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afterburner
2006-Jul-28, 07:30 PM
Firt answer the question: Is the Universe a closed system?

How can we say the conservation of energy law in a closed system exists, if the closed system that we are in, is breaking that very law?

We live in a sphere with a radius of 13.7 billion ligh years. Light further away than that will never reach us, becasue the expansion of the Universe is "faster" than light. Assuming that gravity also propagates at the speed of light, we are effectively in a closed system, since nothing outside the hypothetical sphere will ever be seen or felt. Considering that the rate of expansion is accelerating, the "bubble" that is our universe is shrinking at an accelerating rate as well.

Eventually our "bubble" will be much smaller than it is now. If this bubble/Universe is a closed system (since nothing beyond the "edge" will ever be seen or felt AND it is the only thing we know) and it is losing energy, how can we posibly say that the conservation of energy law is valid?

Assuming that FTL travel is impossible, of course. As well as the fact that our visible Universe is EVERYTHING that exists, since it is the only thing we know.

You cound argue that there used to be more "stuff" in the Universe. In that case, i could argue that there used to be 10^99999999 times more stuff in the universe...

I could be way off on this...what do you think?:eh: :neutral:

Ken G
2006-Jul-28, 08:17 PM
Actually, it is not generally expected that the law of conservation of energy is globally true on the scales of the universe. All laws of physics are local laws, including the conservation of energy.

BigDon
2006-Jul-28, 08:31 PM
Actually, it is not generally expected that the law of conservation of energy is globally true on the scales of the universe. All laws of physics are local laws, including the conservation of energy.

Are you sure about that? Why is here different than there?

Saluki
2006-Jul-28, 08:48 PM
You have defined an open system. You yourself indicate that energy is leaving the "system". In a closed system, no energy leaves. Energy is still conserved. In an open system, you have to account for energy leaving the system in your conservation equations.

Ken G
2006-Jul-28, 08:48 PM
Nothing is different there than here, but the issue is how does one transform one's reference frame from here to there. Reference frames are local, and physical quantities (like energy) are quantified in a reference frame. General relativity tells you how to connect different frames, but there is never conservation of energy when you change reference frames. There have been very technical efforts to try and recover some version of conservation of energy, but it's difficult and frankly over my head. What I would say is, the closest physics comes to having a globally applicable law over the whole universe is the cosmological principle, which says that what you see here is pretty much the same everywhere, and that allows you to extend a local reference frame meaningfully to what is happening over the whole universe. But that by itself does not give you global conservation of energy, nor does general relativity.

Squashed
2006-Jul-28, 10:47 PM
What I would say is, the closest physics comes to having a globally applicable law over the whole universe is ...

If this is true then why does the scientific community seem so reluctant to believe that gravity may operate differently in other parts of the universe? The most likely candidate for revision is the MOND idea but it seems to be generally discounted as a plausible theory/modification of general relativity.

astromark
2006-Jul-28, 11:38 PM
It is wrong to jump to conclusions. We do not know some of the things being said here. The 13.7 billion light year distance is only relative to us. Do you think that if we were halfway out there it might be different. At 6.3 billion l/y do you think they might also see 13.7 in all directions ? or just the 6.3 to where our edge appears and more in the opposing direction. I see this as wrong. We are not in the center. just the center when looking from here. This idea that light may be not reaching us because the source is/has exceeded light speed does seem perfectly logical. I like that. It could lead me to sagest that the universe may be infinite after all. To sagest that is not is I think naive. I am comfortable with GR being the same every where and the rules of physics not changing. BUT ! I can not know this.
yet.

trinitree88
2006-Jul-29, 12:21 AM
Actually, it is not generally expected that the law of conservation of energy is globally true on the scales of the universe. All laws of physics are local laws, including the conservation of energy.

Ken G. I would hope that Lorentz invariance turns out to be globally true on scales of the universe. I can't prove that it is true, but assuming that the universe works that way allows me to assume that any experiment done under controlled conditions anywhere on Earth is reproducible....and hence sensible. If that fails to hold true, what is the sense of anything? This would include not only conservation of energy, but all the conservation laws.Pete

Ken G
2006-Jul-29, 12:49 AM
Lorentz invariance is exactly what I am talking about when I say a local law. I am not saying that our local laws won't hold somewhere else, we just have not done experiments somewhere else. All we have to extend our laws over there is the cosmological principle, which certainly appears to be a very good law of its own. But even the cosmological principle isn't truly a global law, it is a way to splice together local laws so that you can span the whole universe and make predictions about everything you can observe. Physics uses differential equations, which are local. Integral equations are global, but are integrated differential equations and are generally only available when you also have known or assumed boundary conditions. As the boundary conditions are not part of the laws, they have to be added in externally. Like the cosmological principle. That's all I mean, not that I wouldn't expect our local laws to work everywhere else. Occam's razor takes care of that, along with the fact that our observations are entirely consistent with this view. But still, there is not a global principle of conservation of energy, which is where this all began.

Van Rijn
2006-Jul-29, 12:54 AM
It is wrong to jump to conclusions. We do not know some of the things being said here. The 13.7 billion light year distance is only relative to us.


That's relative to any point in the observable universe.



Do you think that if we were halfway out there it might be different.


There is no "halfway out there."



At 6.3 billion l/y do you think they might also see 13.7 in all directions ? or just the 6.3 to where our edge appears and more in the opposing direction.

There is no "edge." You can't get to an "edge" of the universe in three dimensions. It's a bit like asking about the center of the surface of the earth, or asking if London or New York are closer to the edge of a flat earth. But the earth isn't 2D and the universe isn't 3D.

afterburner
2006-Jul-29, 01:14 AM
It is wrong to jump to conclusions. We do not know some of the things being said here. The 13.7 billion light year distance is only relative to us. Do you think that if we were halfway out there it might be different. At 6.3 billion l/y do you think they might also see 13.7 in all directions ? or just the 6.3 to where our edge appears and more in the opposing direction. I see this as wrong. We are not in the center. just the center when looking from here. This idea that light may be not reaching us because the source is/has exceeded light speed does seem perfectly logical. I like that. It could lead me to sagest that the universe may be infinite after all. To sagest that is not is I think naive. I am comfortable with GR being the same every where and the rules of physics not changing. BUT ! I can not know this.
yet.

If someone was "halfway out there", I am almost certain the 13.7 billion light year bubble would exist for the observers on that other planet (although I have absolutely no way of proving this). The main reason being that (this will sound strange in your head, but bear with me)....that.... our Universe is a blubble, and if we follow everything back, we get the singularity and the BB theory. HOWEVER, since we dont know what was before 10^-43 seconds AFTER the BB, we dont know what state the universe was in BEFORE the BB(if there is such a thing(for you loyal BB fans out there))...

IF! there was no singularity, but some "mass", for lack of a better word, and IF this mass was gigantic and "gigantically" dense....then any point on/in that "mass" (would be a singularity, and when "exploded") will be an effective bubble much like our own for observers that happened to be in the "middle" of it all (in fact, they might have also come up with a BB theory, since it is the ony thing they would know, much like us).

Therefor these observers would also be in a 13.7 billion light year bubble (assuming the expansion is uniform throughout the "larger" Universe)...like i said..sounds weird, but hope you got the idea (admittedly my no-way-of-being-proven idea, or mabe its already been though of and dismissed)

:think:
What do you think? Way out in the left field or perhaps? (this or that style):D

astromark
2006-Jul-29, 01:31 AM
Yes, Afterburner I am in agreement with you. That is what I was attempting to convey. . . and No, Van Rijn I can only say that of coarse there is a half way. There is all ways a half way. You just might not ever be able to reach it. As infinity is just a tad to far away to be half way to. I am well aware of the 3 D structure of the universe. You are never going to convince me its other wise. The time component of your more than 3 D universe is irrelevant. I can not be wrong as you can not be right. All we can hope to do is discus this cosmology. We can not be so sure.

RussT
2006-Jul-29, 03:39 AM
Here is a very good breakdown of part of the consideration...Lorentz invariance is exactly what I am talking about when I say a local law.

http://www.mountainman.com.au/aether_3.html

publius
2006-Jul-29, 04:19 AM
Russ,

I disagree with their characterization of frame-dragging. Frame dragging is just gravitomagnetism, the gravitational analog of the magnetic field of moving charges. Well, the GR field is a full tensor field and is more than just two vector fields, E and B, so in the full non-linear solutions, I'm sure gravitomagnestism is more complex than regular magnetism.

You can easily see how their must be a gravitomagnetic effect by a simple thought experiment. Consider two masses with equal electric charge such that the electric repulsion just cancels the gravitational attraction. In a frame at rest with respect to the two charged masses, they are in equilibrium.

Now consider a frame moving with respect to the two. Here, we have a magnetic force in addition to the electric one. That magnetic force is attractive, so the total electric force becomes *less* than the gravitational attraction. And at high relative speeds, the electric field itself will become stronger as the Lorentz contraction "squashes" the equipotentials in the direction of motion.

One has to conclude that gravity must have a magnetic like component as well to maintain the equilibrium between frames. If one frame says there is equilibrium, then all frames must say there is equilibrium.

And indeed, the g field itself (the "gravitoelectric" component) must be modified at high speeds just like the E field is.

And GR does this and preserves that equilibrium. The only reason we don't notice gravitomagnetism like we do regular magnetism is gravity is much weaker and so mass currents do not produce the noticeable forces that electric currents do.

-Richard

Van Rijn
2006-Jul-29, 09:22 AM
I am well aware of the 3 D structure of the universe. You are never going to convince me its other wise.


So ignore observation, eh? Then why do you bother to ask questions in the "Question and Answer" section?



The time component of your more than 3 D universe is irrelevant. I can not be wrong as you can not be right.
All we can hope to do is discus this cosmology. We can not be so sure.

How can we discuss cosmology when you have unswayable beliefs? Anyway, if you have an ATM theory that you want to promote, you know where to take it.

RussT
2006-Jul-29, 09:38 AM
[I disagree with their characterization of frame-dragging.

Yes, I disagree with this as well (I think). That is, as far as I have been able to determine, from everything I have read, it does not appear to me, that baryonic matter interacts with whatever 'space' is made up of at all. Now, that is 'interaction', however, all baryonic matter motion is different than 'interaction'. Baryonic Matters (stars, planets, dust and gases) motion is dictated by two things; the gravity of other baryonic matter and the shape or curvature 'space' is in where the baryonic matter is, in the case of galaxy rotation, and I am not going to say why here and now, for the case of clusters.

The reason I linked this here (I actually hoped you would see this, and maybe relate it to your 'Borehole Gravity'), is that I thought he did an excellent job of showing how all the concepts fit together for mainstream to be at its current thinking and showing how other posiblilities 'might' have room to be fitted correctly, and that the current application of GR is just too mathematical, to be able to tell the 'real', 'physical' story of 'space'.

I got this link, from a link that has all kinds of different ideas, theories, and analysis on 'space', aether, etc, including quite a bit of Einsteins works.
Very good reading, if you can figure out what is pretty much bunk quickly.

http://www.inerton.kiev.ua/links.htm

The site owner's concept is of particular interest, and his resume is nothing less than stunning.

Back to the OP though. He has an absolutely valid point

Which is another reason I linked the 1st site, because he does a good job of showing how and why the current application of GR to the universal expansion and the stress-energy tensor is just mathematical, but since we now know that 'space' is 'something' physical...

More importantly than leaving 'our' space, never to be seen again (actually, the real issue here is leaving via the black holes, especially the massive ones (non-stellar)), how is 'more', 'physical', 'space' being made to be able to expand?

trinitree88
2006-Jul-29, 11:30 AM
Lorentz invariance is exactly what I am talking about when I say a local law. I am not saying that our local laws won't hold somewhere else, we just have not done experiments somewhere else. All we have to extend our laws over there is the cosmological principle, which certainly appears to be a very good law of its own. But even the cosmological principle isn't truly a global law, it is a way to splice together local laws so that you can span the whole universe and make predictions about everything you can observe. Physics uses differential equations, which are local. Integral equations are global, but are integrated differential equations and are generally only available when you also have known or assumed boundary conditions. As the boundary conditions are not part of the laws, they have to be added in externally. Like the cosmological principle. That's all I mean, not that I wouldn't expect our local laws to work everywhere else. Occam's razor takes care of that, along with the fact that our observations are entirely consistent with this view. But still, there is not a global principle of conservation of energy, which is where this all began.

Ken G. Nice semantic clarification. Thanks. Pete:clap:

astromark
2006-Jul-29, 02:08 PM
I most certainly do not have any beliefs. Flexible I try to be. If you look at my contributions I am receptive of new knowledge and will gladly throw away previously held notions as the scientific weight of evidence prevails. Would you please explain your first comment Van Rijn ? I simply, do not understand.

mugaliens
2006-Jul-29, 03:10 PM
Eventually our "bubble" will be much smaller than it is now. If this bubble/Universe is a closed system (since nothing beyond the "edge" will ever be seen or felt AND it is the only thing we know) and it is losing energy, how can we posibly say that the conservation of energy law is valid?

Just because what we see may be limited to a smaller subset of the whole doesn't mean that the whole doesn't exist. While a mass outside our radius might not affect us, it still affects other masses within the sphere of what we see.

The law of conservation holds for any system. Systems are defined by boundaries. If the sphere of our perception is shrinking, it's boundary is changing. If you change the boundary, you've changed the results, and while it may appear that conservation is broken, in reality it remains.

Ken G
2006-Jul-29, 06:15 PM
Let me clarify. The reason energy is not conserved globally in the universe has nothing to do with it being closed or open, it has to do with the way we coordinatize it. By far the best way to coordinatize the universe is to use locally inertial frames that are in free-fall with the local cosmological gravity and move with the average local matter, i.e., the "comoving frame". In this frame, we say that the galaxies are not receding from us, but rather "space is expanding". However, if you shoot a bullet at 99% of the speed of light, it zooms out of our galaxy and passes many others (over billions of years), and as far as I understand these things (not as far as I'd like), it will eventually pass those distant galaxies at slower and slower speeds, relative to the galaxies, though it would still be moving at 99% of the speed of light relative to us (barring other effects). But in the comoving coordinates, which we are using as our global coordinate system to avoid problems like things moving faster than c, the bullet slows down and therefore does not conserve energy. The loss of conservation is due to shifting from our reference frame to the comoving frame, but that is necessary to do global physics. This is my point about the problems with global physics and how it is really just a cobbling together of local (comoving) physics in a way that satisfies the cosmological principle. Hence, no global conservation of energy, just kiss it goodbye, it's not so hard.

Van Rijn
2006-Jul-29, 10:27 PM
I most certainly do not have any beliefs.
Flexible I try to be.


Yet after stating your ATM ideas you wrote "You are never going to convince me its other wise" and "I can not be wrong as you can not be right." In my book, that counts as an unswayable belief.



Would you please explain your first comment Van Rijn ? I simply, do not understand.

You stated that the universe is 3D, despite observation that indicates it is more than that on the cosmological scale, then stated your solid commitment to this belief, as noted above. So given your firm beliefs, why would you ask questions in this section, where mainstream science is discussed?

WaxRubiks
2006-Jul-30, 12:56 AM
isn't the Universe a hypersphere? If so then it has no boundaries but is not infinite. there is no halfway to the edge as there is no edge.

publius
2006-Jul-30, 02:55 AM
The reason I linked this here (I actually hoped you would see this, and maybe relate it to your 'Borehole Gravity'), is that I thought he did an excellent job of showing how all the concepts fit together for mainstream to be at its current thinking and showing how other posiblilities 'might' have room to be fitted correctly, and that the current application of GR is just too mathematical, to be able to tell the 'real', 'physical' story of 'space'.



Russ,

I have been looking at this Borehole gravity business of Cahill some more. It is indeed something of fluid-flow model, or his "velocity field" business looks like a fluid-flow concept. g, the gravitational acceleration, is just the total (or directional) time derivative of that velocity field, dv/dt. Using '@' signs for partial derivatives, you can write the total derivative of a field, 'A', along some direction defined by a velocity vector 'v', is

dA/dt = @A/@t + (v dot del) A. Now, let v be the field itself, and take the derivative along v itself, and you get

dv/dt = @v/@t + (v dot del)v. That is what he defines the gravitational acceleration to be. In his framework, the mass distribution determines this velocity field (whatever it may really be), and 'g' is then the the above operator on that field. In the irrotational case (curl v = 0, which implies curl g = 0), this velocity at a point would be just the velocity of a particle free-falling from infinity that passed through that point, and so its square, v dot v, would be the Newtonian gravitational potential.

Now, what he adds is a self-interaction term (the constant is the strength of this self-interaction, and he says this constant is just the fine structure constant). This has the effect of "gravity making gravity". Mass makes a gravitational field, and that field itself makes more gravity, amplifying the total field.

This self-interaction is actually in GR, but it is very weak and disappears in the weak field Newtonian limit. But Cahill has it being relatively strong in the weak field, but cleverly "hiding" so that it is not apparent outside mass distributions, just inside. So roughly, for a spherical mass distribution of constant density, the self-interaction makes (alpha/2)M more worth of field.

Being non-linear, the Borehole field won't allow superposition in general. That is, the field at a point due to two masses is not equal to the sum of the fields of the individual masses, but it may be almost that in the exterior limit.

It may also suffer from conservation of momentum problems, unless the non-superposition can preserve it somehow.

For example, (and this something I'm not certain of), the Borehole field will be non-zero inside a hollowed-out spherical shell. So a little test mass inside will be attracted to the center. But in Newtonian gravity, that is of course zero, but more importantly, the net force of the field of the test mass itself on the sphere is also zero. And it has to be so the two forces are equal and opposite.

So, if the hollow sphere exerts a force on the test mass, we have a big problem is the test mass doesn't exert an equal an opposite force on the sphere. If the field outside a spherical test mass is still Newtonian, then it won't. The non-linearity and non-superposition might save this, but I'm not sure.

-Richard

Ken G
2006-Jul-30, 02:22 PM
That's a nice analysis publius, thanks for saving me from having to read it! I note that the whole fluid-model business is a complete red herring. Of course the acceleration is the total time derivative of some fluid flow, where the fluid flow describes the motion of the coordinates followed by force-free test particles which are moving with the flow. This is nothing but the equivalence principle, it's not saying a thing. What is saying something is how you get the fluid flow, what are the equations that generate it. His equations are just the same as Newton's, plus an additional term proportional to alpha. That is the crucial part-- there is nothing at all special about all this fluid business, it is just saying that gravity is an acceleration and no more. And I agree with publius that just tacking on a term and saying it agrees better with certain observations is an obvious example of "fixing" a theory for a specific case, without considering how it will mess up a lot of others, including well-worn conservation laws. Pseudoscience does that all the time.

publius
2006-Jul-30, 11:32 PM
Ken,

I caution I do not know if this does violate conservation of momentum, only speculating about the test particle inside a hollowed out sphere. The non-superposition may save it, and say there would be equal and opposite forces.

What would be required is thorough analysis of his governing equations (beyond my capabilities) to see how they hold up to momentum conservation. Being non-linear, that will be difficult. The basic question is do all solutions to Cahill's differential equations preserve conservation of momentum. If they do, then it passes that test. If they don't, then it fails.

I think the hollowed-out sphere -- and what I mean is a constant density sphere with a relatively small hollow region in the center, because that will "amplify" things more than a thin sphere -- would offer the best way to test this in a lab. With Newton, we have a prediction of zero force, but with Cahill, we have non-zero. This "move/no-move" might be easier to test than small differences in force/acceleration. Or the non-zero g in the center might alter the period and motion of some mechanical oscillator in some detectable way.

Am I still intriqued by this, basically because his constant looks like the fine structure constant both with the borehole deviations and the globular clusters.

And then in that second paper I posted, he derived a flat rotation curve. What he did was solve a non-linear differential equation for his "velocity field" under some conditions I'm still not completely clear on, get 'g' and then calculate the circular orbital speed as a function of radius from that. Calling that orbital speed 'u' to avoid confusion with his "v field", that equation boils down to this:

u(r)^2 = A [1/r + B*(1/r)^c], where c is small, c << 1. c is actually alpha over 4 or 8 or something.

The second term then falls off very slowly with r, just a tad less than r^0, a constant.

-Richard

afterburner
2006-Jul-31, 03:56 PM
What about conservation of energy on small scales?

We say force carriers spontaneously appear and dissapear. This effect is commonly described by "borrowed" energy/time, which is later "paid back".

So if every atom in this Universe is doing that...then thats alot of "borrowed" energy. So why is it that I can't borrow energy from the Universe? *puts hand to ear* For the purpose of attracting and repelling things (besides my atoms)? I swear ill pay it back later :eh: ...

This is a quote from Wiki, I dont know how correct it is..."they [virtual particles] are unobservable by their very definition, and furthermore do not respect some of the most fundamental laws associated with physical particles."

So does the Universe have any laws at all or no?
Or are virtual particles simply a figment of our imagination, which we use (very successfully) to model forces and predict certain situations?

Ken G
2006-Jul-31, 05:01 PM
What he did was solve a non-linear differential equation for his "velocity field" under some conditions I'm still not completely clear on, get 'g' and then calculate the circular orbital speed as a function of radius from that. Calling that orbital speed 'u' to avoid confusion with his "v field", that equation boils down to this:

u(r)^2 = A [1/r + B*(1/r)^c], where c is small, c << 1.
That looks quite bogus to me. His ability to do math looked pretty good, but I don't believe you can get a result like this from his equations. You are not going to get alpha in the exponent by having it as a coefficient of the divergence of the velocity field. Let's look at his equation in spherical symmetry, so that his D matrix is just dv/dr times the identity matrix. Then we find his "dark matter" effective rho obeys
4 Pi G rho = (3 alpha/4) * (dv/dr)^2 .
That's a pretty normal kind of differential equation term, with a coefficient that is not going to show up in the exponent of r. Indeed, I find that all this term does is take the normal coefficient of the (dv/dr)^2 term from the div g term, and turn 1 into 1-3*alpha/4. That's not going to do anything significant at all, on any r scale, if alpha is small. His solution looks just plain wrong to me.

Ken G
2006-Jul-31, 05:21 PM
So does the Universe have any laws at all or no?
Or are virtual particles simply a figment of our imagination, which we use (very successfully) to model forces and predict certain situations?
It is arguable that this describes all particles, not just virtual ones, but observable properties are certainly on a different footing from purely theoretical ones. But your point is well taken, conservation of energy is not an absolute law of physics (what is? They are all idealized) as it is violated at both the shortest and longest time scales we encounter in the universe. But it works great at timescales of interest for human endeavors.

five_distinct
2006-Jul-31, 05:30 PM
If someone was "halfway out there", I am almost certain the 13.7 billion light year bubble would exist for the observers on that other planet (although I have absolutely no way of proving this). The main reason being that (this will sound strange in your head, but bear with me)....that.... our Universe is a blubble, and if we follow everything back, we get the singularity and the BB theory. HOWEVER, since we dont know what was before 10^-43 seconds AFTER the BB, we dont know what state the universe was in BEFORE the BB(if there is such a thing(for you loyal BB fans out there))...

IF! there was no singularity, but some "mass", for lack of a better word, and IF this mass was gigantic and "gigantically" dense....then any point on/in that "mass" (would be a singularity, and when "exploded") will be an effective bubble much like our own for observers that happened to be in the "middle" of it all (in fact, they might have also come up with a BB theory, since it is the ony thing they would know, much like us).

Therefor these observers would also be in a 13.7 billion light year bubble (assuming the expansion is uniform throughout the "larger" Universe)...like i said..sounds weird, but hope you got the idea (admittedly my no-way-of-being-proven idea, or mabe its already been though of and dismissed)

:think:
What do you think? Way out in the left field or perhaps? (this or that style):D

Hmm... I've seen someone (maybe on here) post a link to his site with a very similar idea to what you're talking about here (with pictures and everything!). I wish I could remember the name of his hypothesis.

Ken G
2006-Jul-31, 06:09 PM
Hang on, that entire idea is absolutely standard mainstream cosmological-principle Big Bang theory. Not left field, I'd say batter's box.

five_distinct
2006-Jul-31, 07:13 PM
Hang on, that entire idea is absolutely standard mainstream cosmological-principle Big Bang theory. Not left field, I'd say batter's box.

What, that our universe is a bubble within a larger entity? Are you sure?

Ken G
2006-Jul-31, 07:18 PM
As there was no implication that the "bubble within a larger entity" had any kind of physical boundary, but rather merged seemlessly with the rest of the unobservable universe, then yes, this idea is entirely consistent with the cosmological principle. Note that this is a pedagogical view only-- it is not observable, by definition.

five_distinct
2006-Jul-31, 07:21 PM
Fair enough. Maybe I missed something in that post, or am remembering something wrong then. This was earlier this year that I remember seeing the site.

Thomas(believer)
2006-Jul-31, 08:02 PM
Do I read well here, that the conservation law of energy doesn't hold globally? Where is the energy going to or comimg from then? For me this law always has been the most fundamental one. I'm not going to kiss this goodbye, because I'm sure something bad will happen to me.:evil:

Ken G
2006-Jul-31, 11:44 PM
Lol. What happens to the energy in the following situation: a stationary observer emits a photon, which travels to another observer that is moving away at high speed. If the first observer asks the second what he/she perceives the photon energy as being, based on past experience (without actually absorbing the new photon), he/she will report that it is redshifted. If the first observer is using the word of the second as his/her global coordinate system (which is just what we do in cosmology), then where did the redshifted energy go? Energy conservation is a local concept.

publius
2006-Aug-01, 12:05 AM
That looks quite bogus to me. His ability to do math looked pretty good, but I don't believe you can get a result like this from his equations. You are not going to get alpha in the exponent by having it as a coefficient of the divergence of the velocity field. Let's look at his equation in spherical symmetry, so that his D matrix is just dv/dr times the identity matrix. Then we find his "dark matter" effective rho obeys
4 Pi G rho = (3 alpha/4) * (dv/dr)^2 .
That's a pretty normal kind of differential equation term, with a coefficient that is not going to show up in the exponent of r. Indeed, I find that all this term does is take the normal coefficient of the (dv/dr)^2 term from the div g term, and turn 1 into 1-3*alpha/4. That's not going to do anything significant at all, on any r scale, if alpha is small. His solution looks just plain wrong to me.

Ken,

I'm going to go over this "flat orbital" solution a bit more. I'm having to dig out some of my old texts to refresh on the spherical forms of the vector operators (all the basic "del" operations). I can still halfway cut it in Cartesian from memory, but forget about non-Cartesians. :)

There are two papers to look at. The first is the original "borehole" paper, where he took the abrasive tone, and the second is the "flat orbital" paper.

Borehole: http://arxiv.org/PS_cache/physics/pdf/0512/0512109.pdf

flat orbital: http://www.mountainman.com.au/process_physics/HPS28.pdf

He's got a different spherically symmetric form for the "dark matter" density (the result of the self-interaction, or "gravity making gravity") than you have above.

The thing that makes this complicated is he is dealing with his v-field, which for normal Newtonian gravity is just the escape velocity as a function of r. So that adds a bunch of complexity. Rather than solve for 'g', you solve for the escape velocity, in vector form. In the full form with the self-interaction added, you have to do this, as he writes the self-interaction in terms of the v-field.

As an example, for normal Netwonian gravity, the escape velocity, which is v(r) is just 1/2v^2 = -Phi(r) = GM/r --> v(r) = Sqrt(2GM/r), or
call that v(r) = K*r^(-1/2).

You can verify that (v dot del) v = g(r) is just the expected -GM/r^2, using del = d/dr, so (v dot del) v is just v*dv/dr.

So using this form, @v/@t = 0, and (v dot del)v = v*dv/dr, we can write his modified div g equation as

div (v*dv/dr) + self_interaction(v) = -4piG*rho.

Now, in spherical coordinates, the div is going to be
1/r^2 * d/dr (r^2 * ) for spherical symmetry with no theta or phi dependence.....I think.

That is going to make a mess, and his flat curve solution is supposedly a solution to that mess.

I'll post back when I get through it. And I note we have this borehole business spread about several threads -- it might be a good idea to put it all in one thread, in ATM maybe. But we have no real "defender" of this, certainly not me, so it won't be exactly the same ATM format.

-Richard

publius
2006-Aug-01, 12:27 AM
Ken,

I think I see what's going on, and what Cahill is claiming. And I note it would've been nice if Cahill had actually simply said the following.

Reference the first "borehole" paper, where he derives some equations for g for a constant spherical mass distribution of some radius R:

g(r) = -G(1 + a/2)M/r^2; r > R

g(r) = something_different; r < R

These are just "first order" approximate solutions to the full
div(v*dv/dr) = mess, ignoring higher order terms, and only valid in the "near field". IOW, it's almost inverse square outside of the mass, close by. But at high r, the higher order terms modify it. I did not appreciate the above for r > R was only a "first order" solution.

And his flat curve solution is a full solution. The additional
K*(1/r)^(small_number) "kicks in" at high r. For a small mass, K will be small, and this will represent a small "residual" acceleration that doesn't fall off as expected once you get far from the source mass. And yep, I think I remember seeing something about Cahill and the Pioneer anomaly, too when I was searching about this.

For a very large mass distribution, say a galaxy, this K will be large enough to give appreciable and flat contributions at large r.

Once I get my spherical dels and 1/R^2 straight, I'll see what the full equation is, and if the flat-curve solution appears to be a correct solution.


-Richard

Ken G
2006-Aug-01, 12:37 AM
In spherical, del is 1/r^2 * d/dr (r^2...) .
And his argument is starting to get slippery if it is not the simple expression involving alpha he mentioned before....

publius
2006-Aug-01, 03:09 AM
Ken,

Okay. This is his self-interaction term, as he writes it in both papers:

a/8 *[ (trD)^2 - tr(D^2) ], where D is a tensor, defined by

D_ij = 1/2[@v_i/@x_j + @v_j/@x_i ], where I'm using the '@' sign (akwardly) for the partial derivative symbol, and the '_i' and _j' refer to subscripts.

For a spherically symmetric v, he writes this term in spherical coordinates as:

a/2 * [ v^2/2r^2 + v/r *dv/dr ]

That is his spherical self-interaction term, and he uses this in both papers. Now, divide that by 4pi*G, and that's his "dark matter" density to move to the right hand side.

But let's just leave it on the left. We've got
(v dot del)v = 1/2 del(v^2) = v*dv/dr (in the r direction), and that is just 'g', so his full sphericall symmetric div g equation is:

div( v*dv/dr) + a/2 * [ v^2/2r^2 + v/r *dv/dr ] = -4piG*(rho)

with rho being a function of r alone as we're spherically symmetric.

Now the divergence is 1/r^2 *d/dr [ r^2 * v* dv/dr ], which after a mess is:

v*(d^2v/dr^2) + 2/r *(v * dv/dr) + (dv/dr)^2. Now switching to ' symbols to mean d/dr, the whole mess with the a/2 terms is then:

vv" + (2+a/2)/r(vv') + (v')^2 +a/2*v^2/2r^2 = -4piG(rho).

That mess is the spherical, non-linear equation to solve for v(r), and this is what his "flat curve" equation is supposedly a solution to. Solutions to that are the full general solutions for any spherically symmetric mass distribution.

Smoke that mess over -- I could've screwed it up.

-RIchard

publius
2006-Aug-01, 03:48 AM
Continuing:

Now, he claims that a "two parameter exact" solution to the full non-linear differential equation is the following. And although he doesn't make it clear, this has to be for a point mass, ie rho(r) is a delta function that integrates to M.

v(r)^2 = K * [ 1/r + 1/R *(R/r)^(a/2) ]

Now, this 'R' is some parameter that apparently behaves as some "characteristic radius" of some sort. If 'a' were zero, this would be the familiar Newtonian potential with and additive constant. So 'K'
has to be Sqrt(2GM), (but "his" G, not the Newtonian G_n of G(1+ a/2) in this scheme of his).

g(r) = vv' is then:

g(r) = K^2/2 * [ 1/r^2 + a/2R * 1/r * (R/r)^(a/2) ]

For r << R, this looks inverse square. But as r gets large, the inverse linear term dominates, until the (R/r)^(a/2) term finally falls off at very large r. Other proposed modifications to gravity I've read about have it falling off as inverse linear on large scales, and this gives that behavior.

-Richard

Celestial Mechanic
2006-Aug-01, 04:35 AM
Just a bit of clarification about virtual particles.

Energy and momentum are conserved whenever a virtual particle is emitted or absorbed. The difference between a physical and a virtual particle is that for a particle of mass m a physical particle has:

E2 - p2c2 = m2c4.

A virtual particle does not satisfy this equation. The farther away from equality the quicker that it must be absorbed.

Hope this helps.

Ken G
2006-Aug-01, 05:32 AM
OK, thanks for that summary publius, your math looks right and so does his. I oversimplified it above, so it was my result that was "bogus". Nevertheless, I will argue that his theory still won't hold water, to mix the metaphor.

I think we can look at what Cahill is doing in the following way. Einstein's approach to gravity can be considered in the Newtonian limit to essentially be an effect on the force-free coordinate system that free-falling objects will follow. That coordinate system can be thought of as flowing with a v that obeys the acceleration of gravity. So we can think of each coordinate intersection as starting at infinity with some finite speed (to be determined), and then falling inward to the point in question where you have some particle you want to follow. Whatever is the particle speed then determines what speed you had to give that coordinate point at infinity to have the right speed when it reaches the particle in question. From this point on, the particle will simply follow that coordinate intersection. So the idea of a "v flow" is so far just the Einstein idea of the equivalence principle, applied to Newtonian gravity (and we'll never make it general relativistic here, we don't need that bit for galactic rotation curves.)

The innovation is then to tack on an additional behavior to that v field that is not Newtonian. But there's a pretty significant flaw in his approach. In Newtonian, if you have no mass sources anywhere, then a perfectly allowable solution is that the free-fall coordinates just "coast" at constant speed. This is what you want-- inertial motion with v'=v"=0. But note that in Cahill's formalism, this is no longer an allowable solution. Even in the absence of any mass sources, his v field behaves strangely, approximately like the solution to 2v'/r +a/4 * v/r^2 = 0 , to borrow your expression. That is a power law v(r) ~ r^(-a/8) with a very low power, so is close to coasting but not quite. So even in the absence of all real mass, you still have a 'dark matter' effect that causes flows of test particles to decelerate. I realize that the v field is not explained as a test particle flow, it is a make believe construct, but my point is that the test particles do follow v because they accelerate in the same way v does, so all you have to do is choose a v law that coincides with the particles at any time and it will always follow them.

So what is my point-- it is that Cahill includes dark matter as a fundamental aspect of gravity, even in the absence of any true gravity sources. It's not just an augmentation or self-interaction of gravity that starts from true sources, it's there even if true sources are completely absent. So it has much more the flavor of the cosmological constant than it does to dark matter-- it is gravity that exists even in a complete vacuum, but is not antigravity and it causes power-law deceleration rather than exponential increase. This also means that the K in the above formula for g(r) should not multiply both terms-- it is subsumed into R in a way that means it's not really present at all in the second term and can be zero.

So could it be right? I really doubt it. To explain galactic rotation curves, it means this effective gravity from space itself has to make its presence known on kiloparsec scales for flow speeds like a hundred km/s. Since the effect is entirely independent of any actual mass distributions, it is very unclear how it 'knows' where the center of the galaxy even is. Indeed, it predicts galactic rotation curves should apply to any test particle flow anywhere, even out in the middle of nowhere with no galaxy there. It seems to emerge immediately from the symmetry of the chosen coordinates. Since it is not anchored to anything physical at all, it is at best an incomplete theory, and at worst, it's just wrong.

I think what is happening here is we have a theory that claims to work in two cases, but in fact it is two separate theories that work in two separate situations. Each theory then only works in the situation it was jury-rigged to work in, and the use of alpha is just a coincidence. The ramifications of the theory do not appear to generalize well to any other situation, most notably the "middle of nowhere" problem.

RussT
2006-Aug-01, 06:37 AM
KEN G, a very, very, exacting analysis, which clarified the key points extremely well:clap:

How all this relates to the a=1/137 should prove to be very intersting as well.

publius
2006-Aug-02, 04:34 AM
Ken,

Let me second Russ, that's pretty darn good.

You noticed what we can call the "spontaneous gravity" aspect of this, which is one of the big "eyebrow raisers" with me. But Cahill delves into it and seems quite hunky-dory with it, touting it almost. I think his idea is this all has some "quantum roots" in some underlying idea of a quantum vacuum making 3-space, and so the spontaneous gravity would be some sort of quantum "hiccup", that produces an initial central field that "seeded" clumping of matter.

As you noticed, v = constant is not a solution to the v equation with zero source term. But v = 0, the trivial solution, does remain. And so I would argue that while the v equation admits the "spontaneous gravity" solution, it does not *require it*.

I see the v-field as something like a potential. g comes from derivatives of it, and it plays that role, but as non-linear thing, it just doesn't admit adding arbitrary constants.

In the Newtonian case of a = 0, that v is just the escape velocity, (but directed as the velocity of something freefalling from infinity with an initial velocity of 0), and depends on the difference in potential -- the arbitrary constant just cancels. And that is independent of any initial velocity -- you just add that in to the freefalling solution.

So that's how I view the v-field. v = 0 is the g = 0 case.

-Richard

Ken G
2006-Aug-02, 04:52 AM
As you noticed, v = constant is not a solution to the v equation with zero source term. But v = 0, the trivial solution, does remain. And so I would argue that while the v equation admits the "spontaneous gravity" solution, it does not *require it*.

Yes, I figured you would see that, I have been thinking about that too. But what bothers me is that the way to tell which v you want comes from a boundary condition that you are free to set. So what physics is used to set that boundary condition? I fear that this may just be an example of the idea that if you have one free parameter, you can fit one situation, pretty much no matter what your theory is. So the situation he chooses is to fit the effects of dark matter on spirals. Then he has another theory to fit the borehole data. Neither of those are real physics until you can show that you have a general prescription for choosing the boundary condition. Then the physics is in that prescription, not so much in the velocity law itself.

In the Newtonian case of a = 0, that v is just the escape velocity, (but directed as the velocity of something freefalling from infinity with an initial velocity of 0), and depends on the difference in potential -- the arbitrary constant just cancels.
Actually, I think that the v is not necessarily 0 at infinity, I think you can use any v at infinity that you want and it will still generate the same g field so long as you just let the coordinates "fall" from infinity with whatever speed they started with. So that's an ignorable difference, and you may as well choose v=0 at infinity, or you might want to choose v at infinity to match whatever test particle you are interested in once the coordinate 'gets to' where the test particle is (then the test particle moves according to the v field, so the v field is more than just a way to get g, it's the motion itself). But the added term breaks the ignorability of v at infinity, so you have to specify it yourself. How do you do that in general? You can make it conform to the symmetry of the actual mass sources, but note that in the limit as those sources are gradually removed, there is no change in the new gravity terms. That can't be right, it's the "middle of nowhere" problem-- in the limit that you gradually remove the mass, you end up with a solution that doesn't change even as the last of the mass is being removed, so it is rooted in nothing. If he is going to maintain that the dark matter part is just a kind of echo of the true mass distribution, he has to be able to show that his correction goes to zero as the mass goes to zero, and I don't think it does, though I haven't looked at a formal limiting procedure.

publius
2006-Aug-02, 06:55 AM
Ken,

This velocity question made me think of something bigger, and it may be the same thing in a more important form: Is this thing even Galilean invariant?

To look at that, we'd have to go the full time dependent form, and imagine our source mass moving at some constant velocity. Does the v-field solution remain the same, just in simple x' = x - vt coordinates?

I have a feeling it won't, since the thing is so non-linear, and this is the same thing as adding a velocity at infinity. This would mean the velocity field has to have a preferred frame moving with the source. But suppose our source has bits of mass in all sorts of relative motion. What frame do we choose?

Now, if the g field remains Galilean invariant, even though v doesn't, that might save it, if we continue to hold that is g, not v that really matters.

-Richard

Ken G
2006-Aug-02, 03:52 PM
You're right, this is yet another problem with his symmetry requirements. As long as there is any mass at all at some point, then you get his rotation curve around it, no matter how small that mass actually is. If dark matter is some kind of echo of the actual mass, its effects must scale with the amount of mass, and must move with that mass, or it's just not physical. You have unearthed this latter problem, that if the mass is moving, how does he tell his symmetry center to move with the mass? You'd think that you would want to just add the velocity directly to the v field, but his dark matter is invariant to any constant addition to the v field so there's no way to tell it to "move". Good point. But I suspect the Galilean invariance would be easy to fix, working from the analogy of a potential. If we write that the acceleration of gravity is the gradient of a potential, then this is also not Galilean invariant, but we know how to deal with that we just transform to the frame where the mass is stationary and then transform back, and there's a way to put that into the equations. He'd probably need to do the same, it would be OK I would imagine.

Ken G
2006-Aug-02, 10:30 PM
Come to think of it, the easy way to fix potentials is to use as your coordinate the vector displacement from the center of the potential, instead of some rigid coordinate system. I imagine Cahill has the same idea in mind for his "r" coordinate, and I think that would make it Galilean invariant if you also add the reference velocity directly to the v field. That part is not really a flaw, if I'm right.

publius
2006-Aug-03, 05:15 AM
Ken,

Doing some Googling, I ran across a response to one of Cahill's papers about this very point. The author put it in fluid flow terms, which how Cahill describes it. The analogy was to imagine a big tank of fluid with a drain hole. The acceleration of the flow is invariant to any Galiean observer moving with respect to the tank (and you just add the velocity of the tank to the velocity field). We'll forget about any relatavistic Lorentz observers who will see length contractions and time dilations and might measure a different acceleration! :) However, things are different if the drain hole itself is moving with respect to the tank. So an observer moving with the moving drain hole, and seeing no relative velocity of the source would see a different acceleration than observer stationary with repsect to a stationary drain hole. The solution depends on more than just the relative velocity of source and observer.

This is sort of a "medium of propagation" type thing, just like sound, where things depend on both the velocity of the observer and the source with respect to the medium. You hear a different pitch moving with a source moving with respect to the air than you do when you're both stationary in the air. And that "medium" here is defined by our velocity at infinity, which sort of defines the "sides of the tank".

-Richard

Ken G
2006-Aug-03, 05:32 AM
But doesn't this assume that one is using a standard convention, like v=0, at infinity? I shouldn't think that would be necessary, I would just get the v field in the frame of the mass source, and then transform it to other frames in the needed way, analogous to a moving tank rather than a moving drain. In other words, I would not use the fluid dynamical picture of a stationary tank to constrain my v field (thus avoiding your apt "medium" analogy). I'm happy if his approach can get a physically meaningful result in the frame of the source, and then just do the obvious transformations into other frames-- there's no need for the "moving drain" problem. Or another way to put this, since he cooked up the v field, he is free to give it the transformation properties of a moving tank rather than those of a moving drain. I still see the main problem as being that the drain can be vanishingly small and still produce the same flow.

publius
2006-Aug-03, 06:38 AM
Ken,

Just for fun, and to refresh my memory, let's look at the Galilean picture of a classic Newtonian gravity field. Back when "I used to be good", I loved to write partial derivative symbols, and had a ball once with an assigment to transform the Laplacian from cartesians to spherical. Not only did you have the coordinates, but the variation of the spherical unit vectors with phi and theta as well. That was a mess of partial derivatives. Yep, I used to be good. Not any more. :( :boohoo:

Rather than fooling with primed vs unprimed coordinates, let's just use 'r' and 'x', and we'll r and x to be in the same direction, with our relative velocity aligned with them as well. y and z will be the same for both, and we'll only worry about the x and r coordinates, and only worry about the potential and gravity on those axes.

Suppose a point mass is moving along the x axis at velocity v. r will the distance in the moving frame, so x = r + vt. In the rest frame, the potential V is just 1/r (say we're in units where GM = 1, and we'll forget about sign, as really r = sqrt(r^2)

g(r) is just dV/dr (where is 'd' is partial) = 1/r^2.

Now, in the stationary frame the potential is V = 1/(x - vt).

g(x) = dV/dx = 1/(x - vt)^2.

Same thing. Now if we play with the partials, it would go like this for a transform from the moving frame to the stationary frame:

dV/dx = dV/dr*dr/dx + dV/dt *dt/dx. dV/dt = 0 and dt/dx = 0 [in the moving frame, and this would actually be t' -- that's important to keep straight so you won't get mixed up -- t = t', but you've got to keep track of explicit dependence, and I started out not wanting to mess with primes] And dr/dx = 1, so that's just dV/dr = 1/r^2 = 1/(x - vt)^2.

Now going the other way, from the stationary frame to the moving frame:

dV/dr = dV/dx*dx/dr + dV/dt*dt/dr

Here, dV/dt is not zero, but but dt/dr is, so that's just dV/dx.

Gets a bit tedious, don't it?! :lol: Anyway, from this it looks like g = del V is indeed Galilean invariant. You can even make Maxwell Galilean invariant (but it won't agree with reality, of course). You do that by replacing the partial time derivative with total (directional) time derivativis, ie curl E = -dB/dt = -@B/@t - (v dot del) B

Where v is a velocity vector of a moving frame. In the potentials, the expression for E gets replace with the total derivative of A as well. For v << c, this works fairly well.

Only problem is the wave equation gets v in it, and just says the moving observer will see his velocity added to the wave speed. Which ain't what happens, of course. That is indeed what happens with sound, and the sound wave equation so modified would describe exactly what happens (save for relativistic velocities where length contraction and all that kicks in, but even there, different observers see different wave speeds. But you can show that as rest frame wave speed goes to c, all observers measure it as c, and that means that wave is EM like)

You can still do the total derivative trick, but using a Lorentz transform rather than a Galilean (ie something different from @/@t + v dot del), but you find that is superfluous for the wave equation and even the whole Maxwell set (when you transform the sources as well, I think)

If you do the Lorentz transform form of the total derivative with the sound wave equation, you'll see the additional terms depend on the wave speed vs c. In Maxwell, those are equal and the extra terms cancel. But when wave speed is less than c, you get something different, that gives what the moving observer would see with the time dilation and length contraction.

Sorry about this tangent..................

-Richard

publius
2006-Aug-03, 06:58 AM
But doesn't this assume that one is using a standard convention, like v=0, at infinity? I shouldn't think that would be necessary, I would just get the v field in the frame of the mass source, and then transform it to other frames in the needed way, analogous to a moving tank rather than a moving drain. In other words, I would not use the fluid dynamical picture of a stationary tank to constrain my v field (thus avoiding your apt "medium" analogy). I'm happy if his approach can get a physically meaningful result in the frame of the source, and then just do the obvious transformations into other frames-- there's no need for the "moving drain" problem. Or another way to put this, since he cooked up the v field, he is free to give it the transformation properties of a moving tank rather than those of a moving drain. I still see the main problem as being that the drain can be vanishingly small and still produce the same flow.
Ken,

I may not have this straight in my head, and worse, it's sort of hard for me to put in words what I sort of vaguely intuit without trying to do it the hard way with the math:

Being non-linear, this thing does not allow superposition. That is, the field due to two mass sources is not the sum of the fields of the sources alone. When those sources get to moving relative to each other, that causes the problem.

Using the drain and tank analogy. If we have two drains in relative motion, one of them has to be moving with respect to the tank. Because of the non-superposition, we can't just say okay we'll just have a separate tank for each drain, then add the field of each together. We have to have one tank for both sources. Now, how do we choose which one is the one "really moving" with respect to the tank, or our velocity at infinity.

With regular, nice and linear Newton, as you can sort of intuit without working out the transform messes, it doesn't matter which frame you choose for two sources in relative motion. The potential and the field and is the same (although it may be "moving" or time varying depending on your frame).

Now, could that be fixed. It is beyond my abilities, but I wonder if this could be fixed somehow, and still get this "alpha behavior" with the small inverse linear behavior.

I'd love to "get a fit" for Cahill's "two parameter" solution to the sun, and see what that says for solar system orbits. Does it predict the Pioneer anomaly? Does Jupiter (or Mercury) look pretty much the same (or close to what Newton predicts -- ignoring GR effects) so there is no obvious difference close in? Assuming it passes this basic test of not saying Jupiter's orbit would blow up or be ridiculously different, then we might worry about trying to fix this "medium" problem, and see how far out some orbit would have to be for the inverse linear term to dominate, and see if anything is out there that could test it.

-Richard

RussT
2006-Aug-03, 08:39 AM
You noticed what we can call the "spontaneous gravity" aspect of this, which is one of the big "eyebrow raisers" with me. But Cahill delves into it and seems quite hunky-dory with it, touting it almost. I think his idea is this all has some "quantum roots" in some underlying idea of a quantum vacuum making 3-space, and so the spontaneous gravity would be some sort of quantum "hiccup", that produces an initial central field that "seeded" clumping of matter.

Publius, you were way closer to what is really going on back here.

In re-reading both papers, Cahill comes right out and says that the Fine Structure Constant "IS" the whole 'background' gravitational field, all of the darkness of space, and that the Fine Structure Constant can also be seen in the Borehole evidence and is 'extra DM' in the earth.

So, in your paragraph I quoted above, don't think quantum for the 'hickup', think MASSIVE GR (this also solves the 'middle of nowwhere' problem). This is Q&A so I can't go into more detail here!

Where Cahill is not really understanding exactly what he is dealing with is in his second paper you linked, when he defines which type of black hole systems are influencing the 'background' field!

Ken G
2006-Aug-03, 04:38 PM
Being non-linear, this thing does not allow superposition. That is, the field due to two mass sources is not the sum of the fields of the sources alone. When those sources get to moving relative to each other, that causes the problem.

Even before there is movement, this causes the problem. For if you have superposition, then you can reduce the mass by a factor of a million, and get a million less effect. But here you can reduce the true mass source by a factor of a million and nothing happens in the "far field" where the galactic rotation curve is set. But you are also right that one way to get a movement from point A to B is to cancel the mass at A and add a mass at B, and his equations won't allow that either. So it's basically the same objection we are both having.

Using the drain and tank analogy. If we have two drains in relative motion, one of them has to be moving with respect to the tank. Because of the non-superposition, we can't just say okay we'll just have a separate tank for each drain, then add the field of each together. We have to have one tank for both sources. Now, how do we choose which one is the one "really moving" with respect to the tank, or our velocity at infinity.
Yeah, that's tricky too, it seems that a lack of superposition is the big killer.

Ken G
2006-Aug-03, 04:50 PM
In re-reading both papers, Cahill comes right out and says that the Fine Structure Constant "IS" the whole 'background' gravitational field, all of the darkness of space, and that the Fine Structure Constant can also be seen in the Borehole evidence and is 'extra DM' in the earth.

But Russ, for the reasons above, it is now clear that Cahill's "theory" is baloney.

RussT
2006-Aug-03, 11:50 PM
Ken G;

To keep this simple and because we are in Q&A, please re-read Cahill's second paper that deals with the relation of the black holes and their association with both their accompaning baryonic matter, and then that association to the 'Background' field'.

publius
2006-Aug-04, 01:26 AM
Ken,

From a link in the MOND thread, I found this paper:

http://arxiv.org/PS_cache/astro-ph/pdf/0606/0606197.pdf

It puts constraints on deviations from inverse-square bases on solar system orbits, which was something I was interested in. The maximum radial deviation from inverse square would have to be 4 orders of magnitude less than the Pioneer anomaly for Mars, for example.

They mention another modification to gravity which gives a Yukawa like term to the potential:

V(r) = -GM/r(1 + a*exp(-r/R) ), where a is a strength and R is some cut-off radius.

Again, here one must choose the G to be G_n/(1+a). And this gets to looking inverse linear at long distances to. This might be tweaked to fit the Borehole data.

Anyway, they give a plot putting constraints on a and R above, with the solar system excluding certain regions is a-R space.


-Richard

Ken G
2006-Aug-04, 01:47 AM
That's a pretty standard MOND potential, and those have been pretty much beaten to death I would say. It's the first place they looked, it didn't work, time to move on. I can't explain the borehole data, but I'm pretty confident there is no MOND theory that works from the boreholes, to the solar system, to the galaxy rotation curves, to the galactic cluster data. Dark matter has the advantage of being able to respond differently on all these scales, whereas MOND theories must be universal, so it's a pretty big advantage for dark matter. An unfair one, perhaps, but a MOND theory with a free parameter for every situation is not much of a theory, more like a Frankenstein monster of cobbled-together bits.

Ken G
2006-Aug-04, 01:50 AM
To keep this simple and because we are in Q&A, please re-read Cahill's second paper that deals with the relation of the black holes and their association with both their accompaning baryonic matter, and then that association to the 'Background' field'.
I'll read it on two conditions, (1) that you first give me some reason to suspect that Cahill's theory can be made to explain why the "background" effects share the symmetry of the true mass distributions while not vanishing continuously as the true mass distribution is made to gradually vanish, and (2) that you remind me of that link.

RussT
2006-Aug-04, 02:11 AM
flat orbital: http://www.mountainman.com.au/process_physics/HPS28.pdf

This is Q&A, so I can't go into to much detail. In addition I am not ready to fully divulge what all of this really means to cosmology as a whole.

If you pay particular attention and think through what he is saying about DM and the 'stars and planets', that will partially answer your question.

By the way, I need to slightly amend part of the post above where I said...Don't think of it as a quantum 'hickup', but as Massive GR...and then in parenthesis said...and that solves the 'middle of nowhere' problem.

The amended version is...we can see and know what happens in the 'middle of nowhere', but can not yet be able to know 'how it manefests there'.