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#91
November 22nd 06, 11:33 PM
posted to sci.astro,alt.astronomy,sci.physics.relativity
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Max Keon's "anisotropic gravity".
"Max Keon" wrote in message u... "George Dishman" wrote in message oups.com... "Max Keon" wrote in message u... .... This diagram depicted the motion of a star following a curved path through the outer reaches of a galaxy traveling relative to the background universe, and that is generating a gravity anisotropy. Let's stick with the problem at hand and keep the description as the motion of a planet at the mean radius of Mercury but in a circular orbit. That's what we've done the numbers for so far. The principles are the same. I'll add in the Sun (but not to scale): -------force direction- _ _ - - 0 -motion direction - - _ _ - - | - - | | gravity pulls towards Sun v Sun The force will always deflect the "0" mass in the direction initiated by the gravity source, toward the gravity source. No, you have shown the force pointing horizontally whereas the gravitational force is vertical. Remember we are talking about the anisotropy generated by all the other matter in the universe, -------- I'll snip the ball and head wind scenario because it only causes confusion. -------- Now consider the same scenario in the realm of the star. The star is slowed by the "head wind", and it's pulled down to the gravity well of the galaxy. Right. Now note that the slowing due to the "head wind" reduces the energy of the star just as the air reduced the energy of the ball. The amount of energy lost is equal to force times distance and that is true _regardless_ of the mechanism. So, in your physics, you can just ignore questions that you can't answer? This isn't "my physics", this is Newton's physics which you should have learned in your early teens. I find it hard to believe you aren't familiar with these absolute basics. You still have no idea then how energy is transferred between the star and the universe, or Mercury and the universe, or anything else and the universe? Mercury and the Sun? You showed how energy is transferred in the above diagram by drawing it like this: -------force direction- _ _ - - 0 -motion direction - - _ _ The amount of energy transferred is the product of the force and the distance moved parallel to the direction of the force. Since they are pointing in opposite directions, their product is negative so it reduces the kinetic energy. You haven't shown anything that tells me where the energy goes - that is your problem, not mine. The pull of gravity into the galaxy is an exchange between gravitational potential and kinetic energy just as before. This is the key to the analysis of Mercury. During your one second interval, the planet moves 48 km or 4.8e4 m. You said the anisotropic force resisting the motion produced an acceleration of 3.188e-9 m/s^2 so every kg feels a force of 3.188e-9 N and loses 4.8e4 * 3.188e-9 = 1.53e-4 J of energy (I'll talk about where that goes later). But you only ever talk about it. For once, just show me how the energy is transferred. Your own diagram shows it Max, don't you even understand what you drew? The arrow pointing from left to right marked "force direction" takes away the energy to a destination you have not specified. You said at the beginning: I had imagined that I could justify the universe generated gravity anisotropy if the fall rate to the Sun was treated with a simple logic that would still fulfill the requirements. That is right, but the requirement should be that, after losing 1.53e-4 J/kg to the anisotropy, What "losing"? Where did it get lost to? How should I know Max, _you_ drew the diagram and wrote the equation (though I helped simplify your algebra). All I have done is shown you what the consequences would be. any change of orbit should conserve what remains. The inwards motion due to the Sun's gravity conserves total energy by transferring between kinetic energy and gravitational potential energy. That requirement plus not changing the eccentricity, is what lets you calculate the new orbital speed and radius. The answer is that the radius reduces by 7.69 mm in the second, the potential energy reduces by 3.06 J/kg, the kinetic energy increase by 1.53e-4 J/kg and that gives the right total change. The answer is that Mercury is pushing into the head wind generated by its velocity relative to the universe and is deflected .. Nope, you drew this: -------force direction- _ _ - - 0 -motion direction - - _ _ Newtonian physics 101: f = m * a or a = f / m The acceleration is a vector which is equal to the force vector divied by the mass which is a scalar. That means the acceleration is in the same direction as the force, there is no deflection. You drew the direction of the force Max, that's the way the planet is accelerated, not sideways. Forget "head winds", they are don't do what you think and there is no wind in space. I'll snip your misleading 'headwind' stuff, as you said above: -------- I'll snip the ball and head wind scenario because it only causes confusion. -------- The reason it does that is because you haven't got the behaviour of simple ballistics right either. Maybe you should learn that first. You have been saying that it is an unrealistic amount but it means a change of 58 km per orbit which is a millionth of the change between aphelion and perihelion so I don't see why you think it unreasonable on logical grounds. It can be as logical as you like, but it has nothing to do with what the zero origin concept **predicts** (here) What I have shown you is the correct prediction from the equation you gave me _assuming_ the acceleration of the Pioneer craft is a measure of the effect of the "rest of the universe" as you claimed. You obviously need to go back and revise your basic Newtonian mechanics if you are going to understand it but if that equation is the theory you call the "zero origin concept" then then inward spiralling of the planets would be the consequence. Either the "rest of the universe" has a much smaller effect than you thought or your theory is simply wrong. And how might this force be applied? You drew the picture Max, you tell me. George 
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#92
November 27th 06, 03:23 AM
posted to sci.astro,alt.astronomy,sci.physics.relativity
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Max Keon's "anisotropic gravity".
George Dishman wrote: Max Keon wrote: George Dishman wrote: Max Keon wrote: This diagram depicted the motion of a star following a curved path through the outer reaches of a galaxy traveling relative to the background universe, and that is generating a gravity anisotropy. Let's stick with the problem at hand and keep the description as the motion of a planet at the mean radius of Mercury but in a circular orbit. That's what we've done the numbers for so far. The principles are the same. I'll add in the Sun (but not to scale): -------force direction- _ _ - - 0 -motion direction - - _ _ - - | - - | | gravity pulls towards Sun v Sun The force will always deflect the "0" mass in the direction initiated by the gravity source, toward the gravity source. No, you have shown the force pointing horizontally whereas the gravitational force is vertical. Remember we are talking about the anisotropy generated by all the other matter in the universe, -------- I'll snip the ball and head wind scenario because it only causes confusion. -------- Now consider the same scenario in the realm of the star. The star is slowed by the "head wind", and it's pulled down to the gravity well of the galaxy. Right. Now note that the slowing due to the "head wind" reduces the energy of the star just as the air reduced the energy of the ball. The amount of energy lost is equal to force times distance and that is true _regardless_ of the mechanism. So, in your physics, you can just ignore questions that you can't answer? This isn't "my physics", this is Newton's physics which you should have learned in your early teens. I find it hard to believe you aren't familiar with these absolute basics. You still have no idea then how energy is transferred between the star and the universe, or Mercury and the universe, or anything else and the universe? Mercury and the Sun? You showed how energy is transferred in the above diagram by drawing it like this: -------force direction- _ _ - - 0 -motion direction - - _ _ The amount of energy transferred is the product of the force and the distance moved parallel to the direction of the force. Since they are pointing in opposite directions, their product is negative so it reduces the kinetic energy. You haven't shown anything that tells me where the energy goes - that is your problem, not mine. But I explained where the energy goes. Mercury's loss of momentum in its orbit direction is shifted into the plane perpendicular to that motion, driving Mercury toward the Sun by exactly what was lost from its orbit velocity. Nothing else needs to be explained. But Mercury must initially be drawn to the Sun by much more than the 3.188E-9 meters in each second generated by the anisotropy. If that remained unchecked, I would agree that it would hit the Sun in less than 1 million years. But, for reasons I have already given, it only falls so far before it's inward motion comes to a halt. ------ ------ any change of orbit should conserve what remains. The inwards motion due to the Sun's gravity conserves total energy by transferring between kinetic energy and gravitational potential energy. That requirement plus not changing the eccentricity, is what lets you calculate the new orbital speed and radius. The answer is that the radius reduces by 7.69 mm in the second, the potential energy reduces by 3.06 J/kg, the kinetic energy increase by 1.53e-4 J/kg and that gives the right total change. The answer is that Mercury is pushing into the head wind generated by its velocity relative to the universe and is deflected .. Nope, you drew this: -------force direction- _ _ - - 0 -motion direction - - _ _ Newtonian physics 101: Your physics book is 300 years old. It's time to move on, don't you think? ------ ------ I'll snip your misleading 'headwind' stuff, as you said above: I snipped the ball within an air head wind because it was totally unrelated to the "head wind" generated by the background universe as matter moves through it. But you have snipped a crucial element in how the velocity slowing can be accounted for. If the force of the anisotropy slows Mercury's orbit velocity, some explanation is required as to where the energy has been transferred. The problem is quite easily resolved in the fact that Mercury's velocity reduction in its orbit direction has been transferred to the perpendicular plane pointing at the Sun, as a velocity increase of the same magnitude. All energy has been accounted for. If there is nowhere for the orbit velocity reduction to be transferred, there is also no way to explain how its orbit velocity is slowed. Such a situation would develop when Mercury's fall to the Sun has increased its orbit velocity per orbit radius enough for centrifugal forces to counteract the effect of the anisotropy. It would then be pointing squarely into the head wind, and as a consequence, there would be no residual drive in any direction. So, the easy answer is that it doesn't slow at all, or at least the slowing rate is vastly reduced. Mercury would then just rock back and forth within the influence of the background universe as it negotiates its orbit path, as if it was floating in the middle of a gigantic spring. The compression against the force from the universe in any specific forward direction would recoil back when Mercury's motion is reversed during the other half of its orbit cycle. The compression and expansion wave generated from an 88 day orbit cycle in a spring extending over 13 billion light years isn't going to be too hard to maintain. There is no point trying to explain any of this using Newtonian physics, or anyone else's physics where a gravity anisotropy is assumed not to exist. ----- Max Keon
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#93
November 27th 06, 08:36 AM
posted to sci.astro,alt.astronomy,sci.physics.relativity
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Max Keon's "anisotropic gravity".
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#94
November 29th 06, 09:32 AM
posted to sci.astro,alt.astronomy,sci.physics.relativity
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Max Keon's "anisotropic gravity".
George Dishman wrote: wrote: George Dishman wrote: Max Keon wrote: The amount of energy transferred is the product of the force and the distance moved parallel to the direction of the force. Since they are pointing in opposite directions, their product is negative so it reduces the kinetic energy. You haven't shown anything that tells me where the energy goes - that is your problem, not mine. But I explained where the energy goes. Mercury's loss of momentum in its orbit direction is shifted into the plane perpendicular to that motion, ... The definition of the word "force" in physics is rate of change of momentum. Your diagram shows the direction in which the momentum is being changed as parallel to the direction of motion, not perpendicular to it. To match your words, the diagram must look like this: _ _ - - 0 -motion direction - - _ _ | | | force direction V In that case there would be no energy loss since the force changes the direction of the momentum but not the magnitude. That's what I've been saying all along. There is no energy lost. But Mercury's orbit velocity does reduce by exactly the amount that it's now driving toward the Sun. Mercury would in fact be accelerating toward the Sun at the rate of 3.19E-9m/sec^2, and it would very quickly hit the Sun if that was to continue unchecked. But it doesn't because, as it falls, centrifugal forces increase to the point where the inward motion is counteracted. And it will stay there because no energy is being lost from the system. This diagram simply shows the normal gravitational force in a stable circular orbit. .. driving Mercury toward the Sun by exactly what was lost from its orbit velocity. Nothing else needs to be explained. But Mercury must initially be drawn to the Sun by much more than the 3.188E-9 meters in each second generated by the anisotropy. If that remained unchecked, I would agree that it would hit the Sun in less than 1 million years. But, for reasons I have already given, it only falls so far before it's inward motion comes to a halt. Your explanation contradicts the diagram. Considering symetry in a two-body problem with the motion of one body being directly away from the other, the diagram is correct and your words are wrong. The answer is that Mercury is pushing into the head wind generated by its velocity relative to the universe and is deflected .. Nope, you drew this: -------force direction- _ _ - - 0 -motion direction - - _ _ Newtonian physics 101: Your physics book is 300 years old. It's time to move on, don't you think? If you want to provide a replacement for Newton's equation f = m * a then by all means start again. Unless you do, I have to assume you are adopting basic mechanics. I'll snip your misleading 'headwind' stuff, as you said above: I snipped the ball within an air head wind because it was totally unrelated to the "head wind" generated by the background universe as matter moves through it. But you have snipped a crucial element in how the velocity slowing can be accounted for. If the force of the anisotropy slows Mercury's orbit velocity, some explanation is required as to where the energy has been transferred. You seem to feel it is necessary, I am treating it as a separate question. You have shown a force acting on the planet and we can analyse that while leaving the question of where the energy goes for another day. You can't explain where the energy goes, so you just ignore the question? Surely that's not the way it's supposed to work. Anyway, it doesn't actually go anywhere, it just changes direction. The problem is quite easily resolved in the fact that Mercury's velocity reduction in its orbit direction has been transferred to the perpendicular plane pointing at the Sun, Your diagram shows the force parallel to the motion, not perpendicular. You need to decide which direction it points before we can continue. Mercury is deflected toward the Sun by exactly what has been subtracted from its orbit velocity. http://www.optusnet.com.au/~maxkeon/merc-un.gif The set of oscillating rings, that reduce in oscillation magnitude per distance from the orbiting Mercury (red circle), depict how Mercury's motion will be perceived from the background universe. The scale is very compressed. Over each 88 day orbit cycle, one compression-rarefaction wave will be sent off throughout the plane of the orbit, into the universe. There will be around 18 of them between here and our nearest neighbor. As you can imagine, the force in the anisotropy generated by the universe is very flexible over the time span of Mercury's orbit. There is virtually no direct link between Mercury and any part of the universe. The smallest of the oscillating rings should probably be centered more around the Sun than is shown, and that would drive Mercury toward the Sun to an even greater degree. Mercury's fall to the Sun would be halted when centrifugal forces counteract the inward force. It could then cycle around within the background universe as if it was located in the center of a gigantic elastic web, and would lose very little energy in the process. The black line through the orbit axis is the perpetual wall of resistance from the universe generated gravity anisotropy. But over Mercury's orbit cycle time, that can be of very little consequence. ----- Max Keon
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#96
December 1st 06, 11:38 AM
posted to sci.astro,alt.astronomy,sci.physics.relativity
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Max Keon's "anisotropic gravity".
George Dishman wrote: wrote: George Dishman wrote: wrote: The definition of the word "force" in physics is rate of change of momentum. Your diagram shows the direction in which the momentum is being changed as parallel to the direction of motion, not perpendicular to it. To match your words, the diagram must look like this: _ _ - - 0 -motion direction - - _ _ | | | force direction V In that case there would be no energy loss since the force changes the direction of the momentum but not the magnitude. That's what I've been saying all along. No it isn't, it is at right angles to what you have been saying. Pioneer's trajectory curve toward the Sun would initiate the direction through which its momentum loss along the pointing direction will be sent. The momentum loss must go somewhere, and that's where it goes. It can't simply disappear. This would have the Pioneer anomaly perpendicular to its trajectory. And how could you possibly know that it's not? If it's falling perpendicular to its trajectory, in the direction of the Sun, at the rate of 8.4E-10m/sec^2 and its velocity is reduced by the same amount, it doesn't matter which way you look at it, it's still falling to the Sun at much the same rate. This is what you said befo -------force direction- _ _ - - 0 -motion direction - - _ _ ------ ------ If the force of the anisotropy slows Mercury's orbit velocity, some explanation is required as to where the energy has been transferred. You seem to feel it is necessary, I am treating it as a separate question. You have shown a force acting on the planet and we can analyse that while leaving the question of where the energy goes for another day. You can't explain where the energy goes, so you just ignore the question? Surely that's not the way it's supposed to work. Actually it is the way science works. It characterises nature mathematically without bothering about explanations. How can one possibly describe something mathematically when its existence is firmly rejected, even though the Pioneer anomaly very strongly supports it? The problem is that it would throw your neatly packaged little world into chaos. Which will be of no concern whatever in the end. WE REJECT REALITY AT OUR PERIL. You of course already know that because you're a scientist. Those are left to philosophy, although often they work hand in hand. Anyway, you haven't provided the explanation for the mechanism that is supposed to create the anisotropic force so the problem is yours. I don't care since the force doesn't exist, it is a figment of your imagination. Dark matter and dark energy are not figments of your imagination then? First of all you postulate their existence, then you just push the stuff around until the math gives you the answer you want. That's fair enough I suppose, but you really need some sort of prediction to go with it. A gravity anisotropy offers a far more logical explanation for the anomalous galaxy rotation rates. The fact that centrifugal forces between a galaxy center and the stars in its outer region alter according to 1/r instead of 1/r^2 is a fairly good indication that a gravity anisotropy exists. The anisotropy drives the star inward toward the galaxy center, until centrifugal forces build to counteract the fall. The anisotropy is directly proportional to velocity, while the centrifugal force alters according to v^2, thus the fall is halted. The star will now be traveling faster than it should be for the orbit radius. There is obviously a good reason why it stops falling where it does. It would be some amazing coincidence if the balance of dark matter and dark energy in the universe just happened to fall into place to satisfy that same criteria, don't you think? Anyway, it doesn't actually go anywhere, it just changes direction. The problem is quite easily resolved in the fact that Mercury's velocity reduction in its orbit direction has been transferred to the perpendicular plane pointing at the Sun, Your diagram shows the force parallel to the motion, not perpendicular. You need to decide which direction it points before we can continue. Mercury is deflected toward the Sun by exactly what has been subtracted from its orbit velocity. Then you are predcting Pioneer would be deflected in a direction perpendicular to its trajectory, or maybe perpendicular to the line between it and the Sun. Either way your prediction is wrong. Oh, and since the perpendicular to a line is a plane, that doesn't identify a unique direction and by symetry it could equally be to the left or right, so your new suggestion is impossible. It's not actually a new suggestion though. It was born out of necessity because all energy must be accounted for, and that's the only possible way to do it. I too needed to explain where it went. The time span required to shift the relevant mass of the universe, which is providing the local basis for the anisotropy, completely rules that out as an energy absorber. Gravitational radiation is all there is in that direction. But it should be obvious where the momentum shift would be. It's going to take the line of least resistance, being the inside edge of its trajectory curve, toward the Sun. That's where it was already pre-directed. ----- Max Keon
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#97
December 1st 06, 02:32 PM
posted to sci.astro,alt.astronomy,sci.physics.relativity
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Max Keon's "anisotropic gravity".
wrote in message ps.com... George Dishman wrote: wrote: George Dishman wrote: wrote: The definition of the word "force" in physics is rate of change of momentum. Your diagram shows the direction in which the momentum is being changed as parallel to the direction of motion, not perpendicular to it. To match your words, the diagram must look like this: _ _ - - 0 -motion direction - - _ _ | | | force direction V In that case there would be no energy loss since the force changes the direction of the momentum but not the magnitude. That's what I've been saying all along. No it isn't, it is at right angles to what you have been saying. As your diagram got lost in the trimming, I repeat it here purely for reference: -------force direction- _ _ - - 0 -motion direction - - _ _ Pioneer's trajectory curve toward the Sun would initiate the direction through which its momentum loss along the pointing direction will be sent. Momentum isn't something that gets 'sent', it is a property of the body. The momentum loss must go somewhere, and that's where it goes. It can't simply disappear. Of course, just as the momentum lost due to the basic gravitational force exerted by the Sun must go somewhere if momentum is to be conserved. The momentum lost by the craft is transferred to the Sun. The same applies to the planets as a result of their orbital motion. That causes the Sun to move a small amount which is how most extra-solar planets have been found. This would have the Pioneer anomaly perpendicular to its trajectory. And how could you possibly know that it's not? Because the trajectory is almost radial, about 11 degrees off during the time the anomaly was measured, while the anomaly is directed towards the Sun within a few degrees. If it's falling perpendicular to its trajectory, It is not falling perpendicular to its trajectory, it is falling in a direction almost opposite direction to its trajectory. Here is the Sun 'S' and Pioneer 'P' with a line showing its velocity 'v': v\ P S The Newtonian gravitational acceleration 'g' points towards the Sun for Pioneer and towards the craft for the sun: v\ P | g| | | g| | S If the anomaly 'a' is like a drag or friction then it opposes the trajectory. To conserve momentum there would be a change of speed of the material producing the drag or possibly the Sun. The latter would like this: v\ P |\a g| | | |g a\| S Instead, I thought you were saying that your theory was a modification to gravity that meant speed increased the effective force when the objects were moving apart. Certainly you said we had to take the component of the speed in that direction. I can show that as a longer line like this: v\ P | g| | | | | g| | S It is clear in this case that momentum is conserved because the longer line on Pioneer is matched by the longer line on the Sun. In order to calculate the effect we can treat that longerer line as the sum of the standard Newtonian acceleration and a separate anisotropic acceleration. If I remove the Newtonian part from the diagram for clarity, it looks like this: v\ P |a |a S Momentum is still conserved in both this and the version above and they are both compatible with the observations. ... WE REJECT REALITY AT OUR PERIL. Exactly. The Pioneer anomaly could be directed exactly opposite its trajectory or it could be towards the Sun but it certainly is not perpendicular to either. It is the final diagram above that I have been discussing for the last several weeks but the version with the anisotropy opposing the velocity would be very similar for the planets or a star in the galaxy. snip incorrect conclusions and philosophy Then you are predcting Pioneer would be deflected in a direction perpendicular to its trajectory, or maybe perpendicular to the line between it and the Sun. Either way your prediction is wrong. Oh, and since the perpendicular to a line is a plane, that doesn't identify a unique direction and by symetry it could equally be to the left or right, so your new suggestion is impossible. It's not actually a new suggestion though. It was born out of necessity because all energy must be accounted for, and that's the only possible way to do it. Tough luck, the anomaly is not perpendicular to the direction of motion and kinetic energy is observed as being lost from Pioneer at a rate of 2.6mW due to the anomaly. That is over 80kJ per year. I too needed to explain where it went. That remains your problem, not mine. The reality is that Pioneer is losing kinetic energy because of the anomaly. But it should be obvious where the momentum shift would be. It's going to take the line of least resistance, being the inside edge of its trajectory curve, toward the Sun. That's where it was already pre-directed. That is gibberish, momentum is not a substance. George
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#98
December 23rd 06, 01:20 AM
posted to sci.astro,alt.astronomy,sci.physics.relativity
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without dark matter
Merry Christmas!
without dark matter for a pure space! The Law of Universal Gravitation and Separation First of all, I would like to present this fundamental working hypothesis. It is the universal force which unifies gravity and separator into one. No one ever knows about the separator force. So now, I give a definition of the law of universal separation at first. Fs = - Sc Ea Eb / r^2 F ; the separation force Ea ; Energy which belongs to the point a Eb ; Energy which belongs to the point b r ; the distance between a and b Sc ; the separation constant This force will not be detected on the Earth. It can be negligible enven in the solar system. But it will work in the galactic scale. Next step is to unify Gravitation and Separation into one law. Fg = G Ma Mb / r^2 -----------------Then, Fg+s = G Ma Mb / r^2 - Sc Ea Eb / r^2 And assume that Sc = G / c^4, because E = mc^2. Fg+s = G Ma Mb / r^2 - (G / c^4) Ea Eb / r^2 One step forward by using the complex number formula. F = G ( Ma + i Ea / c^2 ) ( Mb + i Eb / c^2 ) / r^2 F = G Ma Mb / r^2 - (G / c^4) Ea Eb / r^2 + i ( G Ea Mb / ( r^2 c^2 ) + G Ma Eb / ( r^2 c^2 )) The real part is Re( F ) = Fg+s, but I don't know how to deal with the imaginary part ; Im( F ) = G Ea Mb / ( r^2 c^2 ) + G Ma Eb / ( r^2 c^2 ). So an existing substance is to be described as S = M + i E / c^2 . In a certain independent area, if M + E / c2 = constant ( in other words when M desceases by ΔM, E will increases by ΔE = ΔM c^2 ), then abs( S ) = will be minimum when M = E / c^2, because abs( S ) = root( M^2 + E^2 / c^4 ). The Two or One dimensional galaxy The spherical harmonics are the angular portion of the solution to Laplace's equation in spherical coordinates where azimuthal symmetry is not present. And there are three types of galaxies. elliptical galaxy e.g. NGC4881 Three Dimension GM(r)m/r*r = mv*v/r spiral galaxy e.g. NGC4414 Two Dimension G'M(r)m/r = mv*v/r barred spiral galaxy e.g. NGC1300 One Dimension G"M(r)m = mv*v/r Emissions like strong lights will distort the gravity. http://www.geocities.jp/imyfujita/galaxy/galaxy01 [...] Iori Fujita
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