All the lepton masses from G, pi, e

**CarlB** · Mar18-07, 05:52 AM

Continuing the story, we return to the case of the baryons.

The Koide formula gives the masses of the electron, muon, and tau by the formula:
$LaTeX Code: \\sqrt{m_n} = \\mu_v + \\mu_s \\cos(2n\\pi/3 + \\delta)$

where the constants are defined as:

$LaTeX Code: \\begin{array}{rcl} \\mu_v &=& 17.716 \\sqrt{MeV},\\\\ \\mu_s &=& \\mu_v \\sqrt{2},\\\\ \\delta &=& 0.22222204717 \\end{array}$

The above formula has one degree of freedom removed with the square root of 2. This is the formula that Koide discovered in the early 1980s. The angle \delta, is surprisingly close to 2/9, and this post is devoted to the application of this angle to the baryon resonances.

The authors model of the leptons supposes that they are composite with three elementary objects in each (simplifying here a bit), and that these elementary objects are held together with a force similar to the color force. That is, the claim is that the electron, muon and tau are color singlets and the generation structure arises from a similar effect.

The author extended the above formula to the neutral leptons, the neutrinos, by jumping to the conclusion that the above numbers have something to do with quantum numbers. The justification for this is beyond the scope of this post, but the formula published a year ago was:

$LaTeX Code: \\begin{array}{rcl} \\sqrt{m_{\\nu n}} &=& 3^{-11} (\\mu_v + \\mu_s\\cos(2n\\pi/3 + \\delta + \\pi/12))$
with the same constants given above.

A short form reason for the pi/12 is that it appears when you convert a 3x3 array of complex multiples of primitive idempotents of the Pauli algebra into a 3x3 array of complex numbers that preserves matrix addition and multiplication. In short, the equation that relates the 0.5 with the pi/12 is:
$LaTeX Code: \\begin{array}{rcl} P_x &=& 0.5(1 + \\sigma_x),\\\\ P_y &=& 0.5(1 + \\sigma_y),\\\\ P_z &=& 0.5(1 + \\sigma_z),\\\\ (0.5\\exp(-i\\pi/12))^4 P_xP_yP_xP_x &=& (0.5 \\exp(-i\\pi/12)) P_x, \\end{array}$
That is, one can eliminate the nastiness of the product of these three projection operators, and turn them into just another complex multiplication if you multiply each by some numbers that have to do with the sqrt(2) in the Koide formula, and the difference between the charged and lepton delta angles. (Note, I haven't checked the above with care. If you play around with it, you can fix any errors.)

There are hundreds of baryon resonances / excitations and understanding their masses is an ongoing project. We will use the word "resonance" to mean a set of baryons that all have the same quantum numbers. We will use the word "excitations" to distinguish between baryons that have the same quantum numbers.

Other than lepton number, the leptons all have the same quantum numbers. So our analogy between the leptons and the baryons, is between generations of leptons, and excitations of baryons. This makes a certain amount of sense in that leptons do not have excitations other than the generation structure. One could also imagine looking at the generation structure of the baryons. For such an analogy, one would want to compare stuff like (ddd,sss,bbb) and (uuu,ttt,ccc). Unfortunately, these more charmed states do not have very good data.

When a baryon has only two or fewer excitations, we suppose that the others are either yet to be detected, or are hidden as we will discuss later. For this program, a worse situation is when an excitation comes in a multiplicity greater than 3. For the baryons, this happens only one time, with the $LaTeX Code: N_{1/2+}$ . The fourth state carries only one reliability star. In the PDG data, this means that "evidence of existence is poor". Accordingly, we will ignore this state.

With the leptons, we saw that the angle 0.22222204717 had something to do with the difference between the charged and neutral leptons. We suppose it has to do with the Weinberg angle. The charged leptons and the neutral ones differed by pi/12 = 15 degrees. We therefore speculate that the excitations of the baryon resonances will carry this same relation, that is, that they will have angular dependency of the form:
$LaTeX Code: \\cos(2n\\pi/3 + \\delta + m\\pi/12)$ .
where m depends on the resonance.

Adding 8 to m is the same as subtracting 1 from n, so we need only consider 8 different values of m, for instance, from 0 to 7. Since the cosine is an even function, we cannot distinguish between positive and negative angles. This causes a reflection in the data. Consequently the algorithm for finding the angle from the mass data will return angles from 0 to 60 degrees rather than 0 to 120 degrees. As a result, the cases for m > 3 are folded over those same 60 degrees and we will bin the calculated values into the following 8 bins

$LaTeX Code: \\begin{array}{rcr} \\delta + 7\\pi/12&==& 2.27\\\\ \\delta + 0\\pi/12&==&12.73\\\\ \\delta + 6\\pi/12&==&17.27\\\\ \\delta + 1\\pi/12&==&27.73\\\\ \\delta + 5\\pi/12&==&32.27\\\\ \\delta + 2\\pi/12&==&42.73\\\\ \\delta + 4\\pi/12&==&47.27\\\\ \\delta + 3\\pi/12&==&57.73\\end{array}$

For example, in the first line, $LaTeX Code: \\delta + 7\\pi/12$ gives 117.73 degrees. Adding 1 to n is the same as subtracting 120 degrees from the angle, so this is the same as -2.73 degrees and is indistinguishable from +2.73 degrees because the cosine is an even function.

Note that the first and last bins are very close to 0 and 60 degrees. An angle of 0 degrees corresponds to a degenerate case with two excitations at the lower mass value while the angle 60 degrees puts the degeneracy at the upper mass value. These degeneracies would correspond to excitations of baryons that only appear with two masses.

The above set of bins have gaps of length 4.54 and 10.46 degrees. The rms average for a random value in a bin of length D is:
$LaTeX Code: \\frac{2}{D} \\int_{x=0}^{D/2} x^2 dx = D^2/12$
The 4.54 degree gap will be hit 4.54/15 of the time, while the other will be hit 10.46/15 of the time. The average rms is therefore:
( $LaTeX Code: (4.54^3 + 10.46^3)/12*15)^{1/2} = 2.622$

I will present the PDG data in the next post.

**CarlB** · Mar18-07, 06:28 AM

Here's the results of the calculations:
$LaTeX Code: \\begin{array}{lccccc|l} Bin/Set & mu_v & mu_s & \\delta&Error & L_{IJ} &Notes \\\\ \\hline m=7 & & & 2.27 & & & Hidden \\\\ \\hline m=0 & & &12.73 & & &\\\\ e,\\mu,\\tau & 17.716 &25.05 &12.73 & & & ****, ****, ****\\\\ N_{1/2-} & 41.89 & 3.92 &12.97 &+0.24 & S_{11} & ****, ****, *\\\\ \\Lambda_{3/2-} & 42.77 & 5.58 &12.67 &-0.06 & D_{03} & ****, ****, *\\\\\\hline m=6 & & &17.27 & & & \\\\ \\Sigma_{3/2-} & 41.52 & 2.45 &16.08 &-1.19 & D_{13} & ****, ***, **\\\\ N_{3/2-} & 41.95 & 3.87 &19.22 &+1.95 & D_{13} & ****, ***, **\\\\ \\hline m=1 & & &27.73 & & & \\\\ \\Sigma_{1/2-} & 42.33 & 2.60 &23.03 &-4.70 & S_{11} & ***, **, *\\\\ \\hline \\end{array}$

$LaTeX Code: \\begin{array}{lccccc|l} Bin/Set & mu_v & mu_s & \\delta&Error & L_{IJ} &Notes \\\\ \\hline m=5 & & &32.27 & & & \\\\ \\Delta_{1/2-} & 43.51 & 3.36 &31.43 &-0.84 & S_{31} & ****, **, *\\\\ \\Delta_{3/2+} & 39.80 & 5.16 &35.70 &+3.43 & P_{33} & ****, ***, ***\\\\ \\hline m=2 & & &42.73 & & & \\\\ \\Sigma_{3/2+} & 41.90 & 4.95 &41.57 &-1.16 & P_{13} & ****, **, *\\\\ N_{1/2+} & 36.69 & 6.34 &42.65 &-0.08 & P_{11} & ****, ****, ***\\\\ \\Sigma_{1/2+} & 39.55 & 5.23 &43.22 &+0.49 & P_{11} & ****, ***, **\\\\ \\Lambda_{1/2-} & 40.21 & 2.81 &43.51 &+0.78 & S_{01} & ****, ****, ***\\\\ \\hline m=4 & & &47.27 & & & \\\\ \\Lambda_{1/2+} & 38.74 & 5.46 &47.46 &+0.19 & P_{01} & ****, ***, ***\\\\ \\hline m=3 & & &57.73 & & & Hidden\\\\ \\hline \\end{array}$

The units of $LaTeX Code: \\mu_v^2, \\mu_s^2$ are MeV. The next column is the calculated delta value, then the error. The final column are notes. The asterisks are from, the PDG and describe how certain the three states are. I've split the table into two because the PF LaTex editor just couldn't quite handle it as one.

The rms error is 1.88 degrees, somewhat below the expected 2.623. The worst fit is for the $LaTeX Code: \\Sigma_{1/2-}$ , which coincidentally also happens to carry the worst asterisk rating from the PDG. The second worst fit is the $LaTeX Code: \\Delta_{3/2+}$ . While this set is well supported by experiment, it includes the $LaTeX Code: \\Delta(1600)$ whose mass is only loosely constrained. About this particle, the PDG writes: "The various analyses are not in good agreement." Together, these two bad fits contribute 80% of the error among the 13 sets of 3 masses each.

The $LaTeX Code: S_{1/2-}$ excitations, in addition to carrying a high error, are also in a class by themselves. They would seem to fit better in the $LaTeX Code: \\delta + 0\\pi/12$ class, with the $LaTeX Code: N_{1/2-}$ which is also $LaTeX Code: S_{11}$ .

What you're seeing here is ALL the data from the baryons. I suspect that the data will look better when the particle mass ranges are taken into account. My mass calculator does not yet have the software to take errors in the data into account. Before this can be published, it needs to have the errors in the excitations taken into account. Hopefully the bad fits will correspond to loose mass ranges.

**CarlB** · Mar19-07, 06:30 AM

Whoops. I left off the $LaTeX Code: \\Sigma_{1/2+}$ . [edit]Oh, no I didn't![/edit]

Looking at the data, it seems that the really well supported classes are the even ones. Also, there are two low-lying excitations with two masses where one of the two masses is said, in the PDG, to be possibly doubled. These would be the $LaTeX Code: N_{3/2-} N(1700), N(2080)^2 = D_{13}$ and the $LaTeX Code: N_{3/2+}N(1720)^2,N(1900) = P_{13}$ .

I've finished the user interface for an online applet that will compute Koide parameters with error bars, but I've not yet put the math code into it. I've got that code in another program so it's just a matter of putting it in there and ironing out any bugs. But I really need to get back to the day job.

Carl

**CarlB** · Mar22-07, 04:13 AM

Okay, I've got a tool that allows you to compute error bars for these Koide type parameterizations of three masses.

In addition, the mesons are famous for being messy with duplicate masses hard to distinguish. One ends up with six masses instead of three. The hope is that one can split those six mesons into two groups of three that are decent. Accordingly, the tool holds six masses at the same time and automatically steps through the various permutations:
http://www.measurementalgebra.com/KoideCalc.html

Source code is available at the above.

Using new tools is difficult. I've initialized the above program with the data for the electron, muon and tau. So all you have to do to calculate your first error bars is hit the "KOIDE" button. Also, I've set it up to give the angles in degrees rather than radians.

**arivero** · Mar23-07, 06:52 AM

Originally Posted by CarlB

In addition, the mesons are famous for being messy with duplicate masses hard to distinguish.

Indeed that is the idea, isn't it? We have six mesons with charge +1, due to combinations of three families of antidown quarks and two families of up quarks. But they have spin 0; then we have the same number of degrees of freedom that in the case of leptons of charge +1.

Have you spotted now some interesting pattern in the mesons, Carl? I tried some pages ago, up in the thread, and I was not very happy.

**CarlB** · Mar24-07, 07:17 PM

Dr Rivero,

Right now I'm busily working on the theoretical side rather than the phenomenological side. I wrote up the software so that one could divide up a set of six mesons into two groups each with decent Koide numbers. But I don't think that alone would be very convincing. To do it right, I think you have to split six states into two groups of three in such a way that the other properties of the states make sense.

There are four S=C=0 mesons for which only three excitations exist, the pi, the omega, the f(1) and the rho(3). Their mass data are:

$LaTeX Code: \\begin{array}{cccclll} \\pi & \\pi &\\pi(1300) &\\pi(1800) &= 134.9766, &1300(100), &1812(14)\\\\ \\omega & \\omega(782) &\\omega(1420) &\\omega(1650) &= 782.65, &1425(25), &1670(30)\\\\ f(1) & 1285 &1420 &1510 &= 1281.8(0.6), &1426.3(0.9), &1518(5)\\\\ \\rho(3) & \\rho(1690) &\\rho(1990) &\\rho(2250) &= 1688.8(2.1), &1982(14), &2250(?100?) \\end{array}$

The rho is ugly in that two of its states are single star, and they don't give limits on the mass of the last one. I've arbitrarily put +-100MeV. Typing the data into the Koide calculator, the resulting angles are (in degrees, min, typ, max):

$LaTeX Code: \\begin{array}{c|ccc} \\pi & 45.5 & 48.5 & 51.3\\\\ \\omega & 43.8 & 46.6 & 49.5\\\\ f(1) & 36.9 & 38.0 & 39.1\\\\ \\rho(3)& 26.1 & 32.7 & 41.8 \\end{array}$

All but the f(1) are consistent with delta angles, that is, 47.27 and 32.27 degrees. The f(1), which unfortunately has the most accurate mass measurements, is between 32.27 and 42.73. The errors are +1.23, -.67, -4.27, -0.43. RMS error is 2.27, less than what chance would suggest, but not by a lot.

When you read the description of the f(1) resonances, at least they are pretty weird. Their lightest entry shows fairly heavy, maybe they are something more complicated. The PDG commentary on the states is interesting. See page 3:
http://pdg.lbl.gov/2006/listings/m027.pdf

**verdigris** · Mar25-07, 08:12 AM

I think that it would be more useful if you had some physical theory to tune these numbers and equations into.Because without one what use will the numbers and equations be anyway!

**arivero** · Mar25-07, 12:40 PM

verdigris, I can think of some uses: Even without a theory, it is an attack to the ideology of GUT unification at high scales, because in such setup the masses run via renormalisation group and you should not be able to find any simple relationship at low energy. And we find some.

The problem was that the best model builders got in love with this business of Very High Energy unification, and most of the standard material is focused from such approaches. If you neglect this point of view, Carl (and others?) effort is not very out from the hopes of group theory based unification. His adopted Fermion Cube holds spinors in the same way that it was done in the late seventies (or early eighties). The unification for one family need to be extended in a way to produce three generations and no more, and this is the usual problem in model building.

(Note for instance the idea of using spinors, or fermionic creation operators, to build the representation. Using six of them you get 32 fermionic degrees of freedom, with the right charges of one generation, but if you add another two then you get not 96=32*3 but 128=32*4. Thus you need a very exotic symmetry breaking scheme for the extra two generators, or alternatively a different way to produce the generation-wise symmetry).

**Hans de Vries** · Mar25-07, 01:16 PM

General Mass Formula:

We possibly have neglected our own mass formula which
we had already at the start of this thread. New is a possible
link to real physics which I will talk more about in coming posts.

$LaTeX Code: M(a,b)\\ =\\ a\\pi-b/\\pi\\ =\\ 2\\sqrt{ab}\\ \\sinh\\left( \\ln\\left(\\pi \\mbox{\\Large $\\sqrt{\\frac{a}{b}}$}\\ \\right)\\right)$

The inputs are always two small integer numbers and the
output is the log ratio of two masses. Almost all the
elementary combinations are close to actual ratios of
the (base states) of existing particles:

Code:

.
M(0,2) : log(mτ/mp)   --->   0.636619 : 0.638635   =   0.316%
M(1,1) : log(mτ/mμ)   --->   2.823282 : 2.822461   =   0.029%
M(1,2) : log(mτ/mπ)   --->   2.504972 : 2.544107   =   1.562%
M(1,3) : log(mp/mμ)   --->   2.186662 : 2.183828   =   0.129%
M(1,4) : log(mp/mπ)   --->   1.868353 : 1.905472   =   1.986%
M(2,2) : log(mπ/me)   --->   5.646565 : 5.609955   =   0.652%
M(2,3) : log(mμ/me)   --->   5.328255 : 5.331598   =   0.062%
M(3,4) : log(mτ/me)   --->   8.151538 : 8.154063   =   0.030%
M(4,2) : log(mw/me)   --->   11.92975 : 11.96646   =   0.307%
M(3,6) : log(mp/me)   --->   7.514918 : 7.515427   =   0.0068%

me  =  electron
mμ  =  muon lepton
mτ  =  tau lepton
mπ  =  pion (+/-) 
mp  =  proton
mw  =  W-boson

The possible relation with real physics would be this:

$LaTeX Code: M(a,b)\\ =\\ \\sum_{k\\ =-\\infty}^\\infty\\ \\mbox{\\huge J}_k(2\\sqrt{ab})\\ \\left(\\pi \\mbox{\\Large $\\sqrt{\\frac{a}{b}}$}\\ \\right)^k$

$LaTeX Code: e^{ieA \\sin(\\omega t)}\\ =\\ \\sum_{k\\ =-\\infty}^\\infty\\ \\mbox{\\huge J}_k(eA)\\ \\ e^{ik\\omega t}$

The later expression describes the non-pertubative interaction
factor with a sinusoidal field.

Regards, Hans

**CarlB** · Mar26-07, 01:18 AM

Originally Posted by verdigris

I think that it would be more useful if you had some physical theory to tune these numbers and equations into.

There is a physical theory behind this. It's based on Schwinger's measurement algebra. The equations are rather distant from what you can get with the usual physics and so is the theory. I've written most of a book dedicated to explaining the principles:
http://brannenworks.com/dmaa.pdf

I know that no one has read the above book with any degree of care because if they did, they'd have pestered me with questions, complaints, pointed out typos, and suggested improvements. People are too busy to spend a lot of time reading an amateur's textbook.

What I do get are comments from people who don't understand my theory, haven't carefully read my many documents, do not understand much about Clifford algebras, but who want to use the equations for promoting their own physics theories.

I don't blame them for this. It is very similar to what I did with Koide's original equation. After a couple hours of playing with it I saw that it had an easy derivation in terms of the principle idempotents of Clifford algebras and ran with it. The new interpretation put his equation into eigenvalue form and I posted it here. But I've ignored completely Koide's work in explaining the equation.

There are something like 30,000 physicists on the planet. I think that the papers of only about 1000 of them are carefully read and studied. The other 29,000 write stuff that is ignored.

**Hans de Vries** · Mar26-07, 04:03 AM

A mass relation for all six principal charge 1 particles:

$LaTeX Code: \\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: $LaTeX Code: Y\\pi -X/\\pi$ , 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 = $LaTeX Code: 3\\pi -6/\\pi$
accuracy: 0.0000677

2) Electron mass ratio with 2VeV (Vacuum exp.Value): log(963699)
13.77853 = experimental
13.79810 = calculated = $LaTeX Code: 5\\pi -6/\\pi$
accuracy: 0.00142

3) Electron-Muon mass ratio: log(206.7682838)
5.331598 = experimental
5.328255 = calculated = $LaTeX Code: 2\\pi -3/\\pi$
accuracy: 0.000627

4) Proton-Pion mass ratio: log(6.72258237)
1.905472 = experimental
1.868353 = calculated = $LaTeX Code: \\pi -4/\\pi$
accuracy: 0.01986

5) Electron W-boson mass ratio: log(157387)
11.96646 = experimental
11.92975 = calculated = $LaTeX Code: 4\\pi -2/\\pi$
accuracy: 0.00307

Try it yourself!

Regards, Hans.

$LaTeX Code: \\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** · Mar26-07, 07:32 AM

Originally Posted by Hans de Vries

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 $LaTeX Code: m-n/\\pi$ 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 $LaTeX Code: n/\\pi$ . We find that:

$LaTeX Code: 639*\\pi - 1/\\pi = 2007.16$

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

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

$LaTeX Code: 639*\\pi - 1.5/\\pi = 2007.00024$

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 $LaTeX Code: n\\pi - m/\\pi$ for small values of n and m, one finds that they are mostly separated by $LaTeX Code: 1/\\pi = 0.3183$ . When one assigns random numbers to the nearest one of these, the worst one can do is half the interval, or $LaTeX Code: 1/(2\\pi) = 0.1591$ 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** · Mar26-07, 10:54 AM

Originally Posted by CarlB

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:

Originally Posted by Hans de Vries

1) Electron-Proton mass ratio: log(1836.1526726) (natural log)
7.515427 = experimental
7.514918 = calculated = $LaTeX Code: 3\\pi-6/\\pi$
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~~ · Mar26-07, 11:09 AM

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

**Hans de Vries** · Mar26-07, 12:38 PM

Originally Posted by whatta

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** · Mar26-07, 02:14 PM

Originally Posted by Hans de Vries

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 $LaTeX Code: e^{2007}$ in the form $LaTeX Code: e^{n\\pi - m/\\pi}$ 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 $LaTeX Code: e^{12.5} < R < e^{13.0}$ , 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 $LaTeX Code: \\Delta^+$ ? 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