kristalris Posted January 13, 2013 Posted January 13, 2013 There's no "standards of proof". You either get the same number on both sides of the equation or you don't, meaning it's illogical. If you have some equation that says "mass=", before you even get a number your units better be in the same units as mass. Can two particles always make "mass=y+z-v+r...=Xkg" true? Of course there are standards of proof. And yes even physics can't do without different standards of proof. Because this will become off topic in this thread I'll answer more in detail in my own thread that also deals with correct scientific procedure. My idea and this thread are linked insofar that my idea provides a more fundamental answer that the idea in this thread lacks as I stated earlier. Further more my idea provides a testable result. If that is positive it will strengthen Przemyslaw's idea, if my idea is busted it will not bust Przem's idea. (This also works vice versa BTW) Przem's idea provides my idea something I lack: a possibility to couple it to the standard model. For the reasons I'm going to state in my thread: you can take the SM as a fact and see where Przem's idea is in conflict with it. Say that there are only qualms with quarks. Does this then bust his idea? No of course not. You must also investigate if Przem's idea can be altered in order to fit the SM. Above that you should also take Przem's idea as a fact and see in how far something could be changed in the SM. I.e logic / mathematics dictate that you take what you put into it as an absolute truth. This even-though absolute truth doesn't exist. What you can say is that the SM is probably fully correct and extremely probably to a great extent correct. Apart from that its correctness is always also dependent upon explicit or implicit assumptions that might be incorrect given a changed problem. So the correct standard of proof for Prezm's idea is: does his idea have sufficient probability in being at least in part correct to warrant further cost and effort to be put into it? Cost for gain. I.e what is it's potential compared to other potential investigations? Or in other words: does it prove to be worth the effort? That is the correct standard of proof for this question. I.e. this standard is relative and not absolute what you make it out to be.
swansont Posted January 13, 2013 Posted January 13, 2013 Or maybe they're smarter than you think. Scientist can speculate, taking into consideration, idea that quarks might simply not exist. Actually there is no evidence of their existence. Even worse, they're creating logic problems and inconsistency. In mine opinion proton-proton collision producing anti-proton is proof that up and down quarks, and anti-up and anti-down quarks don't exist. Up quark and down quark idea has been introduced after shock made by splitting particle that had to be not able to split - proton. And produced smaller, less massive, maintaining the same or opposite electric charge particles. How to split +1 and -1, that were supposed to be fixed-point integers, if we know that their result (electric charge) is also +1 and -1?! Quarks were temporary patch to fix obvious logic problem.. If quarks don't exist we don't have logic problem in f.e. pion+ which is supposed to be made of up and anti-down. But result of its decay has no quarks at all.. quark is +1, antiquark is -1, in terms of quark numbers. Total is zero. I know you understand the concept as applied to charge, because of your model. In post #65 I showed how something that has +1, -1, and 0, electric charge can split, maintaining the same overall electric charge. So math equation matches on both sides. In proton-proton collision we have cloud of particles with +2 electric charge in Standard Model. To what +2 can decay (they're naturally pushing away)? +2 -> +1 + 1 or +2 -> +1 + 1 + 0 or +2 -> +2 + 0 Then see post #65 how +1 can decay: +1 -> +1 + 0 or +1 -> +1 - 1 + 1 or +1 -> +1 - 1 + 1 + 0 Energy that is needed to collide proton-proton is not producing anti-matter from nothing - it's just needed to accelerate particle to such speed that it can overcome natural electric charge pushing away force between them. Anti-particles are created from regular particles, just different configuration of elementary particles. Electron P 5/4 is the smallest anti-matter particle. Add to it enough P 2/1 (one positive and one negative) and it'll become anti-proton. Where enough might mean millions. Positron P 5/1 is the smallest matter particle. Add to it enough P 2/1 and it'll become proton. Electron and Positron that we see and detect in our world might not be (and the most likely are not) P 5/4 or P 5/1! It might be P 5/4+P 2/1 * x giving our world electron. P 5/4 has mass of 2.5x elementary photon. Imagine positron is P 5/1, electron 5/4, neutrino is P 4/2, and proton is P 31/14 P 31/14 + P 31/14 -> P 62/28 (+6 electric charge, Standard Model +2 equivalent) P 62/28 -> P 5/1 + P 5/1 + P 52/26 (equivalent of +2 -> +1 + 1 + 0) or larger than normal unstable proton, positron and neutrino: P 62/28 -> P 53/25 + P 5/1 + P 4/2 (equivalent of +2 -> +1 + 1 + 0) P 53/25 -> P 39/18+ P 5/4 + P 5/1 + P 4/2 (equivalent of +1 -> +1 - 1 + 1 + 0) But combination possible to create is also: P 53/25 -> P 31/17+ P 5/1 + P 5/1 + P 4/2 + P 4/2 + P 4/2 (equivalent of +1 -> -1 + 1 + 1 + 0 + 0 + 0) P 31/17 is anti-particle of initial proton P 31/14 (simply putting their positive and negative elementary count are reversed, one has 14 P, 17 N, second one has 17 P, and 14 N) Have we ever seen the particles and decays? What is your 52/26 particle? The 53/25 particle? 39/18? For the reasons I'm going to state in my thread: you can take the SM as a fact and see where Przem's idea is in conflict with it. Say that there are only qualms with quarks. Does this then bust his idea? No of course not. You must also investigate if Przem's idea can be altered in order to fit the SM. Above that you should also take Przem's idea as a fact and see in how far something could be changed in the SM. You don't have to take the SM as a fact, only the results of the experiments (which are consistent with the SM). If the idea has to be altered, then it does "bust" the idea, but not irretrievably so. However, one must first get the concession that the model has to be altered, and we're not there yet. You also don't go about this by changing the SM in any way that conflicts with any experimental results.
Przemyslaw.Gruchala Posted January 13, 2013 Author Posted January 13, 2013 quark is +1, antiquark is -1, in terms of quark numbers. Total is zero. This is no true answer.. Baryon number is not real: it's math explanation of how to keep electric charge formulas correct on both sides of equation. Baryon number was introduced even before "inventing" quarks. Changing Baryon number from simple -1,0,+1 to B=1/3(q-antiq) was move backward in mine opinion, and completely unnecessarily complicated Quantum Physics theories.. If you have not noticed yet, Baryon number in mine theory is simply B=1/3(Positive-Negative). Or even simpler Electric Charge/3. f,e, Electron-compatible P 5/4 is T=5 N=4 P=5-4=1 B=(1-4)/3=-3/3=-1 Proton-compatible P 31/14 T=31 N=14 P=31-14=17 B=(17-14)/3=+3/3=+1 I know you understand the concept as applied to charge, because of your model. I understand, and see that's not correct although math matches. And I am giving alternative that's either matching in math and logic, and is simpler. Have we ever seen the particles and decays? What is your 52/26 particle? The 53/25 particle? 39/18? P 62/28 would be made in the moment of impact/intersection of particles P 31/14. When radius of one proton core is intersecting with radius of second proton core (when they intersected perfectly in the center of other, which has the smallest probability from all hits in target). You know radius of proton core, you know speed of light, so you can calculate how long such "particle" is existing. I am not even calling it particle - treating it as "cloud of elementary particles". (Giving it special name would be like giving name to two cars that collide on Earth and thinking they're one piece, watching this accident using telescope from Mars) The same is when you are writing Feynman diagrams with Boson W- emitted from f.e. neutron. Neutron has 0 -> +1 (proton) -1 (Boson W-) Then Boson W-, unstable electron-compatible particle, suddenly splits to two stable particles: -1 (electron) and 0 (neutrino) If neutron is P 40/20, proton is P 31/14, electron is P 5/4 and neutrino P 4/2 this equation can be showed as: P 40/20 (neutron 0) -> P 31/14 (proton+) + P 9/6 (boson w-) P 9/6 (boson w-) -> P 5/4 (electron-) + P 4/2 (neutrino) Why 31, you might ask, with P=17 and N=14? Because it's the only particle with two prime numbers below 40. Second one is 37 with P=20 N=17 can't be used, 40-37=3 and it would create incorrect electric charge/incorrect baryon number like here in example illegal equation: P 40/20 -> P 37/17 (stable proton+) + P 3/3 (ILLEGAL, just negative particles, without positive can't create stable particle) You also don't go about this by changing the SM in any way that conflicts with any experimental results. Mine theory is not conflicting with any experimental results - even more, explain how anti-proton is created during proton-proton collision. When I will have better user friendly program simulating universe that I am developing, you will run it, enter particle code, its velocity (or 0) and observe what happens. And it's like watching how electricity is "traveling" between protons, and how "photons are reflecting/refracting" passing through some material, or how particle with higher temperature is cooling down, giving photon to surrounding particles heating them..
swansont Posted January 13, 2013 Posted January 13, 2013 Mine theory is not conflicting with any experimental results - even more, explain how anti-proton is created during proton-proton collision. You haven't actually shown this, though. You still have not matched your particles up to those that have been observed, to see if it's true that you predict what has been seen, and only what has been seen. P 62/28 would be made in the moment of impact/intersection of particles P 31/14. When radius of one proton core is intersecting with radius of second proton core (when they intersected perfectly in the center of other, which has the smallest probability from all hits in target). You know radius of proton core, you know speed of light, so you can calculate how long such "particle" is existing. I am not even calling it particle - treating it as "cloud of elementary particles". How do you tell if a particle will exist longer? Why is the proton (31/14) stable? What about the other particles I asked about?
kristalris Posted January 13, 2013 Posted January 13, 2013 You don't have to take the SM as a fact, only the results of the experiments (which are consistent with the SM). If the idea has to be altered, then it does "bust" the idea, but not irretrievably so. However, one must first get the concession that the model has to be altered, and we're not there yet. You also don't go about this by changing the SM in any way that conflicts with any experimental results. Of course. Did I say or imply anything different? So I fully agree.
SamBridge Posted January 13, 2013 Posted January 13, 2013 (edited) Of course there are standards of proof. And yes even physics can't do without different standards of proof. Because this will become off topic in this thread I'll answer more in detail in my own thread that also deals with correct scientific procedure. My idea and this thread are linked insofar that my idea provides a more fundamental answer that the idea in this thread lacks as I stated earlier. Further more my idea provides a testable result. If that is positive it will strengthen Przemyslaw's idea, if my idea is busted it will not bust Przem's idea. (This also works vice versa BTW) Przem's idea provides my idea something I lack: a possibility to couple it to the standard model. For the reasons I'm going to state in my thread: you can take the SM as a fact and see where Przem's idea is in conflict with it. Say that there are only qualms with quarks. Does this then bust his idea? No of course not. You must also investigate if Przem's idea can be altered in order to fit the SM. Above that you should also take Przem's idea as a fact and see in how far something could be changed in the SM. I.e logic / mathematics dictate that you take what you put into it as an absolute truth. This even-though absolute truth doesn't exist. What you can say is that the SM is probably fully correct and extremely probably to a great extent correct. Apart from that its correctness is always also dependent upon explicit or implicit assumptions that might be incorrect given a changed problem. So the correct standard of proof for Prezm's idea is: does his idea have sufficient probability in being at least in part correct to warrant further cost and effort to be put into it? Cost for gain. I.e what is it's potential compared to other potential investigations? Or in other words: does it prove to be worth the effort? That is the correct standard of proof for this question. I.e. this standard is relative and not absolute what you make it out to be. Nope there's no standards of proof, proof is proof, proof is not proof-1 or proof+1. You have different levels of scientific investigation and confirmation, it goes concept->hypothesis->experiment->theory->more experiments->proven right->scientific law, or proven false->more experimenting or garbage, but if its proven then it's proven, and if its disproved then its disproved. If its results are predictable with a set of equations that fit into the rest of physics it's proven, if it doesn't always fit the results we see in reality, then it's disproved or remodeled to try and account for those differences. Edited January 13, 2013 by SamBridge
Przemyslaw.Gruchala Posted January 13, 2013 Author Posted January 13, 2013 (edited) You haven't actually shown this, though. You still have not matched your particles up to those that have been observed, to see if it's true that you predict what has been seen, and only what has been seen. I told you: to match mine particles with our Universe particle, I need EXACT mass of neutrino. Or exact neutrino wave length. elementary Photon must be P 2/1 and elementary Neutrino must be P 4/2 That's what is constant in mine theory. I thought that I will show mine theory, example math formulas. Then people will help finding relation between known particles and propose experiments for accelerators. And as a team will find final total number of elementary particles in our world proton.. But apparently physicists don't work as a team. And I have to do everything. Even exact ratio between rest mass/wavelength of neutrino and rest mass/wavelength of electron might give final result. How do you tell if a particle will exist longer? Why is the proton (31/14) stable? It's simple concept - stable particles have primary numbers of elementary particles. They're stable, because they can't be divided to smaller without breaking electric charge equation/baryon number. If we will add photon to it: P 31/14 + P 2/1 = P 33/15 (P=18,N=15) it's not stable, so it's emitting what is not needed- P 2/1 (in normal circumstances) Couple such particles (f.e. 137) emitting P 2/1 are creating SM visible or non-visible photon wavelength. If particle is receiving too much photons, more than it can emit, it's growing. Temperature is increasing. Photons are jumping between one particle to another with speed of light. But new one are appearing constantly from annihilation. And we have nuclear fusion process. P 31/14 + P 2/1 + P 2/1 + P 2/1.... etc. after adding 20 photons P 2/1 there is P 71/34 (P=37,N=34) It's Proton with Neutron = Deuterium. Either 71 and 37 are prime numbers, so it's stable isotope. Let's keep adding photons to it: P 71/34+ P 2/1+P 2/1.... again 20x P 2/1 P 71/34+ 20 * P 2/1 = P 111/54 (P=57,N=54) Either 111 and 54 are not prime numbers. So it's unstable. And suddenly is decaying: P 111/54 (tritium+) -> P 102/48 (helium-3) + P 5/4 (electron) + P 4/2 (neutrino) P 102/48 (helium-3) = P 31/14 (proton+) + P 31/14 (proton+) + P 40/20 (neutron 0) P 102/48 - P 2/1 (emitting one photon) = P 100/47 (P=53,N=47, stable, even though it has +6 electric charge (+2 Standard Model) they're not separated by electrostatic opposite pushing away force; stability rule has precedence above it; meaning it's strong force) There is dozen of ways for nuclear fusion.. Two protons P 31/14 colliding: P 31/14 + P 31/14 -> P 62/28, which can decay to P 5/1 (positron) + P 5/1 + P 52/26 (neutron-compatible) Positron from them are annihilating with electrons P 5/4 made in above equation during production of Helium-3 producing another cloud of 5 photons P 2/1. P 52/26 -> P 5/4 (electron) + P 47/22 (proton-compatible unstable) P 47/22 -> P 43/20 (proton-compatible stable) + P 4/2 (neutrino) P 43/20 (proton-compatible) + P 31/14 (proton-compatible) + P 2/1 (photon) = P 76/35 (unstable Helium-compatible) -> P 71/34 (Deuterium) + P 5/1 (positron) swansont, see again post #52 where are attached pictures from application. It's showing how proton P 9181/4589 is "bending" timespace. In 2D to better see effect. Each vertex is representing F=Mass/Radius^2. So it's particle->Total / sqrt(( particle->Position - evaluated_position )^2) Now imagine that it's P 31/14. Imagine "adding" to it photon P 2/1. P 31/14+P 2/1 = P 33/15 It doesn't necessarily mean that they are immediately joining in core. This photon can be orbiting in cone. Slightly but slightly it's influencing also orbiting electron P 5/4, and influencing core, Now start adding more and more photons. Influence is increasing. After some point, electron won't be part of it, and fly away (if it was proton + electron). Matter is becoming plasma. After some point, these photons will be so powerful and strong that they will join with core. And Deuterium will be created, and everything will start again. Imagine that there is P 31/14, and P 33/15. Orbiting photon "see" that P 31/14 has no photons orbiting, so it's jumping to its cone. P 33/15 - P 2/1 = P 31/14... P 31/14+P 2/1 = P 33/15 One particle became cooler, second one has been heated. Edited January 13, 2013 by Przemyslaw.Gruchala
swansont Posted January 13, 2013 Posted January 13, 2013 I told you: to match mine particles with our Universe particle, I need EXACT mass of neutrino. Or exact neutrino wave length. elementary Photon must be P 2/1 and elementary Neutrino must be P 4/2 That's what is constant in mine theory. So you don't really know that the proton is P 31/14, or electron is 5/4 or positron is 5/1? How can a 4/2 neutrino have spin 1/2, while a 2/1 photon has spin 1? Shouldn't a 4/2 be able to split into 2 2/1 photons, according to your model? I thought that I will show mine theory, example math formulas. Then people will help finding relation between known particles and propose experiments for accelerators. And as a team will find final total number of elementary particles in our world proton.. But apparently physicists don't work as a team. And I have to do everything. No, they don't, and yes, you do. The burden of proof is yours. Even exact ratio between rest mass/wavelength of neutrino and rest mass/wavelength of electron might give final result. It's simple concept - stable particles have primary numbers of elementary particles. They're stable, because they can't be divided to smaller without breaking electric charge equation/baryon number. A proton is 31/14, right? 14 isn't prime. How do you get baryon number from your notation?
Przemyslaw.Gruchala Posted January 14, 2013 Author Posted January 14, 2013 So you don't really know that the proton is P 31/14, or electron is 5/4 or positron is 5/1? Yes. P 31/14 is used to show example of math formulas of decaying. If I would start doing math with f.e. P 9181/4589 or millions, it would quickly end up with mess and many mistakes/errors in calcs.. Such calculations are good for computers. At least it won't make mistake, if source code is correct. I have made couple such programs for showing me everything about particle with exact number of positive and negative particles. Electron P 5/4 is the smallest particle that is belonging to electron-compatible particle family. With mass just 2.5x higher than photon P 2/1. Positron P 5/1 is the smallest particle that is belonging to proton-compatible particle family. P 2/1 is elementary photon. What you're calling in Standard Model photon is group of these particles. I showed you image 2 days ago how they can travel close to each other pretending to detecting devices that they're single particle, with same direction, same angular momentum. On these pictures red path was one P 2/1, green path was second P 2/1, blue was third P 2/1. etc. so E= h* v is correct. Just interpretation is not correct (group vs single) Imagine Standard Model photon wave made of three P 2/1 photons. They're traveling in same direction with same attributes, from same source. Origin at bottom, going to center. At center there is collision with something causing one "red path" photon to go different direction. It's flying in right direction. The remaining green and blue photon paths are flying left. It should look similar to this (resize in 3d application is not good way for making multiple waves spiral paths, I wanted make it quickly, there is 3 at night already; they should have equal path distance regardless of frequency/wave length): How can a 4/2 neutrino have spin 1/2, while a 2/1 photon has spin 1? Shouldn't a 4/2 be able to split into 2 2/1 photons, according to your model? What you call in Standard Model neutrino is mine true neutrino. That's why I am interested in knowing it's wave length/frequency/rest mass. I don't think so it's like with SM photon wave made of dozen P 2/1. At the moment I don't think so that Neutrino can split, because it's stable. P=2, N=2 - prime numbers. They're strongly attracting. The same is with single P 2/1, P=1, N=1, They're attracting strongly. If they would split, they would immediately join with other pair of P 2/1 that split too. No, they don't, and yes, you do. The burden of proof is yours. Nobody proved there are quarks, but most of people are learning about them. Nobody proved there are R/anti-R, G/anti-G, "colors" etc. etc. A proton is 31/14, right? 14 isn't prime. But 17 is. P 31/14 (Positive = Total-Negative = 31-14=17). How do you get baryon number from your notation? Baryon number = ( Positive - Negative ) / 3 proton-like P 31/14 has 17 positives: 31-14=17 so 17-14=+3/3 = +1 anti-proton-like P 31/17 14-17=-3/3 =-1
SamBridge Posted January 14, 2013 Posted January 14, 2013 (edited) I think essentially what your math boils down too isn't necessarily that there's just two particles, it's just there there's really only two types of oscillation which can oscillate in not only different frequencies and amplitudes and phases but also in different dimensions, and all those oscillations composite in different ways at different angles to make different particles. This isn't much different than what contemporary theories predict. But the real question still remains: what is it that is waving? Also do you know of a way to make 4 dimensional shapes in blender? I need to run a simulation Edited January 14, 2013 by SamBridge
Przemyslaw.Gruchala Posted January 14, 2013 Author Posted January 14, 2013 I think essentially what your math boils down too isn't necessarily that there's just two particles, If it's wave by itself, then just one just enough, but different phase. One is sin(angle), opposite is cos(angle), then angle+=step*time But if angle is relative to time, then in 4 dimensional universe proton traveling faster would be becoming anti-proton, electron would be becoming positron etc, only neutral would remain neutral such as photon, neutrino. Who would guarantee that something that was proton after acceleration and collision is still proton, and not anti-proton? what is it that is waving? That's question to SM, because in SM it's undefined. In mine theory it's just matter that is moving in wave path. Read mine analogy from the beginning of this thread- car moving on road to work and back saw by telescope from f.e. Mars. Is it car traveling, or is it wave? We know that it's just piece of matter, and wave movement is result of how road looks like and how other cars are currently configured on track - we don't want to hit them, so path is like slalom. That's Ockhams rule: if universe can be described by 4 (or 3) dimensions, explanation made using 4 (3) dimensions is more probable than explanation using 5,6,..10 dimensions. If universe can be described using less number of particles, it's more probable than higher number of particles.
swansont Posted January 14, 2013 Posted January 14, 2013 Electron P 5/4 is the smallest particle that is belonging to electron-compatible particle family. With mass just 2.5x higher than photon P 2/1. But there's only one prime, and it's stable. Also, photon mass is zero. Positron P 5/1 is the smallest particle that is belonging to proton-compatible particle family. Proton-compatible? What you call in Standard Model neutrino is mine true neutrino. That's why I am interested in knowing it's wave length/frequency/rest mass. I don't think so it's like with SM photon wave made of dozen P 2/1. At the moment I don't think so that Neutrino can split, because it's stable. P=2, N=2 - prime numbers. They're strongly attracting. The same is with single P 2/1, P=1, N=1, They're attracting strongly. If they would split, they would immediately join with other pair of P 2/1 that split too. But the neutrino has spin 1/2 and the photon is spin 1. How do you get that from even numbers of particles?
Przemyslaw.Gruchala Posted January 14, 2013 Author Posted January 14, 2013 (edited) How can a 4/2 neutrino have spin 1/2, while a 2/1 photon has spin 1? Shouldn't a 4/2 be able to split into 2 2/1 photons, according to your model? Spin is result of how particle moves and rotates in timespace. If proton-compatible particle but unstable such as Pion+, Boson W+, Muon+, etc and. is decaying it's producing smaller positive that is rotating also in same way as source (just wavelength is higher, frequency lower, because sum of mass in center is smaller), and another neutral particle, it's rotating/spinning in reverse direction, helix path is reversed. If you have particles moving in helix path, and will send photons to it to measure, you will cause helix to flatten, and it'll become linear polarized. On the left is original not measured helix path, in center measured in one axis, on right measured in second axis. (In 3D application I simply used Set Vertex Value Y=0, then Z=0 to flatten helix) Photons from regular neutral particles such as either proton and electron will have both helix path clock-wise and counter-clock-wise, both from protons and electrons. Do you have 3d application? I can attach 3d object, instead of just showing screen-shot. So you will be able to rotate it in viewport. Maya, Max3D, LightWave, Cinema4d? Edited January 14, 2013 by Przemyslaw.Gruchala
swansont Posted January 14, 2013 Posted January 14, 2013 Spin is intrinsic angular momentum. It's not dependent upon motion.
SamBridge Posted January 15, 2013 Posted January 15, 2013 In mine theory it's just matter that is moving in wave path. Oh ok yeah, then it's completely wrong, matter doesn't physically wave otherwise it would radiate all it's energy away from constantly accelerating.
Przemyslaw.Gruchala Posted January 15, 2013 Author Posted January 15, 2013 (edited) Oh ok yeah, then it's completely wrong, matter doesn't physically wave otherwise it would radiate all it's energy away from constantly accelerating. Oh, boy.. But what is "energy" in mine theory? Energy is matter. Matter is energy.. Photon (energy supplier in SM) is also matter in mine theory- P 2/1, made of 2 particles. But they're the lightest elements in the Universe. I gave you example formula of nuclear fusion in post #82 If proton is P 31/14, and we will add to it one photon P 2/1. P 31/14 + P 2/1 = P 33/15 it'll be trying to emit what is not needed. Photon will jump to other proton, or other particle. But if the all surrounding neighborhood particles are also P 33/15 (which means they've the same average temperature), then soon after emitting P 2/1, it will receive other photon back. It'll be such circle of emission of photons and absorbing them back. Until the all particles have equal. Only particles that have too many photons really radiates them for real = too much energy than neighborhood particles. Imagine we have three particles: P 31/14, P 31/14 and P 37/17 P 37/17 has three additional P 2/1 orbiting. What happens? P 37/17 -> P 35/16 + P 2/1 (photon is emitted by hot particle) P 31/14 + P 2/1 = P 33/15 (photon is absorbed by cool particle 1) P 35/16 -> P 33/15 + P 2/1 (photon is emitted by hot particle) P 31/14 + P 2/1 = P 33/15 (photon is absorbed by cool particle 2) At the end we have three particles with equal formula P 33/15 They have exactly the same average temperature - the same number of photons, the same sum of mass.. These additional photons are not in core, core is always stable P 31/14, photons P 2/1 are orbiting Now imagine that we will replace p+ with Deuterium: P 71/34, it's stable, entire 71 particles is in core. If it's surrounded by P 35/16 it'll lose photon P 35/16 -> P 33/15 + P 2/1 and will give it to Deuterium P 71/34 + P 2/1 = P 73/35 Now both have the same number of photons orbiting. Same average temperature. Core is slightly attracted by these additional photons orbiting. The more photons, the more it's attracted. These additional photons are also "irritating" electrons. In 0 K temperature, there is no orbiting photons = electrons can fly without "interference" from these photons. Particle is superconductive. Edited January 15, 2013 by Przemyslaw.Gruchala
SamBridge Posted January 15, 2013 Posted January 15, 2013 (edited) Oh, boy.. But what is "energy" in mine theory? Energy is matter. Matter is energy.. Photon (energy supplier in SM) is also matter in mine theory- P 2/1, made of 2 particles. But they're the lightest elements in the Universe. I gave you example formula of nuclear fusion in post #82 If proton is P 31/14, and we will add to it one photon P 2/1. P 31/14 + P 2/1 = P 33/15 it'll be trying to emit what is not needed. Photon will jump to other proton, or other particle. But if the all surrounding neighborhood particles are also P 33/15 (which means they've the same average temperature), then soon after emitting P 2/1, it will receive other photon back. It'll be such circle of emission of photons and absorbing them back. Until the all particles have equal. Only particles that have too many photons really radiates them for real = too much energy than neighborhood particles. Imagine we have three particles: P 31/14, P 31/14 and P 37/17 P 37/17 has three additional P 2/1 orbiting. What happens? P 37/17 -> P 35/16 + P 2/1 (photon is emitted by hot particle) P 31/14 + P 2/1 = P 33/15 (photon is absorbed by cool particle 1) P 35/16 -> P 33/15 + P 2/1 (photon is emitted by hot particle) P 31/14 + P 2/1 = P 33/15 (photon is absorbed by cool particle 2) At the end we have three particles with equal formula P 33/15 They have exactly the same average temperature - the same number of photons, the same sum of mass.. These additional photons are not in core, core is always stable P 31/14, photons P 2/1 are orbiting Now imagine that we will replace p+ with Deuterium: P 71/34, it's stable, entire 71 particles is in core. If it's surrounded by P 35/16 it'll lose photon P 35/16 -> P 33/15 + P 2/1 and will give it to Deuterium P 71/34 + P 2/1 = P 73/35 Now both have the same number of photons orbiting. Same average temperature. Core is slightly attracted by these additional photons orbiting. The more photons, the more it's attracted. These additional photons are also "irritating" electrons. In 0 K temperature, there is no orbiting photons = electrons can fly without "interference" from these photons. Particle is superconductive. If you're using classical motion to descried anything about particles, you're wrong, there's scientific experiments that confirm they do not have classical motion. A photon can jump, but it does not travel through the intervening space, and that's because it's not making a classical jump, it's making a quantum jump. A particle itself as far as we know also doesn't even oscillate, only it's probability can be described as a wave function, which solves the problem with jumping. The helical motion you describe would mean the particle is constantly accelerating, and acceleration requires the constant application of energy, and if nothing is applying energy to make it rotate which that seems to be the case in you're model, that means a particle is radiating it's energy away by accelerating, which means electrons should just fall into the nucleus. Edited January 15, 2013 by SamBridge
swansont Posted January 15, 2013 Posted January 15, 2013 If P 71/34 is a stable deuteron, doesn't that make P 142/68 an alpha? An alpha is stable, but there are no prime numbers here. But in forming a deuteron from a proton and a neutron, photons must be given off — it's a bound system and must release energy. 31/14 + 40/20 = 71/34 Where's the photon?
kristalris Posted January 15, 2013 Posted January 15, 2013 If you're using classical motion to descried anything about particles, you're wrong, there's scientific experiments that confirm they do not have classical motion. A photon can jump, but it does not travel through the intervening space, and that's because it's not making a classical jump, it's making a quantum jump. A particle itself as far as we know also doesn't even oscillate, only it's probability can be described as a wave function, which solves the problem with jumping. The helical motion you describe would mean the particle is constantly accelerating, and acceleration requires the constant application of energy, and if nothing is applying energy to make it rotate which that seems to be the case in you're model, that means a particle is radiating it's energy away by accelerating, which means electrons should just fall into the nucleus. You are pointing at a problem with the model of Przem that current science has as well. You state that experiments show that there is no classical motion but a quantum jump. The latter is described in a way taking c = max and time is relative. The latter two points are however assumptions.(The first indeed an observation of a strange jump.) On the same basis one can assume speeds > c and that only the clocks slow down and not time. Even given super conductivity it is possible to assume permanent motion in the system if the energy has nowhere to go, and assuming a logical end to the split-ability of particles and physical existence of mass. I.e. acceleration. We readily observe atoms that (must) have been spinning for billions of years.And we also agree that ultimately even an atom will end its life as such. But like I stated earlier you need to get more fundamental than Przem has done. I.e. it is just as speculative to assume a classical Newton explanation with speeds > c and absolute time as assuming things - then - coming from nothing like Krauss et all is on about. The only difference then is that the first assumption when you look at all the evidence and address all the questions is very simple and probable and the latter is extremely improbable because it then needs you to believe in blatant contradictions. I.e. belief in magic, that is less probable even than believing in God. Which is extremely improbable as well, yet not irrational, contrary to belief in magic. That the mathematics adds up does't prove a thing if you haven't gone through the mathematics of the alternate. Concerning the latter the burden of proof lies on everyone dealing with the problem. Especially and the more so true if it concerns a testable alternate. I'm still working on my promised reaction on evidence and proof. But what I've seen so far Przem has provided more than enough to warrant his idea to be further investigated in correct use of public funding IMO. Like he himself correctly complains about: it should be teamwork from a point on-wards And for the reasons I'm yet to provide in my thread. What do you want him to prove that his idea is already perfect? No-one in science can claim that at the moment on this topic, so why should he?
Przemyslaw.Gruchala Posted January 15, 2013 Author Posted January 15, 2013 (edited) If P 71/34 is a stable deuteron, doesn't that make P 142/68 an alpha? An alpha is stable, but there are no prime numbers here. P 142/68 -> P 140/67 (73 & 67 prime numbers) + P 2/1 (photon) But in forming a deuteron from a proton and a neutron, photons must be given off — it's a bound system and must release energy. 31/14 + 40/20 = 71/34 Where's the photon? Reread post #82. I showed there example process of creation of tritium. Hellium-4 is later made. Quoting myself: "Either 111 and 54 are not prime numbers. So it's unstable. And suddenly is decaying: P 111/54 (tritium+) -> P 102/48 (helium-3) + P 5/4 (electron) + P 4/2 (neutrino) P 102/48 (helium-3) = P 31/14 (proton+) + P 31/14 (proton+) + P 40/20 (neutron 0) P 102/48 - P 2/1 (emitting one photon) = P 100/47 (P=53,N=47, stable, even though it has +6 electric charge (+2 Standard Model) they're not separated by electrostatic opposite pushing away force; stability rule has precedence above it; meaning it's strong force)" Do you see now your photon? Let's add more photons, heat it: P 100/47 + P 2/1.... x 20 times P 100/47 + P 40/20 = 140/67 (Hellium-4, Protons=2, Neutrons=2, Positive=73, Negative=67, stable) Edited January 15, 2013 by Przemyslaw.Gruchala
swansont Posted January 15, 2013 Posted January 15, 2013 Reread post #82. I showed there example process of creation of tritium. Hellium-4 is later made. Quoting myself: "Either 111 and 54 are not prime numbers. So it's unstable. And suddenly is decaying: P 111/54 (tritium+) -> P 102/48 (helium-3) + P 5/4 (electron) + P 4/2 (neutrino) P 102/48 (helium-3) = P 31/14 (proton+) + P 31/14 (proton+) + P 40/20 (neutron 0) P 102/48 - P 2/1 (emitting one photon) = P 100/47 (P=53,N=47, stable, even though it has +6 electric charge (+2 Standard Model) they're not separated by electrostatic opposite pushing away force; stability rule has precedence above it; meaning it's strong force)" Do you see now your photon? I don't see the deuteron, which is what I was asking about.
Przemyslaw.Gruchala Posted January 15, 2013 Author Posted January 15, 2013 Let's add more photons: P 140/67 + 20* P 2/1 = P 180/87 (unstable) We're looking at http://en.wikipedia.org/wiki/Isotopes_of_helium and Hellium-5 is decaying to Hellium-4 and Neutron (in normal, not in star core, circumstances): P 180/87 -> P 140/67 + P 40/20 Let's add more photons: P 180/87 + P 40/20 = P 220/107 113 and 107 are primes so stable in this small scale model; but with thousands heavier model (where Proton is made of thousands of elementary particles) it might not be. I have also theory that Neutrinos from Sun are causing large amount of radioactive decay. Their cores are in such state they're almost unstable. Add to them P 4/2 is causing sudden death. Scientists already proved that erruption on the Sun is causing increased radioactivity (in Wormhole with Morgan Freeman one guy was showing this) P 220/107 -> P 211/101 + P 5/4 + P 4/2 Lithium-6: 3 Protons, 3 Neutrons. I don't see the deuteron, which is what I was asking about. P 31/14 - proton+ P 40/20 - neutron (P 31/14 + P 5/4 + P 4/2 = P 40/20) P 31/14 + P 40/20 = P 71/34 (Deuterium+) P 71/34 + P 40/20 = P 111/54 (Tritium+) P 111/54 -> P 100/47 (Hellium-3) + P 5/4 + P 4/2 + P 2/1 P 31/14 + P 31/14 + P 40/20 + P 40/20 = P 142/68 -> P 140/67 (Hellium-4) + P 2/1 If you're using classical motion to descried anything about particles, you're wrong, there's scientific experiments that confirm they do not have classical motion. A photon can jump, but it does not travel through the intervening space, and that's because it's not making a classical jump, it's making a quantum jump. A particle itself as far as we know also doesn't even oscillate, only it's probability can be described as a wave function, which solves the problem with jumping. The helical motion you describe would mean the particle is constantly accelerating, and acceleration requires the constant application of energy, and if nothing is applying energy to make it rotate which that seems to be the case in you're model, that means a particle is radiating it's energy away by accelerating, I have no idea about what acceleration you're talking about. Photon has constant speed v=c. If velocity is constant, acceleration is 0. which means electrons should just fall into the nucleus. And it would immediately decay. Because Neutron compatible particles are decaying to positive piece and negative piece. Or to positive piece, negative piece and smaller neutral piece. P 36/18 -> P 31/14 + P 5/4 P 31/14 + P 5/4 = P 36/18 P 40/20 -> P 31/14 + P 9/6 (Boson W-) P 9/6 -> P 5/4 + P 4/2 Falling photons that orbited core, to core, really happens - but there must be too much photons. It's simply nuclear fusion. From particle that has millions of K temperature, there is made particle with 0 K. All orbiting photons are sucked to core. Of course in star, particle with 0K will immediately receive photons from surrounding hot particles, decreasing their average temperature.
SamBridge Posted January 15, 2013 Posted January 15, 2013 (edited) Let's add more photons: P 140/67 + 20* P 2/1 = P 180/87 (unstable) We're looking at http://en.wikipedia.org/wiki/Isotopes_of_helium and Hellium-5 is decaying to Hellium-4 and Neutron (in normal, not in star core, circumstances): P 180/87 -> P 140/67 + P 40/20 Let's add more photons: P 180/87 + P 40/20 = P 220/107 113 and 107 are primes so stable in this small scale model; but with thousands heavier model (where Proton is made of thousands of elementary particles) it might not be. I have also theory that Neutrinos from Sun are causing large amount of radioactive decay. Their cores are in such state they're almost unstable. Add to them P 4/2 is causing sudden death. Scientists already proved that erruption on the Sun is causing increased radioactivity (in Wormhole with Morgan Freeman one guy was showing this) P 220/107 -> P 211/101 + P 5/4 + P 4/2 Lithium-6: 3 Protons, 3 Neutrons. P 31/14 - proton+ P 40/20 - neutron (P 31/14 + P 5/4 + P 4/2 = P 40/20) P 31/14 + P 40/20 = P 71/34 (Deuterium+) P 71/34 + P 40/20 = P 111/54 (Tritium+) P 111/54 -> P 100/47 (Hellium-3) + P 5/4 + P 4/2 + P 2/1 P 31/14 + P 31/14 + P 40/20 + P 40/20 = P 142/68 -> P 140/67 (Hellium-4) + P 2/1 I have no idea about what acceleration you're talking about. Photon has constant speed v=c. If velocity is constant, acceleration is 0. And it would immediately decay. Because Neutron compatible particles are decaying to positive piece and negative piece. Or to positive piece, negative piece and smaller neutral piece. P 36/18 -> P 31/14 + P 5/4 P 31/14 + P 5/4 = P 36/18 P 40/20 -> P 31/14 + P 9/6 (Boson W-) P 9/6 -> P 5/4 + P 4/2 Falling photons that orbited core, to core, really happens - but there must be too much photons. It's simply nuclear fusion. From particle that has millions of K temperature, there is made particle with 0 K. All orbiting photons are sucked to core. Of course in star, particle with 0K will immediately receive photons from surrounding hot particles, decreasing their average temperature. It seems like you're saying photons comprise electrons. But photons don't have mass, and electrons can't move at the speed of light, electrons aren't photons. Energy isn't directly matter, it's not E=m, it's E=m^2+p^2c^2, which means m=sqrt(E/(p^2c^2)), not exactly the same as a photon. You're also not showing much except conservation of spin states. I could say anything has spins that you're saying, I could saying microscopic magical elves have the spins you're saying and get converted into photons and back because I just label them as having the right spins that can be conserved to equal what those different particles spins have. You need more observable evidence at best, and I observe that photons do not have mass and electrons do, spin does not cause this. Edited January 15, 2013 by SamBridge
Przemyslaw.Gruchala Posted January 15, 2013 Author Posted January 15, 2013 (edited) You're thinking/believing that photons have no mass. Because it's repeated over and over and over again, in schools and universities. If something is as light as bacteria or virus, you aren't counting it while measuring weight of yourself. You're not bothering count it after sneezing, you just lost a few bacteries. But it's producing paradox much worse than grandchild killing grandpa after travel in time: Electron has mass, Positron has mass, but result of their collision is complete massless? XX century Quantum Physics is full of such. Turn on thinking. Let's for a while forget about special relativity: If we will assume that either E=m*c*c is true, and E=h*v is true then: E=h*c/wave length m*c*c = h*c/wavelength m=(h/wavelength)/c Now find the smallest possible frequency, and you will have theoretical mass of photon. or reverse- find the largest wavelength: m=h*v/c^2 Or we can go further: c=v*wavelength so m=h*v/(v*wavelength)^2=h/(wavelength*wavelength*v) If frequency is 1, then wavelength=c/1 If wavelength is 1, then frequency=c/1 Reversed: E=m*c*c= m * v*wavelength * v*wavelength E=h/(wavelength*wavelength*v) * v*wavelength * v*wavelength E=h*c/wavelength E=h*wavelength*v/wavelength = h * v Edited January 15, 2013 by Przemyslaw.Gruchala
swansont Posted January 15, 2013 Posted January 15, 2013 P 31/14 - proton+ P 40/20 - neutron (P 31/14 + P 5/4 + P 4/2 = P 40/20) P 31/14 + P 40/20 = P 71/34 (Deuterium+) Ah, here we go. Now, where's the photon? A deuteron is not simply the combination of a proton and a neutron. It's a bound system, meaning it releases energy and the mass is lower than the sum of the individual particles. You're thinking/believing that photons have no mass. Because it's repeated over and over and over again, in schools and universities. If something is as light as bacteria or virus, you aren't counting it while measuring weight of yourself. You're not bothering count it after sneezing, you just lost a few bacteries. But it's producing paradox much worse than grandchild killing grandpa after travel in time: Electron has mass, Positron has mass, but result of their collision is complete massless? XX century Quantum Physics is full of such. Turn on thinking. Let's for a while forget about special relativity: Let's not, since it has so much experimental confirmation. If we will assume that either E=m*c*c is true, and E=h*v is true then: E=h*c/wave length m*c*c = h*c/wavelength m=(h/wavelength)/c Now find the smallest possible frequency, and you will have theoretical mass of photon. or reverse- find the largest wavelength: m=h*v/c^2 Or we can go further: c=v*wavelength so m=h*v/(v*wavelength)^2=h/(wavelength*wavelength*v) If frequency is 1, then wavelength=c/1 If wavelength is 1, then frequency=c/1 Reversed: E=m*c*c= m * v*wavelength * v*wavelength E=h/(wavelength*wavelength*v) * v*wavelength * v*wavelength E=h*c/wavelength E=h*wavelength*v/wavelength = h * v The equation is E^2 = p^c^2 + m^2c^4 The E=mc^2 of relativity is only true for a particle at rest (which a photon never is)
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