or both might get gravitationally slung out of the galaxy altogether, or anything in between. But we are not concerning ourselves with that in the program. As far as we are concerned, if the orbits of a couple of stars become unstable and get slung out, then so be it, the galaxy is still there.

But more than that, the orbit of a single star placed within the galaxy at some position is always "naturally" stable in the program, because in essence what we are doing is grounding all of the stars in the galaxy into a fine dust and spreading the atoms from one end of the galaxy to the other with the same general density distribution, so there are no other massive bodies to effect the orbit of a single star placed at some position, only a very, very smooth density distribution. I know we can do this because if we were to use only 1000 point masses spread throughout the galaxy with masses that would provide the same density distribution, as I was doing with my first program, then the acceleration at some distance from the center is close to that for what it would be for a fine dust, which is what the integration for a smooth density distribution would give us. If we were to use more points, say 10000 or 100000 point masses, then it gets even closer, so close that the difference becomes neglible. For hundreds of billions of point masses, like the number of stars in a galaxy, spread uniform with the density distribution, the difference between that and a fine dust would be practically imperceptible.

__________________
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 spent yesterday trying to integrate for a ring through a sphere, where I could keep the curved edge for the height instead of cutting each ring off flat using an average height. So I was basically trying to integrate for a wide cylinder with a sphere cap at each end minus a thinner cylinder with sphere caps. Unfortunately, the integration for that still eludes me; I'll have to study it more.

Anyway, today I just decided to go with the cut-off average height over many rings to see if that would be close enough for the check. I figured that for thin disk galaxies, a cut off height would work fine, but I wasn't sure about a sphere with the central height equal to the radius and all, so I just figured I would have to run it over many rings with a very long running time. Well now I wish I had just gone ahead and done that in the first place, because all I did was change a couple of lines and it turned out to be very precise even over just twenty rings, so now I'm also that much more confident about using a cut off height for a thin disk, even over a small number of rings. I ran it for a height of [sqrt(r^2 - r2^2) + sqrt(r^2 - r1^2)] / 2 for each ring with a constant straight-line rotation speed curve from zero at the center to the rim speed, otherwise using all of the same parameters as before, with r = 100000 light years and a rotation speed at the rim of 220,000 m/s, and the density distribution over twenty rings was very good, surprisingly very constant as it should be, rising from about 7.7*10^(-22) kg/m^3 at the center to only about 7.8*10^(-22) at the rim, giving a total mass of M = D*(4pi/3)r^3 = 3.435*10^41 kg with D = 7.75*10^(-22) kg/m^3. So using that total mass, we can get back our rotation speed for the rim with v = sqrt(G*M/r) = 220,165 m/s. It checked out very well indeed. Cool.

__________________
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."

Um, since I got a fairly constant density over the rings when I ran that, I didn't have the program add up the mass; I just found it using basically the average density across the rings. Well, I had the program do it this time, adding it up for each of the rings individually, and since the density and the volume of the rings increases toward the rim, it gave me M = 3.616*10^41 kg, which translates into a rotation speed at the rim of a perfect sphere of 225,874 m/s. Slightly off for a small number of rings, but still not too bad.

__________________
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 let the program run for a while with 40 rings, and got a mass of 3.5243*10^41 kg and a rotation speed at the rim of 222,997 m/s this time. Getting better. For a thin disk, it would be even much more precise than that, I'm sure.

On an interesting side note, when I was performing some integrations yesterday, I ran across a part of one to the effect of log(x) / sqrt(a^2 - r^2). Finding that for x = -r to r, I got [log(r) - log(-r)] / sqrt(a^2 - r^2) = log(r / -r) / sqrt(a^2 - r^2) = log(-1) / sqrt(a^2 - r^2), where a < r, so log(-1) / [i sqrt(r^2 - a^2)]. So then I wanted to find for log(-1) / i. Well, Euler's formula says e^(i pi) = -1, so log(-1) = i pi, and log(-1) / i = pi. I never would have thought in a million years that I would just happen to run across and actually use Euler's formula so casually.

__________________
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, whoops. Sorry, TomT. I don't know what I was thinking. I guess I still had that iteration for the reverse problem stuck in my head, which if I tried to do what you were asking using that, the running time for the program finding for all positions radially and along the circumference would be exponentially exponential. But since you just want to compare the gravity of a set of spiral arms to that of a uniform disk, I can just use one of the earlier programs to find for that. I think Ken N's would be best, because it already has rods situated around the circumference of each ring, so all I have to do, for a rough comparison anyway, is to run it where the height and/or mass of the rods of each ring starts at zero, and then runs up to 1 after half the circumference, then drops to zero again for the other spiral arm, and climbs back to 1 coming back around to the starting point. I'll get to work on that right away.

__________________
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, that was interesting. I didn't even have to run the program for the spiral arms to find the gravity. I realized as I was putting it in that at least for the way I was doing it, it would come out exactly the same as that of a uniform disk with the same average density. If we vary the density along the circumference of the rings from zero to one and then drop to zero on the other side and back to one the way I described, then the gravity that acts directly toward the disk at any position in the galaxy will be balanced by symmetrical points on either side of the circumference. In other words, let's say we found the gravity from some point in the galaxy that acts on our position. The density for that point, which is proportional to the gravity for that point, might be x. Then the density of the same symmetrical point on the other side in the other spiral arm would be 1-x, and the gravity from that point would also be proportional to 1-x. Adding both densities to find for the radial gravity directly toward the center of the galaxy would give us x + (1 - x) = 1, the same as finding for an average density for both points of 1/2, and that would be true for all symmetrical points in the galaxy.

The only difference with the densities of the spiral arms found in this way, then, would be that of the gravity along the circumference itself. For a disk that is uniform around its rings, there would be no tangential acceleration of gravity, since the gravity from both sides cancel out in this direction. But with spiral arms, there will be an acceleration toward the largest part of the spiral arm that is closest to the position we are finding for. So even more interestingly, then, for two disk galaxies of the same total mass, one with a uniform density and the other with a density that varies along the circumference as described, both galaxies will have the same radial acceleration of gravity, but the one with spiral arms will have an additional tangential component for the gravity, giving it a greater gravity overall.

__________________
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."

Ken N,

Here is the iteration program for the reverse problem I am using (you can just copy and paste it for a similar language). For ten rings, it takes about twenty seconds to run on my computer. For more than ten rings, I had to write a separate program to store the array Q(N,N), but I didn't bother including that part since your language would probably do it differently than it did for mine. To find for a varying height, just add lines 116 and 212. For instance, for the check with the sphere, I added 116 H=sqrt(R^2-R2^2)/2+sqrt(R^2-R1^2)/2 and 212 H=sqrt(R^2-(Nn*J*Dx)^2)/2+sqrt(R^2-(Nn*(J-1)*dx)^2)/2 . The densities of the rings are given from the center to the rim in multiples of 10^(-20) kg/m^3 and the total mass is given after the "*" in multiples of 10^9 solar masses.

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

"Let's define another operator, Sz, which we won