Norman Albers Posted April 25, 2007 Posted April 25, 2007 In the paper available in my cache (URL below) on gravitation I offered my take on the Polarized Vacuum (PV) theoretics started (perhaps) by R.Dicke in the 1950's. All the studies I have read to date posit a scalar function of permittivity of the vacuum, which is then expressed in the Scwarzschild metric to yield the distinct radial and transverse responses. I have enjoyed a lengthy correspondence with H.Puthoff who offers the isotropic solution of the gravitational singularity. This is simply the other reasonable possible assumption on physics going into the construction. We either let the same coefficient multiply differential changes in all three spatial dimensions, or we allow radial measure to change differently from transverse. In my study I directly substitute presumed changes of the vacuum polarizability as per dielectric hole theory, into the two distinct Schwarzchild results to be seen when we express light-speed. I think this is a good approach because it echoes what can be seen in the small in my electron study, namely the "thickening" of the polarizability field, but moreso, one in which we see a skewing of a randomly offered dipole population which reduces the nearfield radial population. I told Puthoff that one or the other of our two respresentations is more useful, and that whoever can tie this gravitation persepective in the large, to the vacuum field expectations in the small, wins. The isotropic approach yields solution with no event horizon per se, so Puthoff calls these "dark gray holes" where permittivity blows up as you apporach the center singularity rather than a spheric event horizon characterized by the same asymptotic behavior in [math]\epsilon_o[/math]. Both the PV gravitational singularity and my electron singularity may be described as degenerate event horizons. The dielectric respresentation offers the understanding of this "infinite slowing of light" to be caused by an increase of the local polarizability equalling 3, a quite finite state of affairs on the face of it.
Norman Albers Posted July 26, 2007 Author Posted July 26, 2007 Seeking to represent circular currents as per a nearfield electron, I am working with the Kerr GR solution for the exterior of a rotating axially symmetric mass, for starters. Dealing with radii larger than Planck length I get a form depending upon geometric angular momentum, expressed as the product: [math] ma=-GJ/c^3[/math]. My goal is to unite this with a solution inside the circulating radiation angular momentum pictured as source. Lense-Thirring accomplished this sort of solution in a "low mass density" where they expressed interior and exterior solutions. So also I hope to accomplish, in the realm appropriate to what I am theorizing. In electrons the Schwarzschild radius, or geometric mass m, is very small compared with the angular momentum manifested. Certainly if the geometric angular momentum, is described by the product ma, above, this implies roughly the Compton radius of the electron for the value of a. The Lense-Thirring model is used extensively in analyzing rotating stars shy of black hole intensity (low-field limit). There mass is much larger than its angular momentum. I need the original Lense-Thirring paper if anyone can help me. The other avenue of approach is the Reissner-Nordstrom solution for an electric field density. Magnetic terms such as I have change the relatively simple solution for the radial fields, and the tensor algebra for the off-diagonal terms are complicated, but I can say that a reasonable electric field solution exists in this context, consonant with what I said in the R-N thread elsewhere. Figuring the magnetics inside and their contribution to the metric is the final challenge in this problem.
Norman Albers Posted October 28, 2007 Author Posted October 28, 2007 There are two parameters having dimension of length which appear in the Kerr exterior solution, namely the "geometric mass" m=GM/c^2, which is half the Schwarzschild radius; and then too the "geometric angular momentum" which is the product term: ma=-GJ/c^3 which involves both total "m" amd also angular momentum "J". Both m and a have dimensions of mength. Usually angular momentum is "less than" massive effects; as we approach relativistic large masses, the terms become comparable; in particles, Schwarschild radii are more than twenty magnitudes smaller than angular momentum terms...
Norman Albers Posted January 3, 2008 Author Posted January 3, 2008 Readers are referred to my thread "Reissner-Nordstrom-? metric": http://www.scienceforums.net/forum/showthread.php?p=381686#post381686
Norman Albers Posted January 13, 2008 Author Posted January 13, 2008 I have learned much of the Kerr metric solution, described elsewhere. Right now I am working through the Lense-Thirring model of interior and exterior field. It is a linearized approach where the perturbation is assumed "relativistically small". It will share characteristics with an electron field whose mass terms are small compared with angular momentum expressions.
Norman Albers Posted February 17, 2008 Author Posted February 17, 2008 One might well ask, why be concerned about gravitational effects which are very small compared to electromagnetic energies in these small-scale considerations. I seek to find the common expressions of gravitation and E&M. In large ensembles of matter net charge is often neutral, and is usually considered so in discussions of rotating mass fields. In the small, polar energy can circulate and source the polar response of the vacuum, exquisitely expressed in the Ahoronov-Bohm electron phase-shift. The assumptions made by the Lense-Thirring analysis are of a non-relativistic rotating mass, but importantly this problem introduces a corner diagonal term [math]g_{03}[/math] in the metric. This is the first step in representing angular momentum. As I hope to complete, the electron nearfield will have that term and also a [math]g_{12}[/math] term from a magnetic field cross-term developed when you figure the RHS stress-energy tensor from my assumed circular current field. One has to decide what level of complexity in the metric tensor to solve for. The Lense-Thirring is a first step of learning. The form of [math]g_{ab}[/math] and its derivatives is what builds up the LHS of the Einstein field equations, and to attempt an interior solution the metric tensor is used to raise indices on the Minkowski tensor, and further in reducing the stress-energy tensor to the Laue scalar, [math] {T^a}_a[/math].
Norman Albers Posted February 23, 2008 Author Posted February 23, 2008 If you are expressing an electromagnetic field, even an inhomogeneous one, then the Laue scalar is zero. This is the basic difference in accounting rules, either as identifiable mass, or as energy densities. Relativity allows for our confusion, if we keep to the rules: [math] R_{ab}=C(T_{ab}-\tfrac1 2 g_{ab}T)[/math]
Norman Albers Posted March 9, 2008 Author Posted March 9, 2008 I have finally reproduced the tensor machinery of the Lense-Thirring solution of a nonrelativistic mass (like a large star) spinning at significant but nonrelativistic angular momentum. This is more complicated than I knew and I don't recommend this to the hobbyist! Like a Kentucky man said about tuning pianos, you don't want to feel like you are chasing snakes. As soon as you introduce any new metric term, in this case the [math]g_{03}[/math], there are repercussions throughout the Einstein differential system of equations which must be satisfied. One has new Chritoffel symbols of the first kind, and I did not at first realize that to construct those of the second kind, one must sum over four possibilities. Thus some of the familiar terms from the diagonal metric (Schwarzschild) solution now have additions. After much stirring of the pot I finally came out with an inverse radial dependence for the [math]d\phi cdt[/math] term. I suspect I would have come sooner to the answer had I solved it as per the linearized field equations presented two chapters earlier. Whatever, my goal is to understand the process. 1
Norman Albers Posted March 12, 2008 Author Posted March 12, 2008 The analysis I have of a linearized approach to metric solutions is couched in the language of Cartesian coordinates and I am not yet clear on how to use it here. This will be interesting to learn, but my goal involves using the full solution because I wish to investigate strong angular momentum fields of low mass. Just doing this Lense-Thirring form where we put in a [math]g_{03}[/math] term generates six new Christoffel symbols, where there were (only) nine needed in the Schwarzschild solution. Also one of the Christoffels already present has an added term. So after chasing many snakes, I am left with not even a clear Laplacian expression but I don't care! Knowing that I need to come out with inverse radius for the r-dependence, what I get is: [math] g'' - \frac 2 {r^2} g = 0[/math] and I am not going to complain.
Norman Albers Posted April 20, 2008 Author Posted April 20, 2008 As I build my understanding of the mathematics here, I am impressed by the importance of the many coordinate transforms used at the drop of a simplifying hat. One must keep track of these or be lost! My work today indicates this is true of the degenerate metric form which is quite useful in getting the Kerr rotational solution. Having this form depends upon the coordinate choice, and so one must use caution in finding physical theoretics. To get the axially symmetric Kerr form, one abandons this simplicity.
pioneer Posted April 21, 2008 Posted April 21, 2008 I think we are getting too fancy because we have lost sight of something simple. A simple observation is positive charge has its most stable association with the larger stable mass unit, i.e., proton. This connects positive charge to gravity and GR. Negative charge is sort of once removed and gets involved through its association with positive. Although negative protons do occur, these are not as stable. One may argue that this is because the current universe is anti to this arrangement causing the instability. If we assume this is true, then all cosmology models that assume continuous creation should be forming both with equal probability. We should be able to find anti-galaxies in their own isolated place in space. All it should take is a tiny starting seed of equally likely and equally stable anti-matter to assure this slants the table in its favor. We don't see it. The equally likely assumption is contradictory to continuous creation. The assumption of equally likely and stable narrows cosmology to a one shot deal like BB, to assure the anti can never remain as stable even if it is equally stable on its own. If we want to keep all these other cosmology theories, then the assumption of protons being more stable than anti-protons is more consistent across the board. This results in positive charge assuming a connection to high mass, gravity, and GR. Negative charge is once removed getting involved through its association with positive. Charge is not equal and opposite in all ways, there is at least one distinct difference via its mass association.
Norman Albers Posted April 21, 2008 Author Posted April 21, 2008 Pioneer, if there were particle/antiparticle pairs being produced in an antisymmetric, matter environment, wouldn't they select the matter particle through attrition as the antiparticle recombines with some other particle? There remains no net count. What makes you say antiprotons are not as stable?
pioneer Posted April 21, 2008 Posted April 21, 2008 Say we start with an empty universe and energy-matter-antimatter appears, but in several places at the same time. If there is enough distance to where the anti-symmetric signal will take time to reach the other spot, the local symmetry can favor both equally. Once either takes root, this will set the agenda for that area. The faint distant anti-symmetrical signal becomes to small to turn the tide. The matter or anti-matter area would be neutral in charge and based on gravity so it should be everywhere based on some cosmology scenarios. The only type of cosmology scenarios that can maintain anti-symmetry will require everything forming close enough in the same area to assure the first born remains dominant. This sort of narrows down to BB and BB look-a-likes. This creates a conceptual dilemma, where we need to make a choice as to which range of theories we wish to sever, to be logically consistent across the board. Either way you step on someone's toes. I chose to make matter the most stable state, always preferred, with positive charge preferring the heavier mass due to some symbiotic connection. This preserves all cosmology as is. We can go the other way and preserve the equally stable theory but we then need to eliminate some cosmology. Just there has been a problem getting gravity to hook up in simple way.
Norman Albers Posted April 21, 2008 Author Posted April 21, 2008 I will be thinking about what you say. I say gravitation is described well, to a point, by General Relativity. It is built on the flat-space Lorentz transform. Both light and gravity are possibilities of the vacuum, and we will have unification when our theoretics encompass this from the bottom up.
pioneer Posted April 21, 2008 Posted April 21, 2008 This just popped into my head. If we look at gravity induced fusion, or gravity induced ionization, due to gravitational work, the positive charge is always moving in the direction of gravity contraction or compaction. The electrons are going opposite. In other words, if compressing an atom with higher gravity caused the entire atom to get smaller and smaller, until collapse, then one would stay the charge-gravity is uniform. What happens is positive charge gets more compact while electrons are ionized increasing their resistance to gravity. The electrons are going in the direction of SR instead of GR. It is their connection to positive charge that keeps them from leaving. Positive charge is the middleman between gravity and negative charge. One may even speculate the affinity of positive charge for high mass allows positive charge sharing within the nucleus mass to create what we call the nuclear forces. In other words, the stability of the proton is due to a prime directive of positive charge for high mass. This directive is stronger than the charge repulsion it may create. If look at the hydrogen proton, deuterium and tritium, the last two are easier for fusion. This implies they are higher up the activation energy hill than the proton, requiring less energy to push them up and over. What this suggest is the one mass one positive charge is the most stable, since it starts in the deepest well. The fact that fusion is exothermic even though it leads to about a 2 to 1 mass to positive charge ratio, is that high mass also has a thing for positive charge. In the nucleus although positive may prefer one proton to maximize stability, the neutral mass also prefers positive charge for stability, with the compromise, mutual prime directive. The higher mass association of positive charge will lower the magnetic to charge ratio of the proton relative to the electron. The heavier mass makes it slower at any give temperature. We can still cause the ratio to increase but it requires energy. This suggest this ratio is the native state of positive charge. In atoms, this native state of positive charge appears to rub off on the electrons, requiring they decrease their magnetic to charge ratio. This is done with orbitals which cancel some of the magnetic. The Bohr atom is not the preferred state, since the magnetic ratio is too high. The odd shapes of orbitals is what it takes. Let me explain this last part. The P-orbitals are the most stable arrangement due to 3-D symmetry. This is also reflected in electronegativity numbers with atoms finishing in P-orbitals providing the most stable electron wave addition. That is why we get O-2, or Cl-1, etc.. The S, D, F orbitals are not as perfect. If it was only the electrons, they could do better by recycling the P-orbital schema. But the demands of the lowering magnetic to charge ratio, does not push them them far, allowing some extra mag more in line with their own native state. At the same time, the higher magnetic to charge nature of negative charge rubs off on the positive charge to increase their magnetic to charge ratio making it easier to share. The positive charge gets in motion allowing the slightly higher magnetic to charge to help the mass prime directive.
Norman Albers Posted April 21, 2008 Author Posted April 21, 2008 You would enjoy reading the work of Hans Alfven. He modelled a universe evolving with alternating regions of particle/anti... with annihilating boundaries. For some reason this was seen not adequate. He did a beautiful study of two giant current streams in space, opposite but neighboring plasmas. They become a spiral galaxy...
pioneer Posted April 22, 2008 Posted April 22, 2008 Thanks for the suggestion. First of all, the terms matter and anti-matter are sort of misleading. The mass aspects are essentially the same. The only difference is within which mass gets which charge. If we assume charges are equal but opposite, and with mass essentially the same, the interaction is based on charge. An observation that suggests matter is the stable mass states for charge is comparing the interaction of a proton and electron to a positron-electron pair. If we only look in terms of charge, the hydrogen atom should be an easier way to cancel the charge. With the positron-electron pair, their acceleration toward each other is increasing the magnetic repulsion, using two entities subject to increasing uncertainty relative to position or momentum. With the electron-proton, with the proton heavier, such that the magnetic repulsion is always less, there is also a higher level of certainty for proton position or momentum. Another observation, is although protons within nuclei are matter and the positron is anti-matter, this interaction is not destructive, with the positron able to get involved in the nuclei and act just like it is matter. Simple mass preference can explain all at the same time. If positive charge prefers higher mass the positron finds utopia in the nucleus. It is able to get rid of that unstable mass arrangement without doing damage. Relative to the electron and proton, they can't cancel in spite of the charge because that would violate the prime mass directives of both of them. In terms of the positron and electron pair, in spite of magnetic problems and uncertainty, the positive charge is more stable with higher mass. It tries to do this piece meal with charge attraction sort of confusing the issue.
Norman Albers Posted April 22, 2008 Author Posted April 22, 2008 How do you define 'plus' and 'minus'? It seems at first reading that you are 'justifying the status quo', which I think is what NATURE does.
Norman Albers Posted May 3, 2008 Author Posted May 3, 2008 Doug Sweetser is a bright fellow who has a development of GEM theory, and if I recall correctly he has an assymmetry of charge and gravitation. . . . . . . I am working hard to combine a Kerr exterior solution of the electron as source of angular momentum and charge, yet 'negligible' mass...
scalbers Posted May 10, 2008 Posted May 10, 2008 Interesting that there is searching going on for antiproton decay: http://arxiv.org/abs/hep-ex/9908036
Norman Albers Posted May 11, 2008 Author Posted May 11, 2008 scalbers, can you help me understand the parameters in the abstract? Are they assuming symmetry with the proton?
scalbers Posted May 11, 2008 Posted May 11, 2008 It may be helpful to check the full article PDF file, also available on that web link. At a simple level, any measurable antiproton decay would by itself indicate a symmetry violation, unless it is a more subtle point about a particular decay mode you are looking at. They seem to lay out some plausible decay modes for the antiproton - are these the same modes for the proton? The question I also have about nomenclature is why they are dividing "tau" by "B". Also, here is a general Wikipedia article on proton decay: http://en.wikipedia.org/wiki/Proton_decay
Norman Albers Posted May 29, 2008 Author Posted May 29, 2008 The CP-violation as observed in cobalt nuclei decay gives a strong preference for the angular momentum of the ejected electron to be opposite, in a right-hand sense, to the velocity vector. Do you think this is attributable to the happenstance of the particles existing hereabouts, or is it a characteristic of the vacuum per se? This would pertain to decay of antiparticles we observe.
Norman Albers Posted June 4, 2008 Author Posted June 4, 2008 Do you think the preference of polarization of the matter field extends to the vacuum itself 'hereabouts'?
scalbers Posted June 7, 2008 Posted June 7, 2008 Hard to say, though this title sounds interesting: http://www.springerlink.com/content/x74l73753576n566/ There's a tie in here with Schwinger and what Kalster mentions in another ongoing thread. The proposed Schwinger effect has to do with electron-positron production via vacuum polarization in the presence of an electric field.
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