True, although not necessarily uniquely, we must bear that in mind.
Agreed.

To Ken G

There is only one rotation profile for a given thickness and mass distribution, and only one mass distribution for a given rotation profile, based on the assumptions made. When I constrained things (as a way to simplify the problem) to a constant density and thickness at each ring, things were locked in stone. From then on it's just arithmatic. Looked at from the equations point of view, the forward problem can set the acceleration profile A(j) from a(i,j) and rho(i) in only one way.

......... a(i,j) rho(i) = A(j) , it's a lock

Then given a(i,j) and after some pain ainv(i,j) the reverse problem can only result in one answer for the density distribution ( and thus the SMD) for a given A(j).

........ rho(i)= ainv(i,j) A(j) , it's another lock

I just do the reverse problem in a different way but get the same answers.

With the use of dark-matter spheres ala van Albada and any fixed estimate of the disk mass distribution for a galaxy, and using stepwise values of the spherical shells to make the speeds match, there could again be only one mass distribution for a given rotation profile. Unfortunately, as described before, that is not a valid use of Newton's law. When the fixed estimate of the disk is zero, all mass is in the dark-matter spheres. When the estimate is too large (usually near the origin), the local sphere mass there will be negative.

Yes, I agree, but this just means the uniqueness is determined by the assumptions. As there are many possible assumptions for the mass distribution geometry, there are also many solutions. The most obvious one would be purely spherical, with no disk at all. Any velocity law with a monotonically increasing r*v^2 can be fit with a purely spherical mass distribution, and since yours obeys that, it can be fit spherically. Not that it would be easy to explain why the light comes from a disk, but you did say to ignore M/L issues....

Well we've set the equation to be used to find v(r) as eq 3 p3. But I would sure like to hear some arguments and questions. Where are you Cougar? Anyway let's proceed. To see if this process is any good we have to check against known theoretical results, and I chose the single ring, the constant-density sphere, the analytic solution for the exponential loading with "zero thickness," and a "wire ring" with a gap. Also I talk about the proof for the reverse problem.

Starting with the single ring at Par 2, p7, the dimensioned form of the equation is reduced from the integrand of eq 2-146 of B&T, after a little juggling. And we're into elliptic integrals, with all the table look ups and interpolation. I then put it in dimensionless form by dividing through by the Kepler acceleration at the ring to get eq 5, p7. This is the result for the "wire" ring and blows up at the ring. Figs 7 and 8 show the comparison with my results for a 5% thickness ring. It is seen that my solution follows closely to wire-ring theory until close to the ring from either inside or outside. My solution also goes smoothly almost exactly through the Kepler value (accd=1 for the dimensionless form) in a physically believable way. As the ring gets thinner (results below figs 7,8) the results get closer to those of the wire ring, as expected.

This check is good enough. Score no. 1 for the home team.

But of course that's not enough. Lets try the sphere, in fig 6, p7. We're using rings instead of the standard way of spherical shells. The standard solution is easy to integrate on the back of an envelope. The shells are not. Near the center, the shells make a good approximation, because they match the surface of the sphere well. At the rim, not so. The rings are a poor match there. So use a lot of rings and still expect an error, both in speed and volume. Solving for the dimensionless vd(rd), (it should be vd=rd), it is seen the mass buildup is much faster for the rings than the spheres, but
vd(rd) is the same. And yes there is a small error with 100 rings. Using 5 rings, the solutions are not all that bad and could be used well enough for the buldges at the center of galaxies.

Score no. 2

Now for the theoretical solution of the exponential loading (by Freeman). Here he may be checking me or I may be checking him. If we match then it's a good bet that both are right. Fig 9, p9 shows a good match except at the rim, where I quit and he keeps going to infinity. The mass he uses beyond my rim, when removed, causes a slight speed increase at the rim. Good enough check. He's right and I'm right. By the way, I didn't bother cranking out the speeds here according to his formula (with elliptic integrals), I just picked them off fig 2-17 of B&T.

Score No. 3

For the single ring, I know the acceleration should be Kepler at the ring, but I don't know how to prove it. Does anyone else? I've tried with a computer, putting a test mass at the rim of a bunch of point masses and increasing the number. The acceleration keeps going well beyond Kepler. So I tried this coward's way out. Find the size of a gap in a wire ring that results in the Kepler acceleration at the ring. This we can do on the back of an envelope and a book of integrals, and the answer, p9, par 4, shows the gap is small enough to back up the Kepler result there. Close but no cigar.

Score No. 3 1/2

How about the reverse problem? If I compute the rotation profile for a given density distribution (with fixed thickness distribution), then use that rotation profile and thickness distribution to find the density distribution, the density distributions had better match! In effect this checks the matrix operations discussed before. So take a look at p9 under Appendix C. That's good enough for government work! This required a little adjustment of the location of the rods in the fundamental segment, shifting them a little inside of the cg location.

Score No. 4 1/2

Well with these checks we now have confidence in the equations and coding, and are willing to trust the computer results.

to be continued, with trials for more confidence building

My assumptions are based on what I see and what makes sense to me. NGC4013 on p2 fig 1 looks to me like a disk with thickness increasing toward the center, and an "atmosphere" with more or less exponential density decreasing away from the disk. There are spherical clusters that probably could be treated as a collection of spherical shells. But I don't think it makes sense to use spherical shells to represent galaxies. Part of the reason is that these shells, if centered on the galaxy, would have to build up a large pressure at the center of the galaxy to support them (like stars), and there is no evidence of that sort of pressure. Without that pressure they would have to collapse into the disk or impact it if in orbit, and some "halos" seem to be doing that.

Granted, it is a perfectly reasonable thing to do. I'm just saying the mass could also be distributed in other ways.

But these are all the reasons that such a spherical distribution would need to be noninteracting with normal matter, so that you wouldn't notice its pressure and it would not impact the disk, other than gravitationally.
So we would seem to have two possibilities for fitting v(r): a disk model with normal matter and a maximal amount of mass so that it can "go it alone", or a disk with minimal matter and a spherical distribution of noninteracting matter that not only has a lot more mass than the disk, it also has more mass than the purely disklike model. The latter is certainly very "convenient", since one can always invent physics to fill in holes, and nobody would ever prefer this to the former without some other driving reason. The other driving reason is the M/L ratio, but right now all we are doing is looking at the disklike solution. I must say the calculation you are describing sounds like a perfectly normal calculation to me, nothing heretical there.

There do appear to be some subtleties about ring distributions, however. If there is any error, it is going to be due to these pesky singularities. Obviously you have bent over backwards to try to address them, and it's not your fault they exist-- any calculation of this type is going to run into them. So the validity of your method sounds like a research question to me. So far I see nothing that would argue against publication of your paper, it might be that your claims are made in a somewhat polemic way so you are getting pigeonholed in with the "woo woos" that I originally thought you were, but see no evidence of that now. All that remains is the issue of whether or not you are right, and if a referee cannot explicity show where you went wrong, they should recommend publication and let the community hash it out.

Let me clarify one point: for a truly ringlike mass distribution, the acceleration would increase arbitrarily in the outward direction as you approach from the inside, and arbitrarily in the inward direction as you approach from the outside, but you're saying that right at the ring the acceleration is just the same as if all the mass were at the center? That does sound like something that would be interesting to prove. Might not be so easy though, I see why you have so far not solved it. I make one comment, which is that if you are on an infinitely thin spherical shell, the acceleration is half the Keplerian value. So it would be enough to argue that restricting the spherical shell onto a ring would double the acceleration. That may not help the proof, but at least it does give one example of how restricting the mass to the plane of the galaxy can noticeably enhance its gravitational effectiveness, which is an imporant element of the "bartender" explanation.

To Ken G

I made a big point about avoiding singularities, and I thought I didn't have any in my calculations. Can you show me where I fouled up?

About the non-interacting matter that would not allow us to feel the pressure to hold up the spheres, but would allow us to detect the gravity effects of spheres. Thats the stuff that magic is made of. The mass, if any, would cause gravity effects, and one of those effects is pressure. The only way out, in my opinion is to have that mass in orbit around the disk. This also must be undetectable, even though there should be impacts with the disk.

Unless we have good evidence to have a new kind of matter it seems to me the best approach is to assume there is none. I believe I have shown there is no need to invent new matter here.

For the single ring, the test-mass acceleration does not change "arbitrarily," it changes PRECISELY as it approaches the ring from either direction and passes through.

Yes, that business about having the exact Kepler acceleration right at the rim still bothers me.

Thanks for the questions! And I have one. How do you add those quotes from other inputs in the neat little boxes?

I did not say you fouled up, I said you made a big effort to avoid them. The singularities I referred to arise from the need to use infinite mass density whenever you have finite mass in a 1D rod, which is hard to avoid, so leaving gaps in the ring seems like a reasonable thing to try. I could think of other things to try also, but your checks seem to indicate that your approach is working.


Pressure is not a gravity effect. But yes, invoking noninteracting matter is not something anybody likes to do.

Your argument may reduce or eliminate the need for "magic" in the galaxy, but note that it actually makes it even more necessary to invoke dark matter in clusters of galaxies. Check out the new thread by TomT.

There's a miscommunication here. When I used the word "arbitrarily", I meant it in the physics sense that the acceleration can increase without limit as you approach the ring, not to mean it went up in some imprecise way.

There are two ways. You can click on the word "quote" beneath a post, and it will insert around what they said. Or, you can just write those in yourself in your posting. You have to do the latter anyway if you want multiple quotes. There are also some abbreviations people use around here-- I don't know them all, but IMHO is "in my humble opinion", IIRC is "if I recall correctly", and AFAIK is "as far as I know".

[Note added on edit: As you can see, my effort to tell you how to make a quote has of course resulted in a quote! But you can still see what I did by clicking on the "quote" button beneath this post --> ]

Hi Kens,
Please don't let my question on dark matter between galaxies interfere here. I was just curious on that subject, and my mind is racing ahead as usual.
TomT

Continuing with confidence in computer results now that checks with theory have worked out ok, look at p10 fig 10, the trial shapes. I just drew these up as something to try. The discussion above and below the figure says it all, except for an error that I'm sure you would catch. I was certainly as surprised as anybody about the results for the flat disk. The value of vd= 1.7074 means the accleration at the rim is almost 3 tmes that of a constant density sphere of the same mass. The results for a flat disk like this should have been common knowledge in the community, but it wasn't. In my arguments with Prof Goldreich back in the 90's, I asked him about this, but he said he knew a flat disk caused more acceleration than a sphere, but he didn't know how much more, and refused to read my paper or consider anything else than the dark-matter approach. If the community had known this they wouldn't have expected a Kepler result for galaxies. The solution for a flat disk is not easy, like that for a sphere, and I suppose that is the reason it wasn't well known (if at all). I couldn't get it using any of the methods in B&T, and I've never seen it published.

The results of the figure are correct, but I see a big error in the discussion below! What happens if we assume that the disk could be treated as if the matter inside r acted like a sphere and that ouside r had no effect on the acceleration of a test mass at r? Then

......... acc(r)= v^2/r = Gm(r)/r = G rho pi r^2 h/r = G rho pi r h,

and certainly not what I said there! Dividing through by the Kepler acceleration at the rim, the result I show for the speed with that method is correct.

...........accd(rd)= acc(r)/(G rho pi R^2 h/R) = (r/R) = rd

and vd = sqr (rd) = 1 at the rim, same as Kepler. So the assumption of the mass inside r acting like a sphere still is far off, and the curves look quite different. Anyway I must have been pretty sleepy when I wrote that.

The bottom two shapes in fig 10 are just something that is supposed to look like a galaxy, and both have the density increasing with h. The rotation profiles look something like the measured data, and if the two are averaged, the speed would be constant over a large part of the radius. Reducing the tip thickness from 5 % to 2.5 % moves the mass toward the center, and makes the results look more Keplerian, as expected. The blunt rims cause the speed to peak up there, so if the real galaxy has a blunt rim, the density will probably fall off there.

So far it all seems to hang together, but I've changed my plotting format for the reverse results, so I'll try to show that to you next. Look at these in this paper anyway, even though they are confusing for rSMDd. I was trying to use something that stayed readable near the center. My next try is better.

to be continued

[quote=Ken G]
"The singularities I referred to arise from the need to use infinite mass density whenever you have finite mass in a 1D rod, which is hard to avoid, so leaving gaps in the ring seems like a reasonable thing to try."

I agree that there should be no infinite densities. Therefore I declare the rods to have a diameter of 0.1 pc, unless it has to be smaller.

"Pressure is not a gravity effect."

Pressure at the inside of stars (or any large spherical object) is usually caused by gravity.

Thanks for making me think.

(Don't forget the /quote in [ ] at the end of the quote). It's a bit nitpicky I'll grant, but the pressure at the center of stars is not caused by gravity, although it wouldn't be there without gravity. It's like the force you feel from your chair right now, it's not caused by gravity it's caused by the cushion's tendency to push back on anything that squishes it. Pressure in a star is usually caused by density and temperature of an ideal gas, which in turn generates an ability to withstand gravity. Pressure in dark matter would be caused by something similar, presumably (individual particles is the normal picture one assumes, anyway), but none of the gas or stars we observe directly would be affected by the dark matter pressure, since they don't interact. The point being, the normal matter wouldn't have to help hold up the dark matter against its own gravity, so you would never observe the dark matter pressure anyway.

But in effect the rods do have infinite density, since in the calculation you never use their finite diameter, yes? It is analytic, based on the center of the rod, so it's an infinite density. That's fine, if we didn't allow that half of physics would be questionable! I'm just saying I see why you needed to be careful about how you treated the regions near the rods. What is worrying me here is that if one approximates a disk by a series of very thin rods, then one may be able to get a lot more "back for the buck" by making the rods exceptionally thin. So answer me this question-- if you did your same calculation over again but with rods that had half the radius and twice the density, would you still get agreement with v(r)? That kind of test would help convince that there is no issue with lurking singularities.

To Ken G

It all depends on how you want to approximate the rings. For the real ring with thicknesses h and dr, each segment would have a mass of
rho h r dtheta dr and I could integrate that to get the same answer as for the zero diameter rod, since the segment would have "almost zero" thickness. It depends how small you want to make the segments. Math wise, I still can't let the test mass touch the center of the segment. The physics come in when we realize that the test mass would pass smoothly through the ring, with acceleration going through zero instead of infiinity, so the trick is to represent that in the easiest acceptable way.

With a finite number of rods and a finite ring thickness delta r instead of dr, I get an approximate answer which is "close enough" (and actually very good) When the number of rods become infinite and dr goes to zero, the answer is exact. It's like the digital cameras, the more pixels you have, the better the picture, but you only have so much storage.

Would I get the same answer if I use rods with half the diameter and twice the density?

No. But half the diameter and 4 times the density, yes.

The rods only have to be thin enough to stay well inside the volume h rdtheta dr, and have the same mass and h. The answers improve if the rods are near the cg, but for an infinite number of rods and dr going to zero, it makes no difference.

Good point.

I'm sorry to say I cannot attach any of the new plots because the files are too big. If any of you want to give me your email, I can send you a PDF copy of them. If you like send your addresses to admin.darkmatterprize.net .

Anyway that about wraps it up. I hope I have backed up Tom T ok, and that I have at least given the rest of you some new insights. Cougar's bartender probably fell asleep. Thanks for letting me into your forum and I hope to get into some of these again. In the mean time I'll try to get my paper published again.

Bye.

I'm afraid I haven't had time to read the paper carefully enough to either dispute or corroborate your claims, but I will say that the logic of your reasoning as presented seems solid. If I were refereeing your paper there are two issues I'd focus on: 1) does the result depend sensitively on how you set up the calculation, i.e., are we really sure that half-diameter rods with 4 times the density would give the same result, and 2) are the conclusions you reach from your calculation fully supported by your results. Then of course there are the usual editorial issues about clarity of style and presentation of equations, figures, etc. I think this latter issue may not be positively predisposing the referees, because the equations are not expressed in a standard-looking format (like LaTeX), and the figures are a little rough as well. But those issues can be hammered out. The real problem is being taken seriously, because unfortunately few referees are going to actually check the calculation. So impressions are crucial. I think you would stand a better chance of being published by adopting a two-phase approach, where first you take out all the references to dark matter and errors in other calculations, and you just simply make the point that the mass in a galaxy can be a lot less if it is restricted to a disk. I had the impression that you were saying this point has not been made in the literature, and it, by itself, is important. Some kind of bartender-like analytic argument to support this would also help predispose the referee to believing your result. Then, once you have a published paper you yourself may refer to, you could shoot for a second paper where you try to establish what you feel are the ramifications of this finding. That second paper is a tougher sell, and it relates to stickier issues like M/L ratios for normal matter. This two-phase approach separates the less and the more controversial issues, so the latter does not interfere with publication of the former.