athinker 4 Posted May 17, 2011 Report Share Posted May 17, 2011 I've been pushing the pencil on galactic rotation for a couple of days and find this general rule. I don't find this general rule in any papers online about galactic rotation. The rule is; Galaxies with uniform distribution of mass have a rotation curve sloping up from the center to the edge. Galaxies with a central bulge in the disk have a rotation curve sloping horizontaly flat from center to edge and objects with most of their mass concentrated in the center of their rotation disk, such as the Solar System of panets or the Jovian System of moons, have a rotation curve that slopes down from the center to the edge. Does anyone with experience in galactic rotation know of this rule or have I discovered it myself. It's an interesting rule as LSB (low surface brightness) galaxies follow the rule for uniform distribution more that they do for those with a bulge. None of the LSB galaxies in the NASA high resolution galactic rotation study slope up from the center as steeply as a perfectly uniform mass distribution model predicts showing that they have some degree of central bulge. This is borne out by those galaxies that are bright enough to resolve the slight bulges and bars. Still, those that can't be resolved to show a bulge have the rotation curve indicating that there are unresolved bulges. Quote Link to post Share on other sites

ken hughes 1 Posted May 19, 2011 Report Share Posted May 19, 2011 TRANSFERED FROM "DARK MATTER". I am sceptical about the existence of CDM. As you correctly point out, it cannot be justified by the analogy of Neptune, since with Neptune, we simply "created" another planet, an additional known entity, whereas with CDM, we are "creating" an entirely new entity altogether with some very specific and demanding properties. This is a bit like the Aether before Relativity came along and we know what happened to the Aether.Also, this "creation" goes against the guidance of Occam's razor and we have seemingly failed to pursue other explanations which do not demand the creation of new entities or the violation of known fundamental rules of behaviour. The correct interpretation of Occam's razor is not the usual "The simplest explanation is usualy the correct one", or even that "Entities must not be multiplied beyond necessity", but that "For two equally valid theories giving the correct answer, the one with the least number of new creations and which best matches known science, is usually the correct one".I submit that the claim of CDM best matching known science is tenuous to say the least. Breaking Newtons law and having no fit within GR is clearly no good match at all. The other supporting "evidence" from gravitational lensing and the 3 degree background radiation is also suspect and these could be explained in other ways if only we would try. It is simply convenient to match them all up with the CDM proposal. In principle, it is not very scientific to claim that one idea must be true just because other observations can be explained using the same envisaged mechanism. It blocks any effort to challenge the initial proposal and we might even suspect this is deliberate. The challenge then is to come up with the fundamental cause for the rotation speeds that matches known science and then to calculate and confirm the observational speeds. This will have to happen before anyone in the established scientific community will even listen.Coming up with the fundamental cause is simply a matter of applying known science and deducing what it must be. Let's try it;- Firstly, we ask, "Is the anomaly real or is it an illusion?" The answer is it could be either or both, so let's park that one for the moment and come back to it later. Secondly, "What are the available mechanisms?" Science and scientific progress is always about cause and effect but we seem to have lost sight of this in recent years.The answer, I think we would all agree, is that it is the action of gravitation that causes the orbital motion of any entity. The next question is "What causes the motion not to be, or to not appear to be, in accordance with Newton?", and this is where we have invented CDM to explain the deviation. If we now use the guidance of Occam's razor, we cannot invoke CDM and must look elsewhere or in more depth to resolve the issue. The next question becomes "How can Newton's Law result in the observed rotation speeds?" and to answer this we need to use our powers of thinking, of deduction, our intelligence. We must depart from conventional wisdom into the realm of original thought. Even the fact that we are looking for an "alternative" solution seems, these days, to invoke distain from the scientific community but science and scientific progress always involves "alternatives" and if we stifle the exploration of these, we stifle scientific progress itself. We might re phrase the question;- "How can Newton's Law AND the observed rotation speeds, both hold?" After all, this is the best fit with known science if we can find an answer, There is only one way for this to be the case and that involves the effects of time dilation. Now, our first reaction to this idea might be that time dilation is a very tiny phenomenon and would surely have no significant effect on rotation speeds. Well, that may or may not be the case but we should not throw in the towel just because the numbers MIGHT not work. Let's get the principles established first. The theory will stand or fall when we apply the correct math.As encouragement for you to persevere with this proposal, I suggest that although the time dilation decaying to zero at infinity has very small values indeed, the distances involved are nevertheless immense and the sheer scale of observations will make this effect more significant. In other words, the very small time rate differentials accumulate over the vaste distances and will have a noticable effect on rotation speeds. Rotational speeds in any two frames can both conform to Newton within the frames themselves, whilst they may not appear to conform relatively between the frames if the time rates are different in each frame. This way, any speed we observe from the galactic center outwards will be distorted by the time rate field of the galaxy itself. Speeds will be red shifted or slowed down near the centre and outwards, relative to the intergalactic frame. The next question "Is there any radial position that gives us an non distorted Newtonian predicted speed?" The answer is yes and this position is the place at which the time rate is the same as here on Earth and there is no temporal distortion of events relative to our frame when we observe stars at this radial position.If we look at a galaxy of similar size and mass to the Milky Way, then this "Datum" position will be at the same galactic radius as the Solar system's in the Milky Way.The Newton curve, therefore MUST pass through this point on the observed curve, so that we observe the non distorted Newtonian prediction at that radius. This means we have to raise the Newton curve upwards on the graph such that it passes though this Datum point which also has to lie on the flattened observed curve at the datum radius. The shape of the Newton curve will also change slightly, due to the implied increased mass at the galactic centre. We deduce that the effective central mass of the galaxy is much greater than currently envisaged for it to produce these increased rotation speeds. The central, supermassive black hole must therefore be more massive to produce this effect. However, so far, we have only raised the curve and NOT flattened it. The curve is presently higher than observed INBOARD and lower than observed OUTBOARD of the Datum. (See my Utube video "Answer to the galactic rotation anomaly") The flattening of the curve from this raised Newton curve, is due to the red shifted (reduced)speeds INBOARD of the datum position and the blue shifted (increased) speeds OUTBOARD of the datum. We have a hint of this causality from the Pioneer anomaly, in that there is a very small blue shift (increased speed) due to the time rate differential between the probes and the Earth. It is the same phenomenon as galactic rotation distortion, but on a very much smaller scale and therefore with very much smaller effects. Using this logic and the appropriate math, I get the flatteneing effect on the Newton curve to match the observed. If you plot the Newton curve through the datum position with the observed curve from "www.astronomynotes.com, it becomes obvious from just looking at the shape of the curves.Short report submitted to PHYS.Rev.Lett Quote Link to post Share on other sites

athinker 4 Posted May 20, 2011 Author Report Share Posted May 20, 2011 Ken Hughes, That's all very interesting but I think you missed the point of my OP. The rotation curve rule I discovered comes purely from calculating by Newton disks of rotating particles. The rule I found is that if you model a disk of rotating particles of uniform distribution a steep upslope from center to edge of the rotation curve is derived, purely from Newton. If you model a disk with uniform rotational speeds (flat rotational curve) from center to edge you derive a disk with a central bulge in mass distribution. If you model a disk with 99% of the mass in the center of rotation you get a rotation curve that slopes down from center as the solar system does. These are models that conform to observation without invoking modifications of gravity, CDM or, with all due respect, time dilation effects. The reason I asked if this rule of rotation had been discovered was to see how it's reconciled by astronomers in their conclusions about the matter in galaxies. If it has not been discovered then there is something wrong with the conclusions of astronomers. I discovered it after pushing the pencil for only a couple of days. If I discovered it in a couple of days and all of astrophysics haven't discovered it in decades I think they are not doing their homework. If they're not doing their homework then I suspect that their self absolution in claiming that they are correctly interpreting the luminance to mass ratio they observe, from which the conclusion of dark matter issue, is suspect. They didn't do their homework with regard to the possibility and likelyhood that the issue is most easily resolved by concluding that there is more fusion happening in volumes of mass with higher density distribution than is happening in volumes of mass with lower density distribution. The rotation rule I mentioned is a simple calculation but is not mentioned in papers on the subject. Neither is the fusion to density issue. I think they just didn't do their homework. They just pulled these ideas out of their hats. Remember, The Mercury rotation anomaly wasn't solved by an astronomer. It was solved by a German patent clerk, a mathemetician. I'm a machininst, go figure. Considering the dishonest history of astronomy before the astrologers changed the name (like a crook changing his name to escape justice) and considering that the same people and culture of astrologers became "astronomers" and continued to teach "astronomy", then it's likely that there could be some degree of dishonesty in the present culture and study of "astronomy". Quote Link to post Share on other sites

ken hughes 1 Posted May 26, 2011 Report Share Posted May 26, 2011 Athinker, I get what you're saying and maybe you're right, only I understand that Newton's Law is always an inverse square relationship with all the mass INBOARD of the rotation radius. Any distribution of matter inboard can be resolved as equivalent balls of mass, so I can't quite grasp how you get anything but a falling curve (inverse squares) from the centre outwards, whatever the mass distribution? Don't get me wrong, I am not simply defending my own ideas against counter ones. I have an open mind and would be just as satisfied with a different explanation being proved right. Can you help me understand yours a little better? Can we agree that the effects of the inboard disc can be viewed as many inverse square curves superimposed? If so, then I can see that a simple inverse quare curve is innapropriate for a galactic disc and that the summation curve must begin to stretch as radius increases.(ie the fall off in speed reduction becomes less steep with increasing radius and the curve starts to flatten). However, since all such curves do show speed DECAY with radius, then any summation of such curves cannot, under any circumstances show a level or flattened curve and certainly not a rising one. Where have I got it wrong? Are you also suggesting that the very real field of time dilation will have NO effect on rotation speed? Quote Link to post Share on other sites

athinker 4 Posted May 26, 2011 Author Report Share Posted May 26, 2011 Ken Hughes, I think I understand your questions. I proffer that you may be over thinking what is a simple problem. An edge star is orbiting at a certain speed based on the mass "inboard" the orbit of its rotation. Newton treats that mass in this case as a point center. But a star orbiting a disk of mass closer to the center of the disk is orbiting less mass than the outer star of the disk. It's orbiting less mass thus slower but closer to center thus faster. Complicated huh? Not really. Consider the solar system with the sun at the center. As you move away from the sun the effect of gravity of the sun gets weaker. but what if you add more mas in the form of gadzillions of tons of dust in a disk between the farther point and the sun. You're farther from the sun, yes you would be orbiting at a slower speed if you were orbiting only the sun, but there is more mass than just the sun to be considered because of the extra mass in the disk. For the rising rotation curve I mentioned in the uniform mass distribution disk model there is no central mass. The farther you go from the center the more mass inboard of your orbit you are orbiting Therefore the faster you orbit. This is not an idea I just pulled out of my hat. This is what it calculates out to be. As to time dilation. More colloqually known as a relativistic effect. This is an effect that is only significant at speeds very near the speed of light or very near very high gravitational fields like very near the sun or other very large mass. This is not the case here. Quote Link to post Share on other sites

CraigD 398,736 Posted May 27, 2011 Report Share Posted May 27, 2011 For the rising rotation curve I mentioned in the uniform mass distribution disk model there is no central mass.Here’re results of a simple numerical approximation of a galactic rotation curve based on a flat disk of uniformly distributed point masses, combined with one where mass “outboard” of the test body is ignored: “outboard” mass ignored Radius Acceleration Speed Acceleration Speed 0 0.0000000000 0.0000000000 0.0000000000 0.0000000000 1 0.0314512986 0.1773451399 1.9571067812 1.3989663260 2 0.0629096846 0.3547102609 3.1618952994 2.5147148146 3 0.0943822540 0.5321153653 3.8006182779 3.3766632692 4 0.1258761210 0.7095804987 4.3287047234 4.1611078926 5 0.1573984266 0.8871257708 4.7660616950 4.8816296946 6 0.1889563478 1.0647713777 5.0505473418 5.5048418734 7 0.2205571067 1.2425376240 5.3920002951 6.1436147394 8 0.2522079801 1.4204449447 5.6094890220 6.6989485874 9 0.2839163088 1.5985139283 5.8390447697 7.2492346443 10 0.3156895073 1.7767653398 6.0461001761 7.7756672872 20 0.6386941658 3.5740569828 7.3840341994 12.1523941669 30 0.9773212386 5.4147610435 8.1780002354 15.6633332041 40 1.3422168827 7.3272556465 8.7424929769 18.7002598665 50 1.7486817402 9.3506196056 9.1872801272 21.4327787830 60 2.2214421582 11.5449785401 9.5480581859 23.9349846700 70 2.8064350395 14.0160783662 9.8522100413 26.2612776326 80 3.6081022802 16.9896492729 10.1174168371 28.4498391379 90 4.9738091765 21.1575713606 10.3507713875 30.5216222516 91 5.1844575914 21.7206270816 10.3723791202 30.7227358797 92 5.4214371657 22.3332088883 10.3953183962 30.9252209765 93 5.6921472260 23.0080353793 10.4155577827 31.1230922915 94 6.0074620328 23.7634473737 10.4377825660 31.3233389217 95 6.3842369409 24.6272716594 10.4573911560 31.5190761257 96 6.8504771795 25.6446058505 10.4793830329 31.7178304926 97 7.4571357539 26.8950212517 10.5006171298 31.9148846401 98 8.3068835065 28.5319922830 10.5204823324 32.1093019011 99 9.5705125890 30.7811751939 10.5405344593 32.3034504576 100 10.5595847514 32.4955146927 10.5595847514 32.4955146927Here’s the MUMPS code that produced its R1=100 f R=0:1:9,10:10:90,91:1:100 w $j(R,3) x "s (AX,AY)=0 f Y=-R1:1:R1 s DY=Y-R,DY2=DY*DY,X=$j(R1**2-(Y**2)**.5,0,0) f X=-X:1:X s D2=X*X+DY2 i D2 s D=D2**.5,A=1/D2,AX=X/D*A+AX,AY=DY/D*A+AY" s V=R*$zabs(AY)**.5 w $j($zabs(AY),18,10),$j(V,18,10) x "s (AX,AY)=0 f Y=-R:1:R s DY=Y-R,DY2=DY*DY,X=$j(R**2-(Y**2)**.5,0,0) f X=-X:1:X s D2=X*X+DY2 i D2 s D=D2**.5,A=1/D2,AX=X/D*A+AX,AY=DY/D*A+AY" s V=R*$zabs(AY)**.5 w $j($zabs(AY),18,10),$j(V,18,10) w ! :QuestionM Can you post the results of your galactic rotation curve calculation for a uniform flat disk, athinker, and your code or pseudo code? It’s hard to discuss calculations without actually showing them. An edge star is orbiting at a certain speed based on the mass "inboard" the orbit of its rotation. Newton treats that mass in this case as a point center.My results above don’t support this claim. Notice that the orbital speed calculated using the full disk is very different than the “outboard ignored” speed. The mass “outboard” of the test body’s orbit is important. My quick calculations show a galactic rotation curve resembling the observed galactic rotation curve for the inner part of a galaxy. It fails to peak and decrease or remain nearly constant with radius as realistically calculated or observed curved do, because its assumption of a uniformly distributed disk of masses isn’t realistic. ... Consider the solar system with the sun at the center. As you move away from the sun the effect of gravity of the sun gets weaker. but what if you add more mas in the form of gadzillions of tons of dust in a disk between the farther point and the sun. You're farther from the sun, yes you would be orbiting at a slower speed if you were orbiting only the sun, but there is more mass than just the sun to be considered because of the extra mass in the disk.A point of physical accuracy: In a galaxy, most of the mass is not concentrated in a small central body (example: our Milky Way galaxy’s central MBH masses about 4e6 solar masses, about 0.0006% of the whole galaxy’s mass about 7e11 SMs). In a solar system, however, most of the mass is concentrated in a small central body (Example: our sun has about 99.86% of the solar system’s mass). Thus, as observed, the orbital speed of the planets almost exactly match Kepler and Newton’s laws’ [imath]v = \sqrt{\frac{k}{r}}[/imath], with more distant planets moving more slowly than inner ones. The dust in the solar system has a negligible effect on the motion of the planets. Quote Link to post Share on other sites

athinker 4 Posted May 27, 2011 Author Report Share Posted May 27, 2011 CraigD, While I work up my numbers in a form you can understand you might take a look at the "curve" your numbers graph out. I suspect there's an arithmetical error somewhere. I do understand the steep middle part is in tens and the end parts are in singles. Just that the slope is not as teep as I calculate it to be. That said, however, you should agree that a flat distrubution disk rotation curve slopes up as we both calculate and a disk with all mass concentrated in the center like the solar system slopes down? Then you should agree that mass distributed in a combination of both some concentration in the center with some distributed in a disk would cause the rotation curve to fall somewhere between the up sloping curve and the down sloping curve? The determining factor being the relative percentages of mass in core to mass in disk. Just to remind you as I pointed out earlier. the rotation curve of LSB galaxies slopes up. Further My galactic rotation rule predicts that any galaxy who's rotation curve does not slope as steeply up as the ideal model for flat mass distribution(as you say unrealistic) will be found to have a bulge. The steeper the curve the smaller the bulge as a percentage of the total disk. Quote Link to post Share on other sites

athinker 4 Posted May 28, 2011 Author Report Share Posted May 28, 2011 Editing post. Checking for arithmetical error. Done. Still feel there's an error somewhere. I don't like the area result. Might be having an optical hicup. Dyslexia runs in my family. For this flat mass distribution model I assumed circular orbits in a circular disk 200,000 light years in diameter. total mass 200,000,000,000 solar massestotal area 31,415,926,535sqLY mass per sqLY 6.366 solar masses formula [math]V_{cir} = \sqrt{\frac{GM}{r}}[/math]calculator here http://orbitsimulator.com/formulas/vcirc.html flat mass distribution curve r=10,000LY, area 314,159,265 sqLY, mass 1,999,999,997 solar masses, orbital speed 52 km/s r=20,000LY, area 1,256,637,061 sqLY, mass 7,999,751,530 solar masses, orbital speed 74 km/s r=30,000LY, area 2,827,433,388 sqLY, mass 17,999,440,948 solar masses, orbital speed 91 km/s r=40,000LY, area 5,026,548,245 sqLY, mass 31,999,006,127 solar masses, orbital speed 105 km/s r=50,000LY, area 7,853,981,633 sqLY, mass 49,998,447,075 solar masses, orbital speed 118 km/s r=60,000LY, area 11,309,733,552 sqLY, mass 71,997,763,792 solar masses, orbital speed 129 km/s r=70,000LY, area 15,393,804,002 sqLY, mass 97,996,956,276 solar masses, orbital speed 140 km/s r=80,000LY, area 20,106,192,982 sqLY, mass 127,996,024,523 solar masses, orbital speed 149 km/s r=90,000LY, area 25,446,900,494 sqLY, mass 161,994,968,544 solar masses, orbital speed 158 km/s r=100,000LY, area 31,415,926,535 sqLY, mass 199,993,788,321 solar masses, orbital speed 167 km/s Quote Link to post Share on other sites

CraigD 398,736 Posted May 28, 2011 Report Share Posted May 28, 2011 The results of your hand orbit speed calculation treating the entire “inboard only” disk as a single body match my “outboard mass ignored speed” calculation, which treats the entire disk as a uniformly spaced collection of 31592 bodies. My point, though, is that you can’t ignore the “outboard mass”, or any other nearby bodies, when calculating the force of gravity, and thus the acceleration and orbital speed, of a body. My “outboard mass ignored” speed matches yours (their slight discrepancy is because I’ve chosen so few bodies to model the infinity-approaching number of bodies in a real physical situation), but neither are what would really happen in a uniform gas/dust disk – what I label “realistic” in the graph below. To get a realistic galaxy rotation curve, we must use a realistic galaxy mass distribution – which a uniformly distributed disk isn’t. This gets complicated, because the mass distribution itself depends on gravity, so for a model with a minimum of simplifying – and possibly inaccuracy-introducing assumptions – you having to calculate the position over time of a huge number of bodies – not an easy undertaking for a hobbyist like myself. ;) Quote Link to post Share on other sites

athinker 4 Posted May 28, 2011 Author Report Share Posted May 28, 2011 you can’t ignore the “outboard mass”,Why not? What is the law and formula involved in considering the outboard mass in this situation. You didn't answer my previous question regarding "you should agree that a flat distrubution disk rotation curve slopes up as we both calculate and a disk with all mass concentrated in the center like the solar system slopes down? Then you should agree that mass distributed in a combination of both some concentration in the center with some distributed in a disk would cause the rotation curve to fall somewhere between the up sloping curve and the down sloping curve? The determining factor being the relative percentages or ratio of mass in core to mass in disk." You also didn't mention if you had ever heard of this rotation law. Considering your Q and A I guess the answer would be no. Quote Link to post Share on other sites

CraigD 398,736 Posted May 29, 2011 Report Share Posted May 29, 2011 you can’t ignore the “outboard mass”,Why not? Because both the bodies “inboard” and “outboard” of the test body attract it gravitationally. You may believe you can ignore the “outboard” mass because of the shell theorem. You could, if you were using calculating for a spherical rather than a disk shaped cloud (which this version of my program slightly to use bodies in a 3-D sphere rather than a 2-D disk s R1=100,R12=R1*R1 f R=0:1:9,10:10:90,91:1:99,100:1:109,110:10:200 w $j(R,3) x "s (AX,AY,AZ)=0 f Y=-R1:1:R1 s DY=Y-R,DY2=DY*DY,Y2=Y*Y,X=$j(R12-Y2**.5,0,0) f X=-X:1:X s X2=X*X,Z=R12-Y2-X2 I Z'<0 S Z=$j(Z**.5,0,0) f Z=-Z:1:Z s D2=Z*Z+X2+DY2 i D2 s D=D2**.5,A=1/D2,AX=X/D*A+AX,AY=DY/D*A+AY,AZ=Z/D*A+AZ" s V=R*$zabs(AY)**.5 w $j($zabs(AY),18,10),$j(V,18,10) x:R'>R1 "s R2=R*R,(AX,AY,AZ)=0 f Y=-R:1:R s DY=Y-R,DY2=DY*DY,Y2=Y*Y,X=$j(R2-Y2**.5,0,0) f X=-X:1:X s X2=X*X,Z=R12-Y2-X2 I Z'<0 S Z=$j(Z**.5,0,0) f Z=-Z:1:Z s D2=Z*Z+X2+DY2 i D2 s D=D2**.5,A=1/D2,AX=X/D*A+AX,AY=DY/D*A+AY,AZ=Z/D*A+AZ" s V=R*$zabs(AY)**.5 w:R'>R1 $j($zabs(AY),18,10),$j(V,18,10) w ! shows). However, as my previous program showed, the shell theorem doesn’t apply to circularly symmetrical disks, only to spherically symmetrical spheres. What is the law and formula involved in considering the outboard mass in this situation. The usual one for non-relativistic calculations, Newton’s law of universal gravitation,[math]F = G \frac{m_1 m_2}{r^2}[/math].the definition of force, [math]F = M A[/math]to derive the formula for acceleration of a small body by a large one,[math]A = G \frac{m}{r^2} [/math]The derived formula for the centripetal force[math]F = \frac{m v^2}{r}[/math]to, again with the definition of force, derive[math]A = \frac{v^2}{r}[/math]then[math]v = \sqrt{A r}[/math] To reasonably accept my results, you must understand the short MUMPS program I generate it with. Here’s a detailed explanation:s R1=100 set the radius of the disk, R1, to 100f R=0:1:9,10:10:90,91:1:100 calculate for the displayed radius R valuesw $j(R,3) output (write) the radius, right justifiedx "s (AX,AY)=0 initialize the acceleration of the test particle to 0f Y=-R1:1:R1 calculate for bodies in the disk with Y coordinate –R1 to R1 in increments of 1s DY=Y-R,DY2=DY*DY calculate the Y distance from the disk body to the test body,X=$j(R1**2-(Y**2)**.5,0,0) f X=-X:1:X calculate for disk bodies with X coordinates within radius R1s D2=X*X+DY2 calculate the distance between the disk and test bodies square D2i D1 only calculate for non-zero distances D=D2**.5 calculate distance between the bodies D,A=1/D2 calculate acceleration per #1, letting [imath]G=m=1[/imath] ,AX=X/D*A+AX,AY=DY/D*A+AY" add X and Y acceleration components to nets V=R*$zabs(AY)**.5 calculate orbit speed per #2w $j($zabs(AY),18,10),$j(V,18,10) display acceleration and orbit speedx "s (AX,AY)=0 f Y=-R:1:R s DY=Y-R,DY2=DY*DY,X=$j(R**2-(Y**2)**.5,0,0) f X=-X:1:X s D2=X*X+DY2 i D2 s D=D2**.5,A=1/D2,AX=X/D*A+AX,AY=DY/D*A+AY" s V=R*$zabs(AY)**.5 repeat the preceding only for bodies within radius of test body Rw $j($zabs(AY),18,10),$j(V,18,10) w ! ... and display it You didn't answer my previous question regarding "you should agree that a flat distrubution disk rotation curve slopes up as we both calculate and a disk with all mass concentrated in the center like the solar system slopes down?Sorry for neglecting your question! Yes, I agree. Referring again to the shell theorem, inside a uniform density spherical shell, the force on test body “varies linearly with distance from the center”, that is[math]F = k_1 r[/math] where [math]k_1[/math] is a unit-dependent constant. Since circular orbit speed is a function of centripetal force, and acceleration [math]A[/math] for a constant test body mass a constant function of force [math]A[/math], we get[math]v = \sqrt{A r} = k_2 r [/math] where [math]k_2[/math] is another unit-dependent constant. So in this case, the rotation curve not only slopes upward, it is a straight line. My dumb program confirms. I’m pretty sure we could derive a formula for the rotation curve of any explicitly defined density distribution, and that many people have done so. Deriving it, or even searching for someone else’s derivation, is too much work for me at the moment, and way too geekish a way to spend a holiday weekend! ;) Then you should agree that mass distributed in a combination of both some concentration in the center with some distributed in a disk would cause the rotation curve to fall somewhere between the up sloping curve and the down sloping curve?No. I believe that for any case except a “galaxy” with most of its mass in a central black hole, the rotation curve is upward sloping, reaches a local maximum, then slopes downward. This is what any simple model where the density of the disk nearer its center is greater describes. The realistic case, assumes that Newton’s law of gravity is an accurate approximation, has the curve reaches a local minimum and slopes upward again, then reaches another local maximum, and again slopes downward. This is what is observed. The challenge is to find a mass distribution that, when modeled, gives the observed curve. The determining factor being the relative percentages or ratio of mass in core to mass in disk.For simple density distribution models, I sort of agree. The determining ratio is not mass in core to mass in disk, but mass in core to density of disk. And, the galaxy must be spherically symmetrical, not a flattened disk. For more complicated, more realistic models, a single ratio isn’t enough. For a spherical galaxy of uniform mass density except for a point mass at its center, thanks to the shell theorem we can write a simple formula for the rotation curve as a function of the mass of the core:[math]v = \sqrt{k d r^2 +\frac{M}{k r}}[/math], where [math]M[/math] is the central point mass, [math]d[/math] the disks density, and [math]k[/math] is a unit-dependent constant. You also didn't mention if you had ever heard of this rotation law. Considering your Q and A I guess the answer would be no.You guess correctly. Other than your posts, I’ve never heard or read anything suggesting a scientific law for galaxy rotation curves. If we use wikipedia’s definition, “a scientific law or scientific principle is a concise verbal or mathematical statement of a relation that expresses a fundamental principle of science”, and “concise: brief, yet including all important information”, I don’t believe there can be such a law, as the motion of bodies in galaxies can’t be described concisely. This motion is complicated. If we tried to create an equivalent of Kepler’s laws that applied to galaxies, we’d get as far as “1. The orbit of most stars around the center of its galaxy are roughly an ellipses with the center of the galaxy at one of their foci”, then “2+ Then things get complicated”. Not a very satisfying collection of laws! Jay-qu and JMJones0424 2 Quote Link to post Share on other sites

athinker 4 Posted May 29, 2011 Author Report Share Posted May 29, 2011 I suspect you may be misunderstanding a basic point about the spherical shell as it relates to the disk. In the sphirical shell a test partical might be near one side of the shell and one might think it should be drawn towards that side. But it isn't because although it's farther from the "other side" the "other side" is much larger than the near side. The nearness of the near side is canceled by the larger mass of the far side. In the disk this rule governs all particles in the disk plane. Any particles above or below the plane would be drawn towards the plane by both the inboard and outboard mass. But as long as the particles are within a percent or 2 of the plane as is the case here then the pull of the near side of the outboard mass is counterbalanced by the greater amount of outboard mass on the "other side". See illustration. The "near mass" in white the "far mass" in black. Since both pulls are balanced the outboard mass adds no vector to compute in the rotation curve. Maybe your program is considering rotation of particles above and below the plane? It shouldn't make a difference to the total orbital speed of a particular particle. For example a particle in aq galactic polar orbit would accelerate as it moves down toward the plane then slow down as it passes through moves away and over the pole. It's average speed would be equal to what it would be as if it was in a stable orbit in the plane. In effect whatever its circular orbit would be at it's average speed around any mass inside its orbit in the disk. PS, Enjoy your holiday. Quote Link to post Share on other sites

JMJones0424 119,315 Posted May 29, 2011 Report Share Posted May 29, 2011 you can’t ignore the “outboard mass”Why not?Because both the bodies “inboard” and “outboard” of the test body attract it gravitationally. Please excuse my ignorance. I am trying to visualize why the shell theorem does not apply to circularly symmetrical disk. I imagine a spherically symmetrical sphere. Take a plane passing through the center of the sphere. Any point on that plane has an equal gravitational pull from both hemispheres of the sphere above and below the plane, doesn't it, because the net force of gravity is towards the center of the sphere? So... if both of those hemispheres were removed, why wouldn't the shell theorem still apply to the circularly symmetrical disc formed by passing a plane through the center of the spherically symmetrical sphere? Ack, ninja'd, but at least I'm not alone in wondering :) Quote Link to post Share on other sites

CraigD 398,736 Posted May 30, 2011 Report Share Posted May 30, 2011 Please excuse my ignorance. I am trying to visualize why the shell theorem does not apply to circularly symmetrical disk.That’s what I’m trying to help with, by showing empirically with my two gravity simulation program, which are nearly as simple and terse (449 bytes for the 2-D one is pretty small) as I can imagine, without resorting to any calculus. When a 3-D sphere of uniformly spaced point masses is used, the zero net force on a test body inside the spehere predicted by the shell theorem appears. When a 2-D disk is used, it doesn’t. I’ll give a short explanation of this using the calculus of the shell theorem below. My confidence in results derived using calculus is increased by testing it with a simple computer model, though, and I find such confirmation more intuitively convincing to most people that pure calculus proofs. I suspect you may be misunderstanding a basic point about the spherical shell as it relates to the disk. In the sphirical shell a test partical might be near one side of the shell and one might think it should be drawn towards that side. But it isn't because although it's farther from the "other side" the "other side" is much larger than the near side. The nearness of the near side is canceled by the larger mass of the far side. In the disk this rule governs all particles in the disk plane.Unless you’ve actually derived this result, or found it published somewhere – that is, you can write or reference a proof of it like Newton’s for a spherically symmetrical body – you’re not justified in the claim. I don’t believe such a proof exists, as it contradicts the simple empirical model I implemented with my programs, and an application of the calculus below. Here’s a small extension of the shell theorem that shows that there’s a non-zero net force on a test body from an annulus (where mass is proportional to area) like the sketch attached to your post. Recall (see its Wikipedia article for the proof and definition of variables) that the shell theorem holds that the force on a test body inside a hollow spherical shell is[math]F_r = \frac{GMm}{4r^2 R} \int_{R-r}^{R+r} F_r \left( 1 + \frac{r^2 - R^2}{s^2} \right) ds = [/math] [math] \frac{GMm}{4r^2 R} \left( \left( (R+r) - \frac{r^2-R^2}{R+r} \right) -\left( (R-r) - \frac{r^2-R^2}{R-r} \right) \right)= 0[/math] However, for a thin disk, the bounds or the integral aren’t [imath]R-r,\,R+r[/imath]. There must be 2 integrals with bounds [imath]R-r,\,R-r+q[/imath] and [imath]R+r-+q,\,R+r[/imath], where [imath]q=\sqrt{R^2-w^2}[/imath] where [math]w[/math] is half the thickness of the disk:[math]F_r = \frac{GMm}{4r^2 R} \left( \int_{R-r}^{R-r+q} F_r \left( 1 + \frac{r^2 - R^2}{s^2} \right) ds + \int_{R+r-q}^{R+r} F_r \left( 1 + \frac{r^2 - R^2}{s^2} \right) ds \right) = [/math] [math] \frac{GMm}{4r^2 R} \left( \left( (R-r+q) - \frac{r^2-R^2}{R-r+q} \right) -\left( (R-r) - \frac{r^2-R^2}{R-r} \right) +\left( (R+r) - \frac{r^2-R^2}{R+r} \right) -\left( (R+r-q) - \frac{r^2-R^2}{R+r-q} \right) \right)= [/math] [math] \frac{GMm}{4r^2 R} \left( \frac{2r^3 -2rR^2 -2r^2q +2R^2q}{(R+r-q)(R-r+q)} \right) \not= 0[/math] when [imath]r \not= R[/imath] QED (and am I ever glad and amazed I got the LaTeX right on the first try!) :) It would be more satisfying to give this proof in only 2 dimensions, ie: without the shell theorem’s [math]2\pi R^2\sin\theta \,d\theta[/math], but more work than I want to undertake – especially as I’d have to relearn a substantial fraction of the calculus I’ve long forgotten. It would be nice if someone who still knows his or her calculus would take mercy on us, but until then, I’m satisfied with the simple glomming onto the last step of the shell theorem proof I used above. Maybe your program is considering rotation of particles above and below the plane? It shouldn't make a difference to the total orbital speed of a particular particle.No, my little programs are doing nothing but calculating and summing the force vectors of t unit mass test body and all of the uniformly spaced unit mass bodies in either a disk or a sphere. :Exclamati Again, I encourage you, athinker, and any interested readers, to understand these simple programs. If the explanation I gave in post #11 is not sufficient help with this, let me know, and I’ll try rewriting it in more natural English language pseudo code. The usefulness of using simple simulations to empirically reach conclusions is lost if one doesn’t accept the correctness of the simple simulation. Quote Link to post Share on other sites

athinker 4 Posted May 30, 2011 Author Report Share Posted May 30, 2011 CraigD, It might help you to understand to consider that the basic gravity equation is in 2 dimensions to begin with. The shell theroem is an extension of it into 3 dimensions as it was not immediately obvious from Newton's law. See Newtons law of universal gravitation. http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation Note that all factors refer to lines and points. Those are planer referances. The basic principle is that the force of gravity exists between points not external to them. Simply put all the factors describing an orbit are in terms describing motions and forces of points in a plane. The shell theroem postulates that a shell can be bisected into an infinite number of planes so the original planer description holds for the sphere. Newton itself is the proof in 2 dimensions. If there were a net vector by outboard mass there would be a direction and magnitude coeficient for it in any gravity formulation. If your problem is the geometry you should consider that any section of an area is balanced by the remaining section. Consider this; If you were in the middle of a ring of planets one planet thick. You should agree that you would float weightless. If you moved one planet's distance from the center you would remove two planets from one side and add them to the other side. You are one planet closer to the close side and one planet farther from the far side. But the ring is four planets less in the close side than it is in the far side. Quote Link to post Share on other sites

CraigD 398,736 Posted June 1, 2011 Report Share Posted June 1, 2011 It might help you to understand to consider that the basic gravity equation is in 2 dimensions to begin with. The shell theroem is an extension of it into 3 dimensions as it was not immediately obvious from Newton's law. See Newtons law of universal gravitation. http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation Note that all factors refer to lines and points. Those are planer referances. The basic principle is that the force of gravity exists between points not external to them. Simply put all the factors describing an orbit are in terms describing motions and forces of points in a plane. The shell theroem postulates that a shell can be bisected into an infinite number of planes so the original planer description holds for the sphere. Newton itself is the proof in 2 dimensions. Athinker, you need to support these claims. Neither the Wikipedia link you give nor the one for the shell theorem support your claim, which I gather is that Newton first proved that a body in an circularly symmetrical plane with a nonzero mass experience a zero net gravitational force. This text (bolding mine) from the bodies with spatial extent section of the first Wikipedia article,In this way it can be shown that an object with a spherically-symmetric distribution of mass exerts the same gravitational attraction on external bodies as if all the object's mass were concentrated at a point at its centre. (This is not generally true for non-spherically-symmetrical bodies.) For points inside a spherically-symmetric distribution of matter, Newton's Shell theorem can be used to find the gravitational force. The theorem tells us how different parts of the mass distribution affect the gravitational force measured at a point located a distance r0 from the center of the mass distributionClearly contradicts your claim. More importantly that citing natural language in encyclopedia articles, I’ve show with both simple approximations and calculus that your claim is simply false. I believe that, despite my attempts to explain it, you still don’t understand what my little programs do. They simply place non-moving point masses uniformly inside whatever figure I specify – a disk, a sphere, an annulus (“outboard only” disk with a smaller disk removed from its center), a hollow spherical shell, or any other easy-to-describe arrangement I code – then calculates the force (using Newton’s inverse square law) of each of those masses on a test mass, vector adding them all to find the net force. For simplicity, I use only integer coordinates. Because I assume a unit mass of the test particle, force and acceleration are the same, so I label these calculated values “Acceleration”. For convenience, I then calculate the orbital speed around the center of the disk, sphere, or whatever, and label it “Speed”. This is as simple an application of the inverse square law of gravity as can be – no calculus, no arithmetic more complicated than taking square roots to find the components of the force vectors. It makes no assumptions based on the results of any derived rules or theories. It’s also correct, within the limits of precision dictated by the number of point masses I use. If greater precision is needed, more point masses can be used, but for the purposes of showing that the “outboard mass” in a disk has a non-zero gravitational effect on a test body, so can’t be ignored, it doesn’t need to use many or be very precise. Athinker, you can reproduce these results yourself. I’d recommend using a programmable calculator or computer, as many thousands of arithmetic operations are required, but doing them slowly and carefully with paper and pencil would give the same results. What you should not try to do is argue them away with intuitive, natural language based arguments like “The shell theroem postulates that a shell can be bisected into an infinite number of planes so the original planer description holds for the sphere.” That’s simply provable and demonstrably wrong (and a misunderstanding of the shell theorem proof), and I’ve done both. Quote Link to post Share on other sites

athinker 4 Posted June 2, 2011 Author Report Share Posted June 2, 2011 You should note that your bolded text is part of a section referring to bodies orbiting outside of non sphericly symetrical bodies. (though I'm sure it's also true for bodies orbiting inside non sphericly or circularly symetrical bodies.) As to the point of mine that you particulary took exception to (snip) "The shell theroem postulates that a shell can be bisected into an infinite number of planes". Maybe you should look up "The Iterated function" http://en.wikipedia.org/wiki/Iterated_function Which is the function that extends the 2d planer Newtonian into the 3d spherical shell. It is also the function that extends a line into a circle and how calculus is done. I just mention it as you mentioned you may have forgoten some of your calculus, maybe some geometry too. Little physics refresher might be helpfull. After you do that maybe you should take a look at this page.http://www.sparknotes.com/physics/gravitation/potential/section3.rhtml Note the part where it says (note underlined) "The principle of superposition (see Newton's Law) tells us that we need to add up the vector sum of all the forces on m from the particles in the shell. It turns out that it is easier to calculate the sum of the gravitational potentials (since this is a scalar, not a vector) and take derivatives to find the force. We can do this by using U = (latex on page) and summing over all the masses. To do this, consider cutting the shell into rings Quote Link to post Share on other sites

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