allien

Everthing depends on rules except miracles. So scientific proof is difficult. But mathematic, specially statistic (probabilty and possibility) could give an idea about this. Lets say, If you hit television with hand phone with many times, could you get notebook pc at the end.

If you are in timeless region, they could be samething. If time is a parameter than they are different. like; dark before light. (or only dark) light. dark after light. Dark and dark after light is not same with respect to time. Like before and after Jesus.

I think, This is the rihgt way, first micro cosmos than macro cosmos. I started the relation between nucleons and quarks. After that I worked smaller and bigger particles. My tender blue is (minimum part of) neutrino (single couple of positrino and electrino).

Really there is a problem somewhere. we have an event and we try to formulate it. we had an equation and we want to improve it by math. but while we doing this if we are using assumptions, the result could be wrong even math and formulation right. That is why results always must be checked according to observing or genaralization which is done. Lets turn Big Bang. According to most popular model, Radius was increased from nearly zero to E+25 meters. The time passed during this event is E-11 second. Avarage explosion speed E+33 km/sec. So it was much more than light speed. Than all the particles made a break and slow down suddenly. So model could be wrong, time could be wrong or light speed is not the max. speed in the universe. Big elements occured while supernova explosion by neutron capture. If this is the case, why bigger isotopes of the elements are high amounth than the smaller isotopes? when you deeply investigate, as if bigger elements try to reduce nucleons to reach smaller elements. Any way, I want to say that even equation and math is true, the result must be rechecked. Similar like trial-eror method. They must be compatible the thing in our hand.

If we define nothing is absence of everything, than absence of infinity parameters is nothing. If it is nothing, than it has a meaning which is not nothing.

Or Are you saying (antimatter + matter) = nothing?

The thing in the universe which is the most like BB disc is Black Hole. When you evaulated BB with standart quantum model, you could refuse Black hole acception. But I do not take SQM as referance also. I just wanted to show BB mechanism and composition also speculations even most of the scientist assumed them as true. If you want to hear speculation, you could find them in my theory. Universe was infinity volume at the begining. Particles which has infinity small mass are at rest. no motion and no force. Disturbance caused to inequalty and motion and gravitional force occured. Motion (spin) caused them to be negative and positive elementry particles. compression proses continue according to one center where the big bang occured. unique elementry particle --> negative, positive elementry particle --> multi group of neutral elemnetry particles --> quarks (two type as positive and negative) --> neutroniums --> BB disc --> BB --> neutroniums --> neutron --> proton (anti proton) --> quarks (two type as positive and negative)--> electron (positron) --> photon (neutrino) --> multi group of neutral elemnetry particle --> Infinity times later, unique elementry particles which are at rest will be occur again.

Since it has singularity and high dencity we could assume Big Bang disc as black hole. If we assume BB disc as Black hole, pre-BB state is compression proses. Also standart explenation of BB is meaningless according to dencity (pressure) of the enviroment. Even if protons were occured immidiately, they must be turned to neutrons because of high pressure. When we assume BB expanded (uniformly) with light speed, we found that; Time (sec.) ----- Radius(m)-----Dencity -------Composition 0 --------------- E-25 -------- E+130 ------- Unknown (Possibly Elementry Particle) E-11 ------------ 3E-3 -------- E+62 ------- Unknown (Black Hole) 1 --------------- 3E+8 -------- E+29 ------- Unknown (Black Hole) 10 -------------- 3E+9 -------- E+26 ------- Unknown (Black Hole) 100 ------------- 3E+10 ------- E+23 ------- Quarks (Preon stars) 1000 ------------ 3E+11 ------- E+20 ------- Quarks 5000 ------------ 2E+12 ------- E+18 ------- Quarks (Quark stars) 10000 ----------- 3E+12 ------- E+17 ------- Neutron (Neutron stars) 100000 ---------- 3E+13 ------- E+14 ------- Proton (Stars) Dencity(kg/m^3) Proton is never observed or predicted in very high dencity environment because of degeneration pressure. To say that universe occured at E-11 second, you must accept universe dencity as E+62 kg/m^3 and at this dencity main composition could not be proton. Because we assume neigligible amounth of proton in Neutron stars which has E+17 kg/m^3 average dencity.

could you advise me a link that I could read the experiment and observations about this. Or could be also a link that explain this observation. I am sceptical about this subject in general. For example, exsistance of gluon proved with the interaction of electron and positron in high speed. I am not agree with this assumption. Provement must be done with massive particles not energy particles. This only shows the energy inside the charges. I only saying that if the assuption is not proved from experiments than could be changed. If this is a fixed reality than there is 3 quark in nucleons. But this bodies could be different groups of quarks also. So I must make reseach about it. If you help me, I would be pleasure.

Avarage is not used as in statistical meaning. This is because of the small changes on the values in the resource books and links. And it also means getting idea (predicting) from related subjects. Here is my thought with equations; Assumption; All mass for nucleons comes from quarks. There is one (negative) quark difference between neutron and proton. N – P = 1 z-quark mass = 939,56563 MeV-938,27231 MeV=1,29332 MeV Number of z-quarks in Neutron = 939,56563 MeV/1,29332 MeV =726,48 Number of z-quarks in Proton=938,27231 MeV/1,29332 MeV=725,48 For neutron 726 z-quark meaningful. It must be even number. Half of them is negative and the other half is positive. For proton 725 z-quark meaningful. One z-quark positive, the remaining are equal as negative and positive. 0,48 z-quark has no meaning. Z-quarks are minimum part of the mass quanta in nucleons. So it must not be divided into parts. If we take the numbers 726 and 725 as reference, than; Z-quark mass in neutron = 939,56563 MeV/726 =1,2941675 Z-quark mass in proton = 938,27231 MeV/725 =1,2941687 Z-quark mass in neutron is not equal z-quark mass in proton. But, according to symmetry, they must be exactly same. For neutron 728 and 724, for proton 727 and 723 give worse result than before. So best values are 726 for neutron and 725 for proton. So there is a mistake somewhere. Multi quark model could be wrong. Neutron mass or proton mass or both of them could be wrong as value. (there are much or less same according to resources. And the difference of the sources could not cause this amount mistake. It is not likely that could be the reason. But small error could come from this.) There is something else in nucleons which has mass. This is expressed as shield. So we changed assumption as, huge amount of mass for nucleons comes from quarks. Nucleons have shield which also has mass. There is one (negative) quark difference between neutron and proton. And when we use (best results) 726 z-quark for neutron and 725 z-quark for proton equation becomes; N=939,56563 MeV =726*z-quark mass + c1 (c1: neutron shield mass) P=938,27231 MeV =725* z-quark mass + c2 (c2: proton shield mass) c1 not equal to zero and not equal to c2 c2 not equal to zero and not equal to c1 (Both c1 and c2 positive) D.F=1 (3 unknown, 2 independent equation) so there is no exact solution. With the help of nuclear reaction and other element z-quark numbers and shield mass expectations, I estimated z-quark mass as 0,0013887u (+/- 0,0000007u). But even with this range shield mass could not calculated. To make this possible, a definite z-quark mass must be found. So, as mentioned before, neither z-quark mass nor shield mass did not found exactly. Bare (complately ionized) atom mass could be usefull to determine z-quark number and shield mass. If anybody have this kind of experimental result, I would be gratefull to hear, specially for small elements.

Yes I could not testable predictions by math. But the hypotesis has testable predictions by experiments. Here is some of them; Two proton impact could not produce deuterium. A quark provider needed for this. Quark provider could be pion or muon or atom. So Sun reaction is not two proton reaction. If there is no quark provider in the reaction, than 3 or 4 even 5 proton impact causing the reactions and producing deuterium and hellium-3. This could be the answer of "Sun neutrino problem". If tritium bombarded with proton, hellium-3 and anti-proton occured. (if none of the atom do not lost its integrty by impact.) Tritium decays faster in a enviroment that is occupied by elements which could easly decays beta pozitif. There must be change on the plate which protons strike to turn to anti proton. This could be observed as beta minus decay on the atoms of the plate. As I say before, only lattice and osilation of the z-quarks need complex math. If you ask details of any assumption, I could try to explain to you.

I am not agree about that S.M. clearly defines everthing. That is why my fiction is very different from S.M. Quarks in SM assumed as 3, they are not calculated. The assumption behind this accept is the balance of the loads. So I did not restrict my self with assumptions and imaginary particles. I did not take S.M as referance. So the numbers and assumptions in my fiction are; Neutrino is interaction of electron and positron. clue:neutrinoless decay of muon. All decays produce at least 3 load. I assumed the minimum group of this as Z-quark. and defined them as positive and negative. Neutron has one (negative) z-quark different from proton. Since decay of neutron produces one electron and neutrino (electron+positron). With the average value of z-quark 1,29332 MeV, you could put 726 z-quark in neutron and 725 z-quark in proton. (when you devide it with neutron and proton mass) The clues about nucleons sheild are; Electron particles do not drop to nucleus. (I assume as electron is not single rigid particle) Neutron dencity is not so big and also neutron proton mixture. Ionic radius much more than nucleus radius. Free neutron and proton does not interact easly. 6 orbital for neutron is found according to acceptions that; Dual tetrahedron. (minimum corner, symetric according to each corner.) one tetrahedron has 4(+) z-quark and its dual also 4(-) z-quark. This is the root of neutron (first orbital). And radius r1 to any corner (load) Second radius is (r1+r1) 2r1. surface of the sphere increased 4 times and also load increased 4 times for constant surface. so; orbital .....z-quark.... neg. z-quark...... poz. z-quark 1..............8......................... 4............ 4 2............. 32...................... 16.......... 16 3............. 72...................... 36.......... 36 4........... ..128......................64........ 64 5............. 200.....................100....... 100 6............. 288.................... 144........ 144 Total........728..................... 364......... 364 Since neutron has 726 z-quark, this means last orbital is not full. There is 2 z-quark space. 363 (+) z-quark and 363 (-) z-quark.

Theory does not have and need complex math. equations. Only lattice structure of nucleons need mathematical model and formulation. Geometric model for the iner orbital is thought as dual tetrahedron for neutron and proton. Only osilation band is calculated with respect to orbital diameter. Individual atractive force (moment, frequancy) of each z-quark is not calculated. It is impossible to calculate the effect of 726 z-quarks in 6 orbital according to each other. Probably it needs a program. But if we summarise again; Neutrino is accepted as interaction of electron and positron like photon. neutrinoless decays of muon. http://www.en.wikipedia.org/wiki/Muon Disentegration of the nucleus produce electron, positron and energy which is interaction of electron and positron. Same is valid for mass and anti mass interaction. There is no exception about that. Nothing except electron, positron, photon and neutrino observed. (sub atomic particles also decays to them) There are a lot of opposite loads in the nucleons but never interact with each other. So there is a mechanism keeping them seperate. What is the minimum group of this loads. Beta negative decay. Produce; electron + (electron + positron) Beta positive decay. Produce; positron + (electron + positron) No matter (electron + positron) is neutrino or photon. I defined this minimum group as z-quark. So I have two different z-quark as; Negative z-quark; electron + positron + electron Positive z-quark; positron + electron + positron Normally 3 times of electron mass must give the value for z-quark mass as 1,53 MeV. And this z-quark mass is fit to S.M. quark mass value. But this value is not compatible with mass differance between neutron and proton. And energy of the fusions reactions does not fit the value 1,53 MeV. So the differance between neutron and proton is taken as approximate value for z-quark mass. It is 1,29332 MeV. Since gluons do not accepted, there must be something holding nucleus together. The z-quark location at the last orbital is triangle and star. So one z-quarks surrounded with opposite z-quarks. They osilate outward and inward. They do not allow any z-quark to escape from nucleus. But if there was no sheild. They could be expand infinity with osilating. This means disintegrate. Infinity means the distance which electrical atractions have lost their function. I could not find sheild mass and z-quark mass. When one is found the others could be calculated. I had worked with small elements and the value for z-quark mass is 0,0013887u (+/- 0,0000007u) acceptable.

actually this subject very long. but I would try to mention some of them. but first of all I must correct proton z-quark numbers at the end of my massege. it must be; Proton 363 (+)z-quark ........ not 362 (+)z-quark 362 (-)z-quark 725 z-quark Center positive when you investigate matter-antimatter interaction all the thing produced same as electron, positron, photon and neutrino. photon has not anti particle, for me also neutrino. my neutrinos are electron neutrino and electron anti neutrino (which is positron neutrino). since no anti particle all same as neutrino. In every decay there are plenty of charge produced. This means there are alot of charge in nucleons, no matter it is matter or anti matter. nowadays neutron is defining as (3 layer) onion structure. this is also compatable with my theory. (I define 6 layer and a sheild) Ionic for of the atom is not bare. Other wise they could easly interact with each other. and free neutron has small negative charge which could be say exsistance of sheild for the nucleus. Same for proton, this prevent electron particles droping to nucleus. Standart quark mass is close to my z-quark mass. and there is not strict restriction for only 3 quarks in the nucleons. This is an assumption. If you balance the loads, you could increase them. quark mass allow you to do it even in S.M. Produced energy with my z-quarks are not well fitted with predicted energy. but I could say, if there is proton in the product side this reduce conversion. and if there is tritium in the input side conversion is high. probably because of that, proton causing death energy (neutrino) much and tritium sheild increase the energy produced. In my theory only two proton impact could not produce deuterium. there must be a quark provider in the reaction like pion or muon. The deuterium neutron is same as anti-proton as z-quark numbers. but it could be different geometrically. I mean center negative z-quark could be found at the last orbital. at this time it is defected neutron with negative load. Negative z-quark could be found at the center. At this time it is anti-proton. If it is anti-proton, then deuterium proton changes with anti-proton in a sequence. I mean nucleus also osilate as proton and anti-proton. so I could not proof the fiction, I have no ability of doing this. I collected some clues and got them together. Maybe this could grow with your corrections and experiments The fiction is not only about Micro Cosmos, also about Macro Cosmos. There are also some problems there. The elements untill Germanium is formed by fusion in a surrounding, which has similar ratio of proton and neutron or proton dominant enviroment. Remnant elements (li,Be,B) are produced at neutron dominant enviroment (atleast big, stable isotopes of the elements). Also argon is exception (Argon40 > Argon36). When we ignore this exceptions, smaller isotopes of the elements have great ratio against bigger isotopes of the same elements. For the heavier elements than germanium, this turns opposite. Only zirkonium, But this does not upset to genaralization. Zinc Zn-64 abundance: 48,6 ......... Zn-64+ Zn-66= 76,5 Zn-66 abundance: 27,9 Zn-67 abundance: 04,1 Zn-68 abundance: 18,8 ......... Zn-68+ Zn-70= 19,5 Zn-70 abundance: 00,6 Germanium Ge-70 abundance: 20,8 ......... Ge-70+ Ge-72= 48,3 Ge-72 abundance: 27,5 Ge-73 abundance: 07,7 Ge-74 abundance: 36,3 ......... Ge-74+ Ge-76= 43,9 Ge-76 abundance: 07,6 Slenium Se-74 abundance: 00,9 ......... Se-74+ Se-76+ Se-77 = 17,9 Se-76 abundance: 09,4 Se-77 abundance: 07,6 Se-78 abundance: 23,8 ......... Se-78+ Se-80+ Se-82 = 82,1 Se-80 abundance: 49,6 Se-82 abundance: 08,7 http://periodictable.com/Isotopes/032.74/index.html So there are atleast two types of occuring prosses. One of them is fusion. But the other is not fast (or slow) neutron capture prosses. Because small stable isotopes are less abundance than big ones. If neutron capture prosses was right, abundance must gradually decrase when we go higher isotopes. But it is opposite nearly for all elements. Iron-56 catches 18 neutron to form Se-74. Iron-56 catches 26 neutron to form Se-82. So catching 18 neutron is more possible than catching 26 neutron by iron. But Se-82 is ten times much than Se-74.

My particle is Zerre, origin of the mass. But I will start from quarks to make it easy. My quarks are Zerre quarks. I called them as z-quarks. What is the type and mass of the z-quarks? When we investigate decays and N.R. all the thing produced is; electron, positron, photon (electron-positron interaction) and neutrino. when you investigate decays whithout neutrino you see again, electron, positron, photon (electron-positron interaction) http://www.en.wikipedia.org/wiki/Muon neutrino-less decay modes. So type of quarks according to product side; electron - positron - electron or positron - electron - positron electron - positron couple could be photon or neutrino. Which means neutrino also interaction of electron and positron That is why, neutrino defined as irregular and photon is defined as regular energy. So (1/3) of the quark mass defines net load. and (2/3) of the quark mass polarised. So quarks are negative or positive according to their effective loads. when quarks decays, they produce; electron + positron + electron or positron + electron + positron or electron + photon or positron + photon or electron + neutrino or positron + neutrino All of them shows the decay alternatives of positive or negative quarks. What is the z-quark mass? Three times electron? No... Because electron is not taken as rigid particle. Mass of the z-quark is found from the acception of that; there is one (negative) z-quark differance between neutron and proton. Neutron mass : 939,56563 MeV Proton mass : 938,27231 MeV Z-quark mass : 001,29332 MeV z-quark mass is 1,3MeV. So I define the loads in quarks as quark electron and quark positron. quark electron mass : (1,3/3)= 0,43MeV quark positron mass : (1,3/3)= 0,43MeV standart electron mass --------: 0,51MeV The charge of the quark electron and quark positron could not found. They could be same charge like ordinary electron or could be propotionaly less than electron. (Even they could have bigger load than electron.) in S.M. down quak double of up quark as mass and charge. in z-quarks both mass and loads are same quantity. if you assume nucleons made from only z-quarks, you found as; 726,49 z-quarks in neutron and 725,49 z-quarks in proton. 0,49 says that z-quark mass is wrong or there is something else in the nucleons. Since gluons do not accepted according to the theory, I define this as sheild of the nucleus. Neither z-quark mass nor sheild mass did not found exactly. But as a range, z-quark mass is 0,0013887u (+/- 0,0000007u). Neutron --decay---->Proton+ Neg. z-quark ------ (Neg.)z-quark-> electron + (electron+positron) 363 (+) z-quark ----- 363 (+)z-quark 363 (-) z-quark ----- 362 (-)z-quark 726 z-quark --------- 725 z-quark in multi quark model, neutron structure is defenied as dual tetrahedron in the center. 4(+)z-quark and 4(-)z-quark. and others are replaced according constant surface. z-quarks are osilating inward and outward. there are six orbitals in neutron. proton structure is also defenied as dual tetrahedron but there is positive z-quark in the center. so neutron root is 8 z-quark. (inner orbital) proton root is 9 z-quark. (inner orbital) center positive z-quark differantiate the orbitals cause 12 orbitals for proton. osilation of z-quarks in proton is smaller than the z-quarks in neutron. this makes proton smaller. Positive and negative z-quarks are anti particles of eachother. Proton................................................. Anti-proton 362 (+)z-quark...................................... 362 (+)z-quark 362 (-)z-quark...................................... 363 (-) z-quark 725 z-quark ......................................... 725 z-quark Center positive ..................................... Center negative So Universe do not consist of four types of quarks. They are only two as negative and positive.