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**CarlB** · Mar28-07, 06:34 AM

Hans, I've been playing around with the numbers and now I see that I was too hasty to judge your work as numerology. (I didn't say so, but that was what I was thinking.) In fact, now that I understand the method better, I think I can contribute to this exciting branch of phenomenology. Before, I just had trouble seeing what the heck an exponential could be doing in the mass spectrum.

First, I guess I should mention some theory that drove me in this particular direction. The fact is that there is a lot of periodicity in the elementary particle masses having to do with 3s. 3 is sort of midway between pi and e. The next important number smaller than e is 5/2. Also, 5/2 is the average of the two smallest prime numbers, which suggests that p-adic field theory could be important here, just like the string theorists say. And I like to think of QFT as a probability related theory so equations such as

$LaTeX Code: 5/2 = 1 + \\sum_n3^{-n}$

naturally led me to explore the use of 5/2 in the elementary particles. Enough for the theory (maybe it still needs some work); so here's my formula:

$LaTeX Code: m_{N,M} = (5/2)^{3(N - M/(4\\pi^2))}.$

This formula works very well for the charged leptons, with the electron, muon and tau taking N=0,2,3, and M= 0, 2, and 1, respectively. I need not point out how suggestive these small constants are! The experimental and calculated exponents are:

$LaTeX Code: \\begin{array}{rcc|l|l|l|} &N&M& 3(N+M/(4\\pi^2)) & \\log_{5/2}(m/m) & accuracy \\\\ \\hline \\mu/e &2& 2& = 5.848018& 5.818676 & 0.00501\\\\ \\tau/e &3& 1& = 8.924009& 8.898993 & 0.00280\\\\ \\tau/\\mu&1&-1& = 3.075991& 3.0803163& 0.00141\\\\ \\hline \\end{array}$

The above makes a grid with 4x3 = 12 boxes, three of which are filled. This is suggestive, especially when you look at those tight accuracy figures and the very convincing division by pi. But I don't think that this is nearly enough to write a paper on.

The real test is to see if we can put the other charged particles into the same formula. Admittedly, one might suppose that the color force, being a sort of charge, would contribute something to the mass of a quark or baryon, who reallly knows? In phenomenology, it makes sense to boldy go where no sane man has gone before and simply see what particles naturally fit together.

$LaTeX Code: \\begin{array}{|ccc|c|c|c|c|c|c|c|c|c|c|c|c|c|} \\hline &&&&&&&&&&&&&&&\\\\ &&M=&0&1&2&3&4&5&6&7&8&9&10&11&12\\\\ \\hline &N=0&&&&e&&&&&&&&&&\\\\ \\hline &N=1&&3d&&&&&&&&&&3u&&\\\\ \\hline &N=2&&\\pi^+&&&&\\mu&&&&&&&&\\\\ \\hline &N=3&&&&&\\tau&\\Omega&&&\\Xi&\\Delta&\\Sigma&\\Lambda&& p\\\\ \\hline \\end{array}$

In the above, I'm not so sure about the up and down quark. The masses aren't known very accurately. I've made the assumption that the masses that one needs to use are at the low end of the PDG figures. And since the quarks also have the color force (as well as E&M), I've multiplied their masses by 3.

Only one meson fell into the fit, but it is the most fundamental one. I guess this is evidence that the mesons really are a mess. But I could fit all of the basic (i.e. no charm, bottom or top) low lying baryons. Furthermore, no two particles ended up in the same slot, which is cool.

Some of the fits that I checked are remarkably good. For example, the Lambda/e accuracy is 0.00013 and the 3u/Omega is 0.00041. That last ratio really is stunning accuracy given that the mass of the up is listed as 1.5 to 4.0 MeV in the PDG! I'm not sure about how this happened, I'm just crunching numbers on my calculator.

Anyway, the formula fits all the principle baryons, all the charged leptons, the lightest quarks, and the lightest meson.

Originally Posted by Hans de Vries

There are only 30 grid positions which are the same for each result. 6 of the 30 are hits. I should have discounted the input space only once and not 6 times. This brings the prediction back to 61-log2(30)~56 bits

I'm kind of a dummy and I don't really see how your calculation here makes any sense. But since I've got 12 hits in 52 grid positions, it does look like this should be okay. In addition, all the fermions are bunched together on one side of the diagram so it kind of makes me wonder if it wouldn't be useful to define a "baryon bit" and use three quantum numbers to define the mass exponent.

Originally Posted by Hans de Vries

The expression itself, discounting the small integers, is about 10 bit, which is three times a basic operation like +-x/ (two bit each) plus the use of a single elementary constant (pi) which we presume to be in small group together with the small integers. (3-4 bit) This leaves us with ~45 bits prediction which is three to four times more as the alpha result.

Since I've got 12 masses listed I bet that I should be really positive. Since I was employed for 15 years designing digital logic, and one of my specialties was coding theory, it really bothers me that I don't understand your bit calculations. Maybe I should hit the books -- I learned theory back in the stone age when we mostly used our fingers. Heck, I still remember punch card codes.

I find it hard to believe that you could send someone your expression in just 10 bits. Of course when mapping expressions to bit sequences, it makes sense to code the important expressions in smaller number of bits.

My HP calculator is fairly efficient general purpose scientific calculator. It has over 32 keys so pressing anything that can be coded in one key stroke costs 6 bits. Looking up a data value takes two keystrokes or 12 bits. Pressing the <enter> key costs 6 bits.

Originally Posted by Hans de Vries

It was like this until I decided to put the Vacuum expectation Value at the origin of the grid.

Yes, the factor of 2 was brilliant! I'd have never had the idea to triple the quark masses if you hadn't led the way.

enjoy,
Carl