Subatomic Particle Mass/Stability Spectrum
in [Physics]
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From: Robert L. Oldershaw on 13 May 2010 17:51 On May 13, 4:22 pm, "Al.Riv...(a)gmail.com" <al.riv...(a)gmail.com> wrote: > > Still, it is a bit confusing that you keep speaking of spin and > angular momentum. All the particles you are using in your postdiction > have spin either 1/2, 1 or 3/2, while your sequence goes from 1/2 up > to 7. ----------------------------------------------- As I duly noted in a recent post, when you go from the basic [sqrt n] multiplier to the more refined [sqrt {j(j+1)/a}], then all you need are n and a values in the 1/2, 1 or 3/2 range. See if you can figure out how to work this. Don't forget that most of the particles are UNstable. I will read your post later tonight, but it is clear that you are not discussing these matters objectively and scientifically. You will defend your substandard paradigm to the end, but I don't think you are going to like that end. RLO www.amherst.edu/~rloldershaw <----NEW PARADIGM HERE - NO CHARGE
From: Al.Rivero on 13 May 2010 18:43 On 13 mayo, 23:51, "Robert L. Oldershaw" <rlolders...(a)amherst.edu> wrote: > On May 13, 4:22 pm, "Al.Riv...(a)gmail.com" <al.riv...(a)gmail.com> wrote: > > > Still, it is a bit confusing that you keep speaking of spin and > > angular momentum. All the particles you are using in your postdiction > > have spin either 1/2, 1 or 3/2, while your sequence goes from 1/2 up > > to 7. > > ----------------------------------------------- > > As I duly noted in a recent post, when you go from the basic [sqrt n] > multiplier to the more refined [sqrt {j(j+1)/a}], then all you need > are n and a values in the 1/2, 1 or 3/2 range. {j(j+1)/a} = 1/2 3/4 4/3 3/2 2 5/2 15/4 4 15/2 = 0.5, 0.75, 1.33, 1.5, 2, 2.5, 3.75, 4, 7.5 Now you do not predict Xi (1530) anymore, so I hope that you agree with me that such prediction, in your previous model, was substandard. In fact, given that this model predicts a series very different from the first one, I hope you will agree that the two models differ and at least one of them is not the right model. > I will read your post later tonight, but it is clear that you are not > discussing these matters objectively and scientifically. I expect you will point out to me that parts on my answer show why it is clear so.
From: Robert L. Oldershaw on 13 May 2010 21:43 On May 13, 6:43 pm, "Al.Riv...(a)gmail.com" <al.riv...(a)gmail.com> wrote: > > {j(j+1)/a} = 1/2 3/4 4/3 3/2 2 5/2 15/4 4 15/2 > = 0.5, 0.75, 1.33, 1.5, 2, 2.5, 3.75, 4, 7.5 > > Now you do not predict Xi (1530) anymore, so I hope that you agree > with me that such prediction, in your previous model, was > substandard. ------------------------------ Jeez, you are a SLOW STUDY! For 8 of the 10 major mass/stability peaks (100-1860 MeV) I use: (1) ONLY (n) VALUES OF 1/2, 2/2, 3/2 AND 4/2. (2) ONLY (a) VALUES BETWEEN 0 and 4/3 (note: a couple exceed 1 because they are unstable paticles). THE AVERAGE AGREEMENT IS 99%. THE SUBSTANDARD MODEL CHEATS TO GET FUDGED RESULTS AT THE 95% LEVEL. The peak at ~1500-1535 MeV IS RETRODICTED AT THE 98.6% LEVEL USING n = 2/2 and a = 2/5, YIELDING M = 1509 MeV AS A 1ST APPROX. DO YOU HEAR ME NOW? RLO www.amherst.edu/~rloldershaw <--- unification of hep+QM+GR
From: Al.Rivero on 14 May 2010 03:20 > (1) ONLY (n) VALUES OF 1/2, 2/2, 3/2 AND 4/2. >>> As I duly noted in a recent post, when you go from the basic [sqrt n] >>> multiplier to the more refined [sqrt {j(j+1)/a}], then all you need >>> are n and a values in the 1/2, 1 or 3/2 range. 4/2 is not in the 1/2-3/2 range > (2) ONLY (a) VALUES BETWEEN 0 and 4/3 (note: a couple exceed 1 because > they are unstable paticles). 0 is not 1/2-3/2 range :-DDDD In fact, it is excluded, so lets take it as a typo. are you telling me that you can choose freely the value of a, in this interval? >>> As I duly noted in a recent post, when you go from the basic [sqrt n] >>> multiplier to the more refined [sqrt {j(j+1)/a}], then all you need >>> are n and a values in the 1/2, 1 or 3/2 range. > DO YOU HEAR ME NOW? Your third sequence of numbers in one week? Yes I hear. Pity that you dont ever read your own posts before to press the "send" button, it could be a lot more easy to follow you. It could be more useful if you write the final series, if it is only 8 values, it is not goint to take you a lot of time, and we will see what values of 'a' do you choose for each particle.
From: Robert L. Oldershaw on 14 May 2010 13:06
On May 14, 3:20 am, "Al.Riv...(a)gmail.com" <al.riv...(a)gmail.com> wrote: > plenty of nothing ------------------------------- What we are doing is a FIRST APPROXIMATION of subatomic particles in the 100-1860 MeV mass range using Kerr black holes as our model and Discrete Scale Relativity to give us the correct values for G and (hc/ G)^1/2, the Planck mass. Why this mass range? Because it is the most important set of masses for explaining subatomic physics. [Important note: the electron is conspicuously missing because it is still a mystery to me, as well as tout le monde. It's in the to-be-done pile.] I started by driving a ultra-simple mass formula: M = (sqrt n)(674.8 MeV) See: http://arxiv.org/ftp/arxiv/papers/1002/1002.1078.pdf for details. Then I played around with a slight refinement: M = (sqrt n/a)(674.8 MeV) [note "a" is from the Kerr J vs M^2 relation]. Finally, I got around to doing what I probably should have done in the first place: used J = j{j+1}(h-bar) instead of J = n(h-bar). This gives you: M = (sqrt [j{j+1}/a])(674.8 MeV). For 8 of the 10 major mass/stability peaks (100-1860 MeV) I use: (1) ONLY (j) VALUES OF 1/2, 2/2, 3/2 AND 4/2. (2) ONLY (a) VALUES BETWEEN 0 and 4/3 (note: a couple exceed 1 because they are unstable paticles). THE AVERAGE AGREEMENT IS 99%. THE SUBSTANDARD MODEL CHEATS TO GET FUDGED RESULTS AT THE 95% LEVEL. The peak at ~1500-1535 MeV IS RETRODICTED AT THE 98.6% LEVEL USING n = 2/2 and a = 2/5, YIELDING M = 1509 MeV AS A 1ST APPROX. DO YOU UNDERSTAND ME NOW? Or are you paid to NOT understand me? RLO www.amherst.edu/~rloldershaw |