Talanum46 Posted December 13 Report Posted December 13 (edited) If the particle quantum numbers are not recorded in the particle but in the field of the particle, then there must be a "Field Descriptor" somewhere encoded that records all the quantum numbers that particles of the field has in common. This still leaves the speed to be recorded in the particle, otherwise how are you going to couple a particle velocity to a particular excitation? For an Electron Field this means the Field Descriptor must be a function of type (regular or anti-) and of handedness (Left or Right). Just where is the Field Descriptor recorded, and how is it instantiated? Then spacetime must refer to this descriptor each time a particle's quantum numbers are to be computed about. Edited December 14 by Talanum46
Talanum46 Posted December 13 Author Report Posted December 13 If the quantum numbers are instead encoded into the particle, they are present right there at the interaction to be manipulated mechanically.
Talanum46 Posted December 13 Author Report Posted December 13 (edited) For a boson to interact with a particle it must have a key quantum number to couple to the target particle. For the Weak Interaction this key quantum number is Weak Isospin. Then my decay of pi-minus description needs to include that a virtual W-minus boson couples to the pi-minus (-1 isospin) and cause it to transfer -1/2 isospin to the electron and electron anti-neutrino. Edited December 13 by Talanum46
Talanum46 Posted December 14 Author Report Posted December 14 This ref. shows that masses are needed to be given numbers (names). This is equivalent to giving them quantum numbers. Ref. timestep 1:03:52:
OceanBreeze Posted December 14 Report Posted December 14 (edited) 17 hours ago, Talanum46 said: For a boson to interact with a particle it must have a key quantum number to couple to the target particle. For the Weak Interaction this key quantum number is Weak Isospin. Then my decay of pi-minus description needs to include that a virtual W-minus boson couples to the pi-minus (-1 isospin) and cause it to transfer -1/2 isospin to the electron and electron anti-neutrino. While I don’t agree with everything you wrote in this thread, I will say that your posts are improving. My suggestion: Learn everything you can about what is known about particle physics according to the Standard Model. Once you have that well understood, you will find there is considerable room for improvement of our knowledge, very likely including physics not included in the present Standard Model. As long as you follow a reasoned approach and avoid wild speculation, your posts will be informative and welcomed on this forum. In reply to your statement: “Then my decay of pi-minus description needs to include that a virtual W-minus boson couples to the pi-minus (-1 isospin) and cause it to transfer -1/2 isospin to the electron and electron anti-neutrino.” Pions are weakly interacting particles which can decay into two or three other particles. The specific particles involved depend on the type of pion and the type of lepton. The most common decay mode is the + pion decaying into a muon and a neutrino (π⁺ → μ⁺ν), less commonly the - pion decays into an electron and a neutrino (π⁻ → e⁻ν). These decays occur through the weak interaction. Not mentioned earlier is weak isospin; as you have correctly mentioned, weak isospin interactions require virtual gauge bosons (W +,W−, W0) to mediate transformations between fermions with half-integer weak isospin charges. Keep on researching! Edited December 14 by OceanBreeze
Talanum46 Posted December 14 Author Report Posted December 14 (edited) That picture is wrong. A d quark does not transform into a u quark and emit a W-minus. Although it is a good first guess and economical and straightforward and I cannot blame them for guessing it, it does not reflect good physical logic: I propose that quarks can't change flavor and this is in line with us not seeing particles transform: electrons stay electrons and does not change into positrons. No, what happens here is only that quarks swop position: a u-anti-u starts existing, then the u and d swop positions resulting in a proton and anti-ud, by the strong interaction. Then a virtual W-minus binds with the anti-ud and transform it into an electron and electron anti-neutrino by rearranging quantum numbers onto circles in Riemann Spheres and anti-Spheres (actually half-Riemann spheres, half-anti-spheres). A W-minus cannot bind with a d quark since the d quark does not have the correct Weak Isospin. I have studied until boredom. Edited December 14 by Talanum46
OceanBreeze Posted December 14 Report Posted December 14 50 minutes ago, Talanum46 said: That picture is wrong. A d quark does not transform into a u quark and emit a W-minus. Although it is a good first guess and economical and straightforward and I cannot blame them for guessing it, it does not reflect good physical logic: I propose that quarks can't change flavor and this is in line with us not seeing particles transform: electrons stay electrons and does not change into positrons. No, what happens here is only that quarks swop position: a u-anti-u starts existing, then the u and d swop positions resulting in a proton and anti-ud, by the strong interaction. Then a virtual W-minus binds with the anti-ud and transform it into an electron and electron anti-neutrino by rearranging quantum numbers onto circles in Riemann Spheres and anti-Spheres (actually half-Riemann spheres, half-anti-spheres). A W-minus cannot bind with a d quark since the d quark does not have the correct Weak Isospin. I have studied until boredom. No, that picture is not wrong. It is a correct depiction of Quantum Field Theory. The down quark does transform into an up quark. During this transition, the down quark emits a W¯ boson which allows for charge conservation. The W¯ boson is unstable and decays into other particles; in this case an electron (charge -1) and a neutral anti-electron neutrino. I suggested you learn all you can about currently accepted particle physics before you start spouting off your own unfounded “theories”. Oh well, it seems you just can’t restrain yourself. This thread will most likely need to be locked soon to prevent you spreading more misinformation on this forum. You are walking on egg shells now.
Talanum46 Posted December 14 Author Report Posted December 14 (edited) How does a d quark (Isospin = -1/2) bind with a W-minus (Isospin = -1)? It can't because Isospin is not the same. The two particles must overlap exactly in Isospin. It can react with an anti-ud (Isospin = -1). Edited December 14 by Talanum46
OceanBreeze Posted December 15 Report Posted December 15 16 hours ago, Talanum46 said: How does a d quark (Isospin = -1/2) bind with a W-minus (Isospin = -1)? It can't because Isospin is not the same. The two particles must overlap exactly in Isospin. It can react with an anti-ud (Isospin = -1). Remember to distinguish between Isospin and Weak Isospin. Weak isospin is a quantum number relating to the electrically charged part of the weak interaction. T3 is the third component of the weak isospin, which determines the strength of a particle's interaction with the weak nuclear force. Particles with half-integer weak isospin can interact with the W ± bosons; particles with zero weak isospin do not. Do your own research!
Talanum46 Posted December 15 Author Report Posted December 15 (edited) How does that transformation happen? For example how does the charge of -1/3 e transform into 2/3 e? It means an added node on a solid circle must transform into two added nodes on a left-out circle. It can't transform like this. How do you specify a mechanism in the W-minus to do this? See attached paper for these details. My way the nodes and left-out nodes just have to rearrange - this is more physically logical. Defining Particles 2 computerised computation.pdf Edited December 15 by Talanum46
Talanum46 Posted December 16 Author Report Posted December 16 (edited) If such a creation of a node is required then energy can't be conserved in the interaction. Edited December 16 by Talanum46
OceanBreeze Posted December 18 Report Posted December 18 On 12/15/2024 at 6:07 PM, Talanum46 said: How does that transformation happen? For example how does the charge of -1/3 e transform into 2/3 e? It means an added node on a solid circle must transform into two added nodes on a left-out circle. It can't transform like this. How do you specify a mechanism in the W-minus to do this? See attached paper for these details. My way the nodes and left-out nodes just have to rearrange - this is more physically logical. Charge conservation has already been explained to you: When a down quark transitions into an up quark, it involves the weak nuclear force, one of the four fundamental forces in nature. This process is mediated by the emission of a W¯ boson which is the force carrier for the weak interaction. The emission of the W¯ boson allows for charge conservation. The down quark, with a charge of -1/3, emits a W¯ boson (which has a charge of -1) and becomes an up quark (which has a charge of +2/3). The overall charge before and after the interaction remains the same. (-1/3 e). (I presume you can add -1 to + 2/3) Energy conservation is more difficult to comprehend: The mass energy of the W− boson is about 80 GeV, so it cannot possibly emerge from the nucleus as there are only a few MeV of energy available. The Heisenberg uncertainty principle says the uncertainty in the energy (ΔE) times the uncertainty in the time (Δt) is greater or equal to ћ. (ΔEΔt ≥ ћ ) Here ћ = h/2π = 1.05 × 10-34 J-s. h is of course the Planck constant (6.63 x 10-34 J-s). As you should already know, the uncertainty principle states that some pairs of physically observable parameters cannot be precisely measured simultaneously to within arbitrary accuracy. So, in accord with energy-time uncertainty, the W− boson is unstable and will quickly decay into other particles, such as a lepton (electron with charge -1) and a neutrino (charge 0). To summarize, In the beta-minus decay process: electric charge is conserved the number of quarks minus the number of antiquarks is conserved the number of leptons minus the number of antileptons is conserved flavor changing of quarks or leptons is allowed If you want a complete a derivation of the energy-time uncertainty, I suggest you do your own research first. If you do that, then and only then will I engage with you further on this subject. If you continue to make unfounded and frankly ridiculous assertions, about strange nodes on circles and Riemann spheres, this thread will be closed.
Talanum46 Posted December 18 Author Report Posted December 18 (edited) 1 hour ago, OceanBreeze said: flavor changing of quarks or leptons is allowed No. 1 hour ago, OceanBreeze said: The emission of the W¯ boson allows for charge conservation. The down quark, with a charge of -1/3, emits a W¯ boson (which has a charge of -1) and becomes an up quark (which has a charge of +2/3). The overall charge before and after the interaction remains the same. (-1/3 e). You will run into problems as soon as you try to represent the charge (explain it, or try to encode it into the particles). You cannot encode it into the particle by giving it a number in your mind or on paper, it must be encoded physically: by actualized numbers. You can't couple an abstract quantity to a particle: physics reads the particle, not your mind or a paper. Edited December 18 by Talanum46
OceanBreeze Posted December 18 Report Posted December 18 2 hours ago, Talanum46 said: No. You will run into problems as soon as you try to represent the charge (explain it, or try to encode it into the particles). You cannot encode it into the particle by giving it a number in your mind or on paper, it must be encoded physically: by actualized numbers. You can't couple an abstract quantity to a particle: physics reads the particle, not your mind or a paper. sigh. At least nobody can say I didn't try to reason with you and give you a chance to drop your presumptuous assertions. Enough is enough. Thread Closed. Warning Issued Suspension being considered.
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