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To Ken G
1. That bet sounds fine. Don't expect me to cave for things like dark-matter reports in the galaxy collisions, or other phony stuff, and I will expect you to defend to the end. Should be interesting. 2. You know MOND is still alive and well in some circles. Can you prove it wrong? They simply said that Newton's law had to be modified to get speed matches for galaxies and thus avoid dark matter, but their mods had to be special for nearly every case. Does that prove it wrong? I think so. 3. I'm back to reading Binney and Tremaine to find out how surface light is measured so I can talk with some knowledge on it. Pages 8,9,16,17 have good basic info. They conclude (p16) that our solar neighborhood of the galaxy has an SMD of 75 msuns/pc^2 (made up of all components, stars and non-star ordinary matter) and a surface brightness of 15 Lsunvs (sun visible light intensities/ pc^2) They assume all the light comes from stars alone. Dividing 75/15 they get 5 msuns/Lsunv as a mass/light ratio. (So that's where the 5 came from). They then use 5 to estimate the mass of the Galaxy, assuming 5 as constant over the entire radius. This is where I think the problem lies. I believe the average M/L ratio is closer to 10 for most galaxies, and that the M/L varies a lot for different galaxies and over radius in a given galaxy. Because of the possible situation where there is very low average light, I think it should be a L/M ratio to be more realistic. More later. Ken Nicholson Last edited by knicholson : 06-May-2008 at 06:24 AM. Reason: typos |
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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." Last edited by grav : 07-May-2008 at 01:16 AM. |
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To Grav
Set your disk data as follows: rmax=10000 pcs, h=100 pcs, rho = 1 msun/pc^3 (density), total mass =pi*rmax^2*rho*h 20 rings with rods at the mid radius rr(i) of each ring i, rr(i)= dr*i-dr/2 and read the rotation speeds at the outer radius rs(j) of each ring j, rs(j)=j*dr radial width dr = rmax/20 for all rings mass of each ring becomes mr(i)= 2*pi*rr(i)*dr* h*rho mass of each rod in ring i becomes mrod(i)= mr(i)/120 length of all rods = h, diameter=0 (not important and not used) position your rods around the half-ring mid radii, theta=1.5,4.5,.....,178.5 deg, then find the accelereration caused by a single rod as a function of distance from the rod along a line perpendicular to and bisecting the rod, (if in doubt see my paper), note the acceleraton blows up if the test mass touches the rod, so that never happens, then find the acceleration of a test mass at rs(j) on the theta =0 axis caused by TWO rods in ring i at positions + - theta around the ring then add up the effects of the rods by summing the effects of the 60 pairs to get acceleration a(i,j) of a test mass at radius rs(j) caused by ring i INCLUDING THE CASE i=j, then add up the effects of all the rings (i=1 to 20) on the test mass at locations rs(j), (j=1 to 20), to get A(j) the total accelertion of the test mass at all locations rs(j) That will get you through the forward problem. Notice you should now have the effects of each ring (at density of 1) on a test mass at each speed radius j, and you are ready for the reverse problem. I believe all will become clear after we get you through the reverse problem. Good luck, Ken Nicholson Last edited by knicholson : 07-May-2008 at 06:14 AM. Reason: continue |
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__________________
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." |
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Thanks again, Ken N. Sorry it's taken me so long to respond. I set up the program in the way you stated, but I was a little skeptical about running the program for the same number of points for each ring, since the rods would become more and more dispersed for the rings at a greater radius using the same number of rods per ring and all. I generally think of the rods as all being of equal mass and distributed equally throughout the disk, encompassing equal areas and volumes each, so I wrote programs for both ways. The second one uses the same 1200 rods in all, so has the same running time, but the number of rods in each ring is proportional to the circumference and area of the ring, so starts with 6 rods for the first ring instead of 120, 18 for the second, 30 for the third, etc. As it turns out, there's not too much difference between them, though, but yours is more precise toward the center of the disk, because of the greater number of points used there, and the other is more precise toward the rim. By the way, I noticed you compare your thickness to the radius, not the diameter as I was doing, so the ratio of diameter to thickness would be 200 for what I was using, and the rotation speed for that would be just a touch over 2 times that of a sphere at the rim, then, not 1.9 as I said earlier.
Also, I guess you know that 120 rods per ring for 20 rings is nowhere near precise enough for finding the rotation speeds. I was actually getting a negative acceleration for the inner rings, for example. Ten times as many rods for ten times as many rings begins to come somewhat close to the actual rotation speeds, though, from center to rim. I used the program for the last integration along the x axis thing to compare results. That one is still the quickest and most precise by far. It only takes a few minutes to work through the first two-thirds of an integration with some specified mass distribution and height variance for a disk, and a few more minutes to change a couple of lines in the program to accommodate it, then let it run, gaining a result with amazing preciseness. Of course, I've already got the one for uniform density and height. In terms of using rings for the reverse problem, I was thinking I could just find for a disk with a radius equal to the outer rim of a ring with that program and subtract it from a disk with a radius equal to the inner rim of the ring, using the height and density for each ring, to find for a series of rings, using the rotation speeds for the resulting mass distribution. But then, that would require two integrations per ring all the way from the center to the radius of the ring, for each ring, so it might actually turn out to be less precise when running over the same number of points as your program would for the reverse problem, I don't know yet. I guess I will write a program for that next and compare the preciseness of the results over the same number of points and run time. I will want the most precise program possible over the shortest time when I run the iteration for the reverse problem. I will do that next, and then I suppose I will use that last integration across x program to draw up a few quick graphs for the rotation speeds from center to rim for a few thicknesses of uniform density, uniform height disks and post the images for those. I can't believe I haven't even done that yet. Then, once I have determined which of these few programs is the quickest and most precise overall for the rings, I will run the iteration for the reverse problem to see what mass distribution will give the rotation speed curve as it is observed. I will have to do that a few times over varying thicknesses in order to account for larger thicknesses for gas and dust as well. The way I figure it, there may be an infinite number of ways to apply mass distributions for a particular rotation speed at a particular point using different geometries and so forth, but there is only one mass distribution that will generally give the entire rotation speed curve for a particular disk with a specified thickness. I aim to find it. With any luck, the total mass of the rings will then also match that measured for the galaxy. If nothing else, it should at least provide general information about how the mass distribution must fall from center to rim.
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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." Last edited by grav : 09-May-2008 at 02:05 AM. Reason: clarity |
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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." |
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To Grav
If you do it the way I set it up, then I can check your progress. After you do it that way, even if it doesn't look optimum to you, you can improve it later. What you need to do first is to get a simple working program for the forward problem. You'll learn a lot by doing it this easy way first. Ken Nicholson Last edited by knicholson : 09-May-2008 at 02:59 AM. Reason: addition |
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To Mugaliens
I was not being generous. I couldn't lose that bet. I had already shown that all mass could be put in the disk. My purpose was to smoke out those that claimed the extra mass needed could NOT be put in the disk and thus dark matter was required. It's a bet I'd be happy to make anytime, but it's probably immoral now since it's more of a steal than a bet. Ken Nicholson |
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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." |
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To Grav
You're right! I also got the negative acceleration at r=rmax/20. The problem is one I fixed about 10 years back and forgot. The center ring mass is a circle, and the rods must be placed at the centroids of triangles that make up the circle to properly represent the circle. So for the first speed location the rods radius is 2*dr/3 and the rod location is asymptotic to rs(i)-dr/2 at large radii. I use the following: rr(i) = (rs(i)-dr) + (rs(i)-dr/3)/(rs(i)-dr/2) * dr/2 Try again using this and the original data, and accuracy can be improved later. Good work, Ken Nicholson Last edited by knicholson : 09-May-2008 at 05:37 AM. Reason: continue entry |
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There would be a tight correlation, as supermassive black holes are primarily (exclusively?) found in galaxies, and the dark matter distribution is thought to be responsible for the forming of galaxies (it is concentrated near galaxies and galaxy clusters).
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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. |
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McGaugh's papers provide an excellent read for the MOND issue. See this paper and this paper and this paper . In section 4 of this paper it is shown that MOND cannot not be made to fit a hybrid galaxy that combines the rotation curve data for NGC 2403 with the photometry of UGC 128. MOND is unable to produce a fit to this hybrid galaxy (as would be expected since the galaxy is not real) whereas dark matter is able to make an acceptable fit to this galaxy. This is actually a problem for dark matter that it can be fit to a fictitious galaxy. There are too many free parameters to be fiddled with. And then there are the tidal dwarf galaxies that show mass discrepancies unexpected in the DM framework but are expected in MOND. And Globular star clusters too. Another paper on MOND and globular star clusters.
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"The scientist who asks the right question reconnoiters a new patch of the unknown, and may, with luck, bring it within the constricted but expanding boundaries of the known." ~Timothy Ferris (The Red Limit) 1982 |
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Actually, the part in bold was perfectly correct when it was written (it was the "it is all retrofit" part that you are actually taking issue with). However, now that you have pointed to those articles, I would have to say that the bolded statement is no longer correct, thank you. I don't suppose this thread should delve into the issue of how convincing this evidence really is.
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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. |
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I tried that last formula you posted, and this time I got about 4 times the rotation speed for the first ring, so it's getting closer, but I don't trust the formula because I don't know where it came from, except as a clever rig to fit the rotation curve, which it might come close to doing over all of the rings, but I can't be sure about the effects of the individual rings on particular points, which I'll need for the reverse problem. Besides that, I still wanted something just a little more precise, so I went ahead and broke up that "last integration over x" program for a disk I had into individual rings. I then made an array for the acceleration of gravity for each ring as they would act on a particle placed outside of every ring as (i,j), where i denotes the ring and j the position of the particle. It builds the array in about 5 seconds for ten rings, with about 99.99965% accuracy at the inner ring and 99.987% at the rim, so it is still very precise also. It needed to build the array only once, and I could then find for the acceleration and rotation speeds for any of the ten positions by simply having the program add up the accelerations for each of the ten rings for any particular position, using the array, which only takes a fraction of a second. I can also have the program go through the array over and over for the different positions, while giving each of the rings different densities, also in a fraction of a second, using the same array in combination with another array for the densities of the rings, and find for the rotations speeds over various density configurations. That is what I will use for the reverse problem. I will still be crossing my fingers that it will eventually level off at the corresponding rotation speeds and densities for the rings the way I'm hoping it will, though. If not, I will need to try other methods for the reverse problem as well.
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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." Last edited by grav : 11-May-2008 at 07:22 PM. Reason: changed "disks" to "rings" |