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An Alternative To The Nuclear Atom (part 1)
INTRODUCTION.
---------- Stored here for convenience http://members.optusnet.com.au/maxkeon/q1-q6htm.html and here as a FreeBasic compiled Qbasic program http://members.optusnet.com.au/maxkeon/q1-q6x.exe. Norton will give it an immediate tick of approval if it's OK to run. Otherwise, check that the file size is 181760 bytes before using. Whatever the case, before Windows 10 deletes your downloaded file you'll need to click 'Details'. ---------- This simple experiment and what is to follow give an excellent insight into the workings of the atom. The active components for the experiment are, a Na22 radioactive disc, a si3bg Geiger tube, an event counter and a series of lead blocks ranging in length from 5.5mm to 44mm, in 5.5mm increments. The Geiger tube detects only hard beta and gamma rays. http://members.optusnet.com.au/mskeon/0-0-0t.html for more component detail. The Geiger tube and the radioactive disc were placed far enough apart so that the 44mm long lead block could be fitted between them. The different length lead blocks were placed between the source and receiver and the variations in received gamma rays were recorded. If the photon energies from electron-positron annihilations are common, the expected reduction in gamma rays reaching the Geiger tube would be proportional to the increasing lead length. But that's not the case at all. The reduction rate is fairly rapid at the start and reduces substantially at the end. A significant number of supposedly identical gamma rays travel through 88mm of lead as well. Apart from 88, each data point was derived from the average result from 18 one hour duration tests. The curve shape may not yet be precise but it has been clearly evident right from the start. http://members.optusnet.com.au/maxkeon/grf1.jpg Inconsistencies in the contribution from the surrounding universe are fairly inconsequential for all but the 88mm lead test. The total strike rate per hour through the 44mm lead, being the longest in that series, is 28 of which the universe contributes around 10 strikes per hour. Both have significant error margins which becomes insignificant over time. Another cause for concern in this area was the shielding effect of the lead relative to the contribution from the universe. No obvious change was noted when the 11mm lead block position was shifted back and forth between a close proximity to the Geiger tube and a close proximity to the gamma ray source. The blocks were always positioned close to the Geiger tube to minimize the effect. It was going to be fairly inconsequential anyway. http://members.optusnet.com.au/maxkeon/11mmblok.gif Justifying the '88mm' result for the above graph. Note: Increasing the distance between the source and receiver to accommodate the 88mm lead block reduces the received emissions to roughly 25% of the amount received using the spacing required for the 44mm lead block. http://members.optusnet.com.au/maxkeon/88mm.jpg The 88mm lead remains in place throughout this test so that the shielding effect of the lead between the receiver and gamma rays from the universe is of no consequence. The only variant was whether or not the source was in place (swapped after each run (on/off)). With this setup, gamma ray strikes from the universe are greater than ray strikes from my source. And since 50% of the universe is shielded by a rotating earth relative to a changing picture of the universe, the runtime for each test was necessarily one sidereal day. http://members.optusnet.com.au/maxkeon/grf3.jpg 20.24 strikes per day per local source (average1-average2). (average1-average2)/24*4 = 3.37 equivalent strikes per hour for the 44mm between source and receiver comparison. The average strike rate per hour with the lead completely removed was 10. 10 strikes / hr. were necessarily subtracted from all test results for the first graph prior to generating that graph. The gamma ray energies from the Na22 source either vary substantially or some of them are less affected by their surroundings than others. The latter would seem to be the case. The degree of gamma ray interaction with the surroundings depends on the electron_positron link during the process of (presumed) annihilation when the ray was being created. They would be unstoppable if the interaction had been immediate (involving no other charged particles). The maximum travel distance through the lead varies according to how much of the recorded interaction carried in the ray extends beyond the immediate association. The recording will be played back exactly as it was created when any part of the atomic charge structure in the lead interacts with it. (the zero origin concept) http://members.optusnet.com.au/mskeon/zerorign.html The stopping distance exponentially decreases from infinity for 0% accessibility to zero for 100% accessibility, which implies a linear change rate. But it's nothing like that because the zero point on the graph is set at infinity. The tiniest degree of accessibility will infinitely shrink the stopping distance to a measurable quantity. It's an exponential decay mechanism based on zero at infinity so the change rate would be phenomenal. ---------- An Alternative To The Nuclear Atom (part 1) An understanding of the atom is absolutely impossible while the nuclear atom is accepted as reality. The most obvious failing is that it can't possibly explain why liquids are virtually incompressible. The contacting surfaces of the atoms involved are necessarily very rigid and the only way such rigidity can be achieved is if the contents of the atoms are held in an extremely compressed state, and there is direct contact between the surfaces of whatever it is that encloses the contents. The idea that this rigid boundary is somehow generated in the relatively gigantic expanse of atomic space surrounding the nuclei is completely absurd. i.e. For the solar system comparison where the nuclear radius is the radius of the sun the atomic radius would be 1.2 times the average orbit radius of Pluto. In the past I've made several attempts at explaining how the charge interaction between a group of electrons and positrons could create such a boundary. I took a backward step with the last one. They were all based on the fundamental requirement of the Zero Origin Concept that everything in existence is made up entirely of electrons and positrons. 918 positrons and 917 electrons is near enough to the mass equivalent of the proton. My starting point is once again the proton of the hydrogen atom. It necessarily encompasses the entire atomic volume. A point sized electron is a natural companion. IF THE INWARD ACCELERATION RATE WHICH ENCLOSES DIMENSION AROUND A GRAVITY INDUCED BLACK HOLE IS THE SAME AS A CHARGE GENERATED ACCELERATION RATE FOR THE SAME RADIUS, DIMENSION MUST BE ENCLOSED FOR BOTH SCENARIOS. If a total of 1835 electron mass components (918+917-) are housed within an atomic radius of 1e-10 meters and the attractive forces between the component pairs are added together to give a final force acting toward the center of charge the results closely compare. A more precise compare radius is 9.681e-11 meters. The charge acceleration rates are based on the masses of an electron_positron pair accelerating toward each other. These are the results when the two acceleration rates compare. ..00000000009681 Schwarzchild radius = test radius 'r1'. 6.531409295352324D+16 required mass per M = r1*c^2/(2*G) 4.64828013634955D+26 m/sec^2 acceleration rate (GM/r^2) ***** 3.800583429733027D-30 meters^3 total atomic volume 2.071162632007099D-33 meters^3 for each of the 1835 components 7.907542424152498D-12 radius for each component 1.5815084848305D-11 distance between component centers 'r3' Q1*Q2/(4*pi*e0*r3^2) = 9.230279311249665D-07 newtons per pair 8.468781268071567D-04 N total force acting within the 917.5 pairs 4.64806875305794D+26 m/s^2 (total force / 2 electron masses) ****** The mass of every e-e+ pair on the group perimeter is accelerating toward each other and toward the focal point of the entire web of charge interaction, via that web. Considering the focal point of the attraction to be the center of charge, N * number of pairs * r1 = 8.198627145620084D-14 joules http://members.optusnet.com.au/maxkeon/grf4.jpg Unlike gforces, charge interaction is directional. So the opposite charges here are assumed to be directly aligned. Repulsive forces are considered irrelevant for this case. At the junction between the two curves a force is drawing dimension inward toward the center of charge anywhere around the proton event horizon at a rate which is equal to the electron rest energy, and that encloses dimension. So if dimension is enclosed around 1835 components at a 9.681e-11 meter radius, where would it be enclosed around the 207 Muon components (according to mass)? The result is of some significance, as should be expected. The intersect point for a 207 component Muon is 2.55e-12 meters, which is very close to the wavelength equivalent of the electron rest energy. ..00000000000255 Schwarzchild radius = test radius 'r1'. 1720389805097451 required mass per M = r1*c^2/(2*G) 1.764705882352941D+28 m/sec^2 acceleration rate (GM/r^2) ***** 6.945606124699114D-35 meters^3 total atomic volume 3.355365277632422D-37 meters^3 for each of the 207 components 4.310722489647821D-13 radius for each component 8.621444979295643D-13 distance between component centers 'r3' Q1*Q2/(4*pi*e0*r3^2) = 3.105973698181912D-04 newtons per pair 3.214682777618279D-02 N total force acting within the 103.5 pairs 1.764370349955147D+28 m/s^2 (total force / 2 electron masses) ****** The mass of every e-e+ pair on the group perimeter is accelerating toward each other and toward the focal point of the entire web of charge interaction, via that web. Considering the focal point of the attraction to be the center of charge, N * number of pairs * r1 = 8.19744108292661D-14 joules http://members.optusnet.com.au/maxkeon/grf5.jpg The intersect point for a 201 component Muon matches the Compton wavelength. ..000000000002426 Schwarzchild radius = test radius 'r1'. 1636731634182908 required mass per M = r1*c^2/(2*G) 1.854905193734542D+28 m/sec^2 acceleration rate (GM/r^2) ***** 5.9808374587541D-35 meters^3 total atomic volume 2.975541024255771D-37 meters^3 for each of the 201 components 4.141510601314954D-13 radius for each component 8.283021202629908D-13 distance between component centers 'r3' Q1*Q2/(4*pi*e0*r3^2) = 3.364963436172814D-04 newtons per pair 3.381788253353678D-02 N total force acting within the 100.5 pairs 1.856085759250098D+28 m/s^2 (total force / 2 electron masses) ****** The mass of every e-e+ pair on the group perimeter is accelerating toward each other and toward the focal point of the entire web of charge interaction, via that web. Considering the focal point of the attraction to be the center of charge, N * number of pairs * r1 = 8.204218302636024D-14 joules http://members.optusnet.com.au/maxkeon/grf6.jpg The missing mass could well be explained as interaction energy. The proton mass would include some interaction energy as well, which can easily be accommodated. --------------- http://members.optusnet.com.au/maxkeon/muon.jpg The proton and the Muon both carry a single charge imbalance which would be in constant motion as it regenerates throughout the enclosed contents. That's the only tangible property which extends beyond the event horizons of each. The entire process is otherwise directed inward and the contents could not be affected by anything on the outside. Regardless of the Muon's charge it can travel fairly freely through matter, deflecting around the geodesic pathways at half the rate per time given for a gamma ray. Any number of e-e+ pairs will give a similar result, so what's the significance of the 917e- 918e+ proton? Why should that create a stable particle while others don't ?? Why not the Muon? Every combination of e- e+ particles where there's a single charge imbalance has a unique system of interaction. The result from dividing 917 by 918 is unique to that combination only. e.g. Proton ?? 917 e- components 916 e- components 918 e+ components 917 e+ components ..99891067538 to 1 .99890948746 to 1 I have no idea why that combination works. But if no such stable state existed the universe would be nothing like it is now. A constant sized hydrogen atom proton should substantially increase the volumes of the more complex atoms. But that's clearly not the case according to this graph, which is a fair representation of the atomic radii. http://members.optusnet.com.au/maxkeon/grf8.jpg The charge configuration shown below can be duplicated indefinitely and all linked together in the one common bond. The charge field lines of force would all be channeled between the opposite charges. There's no reason at all why the repulsive forces should have any effect here. http://members.optusnet.com.au/maxkeon/cube.jpg The proton could contain 458 of these cubic structures, but one electron would always be missing somewhere. Surrounding electrons would be constantly driven to fill the forever shifting void. When an electron is forced into the proton the charge imbalance is removed and the neutron is born. The package could now shrink to zero if the energy of their diminishing separation could escape. Removing any charge imbalance from within an enclosed group of e-e+ particles so that all charges are equally represented would cause the contents to assume a much closer relationship. The event horizon radius would shrink substantially, BUT IT COULD NEVER SHRINK TO ZERO WHILE THE INTERACTION ENERGY REMAINS TRAPPED WITHIN THE EVENT HORIZON OF A BLACK HOLE. If the separation distance of the e- e+ components in each pair is such that the acceleration rate of each to the other is equivalent to the electron rest energy as previously described, each pair can be considered as being enclosed within a black hole as well. The next set of figures are based on a single e- e+ pair. The e- e+ separation distance which gives an energy equivalent of the electron rest energy is 1.118e-15 meters. The contained group radius for the neutron is necessarily 1e-14 meters. 1.118D-15 Schwarzchild radius = test radius 'r1'. 754272863568.2159 required mass per M = r1*c^2/(2*G) 4.025044722719141D+31 m/sec^2 acceleration rate (GM/r^2) ***** 5.853491887738991D-45 meters^3 total atomic volume 2.926745943869496D-45 meters^3 for each of the 2 components 8.873571880502236D-16 radius for each component 1.774714376100447D-15 distance between component centers 'r3' Q1*Q2/(4*pi*e0*r3^2) = 73.29949135765169 newtons per pair 73.29949135765169 newtons total force acting within the 1 pairs 4.023023674953441D+31 m/s^2 (total force / 2 electron masses) ****** The mass of every e-e+ pair on the group perimeter is accelerating toward each other and toward the focal point of the entire web of charge interaction, via that web. Considering the focal point of the attraction to be the center of charge, N * number of pairs * r1 = 8.194883133785459D-14 joules A residual interaction acting between the enclosed pairs equivalent to the rest energy of 7.45 electrons is required to enclose dimension at the 1e-14 meter neutron radius. 1D-14 Schwarzchild radius = test radius 'r1'. 6746626568264.641 required mass per M = r1*c^2/(2*G) 4.500000078967486D+30 m/sec^2 acceleration rate (GM/r^2) ***** 4.188799858093262D-42 meters^3 total atomic volume 5.622550289040131D-43 meters^3 for each of the 7.5 components 5.120133044994142D-15 radius for each component 1.024026608998828D-14 distance between component centers 'r3' Q1*Q2/(4*pi*e0*r3^2) = 2.201584769140651 newtons per pair 8.2009 newtons total force acting within the 3.725 pairs 4.50104448687675D+30 m/s^2 (total force / 2 electron masses) ****** The mass of every e-e+ pair on the group perimeter is accelerating toward each other and toward the focal point of the entire web of charge interaction, via that web. Considering the focal point of the attraction to be the center of charge, N * number of pairs * r1 = 8.200903199001594D-14 joules ----- The mass required to enclose dimension by gforces is 3.7e39 times greater than the neutron mass. A strong nuclear force is certainly to be expected. ----- Energy expended in compressing a gas is converted to heat energy and is proportional to e=mc^2. The mass increase when a proton changes to a neutron can be compared with this. The fusion environment provides the compressive force while the electron inclusion maintains it. Unit temperature in degrees K and unit pressure will be magnified by around 1e12 X. The mass increase depends on the unit values at the start. So how does a neutron bind protons together in the more complex atoms? The radius of the proton event horizon is 10000 times that of the neutron event horizon. In a fusion environment, when a proton absorbs an electron and collapses into a neutron state, it could become embedded in the event horizons of any two of the surrounding protons which were involved in the collapse. But only a dimensionless object could fit into the 2D realm of the event horizons. If the protons are forced together by the surroundings the neutron event horizon will extend into both protons. The internal distortion will reduce the e-e+ content across each proton along this line, so the proton event horizons won't be completely closed where the neutron resides. The charge balance is achieved when the two e+ proton charges and the two proton companion electrons assemble together around the neutron event horizon where the proton-proton association can be almost immediate. Those electrons will now be bound to the atom. A channel for some interaction between the contents of both protons will have opened up, so each proton would shrink to some degree. http://members.optusnet.com.au/mskeon/elements.html The contents of this link were posted to some of the science newsgroups around 20 years ago. It seemed to be the most logical path for the develepment of the elements in the proposed zero origin universe. I was a little bewildered when it was treated as some kind of joke while the nuclear atom was considered reality. ----- Max Keon |
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