All the lepton masses from G, pi, e

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In summary, the conversation revolved around using various equations and formulae to approximate the values of fundamental constants such as the Planck Mass and the fine structure constant. The discussion also delved into the possibility of using these equations to predict the masses of leptons and other particles. Some participants raised concerns about the validity of using such numerical relations, while others argued that it could be a useful tool for remembering precise values.

Multiple poll: Check all you agree.

  • Logarithms of lepton mass quotients should be pursued.

    Votes: 21 26.6%
  • Alpha calculation from serial expansion should be pursued

    Votes: 19 24.1%
  • We should look for more empirical relationships

    Votes: 24 30.4%
  • Pythagorean triples approach should be pursued.

    Votes: 21 26.6%
  • Quotients from distance radiuses should be investigated

    Votes: 16 20.3%
  • The estimate of magnetic anomalous moment should be investigated.

    Votes: 24 30.4%
  • The estimate of Weinberg angle should be investigated.

    Votes: 18 22.8%
  • Jay R. Yabon theory should be investigate.

    Votes: 15 19.0%
  • I support the efforts in this thread.

    Votes: 43 54.4%
  • I think the effort in this thread is not worthwhile.

    Votes: 28 35.4%

  • Total voters
    79
  • #421
I'm sure the magnetic moment plays a major role, which reminds me of another electro-
magnetic duality: There are types of two magnetic dipoles with indistinguishable fields,
the vector dipole and the axial dipole. The first is a combination of two opposite magnetic
monopoles and the second is a point like circular electric current.

So, associating the field with a continuous charge/spin distribution, which of the two
types is involved in case of the magnetic moment? Well, the net force in a magnetic
field is quite different:

Force on a Vector dipole in a B field:

[tex]
\begin{array}{l l |l l l| l | l | l}
& & \partial_x \textsf{B}_x & \partial_y \textsf{B}_x & \partial_z \textsf{B}_x & & \mu_x & \\
\vec{F}_{magn} & =\ \ & \partial_x \textsf{B}_y & \partial_y \textsf{B}_y & \partial_z \textsf{B}_y & \cdot & \mu_y & \quad =\ \partial_j B_i\ \mu_i \\
& & \partial_x \textsf{B}_z & \partial_y \textsf{B}_z & \partial_z \textsf{B}_z & & \mu_z
\end{array}
[/tex]

Force on an Axial dipole in a B field:

[tex]
\begin{array}{l l |l l l| l | l | l}
& & \partial_x \textsf{B}_x & \partial_x \textsf{B}_y & \partial_x
\textsf{B}_z & & \mu_x & \\
\vec{F}_{magn} & =\ \ & \partial_y \textsf{B}_x & \partial_y \textsf{B}_y & \partial_y \textsf{B}_z & \cdot & \mu_y & \quad =\ \partial_i B_j\ \mu_i \\
& & \partial_z \textsf{B}_x & \partial_z \textsf{B}_y & \partial_z \textsf{B}_z & & \mu_z

\end{array}
[/tex]

Note the exchange of the indices. The second case is what you would expect:

[tex]\vec{F} ~~=~~\mbox{grad}(\vec{B}\cdot\vec{\mu})[/tex]


So, it should be easy to distinguish between the two cases you would say,
and this rules out the pair of magnetic monopoles. There is an interesting twist
however: The net force between either two vector dipoles or two axial dipoles
turns out to be same:

[tex]
\vec{F}\ =\ \frac{3\mu_o\mu_e^2}{4\pi r^4}\left[ \left\{\left(
\hat{\mu}_1 \cdot \hat{\mu}_2 \right) - 5\left( \hat{r} \cdot
\hat{\mu}_1 \right)\left( \hat{r} \cdot \hat{\mu}_2 \right)\right\}
\mbox{\Large $\hat{r}$} + \left( \hat{r} \cdot \hat{\mu}_2
\right)\mbox{\large $\hat{\mu}_1$} + \left( \hat{r} \cdot
\hat{\mu}_1 \right)\mbox{\large $\hat{\mu}_2$}\ \right]
[/tex]

At least in the case that both are at rest, which reminds me that I should find some
time to work out the general case too really proof the exclusion of the first case.


Regards, Hans
 
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  • #422
Okay, if the algebra works out, (it is usual for me to get 2s and factors of pi wrong the first time), then I should give credit where credit is due. The formula for i^i was discussed at Built On Facts:
http://scienceblogs.com/builtonfacts/2008/08/sunday_function_4.php

So what are the attributes of the complex mapping
[tex]f(u) = u^{\ln(u)}[/tex]
and why would you be interested in this function at the point i?

What are the poles of ln(u)?

I bet Kea can tie this into the Riemann zeta function.
 
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  • #423
Just to fix the idea, the point about the exp is that it is gaussian but a peculiar one. It is a wavepacket with the momenta distribution centered in pi instead of the origin. To be precise, the normalized solution of schroedinger equation, at t=0, for a gaussian packet of mean wavenumber [itex]k_0[/itex] and distribution width a=1.
[tex]
\Psi(x)= ({2\over \pi})^{1/4} e^{i k_0 x} e^{-x^2}
[/tex]
so that at least formally
[tex]
\int_{-\infty}^{\infty} \Psi(x) dx = ({2 \pi})^{1/4} e^{-k_0^2\over 4}
[/tex]

And NOW I see that I do not understand why I am trying to fit the minus sign; Hans "first order" formula uses i^(-ln i). CarlB reversed it only to use the 0th-order.

So it is the reverse situation: we actually have a wavepacket of imaginary average wavenumber [tex]k_0=i \pi[/tex]. I'd say that this kind of beast fits with the definition of instanton.
 
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  • #424
An extra rewrite: [itex]\sqrt \alpha= e /\sqrt {2 \pi} [/itex] so

[tex] e + {2 \pi \over e} = \sqrt{2 \pi} \rm e^{\pi^2/4}[/tex]

(typography problem: should we use \rm for the exponential or for the electron charge?)

While it seems lot uglier, it agrees with the conventions for Dirac monopoles, whose charge g is such that
[tex] e g = 2 \pi n[/tex]
for n any integer.
 
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  • #425
Hans de Vries said:
So, associating the field with a continuous charge/spin distribution, which of the two
types is involved in case of the magnetic moment? Well, the net force in a magnetic
field is quite different:

Witten's http://ccdb4fs.kek.jp/cgi-bin/img_index?197909065
quotes some old (1936-1949) works on "heuristic derivations" of the relationship between electric and magnetic charge when both kinds are present in the same point.

Thoughts of today, if the exponential is actually a gaussian wavepacket:

- there is a imaginary momentum, and this is Euclidan wick rotation.
- such momentum is a multiple of pi, this is periodicity or circular configuration: U(1)??
- self duality is usually the hallmark of a BPS state.

the problem is how has a field theoretical beast, the electromagnetic instanton, descended to the realm of naive quantum mechanics. Of course we "known" the answer: it can only descend for a particular value of alpha. But we do not know what the question is.
 
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  • #426
arivero said:
the problem is how has a field theoretical beast, the electromagnetic instanton, descended to the realm of naive quantum mechanics. Of course we "known" the answer: it can only descend for a particular value of alpha. But we do not know what the question is.

Hmm, interesting that you would put it that way. All the Koide formulas seem to be reduction of a QFT problem to QM methods.

And the paper I'm working on with the hadron mass formulas amounts to the same thing.

Naive QM wise, we can think of the meson as one quark moving in the field of the other, that is, we go to center of mass coordinates in both space and color. As a first approximation, assume the quark is in a 1S state and we will ignore spatial degrees of freedom. What's left is color degrees of freedom, R, G, and B.

Let H be the Hamiltonian for the meson. Since there's no spatial dependency, H is only a 3x3 matrix, with color indices:
[tex]H = \left(\begin{array}{ccc}
H_{RR}&H_{RG}&H_{RB}\\
H_{GR}&H_{GG}&H_{GB}\\
H_{BR}&H_{BG}&H_{BB}\end{array}\right)[/tex]
By color symmetry,
[tex]H_{RR} = H_{GG} = H_{BB} = v,[/tex]
[tex]H_{RG} = H_{GB} = H_{BR} = se^{i\delta},[/tex]
[tex]H_{RB} = H_{GR} = H_{BG} = s'e^{i\delta'}.[/tex]

To get real eigenvalues H must be Hermitian so v is real, s=s' is real, and [tex]\delta' = -\delta[/tex]. The three eigenvectors are (1,1,1), (1,w,w*), (1,w*,w), where w is the cubed root of unity, and the three eigenvalues are:
[tex]s+2v\cos(\delta + 2n\pi/3),[/tex]
for n = 0,1,2.

The above is almost in the form of Koide's formula. The difference is that Koide's formula is for sqrt(H) instead of H. To get that last step, note that, without color, the non relativistic Hamiltonian is in the form:
[tex]H = \nabla^2 + V[/tex]
where V is a potential, and a slightly more complicated equation for relativistic energy. To get this into clean form, we do the same thing Dirac did to get the gamma matrices, that is, we take the square root. The difference is that in our case, we are taking the square root of a 3x3 color matrix instead of the square root of a d'Alembertian (or whatever the 4-d gradient is called).

I'm working on the statistics of the fits for 20 hadron triplets to this formula and should get something done in the next few days. I've discussed various fits here and elsewhere, but this wil be the first time that everything is brought into one paper, along with the statistical fits. This involves a fair amount of computer programming.
 
  • #427
CarlB said:
Hmm, interesting that you would put it that way. All the Koide formulas seem to be reduction of a QFT problem to QM methods.

And the paper I'm working on with the hadron mass formulas amounts to the same thing.

Yes, and Hans' version of the Weinberg angle was also a QM object. In general, this thread is defying the lore of the origin of constants from HEP (from Planck scale GUT groups). Of course, if we don't go high (on energy, I mean) we do not need to create/annihilate particles, and QM should work.

What worries me today is that the we are not looking in a far dark corner; your angles on neutrinos were a "minor" mainstream topic in recent years. The reinterpretation of Hans alpha as coming from self dual objects is not a minor stream, it is a major fluent of the Amazon river. Damn, it is just a "wrong sign version" of expression 1.2 of hep-th/9407087.
 
  • #428
arivero said:
What worries me today is that the we are not looking in a far dark corner...

Why is that a concern? Are you worried that the string theorists will swallow this whole? If they had really understood these kind of observations, we would have discovered it by now.
 
  • #429
Kea said:
Why is that a concern? Are you worried that the string theorists will swallow this whole? If they had really understood these kind of observations, we would have discovered it by now.

More or less, this is the point. It is hard to think that they can overlook a critical value of alpha in a theory whose main paper has almost two thousand citations. They should have discovered it by now. The gate is not very hidden; probably it amounts to consider f(e+g) instead of f(e+ig) as they usually do. And there are some hints that they have tried, notably Poliakov and a lot of stuff on condensates.
 
  • #430
Hans de Vries said:
I'm sure the magnetic moment plays a major role, which reminds me of another electro-
magnetic duality: There are types of two magnetic dipoles with indistinguishable fields,
the vector dipole and the axial dipole. The first is a combination of two opposite magnetic
monopoles and the second is a point like circular electric current.

And I guess you can also compare circular monopole currents against electric charges.
 
  • #431
Or
[tex]
e^{\pi^2/4} - e^{-\pi^2/4} \approx \sqrt{ \alpha + {1 \over \alpha}}
[/tex]
instead of
[tex]
e^{\pi^2/4} \approx (\sqrt \alpha + {1 \over \sqrt\alpha})
[/tex]

Actually it is not so good as starting point
11.706956417... / 11.7065492967 (22) = 1.000034777
11.791761389... / 11.7916621597 (22) = 1.000008415
and a sinh is less atractive, to me, than a clean gaussian. And I doubt it could receive the same corrective series than the original guess. It is just that it is closer to popular shapes.
 
  • #432
At least the new 2008 value for alpha from Gabrielse/Kinogarbagea begins to
solidify the n=3 term which now gives some convidence in the whole series.

[tex]\alpha\ =\ 1/137.035999084 (51)[/tex]

The alpha "radiative" series:

[tex]\mbox{\Huge i}^{~ln\, i }\sum_{n=0}\frac{\mbox{\Huge $\alpha$}^{n-\frac{1}{2}}}{(2\pi)^{b_n}} ~~ = ~~ \mbox{\Huge 1}[/tex]

Where [itex]b_n[/itex] is the binominal series 0,0,1,3,6,10.. with successive increments
of 0,1,2,3,4...

The result of the series after each extra term is:

n=0: ... 0.992 747 158 626 634
n=1: ... 0.999 991 584 655 288
n=2: ... 0.999 999 998 402 186
n=3: ... 0.999 999 999 957 418
n=4: ... 0.999 999 999 957 464
Regards, Hans
 
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  • #433
Re: string theory, writing down simple equations like this puts a crimp in the anthropic theories.
 
  • #434
I apologize if I'm am interupting the flow with an irelevant question.
(I can barely understand what you are doing.)
From what you are doing,

arivero,
Just to fix the idea, the point about the exp is that it is gaussian but a peculiar one. It is a wavepacket with the momenta distribution centered in pi instead of the origin. To be precise, the normalized solution of schroedinger equation, at t=0, for a gaussian packet of mean wavenumber k_0 and distribution width a=1.
Can the numbers of wavepacket that are on the shell be calculated?
jal
 
  • #436
CarlB said:
Re: string theory, writing down simple equations like this puts a crimp in the anthropic theories.

It puts a crimp in GUT almost. The thread is telling that no yukawa couplings are to be predicted from GUT. Moreover, Weinberg angle and fine structure constant comes from Hans, so the "GUT coincidence" can only predict the SU(3) coupling. Furthermore, there is probably a high-low consistence rule: one climbs up from the electroweak scale, get the SU(2) and U(1) couplings to meet, then climbs down the SU(3) coupling until it becomes large and all the QCD nonperturbative stuff (as Marco's) applies, and then surprise, after all this walk we are in a energy scale no far from the original one.

(Of course, it is not fair in a thread on coincidences to discard the One of GUT coupling constants).

jal said:
Can the numbers of wavepacket that are on the shell be calculated?
jal

Yep, when the packet halfwidth is taken as 1, the wavenumber is k0= i pi, or perhaps i pi/4 depending on what normalization do we aim to. It stinks to periodic potential or periodic configuration space, as a first conjecture.
 
  • #437
arivero said:
Yep, when the packet halfwidth is taken as 1, the wavenumber is k0= i pi, or perhaps i pi/4 depending on what normalization do we aim to. It stinks to periodic potential or periodic configuration space, as a first conjecture.

Or perhaps one hidden dimension, cyclic, which is what my Clifford algebra density matrix analysis of the situation requires. That is, using primitive idempotents (projection operators or pure density matrices) you can count the hidden dimensions of spacetime by looking at the weak hypercharge and weak isospin quantum numbers:
http://brannenworks.com/a_fer.pdf

With this sort of idea, the arbitrary complex phase universal to all quantum mechanics becomes a position coordinate in the hidden dimension.
 
  • #438
In the spirit of keeping this thread for numbers and leaving (most of) the standard theory in other hands, I have hitchhiked to the thread

https://www.physicsforums.com/showthread.php?t=240247

to discuss on EM duality and like. Here, let me just to note that a fix of notation, from Dirac 1948:

[tex]e_0^2={1 \over 137} \hbar \ c[/tex]
[tex]g_0^2=4 {137 \over 1} \hbar \ c[/tex]

for n=1 in [tex]{1 \over 2} n \hbar \ c[/tex]

The point is that there was other papers where it is argued that n must be even. Let me call [tex]g_2[/tex] to this forceful even constant. In this case
[tex]g_2^2= {137 \over 1} \hbar \ c[/tex]

As for the 4 pi factor, it seems, looking at Dirac's paper, that it was because some other publications use h instead of \hbar, and this use has propagated until today.

I need to locate the paper where it is argued that n=2 is the minimum case.
 
  • #439
In the spirit of keeping this thread for numbers and leaving (most of) the standard theory in other hands, I have hitchhiked to the thread

https://www.physicsforums.com/showthread.php?t=240247

to discuss on EM duality and like. Here, let me just to note that a fix of notation, from Dirac 1948:

[tex]e_0^2={1 \over 137} \hbar c = \alpha_e \hbar \ c [/tex]
[tex]g_0^2=4 {137 \over 1} \hbar c= 2^2 {1 \over \alpha_e} \hbar \ c [/tex]

for n=1 in
[tex] eg= {1 \over 2} n \hbar c[/tex]

The point is that there was other papers where it is argued that n must be even. Let me call [tex]g_2[/tex] to this forceful even constant. In this case
[tex]g_2^2= {137 \over 1} \hbar \ c[/tex]

As for the 4 pi factor, it seems, looking at Dirac's paper, that it was because some other publications use h instead of \hbar, and this use has propagated until today.

I need to locate the paper where it is argued that n=2 is the minimum case.
 
  • #440
arivero said:
In the spirit of keeping this thread for numbers and leaving (most of) the standard theory in other hands, I have hitchhiked to the thread

https://www.physicsforums.com/showthread.php?t=240247

to discuss on EM duality and like. Here

One can of course consider the continuous spin-density distribution of an electron
field as a distribution of parallel Dirac strings. (Which are basically solenoids)
In one way or another this could lead to charge quantization in the Dirac sense.

Jackson, in section 6.12 mentiones this issue of n versus n/2. There are semi-classical
derivations from Saha (1936) and Wilson (1949) of the Dirac condition which lead to n.


Regards, Hans
 
  • #441
Hans de Vries said:
One can of course consider the continuous spin-density distribution of an electron
field as a distribution of parallel Dirac strings. (Which are basically solenoids)
In one way or another this could lead to charge quantization in the Dirac sense.

It seems that the interaction between "Dirac-Schwinger-Zwinger-Winger" quantization and topological solutions of electromagnetism is a well known candidate to fix the fine structure constant. Ketov 9611209v3 starts his lecures on Seiberg-Witten underlining that "the sole existence of duality symmetry allows one to exactly determine the critical temperature which must occur as the self-dual point where K=K* or sinh(2J/k T)=1". And, more in our electromagnetic constext, Julia and Zee 1975 tell that

If for some reason ... can only take on discrete values ... and if the argument of Schwinger and Zwanziger is relevant for the present case, the one would apparently be faced with a misterious condition saying that the theory will only make sense for some definite value of e^2
 
  • #442
arivero said:
It seems that the interaction between "Dirac-Schwinger-Zwinger-Winger" quantization and topological solutions of electromagnetism is a well known candidate to fix the fine structure constant. Ketov 9611209v3 starts his lecures on Seiberg-Witten underlining that "the sole existence of duality symmetry allows one to exactly determine the critical temperature which must occur as the self-dual point where K=K* or sinh(2J/k T)=1". And, more in our electromagnetic constext, Julia and Zee 1975 tell that

Stimulating to follow this line are Tonomura's video's of real life toplological solutions.

http://www.hqrd.hitachi.co.jp/global/movie.cfm
https://www.amazon.com/dp/9810225105/?tag=pfamazon01-20

Specially movie no. 5 with shows the annihilation of "particle/ anti-particles" (vertices/anti-
vertices) There's a close relation of these vertices with the Dirac's charge quantization.

Concerning introductions in the major QFT textbooks, there's Ryder and more recently
Zee for those interested in the general topic.


Regards, Hans
 
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  • #443
Damn, it is alpha=1

It just dawned on me: the square root of 4 pi is not a normalisation factor nor a volume factor; it is the charge of the electron when the fine structure constant is unity.

So Hans formula is

[tex]
e_{\alpha=1/137...} + g_{\alpha=1/137...} = e_{\alpha=1} \ {\bf e}^{\pi^2 \over 4}
[/tex]

where
[itex]e_{\alpha=1/137...}= \sqrt{4 \pi \alpha}[/itex] is the electric charge of the electron
[itex]g_{\alpha=1/137...}= 4 \pi / e_{\alpha=1/137...}[/itex] is the DSZ charge of the monopole.
[itex]e_{\alpha=1}= \sqrt{4 \pi}[/itex] the charge of the electron in the fixed point of duality (alpha=1)

Note that if we restore units h,c, they appear congruently in the three terms, because the 4pi of DSZ inversion is actually 4pi hbar c, or 2hc

As for the scaling/rotation factor, the two best ansatzes I have got are still
[tex] {\bf e}^{\pi^2 \over 4} = (\sinh {\pi^2 \over 4} + \cosh {\pi^2 \over 4} ) [/tex]
and
[tex]\sqrt {4 \pi} \ {\bf e}^{\pi^2 \over 4} = \int {\bf e}^{\pm \frac 12 \pi x } {\bf e}^{-{x^2 \over 4}} dx = k \int_{-\infty}^{\infty} {\bf e}^{\pm \frac 12 k \pi x} {\bf e}^{-{k^2 x^2 \over 4}} dx [/tex] for any k.

the idea of the second one is that the integral is the square modulus of an Euclidean wavepacket (thus of imaginary wavenumber i k pi). Wick rotating back to Minkowski, the wavenumber gets real, its exponential regains the usual imaginary factor, and then its "Minkoswki" square modulus is
[tex]k \int_{-\infty}^{\infty} {\bf e}^{-{k^2 x^2 \over 4}} dx = \sqrt {4 \pi}[/tex]
 
  • #444
it is h, not pi!

Er did I say "natural units". Of course in natural units 1=hbar = h/ 2 pi. So a new conjecture for the formula is

[tex]

e_{\alpha=1/137...} + g_{\alpha=1/137...} = e_{\alpha=1} \ {\bf e}^{h^2 \over 16}

[/tex]

It has a satisfactory point, all the objects are now quantum objects. But it has a couple problems: it is not clear how units are rebuilt in the exponent, and the 16 is a pity; I had hoped it to disappear too :-(.

In any case, the point is that the pi is not coming from the compactification of a KK circle: it is coming directly from Heisenberg principle when applied to the wavepacket. The wavepacket is actually a regularisation family for a dirac delta; one could even hope that in the cero radius limit the exponential factor could be the same for every regularisation; it is perhaps too much to ask for.
 
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  • #445
For some reason, the semiclassical approximation to QCD shown in this paper seemed to m to be germane to the alpha calculation:


Yang-Mills Propagators and QCD
Marco Frasca
We present a strong coupling expansion that permits to develop analysis of quantum field theory in the infrared limit. Application to a quartic massless scalar field gives a massive spectrum and the propagator in this regime. We extend the approach to a pure Yang-Mills theory obtaining analogous results. The gluon propagator is compared satisfactorily with lattice results and similarly for the spectrum. Comparison with experimental low energy spectrum of QCD supports the view that  resonance is indeed a glueball. The gluon propagator we obtained is finally used to formulate a low energy Lagrangian for QCD that reduces to a Nambu-Jona-Lasinio model with all the parameters fixed by those of the full theory.
http://arxiv.org/abs/0807.4299
 
  • #446
hans de vries said:
at least the new 2008 value for alpha from gabrielse/kinogarbagea begins to
solidify the n=3 term which now gives some convidence in the whole series.

[tex]\alpha\ =\ 1/137.035999084 (51)[/tex]

the alpha "radiative" series:

[tex]\begin{aligned}&. \\ &.~~~~\sum_{n=0}^\infty~\frac{\mbox{\huge $\alpha$}^{n-\frac{1}{2}}}{(2\pi)^{b_n}} ~~ = ~~ \mbox{\huge $ e^{\frac{~\pi^2}{4}}$}~~~~. \\ &. \end{aligned}[/tex]

where bn is the binominal series 0,0,1,3,6,10.. With successive increments
of 0,1,2,3,4...

apparently somebody solved it for 1294 digits :smile:

http://science6.2ch.net/test/read.cgi/sci/1091534329/l50

Code:
[size="2"]
1/α=
137.0359990958 2970048964 7400982482 4649832472 5408221072 8280453419 8236353775 3291508251 5164063679 7696140100
    0562638972 3955307219 1131054898 8411966378 9253597119 3431450591 9378868411 2253831430 9814642191 1084940907
    8132259876 7066274120 1689553474 9369387943 1203244561 3909631074 6327549332 6111965801 5656960227 1639347243
    0757007082 5600347556 4972841179 8177526459 2238581338 1031279567 7161638183 9778886042 0778973428 2353736066
    0555206708 9800179254 4185014141 4229664797 4602586290 4504613346 3047999465 7523213673 2504266689 6696242124
    0527823171 5115527412 2534727726 0326504468 2081772605 6146751666 7986300114 2671896058 3800062750 9682551272
    0001496187 4633319304 9496438649 4087857886 0022706370 1852275955 5736144810 3351280327 2115723542 9277212344
    6351827920 3207784574 9582376454 9389618767 5666772174 2641243315 1604981948 7898184100 7657369800 7939844563
    0773213335 5609920096 1118610374 0331148902 0684994710 0221083527 2203073949 1900453584 3300826440 5980415900
    4846789648 2011040441 7960811223 2721740912 7687806462 3818793279 0079147864 9620502089 5698441703 6636685100
    2052169799 1019507098 3686916619 8730033698 5715452876 0094017710 1259020172 1556975359 8282754919 5804007348
    2935098701 8839649369 8755374879 7396746681 2007580127 6048718855 2240202268 0518075853 2654287399 3572658304
    7624575104 9403524357 1121045001 8139733681 5436362649 7016835615 5476863905 7693063483 1562871355 1・・・ [/size]

Regards, Hans
 
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  • #447
from post #342:

CarlB said:
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:

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

where [tex]\epsilon = 0.22222204717(48) - 2/9[/tex] and I've left off an overall scaling factor. Since [tex]2(n+3m)\pi/3 = 2n\pi/3 + 2m\pi[/tex], 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 [tex]\sqrt{2}[/tex] is what Koide found in 1981, the [tex]\epsilon[/tex] is what I found a year ago.
.

it allows 4 generations if you count m=0
 
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  • #448
Wow...that post you are responding to is two years old.

There is no massless 4th generation of leptons.
 
  • #449
just in case it wasnt clear here is the koide formula

2faa10c9790027937a525b23039fc688.png


zero can obviously be added to both numerator and denominator without changing anything.

http://en.wikipedia.org/wiki/Koide_formula

I don't know if that changes anything or not. the math in the rest of this thread is far beyond me.
 
  • #450
Vanadium 50 said:
Wow...that post you are responding to is two years old.

There is no massless 4th generation of leptons.

Still, from time to time I wonder if there is a massless 3rd generation of leptons. Meaning that if the electron mass is put equal to zero, the masses of muon and tau still are near of Koide's conditions. On the other hand, in the limit where all the e,u,d generation is massless, the pion should be, and it strikes me: in this limit, it should have the same mass that the electron. In the actual, broken situation, it has almost the same mass that the muon. Besides, if it had exactly the same mass that the muon, the charged pion would be an stable particle.
 
  • #451
granpa said:
from post #342:
it allows 4 generations if you count m=0

If you'll look carefully at the formula, you will find that putting 4 into the equation gives you the same masses as if you'd put 1. This is because trig functions repeat every 2 pi.

I'm preparing a submission for Phys Math Central on the subject of Koide's mass formulas for the hadrons. I'm supposed to have something ready by January 10th. The latest cut is here:
http://www.brannenworks.com/koidehadrons.pdf

Carl
 
  • #452
I was using this formula:

2faa10c9790027937a525b23039fc688.png


a mass of zero can be added to numerator and denominator without changing anything.
 
  • #453
CarlB said:
I
I'm preparing a submission for Phys Math Central on the subject of Koide's mass formulas for the hadrons. I'm supposed to have something ready by January 10th. The latest cut is here:
http://www.brannenworks.com/koidehadrons.pdf

Carl

Hi Carl,

I would use some more group notation. Instead of your 6x6=36, I'd vote by 6 x bar6 = sum of irreps.

Second, besides to the "extraordinary" case you put, I'd include more of the "Ordinary" case, meaning the original argument of Koide about his formula for composite particles, along the lines of the original PhysRev.
 
  • #454
there are obviously an infinite number of formulas relating the masses of the electron muon and tauon. the beauty of the koide formula is its simplicity and elegance. especially the fact that Q=2/3 which is exactly halfway between the upper and lower limits of 1 and 1/3.

as wikipedia puts it:
Not only is this result odd in that three apparently random numbers should give a simple fraction, but also that Q is exactly halfway between the two extremes of 1/3 and 1.

the formula involving cos given above seems to me to lack this elegance. it should be obvious that anyone can always find a formula involving cos that will pass through any 3 points quite easily.
 
  • #455
arivero said:
Hi Carl,

I would use some more group notation. Instead of your 6x6=36, I'd vote by 6 x bar6 = sum of irreps.

Hmmm. The only place I've got 36 in the thing right now is the calculation for the square of a set of MUBs. In terms of color, this would be
[tex](3+\bar{3})\times (3+\bar{3})[/tex]
i.e. (R+G+B+/R+/G+/B)^2,
and I guess the symmetry would be, (hems and haws whilst trying to recall Georgi), uh, let's see, maybe
[tex](3+\bar{3})\times (3+\bar{3})
= 8+\bar{8}+6+\bar{6}+3+\bar{3}+1+\bar{1}[/tex]?

Yes, as usual, your comment is brilliant and I will incorporate it.

In a certain sense, what I'm doing is taking the
[tex]3\times \bar{3} = 8+1[/tex]
definition of gluons and claiming that I can compute in qubits with it using 3x3 matrices. [If you can read this, then PhysicsForums let me edit out a silly comment about 6+3-bar.]

Because both can be faithfully represented by 3x3 complex matrix addition and multiplication. From the symmetry point of view, it's a matter of selecting a certain ratio of the coupling constants.

Second, besides to the "extraordinary" case you put, I'd include more of the "Ordinary" case, meaning the original argument of Koide about his formula for composite particles, along the lines of the original PhysRev.

You mean the 1982 article I suppose. I guess I'll have to hike over to the university and read the article. Somehow I don't think I've ever downloaded it. One of the things I'm trying to do is to avoid talking about the preon model. There are two reasons for this. First, it doesn't apply to hadrons, and second, it gets into hot water with Coleman Mandula or (worse) spin statistics.

I think I should be ready to release a first cut in a few days. It still doesn't have "results" and the abstract is obsolete. But I've redone the results section to be make an argument that hangs together better.
 
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