All the lepton masses from G, pi, e

Hans de Vries · Mar 26, 2007

A mass relation for all six principal charge 1 particles: \begin{array} {|ccc|c|c|c|c|c|c|c|} \hline & & & & & & & & & \\ & & \ \ &\ \ \frac{0}{\pi}\ \ &\ -\frac{1}{\pi}\ &\ -\frac{2}{\pi}\ &\ -\frac{3}{\pi}\ &\ -\frac{4}{\pi}\ &\ -\frac{5}{\pi}\ &\ -\frac{6}{\pi}\ \\ & & & & & & & & & \\ \hline &0\ \pi & &2VeV&\cdot&\cdot&\cdot& & &\cdot\\ \hline &1\ \pi & &\cdot&\cdot&\cdot&\cdot& W &\cdot&\cdot\\ \hline &2\ \pi & & p &\cdot&\tau &\cdot&\cdot&\cdot&\cdot\\ \hline &3\ \pi & &\cdot&\cdot&\cdot& \mu & \pi^\pm &\cdot&\cdot\\ \hline &4\ \pi & &\cdot&\cdot&\cdot& &\cdot&\cdot&\cdot\\ \hline &5\ \pi & & & &\cdot&\cdot&\cdot&\cdot& e \\ \hline \end{array} We can put all six principal charge 1 particles in a simple 2D grid.
All grid positions with "." are forbidden via a simple rule that says:

"No two pair of particles may the same mass ratio." The log mass ratio calculation is: Y\pi -X/\pi, where X and Y are the
2D grid's axis. The origin of the grid is 2VeV. (Vacuum expectation Value)Some examples:

1) Electron-Proton mass ratio: log(1836.1526726) (natural log)
7.515427 = experimental
7.514918 = calculated = 3\pi -6/\pi
accuracy: 0.00006772) Electron mass ratio with 2VeV (Vacuum exp.Value): log(963699)
13.77853 = experimental
13.79810 = calculated = 5\pi -6/\pi
accuracy: 0.001423) Electron-Muon mass ratio: log(206.7682838)
5.331598 = experimental
5.328255 = calculated = 2\pi -3/\pi
accuracy: 0.0006274) Proton-Pion mass ratio: log(6.72258237)
1.905472 = experimental
1.868353 = calculated = \pi -4/\pi
accuracy: 0.019865) Electron W-boson mass ratio: log(157387)
11.96646 = experimental
11.92975 = calculated = 4\pi -2/\pi
accuracy: 0.00307Try it yourself!

Regards, Hans.

 \begin{array} {|clc|c|rc|} \hline & & & & & \\ & electron & & e & 0.51099892(40)& MeV \\ & muon\ lepton & & \mu & 105.658369(9) & MeV \\ & tau\ lepton & &\tau & 1776.99(29) & MeV \\ & pion\ \pm & & \pi & 139.57018(35) & MeV \\ & proton & & p & 938.27203(8) & MeV \\ & W boson & & W & 80398(25) & MeV \\ & & & & & \\ \hline \end{array}

CarlB · Mar 26, 2007

Hans de Vries said:

Try it yourself!

I've violated our tradition of ignoring ideas that don't immediately strike us as useful for our own silly ideas and tried it myself.

Your accuracy numbers are exaggerated. You're using the natural logs of the masses. And your formula is linear in those logs. So there's no reason to divide by the logarithm.

You are not fitting to a linear formula like y = ax + b where you would be testing for, for example, b = 0. In such a case it would make sense to divide the error by y because you would want a relative error.

Instead, you are fitting a sort of Diophantine equation and the errors are compared to integers. Another way of putting this is that the way you are calculating errors, the larger the ratio, the more accurate your ratios will be. But the steps between different choices of m-n/\pi are constant, so the expected error does not depend on the ratio.

Here, let me make an error calculation the way you are doing it. The year is 2007, which turns out to have approximately a worst case error for fitting to stuff with a small value for n/\pi. We find that:

639*\pi - 1/\pi = 2007.16

The way you are calculating errors, this would have an accuracy of

0.16/2007 = 0.0000794

Do you really want to claim this for a fit to 2007? By the way, a better approximation for 2007 is:

639*\pi - 1.5/\pi = 2007.00024

Let me put it this way, if you were claiming that the ratios were integers, then the error would obviously be the difference between a ratio and the nearest integer. The worst you could do would be 1/2, and this is the number you should divide your errors by, not for example, 2007 or whatever the nearest integer.

When one describes the points on the real line of the form n\pi - m/\pi for small values of n and m, one finds that they are mostly separated by 1/\pi = 0.3183. When one assigns random numbers to the nearest one of these, the worst one can do is half the interval, or 1/(2\pi) = 0.1591 so this is the number you need to divide the differences by, not the log of the mass ratio.

With these changes, the mass ratios you've listed have errors as follows:

0.3%
12.3%
2.1%
23.3%
23.1%

I think that this is impressive enough as it is. What I would like to see is the results of a computer program that can find these sorts of fits, and see various randomized data thrown at it.

Now the other thing I wanted to point out is that the mass formula you have here is not at all incompatible with the Koide formula or the formulas that I've described. Furthermore, the Koide formula ends up putting an angle of 0.22222204717(48) radians into an exponential in the mass matrix for the charged leptons. This suggests that taking natural logs of masses is likely to be a useful thing to do.

Carl

Hans de Vries · Mar 26, 2007

CarlB said:

Your accuracy numbers are exaggerated. You're using the natural logs of the masses. And your formula is linear in those logs. So there's no reason to divide by the logarithm.

Carl,

These accuracies correctly describe the number of prediction bits
according to information theory. Don't forget that a dynamic range is
needed next to the precision. In terms of floating point numbers:
You need an "exponent" as well as a "mantissa".

Let us simply do the calculation: Take example 1:

Hans de Vries said:

1) Electron-Proton mass ratio: log(1836.1526726) (natural log)
7.515427 = experimental
7.514918 = calculated = 3\pi-6/\pi
accuracy: 0.0000677

The number of bits predicted is -log2(0.0000677) = 13.850 ~ 14 bits.

If I would express the number exp(7.514918) = 1835.2181 which has an
accuracy of 0.000509, as a floating point number, then I would need:

-log2(0.000509) = 10.939 ~ 11 bits for the mantissa,

but I also need another 4 bits for the exponent which would have the
value 10 for 2^10 = 1024 to express the range between 1024 and 2048
in which the value 1835.2181 falls.

To be entirely exact: For the exponent I need -log2(10) = 3.321 bits.

Now 10.939 + 3.321 bits ~ 14 bits or the same number of bits as in the
case of the logarithm!

The total number of predicted bits in the 5 relations is 50. There are
six independent relations which provide a total of 61 bits. Good for
18.4 correct decimal digits.

Now we are also inputing information here which we need to subtract.
These are the grid positions. The information we provide for the six
relations = 6 x( log2(5) + log2(6)) = 29.441 bits.

Subtract these from the 61 bits and we are left with 31.6 bits which is,
by far, the best result I've presented on this entire thread until now...

Now here is something else very interesting. The exclusion rule: Note
that almost all grid positions are forbidden by it and a random placement
would very likely break the rule. Thus, the 29 bits number would be
much lower if the rule is correct.

Regards, Hans

whatta · Mar 26, 2007

this thread needs only one number, 42. this will probably be the page number when it gets locked.

Hans de Vries · Mar 26, 2007

whatta said:

this thread needs only one number, 42.

Dear Whatta,

This thread was initiated to allow, (but also to contain) posts with
the purpose of the archival of numerical coincidences. These
post are allowed (here on this thread) within the following restrictions:

1) The reported numerical coincidences should be independent of
the units used (meters, kg..). Only coincidences with dimensionless
numbers are allowed.

2) The numerical coincidences must be independent of the number
system used, like, decimal numbers, binary, hexadecimal.

3) The reported numerical coincident should have a sufficient
"predictability". That is, It should produce significantly more result
bits as the number of bits used as input.Kind Regards, Hans de Vries

CarlB · Mar 26, 2007

Hans de Vries said:

These accuracies correctly describe the number of prediction bits according to information theory.

I would like to think that information theory is appropriate for the calculation of a fit, but I suspect that even a well accepted calculation like the g-2 value for the electron will fail an information theory test. That is, the amount of bits required to describe the thousands of Feynman diagrams, or the rules to generate those Feynman diagrams along with coupling constants plus etc., etc., etc., will be greater than the amount required to simply code g-2 to experimental error.

But if you insist on using information theory, you need to count the number of bits needed to describe your formula, along with the number of bits used to describe the exponent and the number of bits needed to describe the mantissa. This will blow up the accuracy. If the formulas really were that good they wouldn't be ignored so much. You haven't faulted my calculation for the approximation of e^{2007} in the form e^{n\pi - m/\pi} because it was correct.

The calculation I provided has the useful feature that for masses distributed uniformly over relatively small ratio spaces, for example, if the mass ratio is from e^{12.5} < R < e^{13.0}, it will give approximately the correct probability for getting a hit, in that the error will be approximately uniformly distributed from 0 to 100%. This feels to me like the right way of looking at it.

There are some other things I forgot to mention. The vertical column would make more sense to me if it was turned upside down relative to the pi values. The way it's set up, increasing values of the n cause decreases in masses and that is counterintuitive. So the electron would have the 0\pi rather than 5\pi.

Second, I'm not sure what "principle" means with respect to charge 1 particles. Why not include the \Delta^+? Did you try to fit other charge +1 particles and they didn't fit or what? I guess there are most of a thousand meson / baryon resonances and excitations with charge +1.

Carl

Hans de Vries · Mar 26, 2007

CarlB said:

But if you insist on using information theory, you need to count the number of bits needed to describe your formula.

I see I was too conservative on the input bits side.

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

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.

CarlB said:

There are some other things I forgot to mention. The vertical column would make more sense to me if it was turned upside down relative to the pi values. The way it's set up, increasing values of the n cause decreases in masses and that is counterintuitive. So the electron would have the 0\pi rather than 5\pi.

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

CarlB said:

Second, I'm not sure what "principle" means with respect to charge 1 particles. Why not include the \Delta^+? Did you try to fit other charge +1 particles and they didn't fit or what? I guess there are most of a thousand meson / baryon resonances and excitations with charge +1.

Many, many indeed, and there are only 30 grid positions...

It's more that the particles select them self by fitting. The resonances
and excitations don't fit but the base states do. Hadrons with masses
dominated by a single (fractional charge) quark probably won't fit either
but I have yet to try.

Heck, this formula is one of the subjects of the very first post in this
thread and I never even bothered to try a hadron mass because of
me being convinced that these can only have incredibly complicated
QCD determined mass values.

But then, looking at the infamous 'proton spin crisis', something can be
incredibly complex inside, and then, all the small spin contributions from
the quarks, the gluons, the sea-quarks and all the various angular
momenta add up to a very simple number determined by geometry only.

Regards, Hans

CarlB · Mar 28, 2007

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

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:

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:

\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.

\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.

Hans de Vries said:

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.

Hans de Vries said:

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.

Hans de Vries said:

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

whatta · Mar 28, 2007

was that post sarcastic

Kea · Mar 28, 2007

CarlB said:

...which suggests that p-adic field theory could be important here...

Yes, Carl. A nice paper on this point is

On the universality of string theory
K. Schlesinger
http://arxiv.org/abs/hep-th/0008217

which introduces the concept of a tower of quantizations. All of String theory fits into the prime 3 on the tower, from the point of view of the true M Theory. In other words, the unified theory can address all scales on an equal footing. The classical landscape is Mathematics Itself.

How do you like that, whatta?

kneemo · Mar 29, 2007

Kea said:

On the universality of string theory
K. Schlesinger
http://arxiv.org/abs/hep-th/0008217

which introduces the concept of a tower of quantizations. All of String theory fits into the prime 3 on the tower, from the point of view of the true M Theory. In other words, the unified theory can address all scales on an equal footing. The classical landscape is Mathematics Itself.

T. Pengpan and P. Ramond showed in http://arxiv.org/hep-th/9808190" that the 11D supergravity triplet of SO(9) representations sits at the base of an infinite tower of irreps of SO(9), describing an infinite family of massless states of higher spin. They muse that such higher-spin states describe degrees of freedom of M-theory.

CarlB · Mar 29, 2007

whatta said:

was that post sarcastic

Somehow I forgot to give the calculations for the absolute errors and the masses of the predicted particles. Of course these aren't as pretty as the selected ratios, but are probably more of an indication of how good the fit is:

Calc = 0.5743319321193890 MeV * {(5/2)^{Exp}}

\begin{array}{cccc} Exponent&particle&Mass&Calc/Mass\\ 3(0 + 2/(4\pi^2))&e&0.51099892&0.9778\\ 3(1 + 10/(4\pi^2))&3u&4.5&0.9940\\ 3(1 + 0/(4\pi^2))&3d&9.0&0.9971\\ 3(2 + 4/(4\pi^2))&\mu&105.658369&1.0044\\ 3(2 + 0/(4\pi^2))&\pi^+&139.57018 &1.0046\\ 3(3 + 12/(4\pi^2))&p&938.27203&1.0126\\ 3(3 + 10/(4\pi^2))&\Lambda&1115.683&0.9788\\ 3(3 + 9/(4\pi^2))&\Sigma&1189.37&0.9843\\ 3(3 + 8/(4\pi^2))&\Delta&1232.0&1.0188\\ 3(3 + 7/(4\pi^2))&\Xi&1321.31&1.01845\\ 3(3 + 4/(4\pi^2))&\Omega&1672.45&0.9915\\ 3(3 + 3/(4\pi^2))&\tau&1776.99&1.0005 \end{array}

Getting back to the charged leptons, their masses (and ratios) are exact numbers and one expects that to store them requires an infinite number of binary bits. Our experimental measurements, on the other hand, are inexact, so one can certainly expect that data they contain can be easily compressed.

The charged lepton mass numbers run over a very wide ratio and so the natural way to compress them is by a power series. Koide's formula uses a square root, which is also a compression method and so at first I suspected it as well.

But if Koide were looking for a compression formula rather than physics, he would have done well to begin with the 7th root of masses rather than the square root. Then the masses of the charged leptons are fairly close to 1^7, 2^7, and 3^7, and one could write a generation formula in the form

m_n = (n+f(n) )^7

for n=1,2,3. In fact, I'm quite certain that I could find such a formula, and then bend it around to pick up the baryon masses which form such a convenient linear series. But I think my point here is made.

In the face of how easy it is to find compression algorithms for sparse data, where the Koide formla is more convincing is that it is consistent with exactly three generations and no more. The short form for the Koide formula is:

\sqrt{m_n} = 1 + \sqrt{2}\cos(2n\pi/3 + 2/9 + \epsilon)

where \epsilon = 0.22222204717(48) - 2/9 and I've left off an overall scaling factor. Since 2(n+3m)\pi/3 = 2n\pi/3 + 2m\pi, the formula gives exactly three masses so there are only three generations implied. These are the electron, muon, and tau for n the generation number, 1, 2, 3. The \sqrt{2} is what Koide found in 1981, the \epsilon is what I found a year ago.

So the Koide formula is exact to experimental error, and it's not really in a form that is obviously convenient for simply hiding a compression algorithm.

Mentz114 · Mar 30, 2007

"It is shown here that the rest energies and magnetic moments of the basic elementary particles are given directly by the corresponding Planck sublevels."

http://uk.arxiv.org/pdf/physics/0611100

Enjoy.

CarlB · Apr 2, 2007

This was released by Hans on sci.physics.foundations:

A non-perturbative derivation of the exact value of the SU(2) coupling value g from the standard Electroweak Lagrangian itself.
Hans de Vries, March 30, 2007
http://chip-architect.com/physics/Electroweak_coupling_g.pdf

I wasted two days playing with the charged particle formulas and ended up quite disgusted and angry. Having learned my lesson, I'm leaving this one for others to analyze.

In short, it relates a smaller number of constants but to a higher accuracy and with simpler formulas, which is about what one would expect.

Hans de Vries · Apr 3, 2007

Hi, Carl

I must say you did impress me with your feverish activity the in last few days.
From experience I know that these extreme bursts of mental activity have a
risk of ending up in the type of exhaustion you're describing here. :blushing:

Relax, Rome wasn't build in a day, they say. I was still considering a
response before starting this subject.

CarlB said:

This was released by Hans on sci.physics.foundations:A non-perturbative derivation of the exact value of the SU(2) coupling value g from the standard Electroweak Lagrangian itself.

Hans de Vries, March 30, 2007
http://chip-architect.com/physics/Electroweak_coupling_g.pdf
In short, it relates a smaller number of constants but to a higher accuracy and with simpler formulas, which is about what one would expect.

Before jumping onto this one I should maybe first point out the "somewhat
vague" mathematical relation between the new paper and the numerical
coincidents in these mass ratios: \mbox{\huge $ e^{\left( m\pi-\frac{n}{\pi} \right) } $}\ \ =\ \mbox{mass ratio numerical coincidents} 

Where m and n should be small integer values. This can be written as a
power expansion like this:

 \mbox{\huge $ e^{\left(m\pi-\frac{n}{\pi}\right)} $}\ \ =\ \sum_{k\ =-\infty}^\infty\ \mbox{\huge J}_k(2\sqrt{nm})\ \left( \pi \mbox{ $\sqrt{\frac{n}{m}}$}\ \right)^k 
This now relates to the core of the new paper:

 \mbox{\huge $ e^{iQ \sin(\omega t)}$}\ \ =\ \sum_{k\ =-\infty}^\infty\ \mbox{\huge J}_k(Q)\ \mbox{\huge $ e^{ik\omega t} $} 

The later is the phase a charged particle acquires in a sinusoidal
electromagnetic (electroweak) field. This is a superposition where
the Bessel coefficients can be interpreted as amplitudes:

J₀(x) = amplitude to absorb 0 quanta
J₁(x) = amplitude to absorb 1 quanta
J₂(x) = amplitude to absorb 2 quanta
J₃(x) = amplitude to absorb 3 quanta
...

These Bessel coefficients have the unique property that they
are Unitary for both amplitudes as well as probabilities
for any value of Q:

 \sum_{k\ =-\infty}^\infty \mbox{\huge J}_k(Q)\ = 1, \quad \ \ \sum_{k\ =-\infty}^\infty \left|\mbox{\huge J}_k(Q)\right|^2\ = 1 Regards, Hans

PS: Try the link below. There's a lot on these Bessel coefficients
in regard with frequency modulation on the internet:

http://images.google.nl/images?hl=e...gle+Search&ie=UTF-8&oe=UTF-8&um=1&sa=N&tab=wi

CarlB · Apr 4, 2007

I guess I should probably update three things.

First of all, I've been using the constant 17.716 sqrt(MeV). Squaring to get to the units everyone else uses, this is 313.85 MeV.

Nambu uses 35 MeV in his empirical mass formulas. The relation to my 313.85 MeV is that 35 x 9 = 315 MeV. Of course 9 is a power of 3 and powers of 3 are important in my theoretical stuff, the majority of which is not published. Some links for the Nambu theory are:

http://www.google.com/search?hl=en&q=nambu+mass+formula

Also

http://www.arxiv.org/abs/hep-ph/0311031

refers to it as Y. Nambu, Prog. in Theor. Phys., 7, 595 (1952). I've not yet read much on the theory. I'd like to thank Dr. Koide for noting that my mass formulas reminded him of the Nambu stuff.

The Nambu formula has probably been discussed around here but I haven't found it. I'll drop by the local university and read the articles on it sometime in the next week or so. We should discuss the Nambu formulas here or maybe on another thread. Dr. Koide also mentioned the Matsumoto formula, which I've not yet looked up.

Since the mass I'm using comes from the electron and muon masses, I can calculate the "Nambu mass" to much higher accuracy. The number starts out as 34.87 MeV.

Second, on the analogy between the force that composes the electron, muon and tau, and the excitations of the elementary particles: At first I was thinking that the analogy should be strongest when the three quarks making up the baryon were identical, as in the Delta++. But the spin of a Delta++ has to be 3/2 which is different from that of the electron.

In the theory I'm playing with, the 3 preons inside an electron are assumed to be in an S state and can transform from one to another by a sort of gluon. To get that kind of wave function, one should instead look at the baryons that are made up of three different quarks.

Among the low lying baryons, there are two that are composed of one each of u, d, and s. These are the Lambda and Sigma. The charged lepton Koide formula is:

\sqrt{m_n} = 17.716 \sqrt(MeV) (1 + \sqrt{2}\cos(2n\pi/3 + 0.22222204717(48) ))

and the neutrinos by a similar formula (multiplied by 3^{-11}), but with the angle \pi/12 added to the angle inside the cosine. These are the m=0 and m=1 mass formulas listed above, though the neutrinos are not included above.

To get the analogy between the charged leptons and the baryons as close as possible, one naturally looks for a set of three "uds" baryons that have the same angle as the charged lepton mass formula. Such a triple does exist, it is the \Lambda_{3/2-} D03. The triple consists of the \Lambda(1520), \Lambda(1690), \Lambda(2325). Putting these into the Koide formula gives the form:

\sqrt{m_n} = 42.769 + 5.5856 \cos(2n\pi/3 + 0.22186)\;\;\;\sqrt{MeV}

The angle is close to the 0.22222204717, though it cannot be distinguished from 2/9. The other two constants are related to the 17.716 constant approximately as
\mu_v = 42.769 = (1+\sqrt{2}) \;\;\;17.716
\mu_c = 5.5856 = \sqrt{2}\;\;\; 17.716\times 2/9

Making the assumption that these are exact allows one to "predict" the associated resonances as:

\Lambda(1520) = 1520.408
\Lambda(1690) = 1690.673
\Lambda(2325) = 2323.355

These numbers are well within the PDG estimates. This is the only uds excitation that falls in the m=0 class. The other Lambda and Sigma excitations have some interesting numbers as well, but are not as nicely suggestive.

The suggestion is that \mu_v comes from the internal energy of the particles. Looking at a quark as a system, its internal (square root) energy is the 1+\sqrt{2} number in the charged lepton formulas when you ignore the cosine.

The idea here is that if you ignored the color effects and the energy of the stuff that glues them together, all quarks would weigh the same amount. The "1" is the length of the mass vector that the preons differ in, while the sqrt(2) is the length of the mass vector that they share. This sqrt(2) gets modified by the cosine according to how well they cancel their fields. (And the generations arise from glue effects.)

The \mu_s comes from the color force. The color force between quarks is only 2/9 of the force between the preons. One can provide various unconvincing arguments for why this should be 2/9. Suffice it to say that 2/9 shows up fairly frequently in these formulas.

The third thing I need to mention is that I made an error in a calculation for the delta angles from the baryon excitations. I was making calculations by calculator, this was before I coded it up into Java. There were two excitations that gave particularly bad errors in their delta calculation. The primary change is that these errors decreased considerably and the fit is much better than advertised.

The \Sigma_{1/2-} delta error was -4.7 degrees in the m=1 class, now it is 20.44 and is in the m=6 class with an error of +3.17. The \Delta_{3/2+} error was 3.43. Now the best angle is 34.10 and the error is 1.83 degrees. There are still wide error bands on the calculated angles, but the RMS error is close to halved as these two outliers contributed 80% of the old RMS error.

Eventually I'll write this up in a LaTex article and check the numbers carefully. Right now, I'm amusing myself by alternately pushing from the theoretical and phenomenological sides. Also I should mention that I found and fixed an unrelated minor Java programming error in the Koide calculator.

When I finally get around to writing up the LaTex article, I will try to figure out how Alejandro and Andre wrote the "Gim" symbol in this paper:

http://www.arxiv.org/abs/hep-ph/0505220

and redefine it as a vector, so that mass = |Gim|^2.

There are obvious reasons for expecting powers of e in physics. Powers of 3 are more rare. One way of getting a power of 3 is by exponentiating ln(3). Lubos Motl's blog recently brought the subject of how ln(3) shows up in black hole calculations here:
http://motls.blogspot.com/2007/04/straightforward-quasinormal-calculation.html

Hans de Vries · Apr 18, 2007

The latest Standard Model prediction for the tau magnetic moment is:

1. 00117721 (5)

The theoretical value is six orders of magnitude more accurate
as the experimental one due of course to the short lifetime.

The tau lepton anomalous magnetic moment
S. Eidelman, M. Giacomini, F.V. Ignatov, M. Passera
http://arxiv.org/abs/hep-ph/0702026

Theory of the tau lepton anomalous magnetic moment
S. Eidelman, M. Passera
http://arxiv.org/abs/hep-ph/0701260

Regards, Hans

Hans de Vries · Apr 25, 2007

Another numerical coincident of the vertex correction (magnetic anomaly)
in a quite elementary mass ratio. This time the square of the pion mass delta:

\left|\ \frac{\pi^\pm}{\pi^0} - 1\ \right|^2\ =\ 0.00115821 (26)

So we have as numerical coincidences:

0.001159652________ Electron Magnetic Analomy
0.001159567________ Mass independent Magnetic Analomy
0.001158692_(27)___ Muon / Z boson mass ratio.
0.00115821__(26)___ Pion mass delta square.

The latter two relations are as good as sigma 1.8. Not as good but similar
is this one concerning the proton neutron mass delta:

\left|\ \frac{m_p}{m_n} - 1\ \right|\ \ =\ 0.0013765212 (6)

0.00131419__(41)___ Muon / W boson mass ratio
0.0013765212_(6)___ Proton-neutron mass delta Regards, Hans

http://arxiv.org/PS_cache/hep-ph/pdf/0503/0503104v1.pdf

Hans de Vries · Apr 25, 2007

A pretty amazing coincident isn't it? The relation:

\mbox{\Huge $\frac{m_{\pi^\pm}}{m_{\pi^0}}\ =\ 1+ \left(\frac{m_\mu}{m_Z}\right)^\frac{1}{2}\ =\ 1.0340344(55)$}

following from the previous post is as exact as:

1 : 1.0000067 (42)

Where more than half the error is experimental uncertainty.And secondly. The delta isn't just any value. It's square, which is:

\mbox{\Huge $\frac{m_\mu}{m_Z}\ =\ 0.001158692(27) $}

is to a very high degree equal to the magnetic anomaly. Most notably
the mass independent value (without the vacuum polarization terms),
which is 0.001159567I've used the latter two of following pion mass data:

m_{\pi^\pm} = 139.57018 \pm 0.00035\ MeV
m_{\pi^0} = 134.9766 \pm 0.0006\ MeV
m_{\pi^\pm}-m_{\pi^0} = 4.5936 \pm 0.0005\ MeV

Where the difference is experimentally known better as the two
absolute values. So the error I use is the 0.0005 MeV. The
combined error of the charged and neutral pions is larger as
the error in the numerical coincident!Regards, Hans.

Jay R. Yablon · Apr 26, 2007

Baryons, Mesons, Gluons and QCD Confinement

Dear friends over at Physics Forums:

I have for more than two years been researching the possibility that
baryons may in fact be non-Abelian magnetic sources.

The result of this research are now formally and rigorously presented in
a paper at:

http://home.nycap.rr.com/jry/Papers/Baryon%20Paper.pdf

Among other things, I believe this paper fundamentally solves the
problem of quark and gluon confinement within baryons, and origin of
mesons as the mediators of nuclear interactions. It may also resolve
the question of fermion generation replication.

I think you guys may enjoy playing with the mass formula which I first develop in (3.7). It bears a resemblance to Koide formula.

I would very much appreciate your constructive comments.

Best to all.

Jay
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
Web site: http://home.nycap.rr.com/jry/FermionMass.htm
sci.physics.foundations co-moderator

Hans de Vries · Apr 26, 2007

Jay R. Yablon said:

Dear friends over at Physics Forums:

I have for more than two years been researching the possibility that
baryons may in fact be non-Abelian magnetic sources.

The result of this research are now formally and rigorously presented in
a paper at:

http://home.nycap.rr.com/jry/Papers/Baryon%20Paper.pdf

Thank you for presenting your work over here Jay. It's becoming more
interesting with each new paper.

I think we do share the viewpoint that much of the interesting and
important physics is the result of the special terms, like the spin term in
the Dirac equation for the fermions, and the extra non-Abelian term for
the Bosons. My feeling is that the universe would be just a dull homo-
geneous "soup" without these two terms.

Regards, Hans.

Hans de Vries · Apr 26, 2007

I never did expect numerical coincidences with hadron masses but
the pion shows a second one, again magnetic anomaly related.
Half the mass ratio of the (charged) two quark pion and the electron
is roughly equal to alpha. Now if we divide this value by 2pi then we
find a coincidence with the Muon's magnetic anomaly:

\mbox{\Huge $\frac{1}{\pi} \frac{m_e}{\ \ m_{\pi\pm}}\ =\ 0.001165407 (3) $}

We already had an Electron magnetic anomaly coincidence:

\mbox{\Huge $\left(\ \frac{m_{\pi^\pm}}{m_{\pi^0}} - 1\ \right)^2\ =\ 0.00115821 (26) $}

This adds to the list of Magnetic Anomaly related numerical coincidences:

0.001165920________ Muon Magnetic Analomy
0.001165407__(3)___ Electron / Pion mass ratio /pi.
0.001165892________ Electrom / Pion mass ratio +Vpi-Vmu

0.001159652________ Electron Magnetic Analomy
0.001159567________ Mass independent Magnetic Analomy
0.001158692__(27)__ Muon / Z boson mass ratio.
0.00115821___(26)__ Pion mass delta square.

0.0000063532_______ Muon Mag.Anomaly mass dependence
0.0000063558_(20)__ Electron / W mass ratio (2007)

The third value (+Vpi-Vmu) is closer. What we have done here
is to add the one-loop QED vacuum polarization difference for
a particle of muon mass and a particle of pion mass. A lepton
with the mass of a pion should have a higher magnetic anomaly
as the muon by this amount plus another 4-8% or so for higher
order VP loops which I can't calculate.

The "Mass independent Magnetic Anomaly" is the same for
all leptons. It is calculated here by subtracting the one loop VP
contribution from the electrons magnetic anomaly.

The third numerical coincidence was found earlier and relates
the electron / W mass ratio with the mass dependent part
of the Muon's magnetic anomaly. It was calculated by taking
the difference between the Muon and Electrons magnetic
anomalies and adding the Electrons largest (one-loop) VP term.Regards, Hans

Code:

 One loop vacuum polarization:

 m1/m2 = 273.132044975751,   one-loop vpol = 0.000006389950   (pion +/-)
 m1/m2 = 206.768283800000,   one-loop vpol = 0.000005904060   (muon)
 m1/m2 = 1.00000000000000,   one-loop vpol = 0.000000084641   (electron)

Jay R. Yablon · Apr 30, 2007

Hans de Vries said:

Thank you for presenting your work over here Jay. It's becoming more
interesting with each new paper.

I think we do share the viewpoint that much of the interesting and
important physics is the result of the special terms, like the spin term in
the Dirac equation for the fermions, and the extra non-Abelian term for
the Bosons. My feeling is that the universe would be just a dull homo-
geneous "soup" without these two terms.

Regards, Hans.

Hi Hans, thank you for your encouragement.

As you know I mentioned over on SPF, the most important results in the baryon paper have to do with confinment. I have extracted and consolidated these results and posted them at the link below.

http://home.nycap.rr.com/jry/Papers/Confinement%20Paper.pdf

This paper may provide an exact analytical solution to the problem of QCD confinement and the existence of short-range meson mediators of the nuclear interactions.

As you noted over on SPF, this paper evolves from the non-vanshing three form P = dF which would be the Yang-Mills (non-abelian) "magnetic current."

If I were to put the paper in one sentence it would be:

Baryons are non-Abelian (Yang-Mills) magnetic monopoles.

And if permitted a second sentence:

These Yang Mills monopoles exhibit all the properties of a baryon, including three fermions which via exclusion we connect with quarks, confinement of these quarks and their mediating gluons such that there is never any net flux of gluons or individual quarks across any closed two-dimensional surface of integration through the baryon current density, and emission of mesons which have short range such that these mesons, which account for interactions between baryons, are the only entities for which there is a net flux through the integration surface.

I appreciate any comments which the Physics Forums readers may have.

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
Web site: http://home.nycap.rr.com/jry/FermionMass.htm
sci.physics.foundations co-moderator

arivero · May 1, 2007

Hans de Vries said:

0.001165407__(3)___ Electron / Pion mass ratio /pi.

Well, that is rare, as every you find

! I would expect the quotient between charged and neutral pion to be related somehow to the fine structure constant, and then this natural connection would propagate to the rest of findings in the thread. But again, too much exactitude.

In a related thems, did I tell that I was interested on the history of the calculation of g-2? It seems that a guy, who helped Kino****a to set up the software, is now in the dark side of accidental numeric coincidences.

I have been for the last four months working in a BOINC supercomputing platform, Zivis. I wonder how much time could take to repeat K. et al calculations; I guess that programming time should be higher than computing time.

Jay R. Yablon · May 2, 2007

The so-called Yang- Mills "mass gap"

Dear Physics Forum friends:

I have posted an updated draft of my confinement paper to:

http://home.nycap.rr.com/jry/Papers/Confinement%20Paper.pdf

Last evening, I uncovered and added to the paper, a line of calculation
which explicitly and formally gives a non-zero rest mass to the meson
mediators of QCD without resort to the Higgs Mechanism, naturally
eliminates the propagator poles and therefore allows one to have "real"
rather than "virtual" particles without ad hoc tricks, formally renders
QCD a short-range interaction, and solves the Yang-Mills mass gap
problem. This is in addition to the confinement solution which has
already been posted for several days now.

Interested in feedback.

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
Web site: http://home.nycap.rr.com/jry/FermionMass.htm
sci.physics.foundations co-moderator

arivero · Jul 12, 2007

Gerald Rosen

I believe we have not included G. Rosen in the references for the topic of this thread.

His http://prola.aps.org/abstract/PRD/v4/i2/p275_1 , for instance, gives one of these formulae with exponential relationship between Planck and electron masses. Other papers look for masses of electron and quark, value of the fine structure constant, etc. http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=find+a+rosen%2C+g&FORMAT=WWW&SEQUENCE=

arivero · Jul 17, 2007

arivero said:

Then we use http://wwwusr.obspm.fr/~nottale/ukmachar.htm to get the mass of the electron

 \ln (M_P/m_e) = \alpha^{-1} \sin^2 \theta_W

Note that this remark is mostly an elaborate rewritting of the argument about the self energy of the electron. For instance in a modern text such as Polchiski's string textbook, it appears as
 \delta m \approx \alpha m \ln {1 \over m l} 
without any reference except "known since the 1930s", and then it is explained that l should to be expected to ultimate cutoff, thus the length of Planck, and that \delta m \over m should be expected to be of order O(1).

Thus the formula can be rewritten as
 \alpha^{-1} {\delta m_e \over m_e} \approx \ln {m_P \over m_e} 

And Nottale's statement translates, in standard knowledge, as telling that the electromagnetic contribution to the mass of the electron is about 3/8.

While most of the "rediscoverers" of this relationship are outsiders, "out of the loop" people, in the case of Nottale it is sort of astonishing that he forgets to tell that he is recovering a "known" relationship and not a new one (he does not have, at all, new arguments for the 3/8 adjustment neither).

Another remark nobody does is that the argument can be reverted to predict the GUT-Planck-String scale from the low energy scale. Thus one could to try to predict muon or electron masses from the electroweak scale, then use the cutoff argument to predict the GUT scale, and the to use the seesaw to predict the neutrino scale.

EDITED: Nottale presentation can be found in http://luth2.obspm.fr/~luthier/nottale/ukmachar.htm Also, it is argued in http://luth2.obspm.fr/~luthier/nottale/arDNB.pdf that better results are met if one uses in a very peculiar way the fine structure constant running value instead of the infrared one. Indeed using alpha at m_e, one really "predicts" (except the justification of the 3/8) the value of Newton Constant. In any case, I am ashamed 1) about our inability to note that it was just the renormalisation of electron mass as usual, and 2) that Nottale hides or ignores this fact, after being in contact with QED theory during decades.

arivero · Aug 12, 2007

For numerical "coincidences", hep-ex/0412028 remembers that "the infamous Florida 2000 presidential election recount with the official result of 2,913,321 Republican vs. 2,913,144 Democratic votes, with the ratio equal to 1.000061".

Hans de Vries · Aug 13, 2007

Once in a while one runs into a nice numerical coincident like this one
relating alpha to a Yukawa potential:

\mbox{\Huge $\frac{\alpha}{2\pi}\ =\ \frac{e^{-2\pi r}}{2\pi r}$}

Where r is the quotient of two radii:

r\ =\ \frac{r_c}{r_l}

in which rc is the Compton radius of the electron, muon or tau lepton and
rl is the corresponding cut-off radius for which the electrostatic energy
is equal to the magnetostatic energy of the classical electron, muon or
tau lepton. It's the magnetic moment which is responsible for the magneto-
static energy. One gets the following relations:

\alpha =1/137.05268 -\ \quad \mbox{for the electron}
\alpha =1/137.04743 -\ \quad \mbox{for the muon}
\alpha =1/137.03796 -\ \quad \mbox{for the tau lepton}
\alpha =1/137.03599971\ \quad \mbox{experimental}

The differences stem from the differences of the magnetic moment analomies. Regards, Hans==================================================

PS: For the EM fields the following expressions were used:

 \textsf{E}_x\ = \frac{q}{4\pi\epsilon_o}\ \frac{x}{r^3}, \quad \textsf{E}_y\ = \frac{q}{4\pi\epsilon_o}\ \frac{y}{r^3}, \quad \textsf{E}_z\ = \frac{q}{4\pi\epsilon_o}\ \frac{z}{r^3} 

 \textsf{B}_x\ = \frac{\mu_o\mu_e}{2\pi}\left(\frac{3z}{r^5}x \right), \quad \textsf{B}_y\ = \frac{\mu_o\mu_e}{2\pi}\left(\frac{3z}{r^5}y \right), \quad \textsf{B}_z\ = \frac{\mu_o\mu_e}{2\pi}\left(\frac{3z}{r^5}z-\frac{1}{r^3} \right) 

And for the total energy: \mbox{Energy:} \qquad E\ =\ \frac{1}{2}\left(\epsilon_o\textsf{E}^2 + \frac{1}{\mu_o}\textsf{B}^2\right) 

 E\ =\ \int \left\{ \frac{q^2}{32\pi^2\epsilon_o r^4}\ +\ \frac{\mu_o\mu_e^2}{8\pi^2}\left(\frac{3z^2}{r^8}+\frac{1}{r^6} \right) \right\} dx^3\ =\ \frac{q^2}{8\pi\epsilon_o r_o}\ + \ \frac{\mu_o\mu_e^2}{4\pi}\ \frac{3}{r_o^3} 

For an electron:

 \frac{q^2}{8\pi\epsilon_o} = 1.1535385\ 10^{-28},\quad \frac{3\mu_o\mu_e^2}{4\pi}\ =\ 2.5862051\ 10^{-53} 

and ro is the cut-off radius.

arivero · Aug 14, 2007

This last comment reminders me about Born calculation of the Lamb shift,
in the train cominb back from Shelter Island. He had no serious argument to fix the cut-off in the renormalized equation, so he simply choose the compton length of the electron because it is the cutoff of particle creation.

All the lepton masses from G, pi, e

Multiple poll: Check all you agree.

Logarithms of lepton mass quotients should be pursued.

Alpha calculation from serial expansion should be pursued

We should look for more empirical relationships

Pythagorean triples approach should be pursued.

Quotients from distance radiuses should be investigated

The estimate of magnetic anomalous moment should be investigated.

The estimate of Weinberg angle should be investigated.

Jay R. Yabon theory should be investigate.

I support the efforts in this thread.

I think the effort in this thread is not worthwhile.

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