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geistkiesel
USA
5 Posts |
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Larry Burford
USA
1355 Posts |
 Posted - 01 Dec 2009 : 13:20:32
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No such data exists. Newton and Einstein treat gravitational force as if it has infinite range.
The sizes and shapes of the various types of galaxies, and the observed orbital velocities of stars in them at various distances from their cores, can be interpreted as evidence for a limited range of gravitational force. Based on this interpretation, some alternative theories either postualte or deduce a limited range for gravitational force.
One of them is DRP, which uses a particle based LeSagian model for gravitational force. It deduces that the range of gravitational force is on the order of a few kiloparsecs. |
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Jim
1607 Posts |
Posted - 01 Dec 2009 : 14:18:15
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| It seems you could say all gravity force is local and not go too far wrong. The force centers on mass and is very weak so any large mass has its own domain. |
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MarkVitrone
USA
386 Posts |
Posted - 01 Dec 2009 : 16:57:09
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Larry, I believe Tom had calculated somewhere around 21 kiloparsecs from galaxy data....I can't put a finger on the source at the moment, but that is what comes to mind....
Mark Vitrone |
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dholeman
USA
79 Posts |
Posted - 02 Dec 2009 : 01:55:16
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I forget which author it was (probably Tom) in Pushing Gravity who gave the mean free path of gravitons as being about 1 or 2 kiloparsecs. Tom also discussed it somewhere on this messageboard as well. The mean free path is the distance a graviton travels, on average, before colliding with another graviton, and is thus the limiting factor for the effective range of gravity.
>> The force centers on mass and is very weak so any large mass has its own domain.
Well, the most massive body in the neighborhood dominates all other bodies because the gravitational forces net out in it's favor but each body still 'feels' the gravitation of all the others within range of a couple of kiloparsecs.
No great thing was ever created suddenly - Epictitus |
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Larry Burford
USA
1355 Posts |
Posted - 02 Dec 2009 : 09:16:55
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Tom discusses this in Ch 4 (pg 79 in both editions). He concludes that the mean free path of gravitons appears to be about 2 kiloparsecs, but does not specify a value for the "range" of the force. After traveling two or three mean free path lengths the vast majority of gravitons will have interacted with at least one other graviton and will no longer convey any information about the last mass they encountered.
So, gravitational force should begin to deviate from the inverse square model almost immediately, probably within our solar system. But we will need better measuring devices than we have to detect this small variation. It is likely that some amount of gravitational force can still be detected as far out as 21 kiloparsecs (again, our instruments will influence this). I guess we ought to try to define what we mean by the "range" of gravitational force so that we can be sure we are talking about the same thing. First detectable deviation from inverse square? Maximum distance for detection of any effect? Or some middle ground, like mean free path?
Of course, this sort of thing is just one level above rank speculation. Until we can actually detect gravitons and actually measure the mean free path of their flight, we must admit to ourselves that we do not really know. |
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Jim
1607 Posts |
Posted - 02 Dec 2009 : 13:42:50
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| Even defining gravity is impossible-all that is known for sure is its a predictable force. |
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Larry Burford
USA
1355 Posts |
Posted - 05 Dec 2009 : 10:02:22
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Hmmm. Define. Explain. Describe.
Sometimes you really puzzle me, Jim. Why do you say things that are so easy to disprove? Every dictionary I've looked at has a definition for gravity.
Perhaps you meant explaining, rather than defining? But that is not impossible either. We have several explanations that seem at least a little plausible. Whether any of these explanations are a good match with physical reality remains to be seen, but we will get there.
We also have several really good mathematical descriptions of how gravitational force behaves. These are what allow us to predict ... but gravitational force is observed to behave differently than our predictions under some conditions. And these deviations are not always small, so it is presently not "for sure ... a predictable force".
LB |
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Jim
1607 Posts |
Posted - 06 Dec 2009 : 20:33:30
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| Hi LB, I could guess what you refer to that defys your best models but it would be clearer if you listed these items. I guess you are thinking of all the problems galatic disks display but you may be aware of things I am not where gravity defys your models. A humble statement might include some reference to the fact the models could have shortcomings rather than say gravity is unpredictable(don't you think?). |
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PhilJ
USA
269 Posts |
Posted - 18 Jan 2010 : 19:41:25
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quote:
geistkiesel: Does anyone have information regarding any distance limitation for gravity forces?
Chinese Academy of Sciences conducted highly sensitive gravity measurements during last July's solar eclipse in an effort to detect whether the moon blocks a fraction of the sun's gravity. I'm still waiting to hear what they found.
If TVF is correct about the limited range of gravity, Newton's universal law of gravitation becomes
I used to think that was an inverse cube law for large values of r; on second though it decreases much faster than the inverse cube. (I guess I better hunt down those erroneous posts and fix them.) That being the case, I would expect the outer arms of galaxies to orbit much more slowly than what the inverse square law predicts. Observations show them rotating faster, not slower. If dark energy is the answer, then a limited range of gravity would require more dark matter, not less.
I'm not enough of a mathematician to determine what effect the finite speed of gravity might have on galaxy rotation curves. My limited reasoning is, if the finite speed of gravity tends to move planets farther from the sun by pulling them forward, it should have a similar effect on galaxies. TVF used that postulate to set a lower bound on the speed of gravity. Has anyone applied that same reasoning to the rotation curves of galaxies?
My own Fractal Foam Model of Universes does not suggest a limited range of gravity. My gravitons are ethereal pressure waves (dark energy), not elastic particles. Pressure waves pass thru one another rather than bounce off one another. I do believe these pressure waves exchange momentum with ethereal shear waves, and matter consists of orbiting pairs of shear waves, so gravity blocking is not precluded by my model.
Fractal Foam Model of Universes: Creator |
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Jim
1607 Posts |
Posted - 19 Jan 2010 : 15:34:44
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| PhilJ, The rule used to estimate the speed of stars in a galatic structure is not designed to work on a disk because mass of a disk is not located anyway near the center of the disk. The simple fact is too simple for the advanced minds of astronomers to comprehend but the mass of a disk is scattered throughout the disk structure and every star within the structure adds some small fraction of the total force effecting the the structure. This simple fact also applies to more compact structures like stars and planets so when viewed from outside the rule works very well-it just doesn't work within the structure. Models have limits. |
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Jim
1607 Posts |
Posted - 25 Jan 2010 : 15:04:17
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| Why not estimate the limit of a gravity field by assuming it's force vanishes when it becomes equal to the gravity force of the universe in total? No one yet agrees on a concept of universal gravity but I see it as a force equal to ~1nm/s^2 that is observed and defined as a universal redshift of about that force. This would explain the cosmic redshift as a gravity effect. |
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PhilJ
USA
269 Posts |
Posted - 29 Jan 2010 : 19:46:29
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quote:
Originally posted by Jim
Why not estimate the limit of a gravity field by assuming it's force vanishes when it becomes equal to the gravity force of the universe in total? No one yet agrees on a concept of universal gravity but I see it as a force equal to ~1nm/s^2 that is observed and defined as a universal redshift of about that force. This would explain the cosmic redshift as a gravity effect.
I haven't done the math, but isn't the problem with galaxy rotation curves happening in a much stronger field than what you are referring to?
What is this "gravity force of the universe"? Are you referring to the acceleration of every point away from the origin. That acceleration does not exist in comoving coordinates. In non-comoving coordinats, it is equivalent to a parabolic gravity hill centered on the origin; its slope (rH0^2) is proportional to the distance from the origin (except when relativity turns it into a hyperbolic function). Perhaps the cosmic redshift is partly a gravity effect in non-comoving coordinates. We have to be very careful to specify the coordinate system in which we are arguing. Different coordinates systems define space, time, etc., differently. Disagreements often result from the fact that the same words have different meanings in different systems.
Fractal Foam Model of Universes: Creator |
Edited by - PhilJ on 29 Jan 2010 19:48:36 |
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Jim
1607 Posts |
Posted - 30 Jan 2010 : 13:08:20
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| PhilJ, The point of where gravity is generated is the mass center-is that right? The mass is distributed throughout the universe-right? Therefore there is trillions of trillions gravity centers and the force is more or less the same everywhere with small scale bubbles of much more force where mass concentrates such as galatic disks and stars within those disk structures. As to the math-you can make anything from that depending on what is assumed. In the current standard model you must assume(as you clearly imply)the center of gravity is at the point where the mass is centered. But, is it not true that only works when the observer is outside the observed structure? Maybe you can locate a mass center from the outside but not from the inside. |
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PhilJ
USA
269 Posts |
Posted - 31 Jan 2010 : 19:24:49
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quote:
Originally posted by Jim
PhilJ, The point where gravity is generated is the mass center-is that right? The mass is distributed throughout the universe-right? Therefore there is trillions of trillions gravity centers and the force is more or less the same everywhere with small scale bubbles of much more force where mass concentrates such as galatic disks and stars within those disk structures. As to the math-you can make anything from that depending on what is assumed. In the current standard model you must assume(as you clearly imply)the center of gravity is at the point where the mass is centered. But, is it not true that only works when the observer is outside the observed structure? Maybe you can locate a mass center from the outside but not from the inside.
The Big Bang model has all the gravity in the universe pulling inward toward the center, but every point is the center. But if the universe is homogeneous, homotropic and infinite in extent, then the total gravity at any point results only from local clusters of mass; the rest of the universe contributes nothing to gravity.
The mass of our galaxy is distributed, more or less, in a disk. But, if Newton's law of gravity is correct and the speed of gravity is infinite, then from where we are, the center of the galaxy's gravity must be at the center of the galaxy, about 3.0 x 10^4 ly away.
That's not the same as all the mass being concentrated at the center.
Inside a uniform sphere of radius R, only the mass inside the smaller sphere of radius r contributes to gravity at r. I believe the same is true of a uniform disk. But proofs of that theorem assume that R is small or the speed of gravity is infinite.
Due to our orbit around the galaxy, the direction of the galactic center moves 360 deg westward about every 2.2 x 10^8 years. If the speed of gravity is 3 x 10^10 c, there should be about 10^-6 year (30 seconds) of delay (aberration) for gravity from the galactic center. We should feel the pull of the galactic center's gravity approximately 360/2.2 x 10^14 = (1.63x 10^-12)deg east of the direction where the galactic center is now, as seen from Earth.
Due to the distribution of mass within the galaxy, the aberration should be less for the near side of the galaxy and more for the far side. I invite you mathematicians to calculate the overall aberration of the galaxy's gravity due to speed-of-gravity delay.
Of course, the math is very different if you accept the arguments of Kopeikin and Carlip. Then, the aberration for the galactic center should be 360 deg x 30 / 2.2 x 10^8 = (4.91 x 10^-5)deg ; but some alleged relativistic effect (which I don't understand) cancels that out.
Fractal Foam Model of Universes: Creator |
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Stoat
United Kingdom
863 Posts |
Posted - 03 Feb 2010 : 04:58:10
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Perhaps we could get Joe Keller to adapt his computer program to look at the star S2. This is the most studied; they have a complete orbit of it; of the stars going round the super massive dark object at the centre of our galaxy. This thing is really shifting. I did a quick calculation of its relativistic mass but I had the wrong figure for the mass. I thought it was about six solar masses but it turns out to be fifteen. Now it look as though it has a rosetta orbit but we'll have to wait for confirmation. That means that the very elliptical orbit moves slightly round a radius, on each orbit, and the track of it draws out a flower petal shape. Mass wise it will have to vary by more than a solar mass!!!
It's a baby, 15 solar masses and not in the best place for a toddler. Why hasn't gone nova?
Last but not least, how does the idea of a neutrino ball grab ya? One of these at the centre would be pretty wild, huge radii, and invisible.
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Stoat
United Kingdom
863 Posts |
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PhilJ
USA
269 Posts |
Posted - 03 Feb 2010 : 15:58:49
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quote:
Originally posted by Stoat
Perhaps we could get Joe Keller to adapt his computer program to look at the star S2.
Here's a free gravity simulator that you can install on your computer to do the simulation yourself. Just type in the orbital parameters and it will do the rest.
You have to be pretty sharp if you want to modify the program to accommodate a new theory of gravity, though.
Fractal Foam Model of Universes: Creator |
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Stoat
United Kingdom
863 Posts |
Posted - 04 Feb 2010 : 04:51:52
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Ah, that program was mentioned by a south american guy in Joe's thread, now I'm not sure whether Joe adapted that or wrote his own, I am sure that I didn't look at it, because I'm bone idle.
My first thought with the S2 star problem, was to simply scale things down and look at the orbit. That means taking the dark massive thingy and divided it through by 2.5E 6 to make it an average Sol. Then do the same for S2. That makes S2 a terrestrial mass planet. It's way out in the boonies, like Joe's proposed planet.
I did manage to find one paper for free aout neutrino balls, which I haven't read yet. I did have a quick read of the explanation of the rosetta orbit of S2. We need to turn the space around the super massive, whatever it is, into a neutrino ball. Our S2 planet can then dive through this and ignore the mass above it.
I think this ball has to be invented, because the famous rubber space-time sheet is fossilized round such a huge black hole. I'm not a betting man but I think a loaded new test of relativity will be wheeled out for us to marvel at. |
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PhilJ
USA
269 Posts |
Posted - 04 Feb 2010 : 11:46:21
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A caution to anyone who plays with the free gravity simulator: You can run the simulations as fast as you like. You can watch planets zipping around like a swarm of bee's. Unfortunately, if you run it too fast, the accuracy goes out the window. If you're simulating our solar system, you can eject Mercury and Venus from the galaxy by running the simulation too fast.
Fractal Foam Model of Universes: Creator |
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Stoat
United Kingdom
863 Posts |
Posted - 05 Feb 2010 : 06:52:12
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I never did like Mercury and Venus anyway  I've spent the last hour going crazy looking at a couple of exponential graphs. The main reason why I always try to get Joe Keller to do all the work.
I was looking at MOND. Say we've got c^2 / b^2 b being the speed of gravity. Our answer will be a tiny number. I reckon that it will be 6.626E-34 but we should try as many numbers as we like.
We can write that as 1/eta^2 which will be the reciprocal of 6.626E-34 that's 1.509E 33
With MOND we have F = GMm / r^2 = m* mu(a/a_0)a
We can lose that lower case m which is the mass of a star. The upper case M is the mass of the galaxy.
Milgrom takes a_0 as being 1.2E-10 which he says is the acceleration
a = v^2 / r (here v is the speed of light and r is the age of the universe) Yeah, the universe has to be about four and a half times bigger with this but we'll leave that for now.
We do the maths for the acceleration of a star at the edge of our galaxy find out what a / a_0 is, that comes out at 2.718 as near as damn it. Then we need to look at mu(x) = 1 - e^-x which is
1 - 1 / e^2.718
The answer is, 0.934
Now if we do the same for my speed of gravity the answer comes out at 0.9869
A Newtonian infinite speed of gravity will get closer to one.
Now I think that there's a phase change at the speed of light, which means we flip the sign in the Lorentzian but I'll leave it at that for now.
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what is the observed distance range of gravity?
