All the lepton masses from G, pi, e

[arivero]

Let me put numbers, from PDG 2004, into the formula

$$\Gamma_{\pi^0}= \Big({M_{\pi^0} \over M_{A} }\Big)^3 \; \Gamma_{A}$$

where mass of pi0 is 0,134976 GeV and we expect to find a value for its gamma around experimental
Gp(exp) =0,000 000 007 8 (-0 5, +0 6) GeV
or theoretical
Gp(the) =0,000 000 008 10 (+-0 08) GeV

Well, I found two valid neutral (but no scalar) particles.

First, for the aforementioned Z0, we have M= 91,1876 , G=2,4952 (all the values in GeV) and it results in a prediction
Gp(Z0) =0,000 000 008 1 (+-0.000 000 000 0)

And second, for J/Psi ! We have M=3,096916 G=0,000091 and
Gp(J/P) =0,000 000 007 5 (+-.000 000 000 3)

No bad. It should be nice to have a theory backing it.

Ah, the other neutral boson, the upsilon $$\Upsilon$$, does not fit, but it is possible to make use of MacGregor's alpha pattern and tweak a bit with the data; the best fit is with the least bounded state, upsilon(3), that after multiplication times alpha gives 0.000 000 007 9.


EDITED, VERY TECHNICAL: One argument is that as vectors can not decay to two photons except virtually, they compose the whole decay amplitude. As for Upsilon, perhaps it is decaying using the square anomaly instead of the triangular one.

 

[Kea]

alpha and the golden ratio

The following may not be in the same league as the main contributions to this thread, but I think it's cool. It doesn't work to many significant figures. Now, never mind why I'm doing this, but if we let $\phi = 1.618 \cdots$ be the golden ratio and we try to relate $\alpha$ to the Temperley-Lieb $d$ factor, we come up with $\alpha = 137.08$ from

$$\frac{\sqrt{\alpha}}{2} = e^{\frac{2 \pi}{2 + \phi}} + e^{\frac{- 2 \pi}{2 + \phi}}$$

Cheers
Kea

 

[arivero]

Kea, you can prettify your formula, imho, by writting it as the square of a hyperbolic cosine... and adding a reference to that Temperley-Lieb factor; I am afraid it is not a object of my mathematical bagage. In any case it does not sound as bad, against the measured 137.0359, if it were the first term of a series.

Let me note that the updated Wolfram-Weisstein webpage has a pair of recent approximations, formula 6 and 7. They refer to work from Bailey and http://www.lacim.uqam.ca/~plouffe/.

About alpha from a physics point of view, I do not remember if we quoted already here Adler's program, the unsuccessful idea of looking for an eigenvalue equation for it: href=http://prola.aps.org/abstract/PRD/v5/i12/p3021_1[/URL]) of experimental values of alpha which are interesting to see (they are quoted in Peskin/Schroeder about pages 90-100, depending on edition)

Unrelated to this... I have started a note on the pion relationship, you can see the [PLAIN]http://dftuz.unizar.es/~rivero/research/pion.pdf

 

[arivero]

My last formula in hep-ph/0507144 involved a summatory in all the fermions the Z0 can decay to,
$$\sum_f C_f (|V_f|^2 + |A_f|^2)$$
where C is a coefficicient 1 for leptons, 3 for quarks, and V and A are the vector and axial couplings.

After releasing it, I read again the calculation of the Z0 decay width in "QCD and Collider Physics" (Ellis-Stirling-Weber) and I noticed that, when summed for a complete generation of fermions, this coefficient is very near of the number e.

 

[arivero]

Let me describe the calculation. The A coupling is always $$\pm 1/2$$, but the V coupling depends of charge and Weinberg angle. Table 8.3 in the above quoted book lists the values $$V_u \sim +0.191, V_d\sim -0.345, V_\nu=0.5, V_e=-0.036$$. And as I said, C is 3 for quarks (it receives corrections if the colour coupling constant is enabled, but the book does not worry about this, so neither me) and 1 for leptons.

So the sum is 2.717814
versus e^1= 2.71828... it makes a 0.017% discrepancy.

Now, the book calculation was for (running) sin^2 of Weinberg angle 0.232 and the intermediate rounding was against us (we get 2.7181 if we use this angle without the intermediate rounding). Let's forget about the possible corrections from C, and ask for which value of Weinberg angle is this equality exact. We need then the formulae,
$$A_f=T^3_f$$
$$V_f=T^3_f-2Q_f \sin^2 \theta_W$$
and we get $$sin^2 \theta_W = 0.231948$$

Of course, we are speaking of Z0 couplings for Z0 decay, so the value of angle is to be taken to this scale, in distintion to Hans approximation early in this thread, which was unrenormalized. Check table 10.5 of the pdg review.

Hmm if we believe both Hans estimate plus the exactitude of this relationship, the running of Weinberg angle could be interpreted as a prediction of the mass scale of Z0. Then in turn the mass of W is predicted (via Weinberg angle again!) and then the scale of electroweak vacuum follows.

On other hand, it should be remarked that the minimum of this expression, 2.5, is got for Sin^2 W equal to 3/8, a value known from GUT theories, and that has the additional virtue of cancelling the top [,up, charm] V_f couplings (but not A_f ones, of course).

 

[arivero]

I was worried because in this thread we have got two very high precision inventions for $$\sin^2 \theta_W$$ and they are very different in spirit; the one by Hans about middle way in the thread, the other here in the message above.

But now I think about, this later one is about Weinberg angle as parametrisation of the coupling constants while the first one (Hans will correct me if I am wrong) was in the scent of the mass quotient between Z and W. One number can be calculated directly from the another if SU(2)xU(1) is broken via the minimal Higgs mechanism, but they are different concepts and it is possible to have different unrelated ways to estimate them. Uff.

Now I think about, I haven't put here explicitly the estimate from previous message yet; it is
$$\sin^2\theta_W=\frac 38 \Big(1-\sqrt\frac 23 \sqrt{(\sum_{n=0}^\infty {1\over n!}) - \frac 52}\Big)$$

Or, if you prefer
$$4(1-2 \sin^2\theta_W + \frac 83 \sin^4\theta_W) =\sum_{n=0}^\infty {1\over n!}$$
Or
$$4 ( \cos^4\theta_W + \frac 53 \sin^4 \theta_W) =\sum_{n=0}^\infty {1\over n!}$$

or
$$\ln ( 4 (\cos^4\theta_W + \frac 53 \sin^4 \theta_W)) -1 =0$$

 

[arivero]

I was thinking to start a new thread, but well anyway... this last number, 0.231948... when checked against the aforementioned table 10.5 selects only a particular measurement: 0.23185(18), the one from the Forward/Backward asymmetry!

Why the exclamation marks? because this measurement is a well known problem of standard model fits; depending of the data it stands between 2.9 and 3.5 sigmas of the expected value, no bad enough to claim new physics, but no good enough to consider it in the same bag than the rest. And while the NuTeV anomaly comes from only an experimental group, this one comes from different teams, so it is usually assigned to "unknown systematic error".

Hans' number from posting #44, 0.2231013... , when used for the second column of table 10.5 selects the first three measurements, ie the accepted standard model value.

Perhaps it is worthwhile to remember that the relationship between Weinberg angle (ie the mixing of couplings in Weinberg model) and the masses of Z and W depends on the structure of the Higgs sector; the usual M_W/M_Z is a result for the minimal Higgs.

 

[Jay R. Yablon]

Hans, Please view the attachment (if this works right), I have summarized some possible relationships between your Fine Stucture Constant Work and my Earlier formula for the Tau Electron mass.

Jay.

 

[Jay R. Yablon]

New Paper: Magnetic Monopoles and Duality Symmetry Breaking in Maxwell's Electrodynam

Hello to all:

I wanted to let you know about my new paper just posted at
http://arxiv.org/abs/hep-ph/0508257, titled Magnetic Monopoles and Duality
Symmetry Breaking in Maxwell's Electrodynamics.

This paper summarizes the main direction of my research over these past
eight months.

The abstract is as follows:

It is shown how to break the symmetry of a Lagrangian with duality symmetry
between electric and magnetic monopoles, so that at low energy, electric
monopole interactions continue to be observed but magnetic monopole
interactions become very highly suppressed to the point of effectively
vanishing. The "zero-charge" problem of source-free electrodynamics is
solved by requiring invariance under continuous, local, duality
transformations, while local duality symmetry combined with local U(1)_EM
gauge symmetry leads naturally and surprisingly to an SU(2)_D duality gauge
group.

As regards this thread, I note the extremely accurate formula (no "mere coincidence" in my
view) that Hans has posted for the fine structure constant and the
appearance of 2pi and (pi/2)^2 as a key numeric drivers in this formula. In section 6, I call your attention particularly
to equation (6.7) which brings pi into the fine structure formula based on
phenomenological origins, and to the derivation leading to this including
the Dirac Quantization Condition expressed as (5.22) where both the 2pi and
the a^.5 (the electric charge) are apparent ingredients. Regarding the
(e^pi/2)^2 factor, pi/2 for the "complexion" angle I am using in this work
is what rotates between the electric and magnetic monopoles. And, with some
rejuggling of (5.22), one can clearly get pi^2 factors to show up there.
The exponentials do not yet have a clear phenomenological origin, but, since
running couplings a connect via a log relationship to probe energy u, that
is, ln u ~ a, so that u ~ e^a, one can conceivably get this exponential onto
phenomenological footing as well once a connection is made to probe
energies. The connection to probe energies, I will make the topic of a
follow up paper.

In other words, the complexion angle, which is central to this
paper, is a direct function of the fine structure constant a, but is an angle
of rotation which naturally introduces factors like 2pi and pi/2. I have a
strong suspicion that this might provide a basis for going beyond
"mathematical fitting" of the numbers (I am being careful not to use the
word "numerology" because of its negative connotations) to explain how Hans' Fine Structure Constant
mathematics can possibly be given a phenomenological origin.


I would be interested in any feedback, public or private, that you may wish
to provide.



Jay R. Yablon
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com

 

[Kea]

Hi Jay

Very interesting! I see on page 18 you begin to discuss monopole masses. I assume you know that one of the LHC experiments will be searching for such things in the right ballpark...

http://moedal.web.cern.ch/moedal/

 

[Jay R. Yablon]

Hi Kea:

Yes, thank you, a few people have pointed the LHC plans out to me.

I do want to clarify for the readers of this post that the ~ 2.35 TeV is the predicted mass of the vector boson which mediates magnetic monopole interactions (assuming the Fermi vacuum expectation value applies here), and is not the mass of the magnetic monopoles themselves, which thus far are not predicted.

Jay.

 

[arivero]

Funny. Terry Prachett's "Guards! Guards!", a old book of the Discworld series, contains a formula including Planck Length, alpha, and the quotient between proton and electron mass. It seems thus an old maladie.

I will try to locate the formula and plot it here, but the Spanish translation has typographical problems, so if someone can check the UK one please do.

 

[Hans de Vries]

A short latex style paper on the alpha result discussed earlier on this thread:

A simple radiative series leading to the exact value of the fine structure constant


Regards, Hans.

 

[arivero]

Hans, my problem with your result, and the secret because I did not concentrate on it, is that I am not able to see how standard the method of successive differences is. If could be good if you had time to put some examples using other ansatzes, ie what number will we get if instead of (1) we take $$\sqrt \alpha \approx e^{-\pi}$$? or if we take $$\sqrt \alpha \approx \pi^2$$? Or $$\sqrt \alpha \approx \alpha^2$$? Neither I understand the systematics to get all these $$\pi$$s in denominators along the process of expansion. Do they come from the exponent in the exponential somehow?

 

[Hans de Vries]

Alejandro,

I'll add a second section to the paper demonstrating this successive
difference method. Indeed, currently only the end-result is presented
without any explanation.

Regards, Hans

 

[Jay R. Yablon]

New paper, may contain a solution to the NuTeV anomaly

Hello to all:

I am pleased to announce that my newest paper, "Magnetic Monopoles, Chiral
Symmetries, and the NuTeV Anomaly," has now been published at
http://arxiv.org/abs/hep-ph/0509223.

This paper is a follow up to my earlier publication at
http://arxiv.org/abs/hep-ph/0508257, and takes a closer look at the magnetic
monopoles themselves as fermionic particles. I have reported interim
progress along the way on the sci.+ boards; now you can see the full
picture.

This paper calculates widths and cross sections associated with the
predicted magnetic charge, and determines that there is a very slight
cross-section enhancement at sqrt(s) = M_z ~ 91 GeV due to magnetic
monopoles.

If one were to do experiments and NOT understand the magnetic monopole
origin of this small cross section enhancement, one might instead conclude
that the weak mixing angle had decreased for e/ebar scattering, in relation
to neutino/neutrino-bar scattering, by a small amount. How small? This
paper predicts a reduction of approximately .003, which is right near the
magnitude of the NuTeV anomaly and goes in the right direction as well.

Fundamentally, the NuTeV anomaly is thus seen to be the first experimental
evidence of the existence of the magnetic monopole charges, which have been
a mystery ever since Maxwell's era.

Also, some fundamental connections are drawn between the magnetic / electric
symmetries, and chiral symmetries.

If you want the quick tour, look at equations (9.12) to (9.15) which contain
the final numeric results. Then look at (8.16) through (8.20) which shows
these same results represented in term of the cross section enhancements
from which they were derived.

If you are doing NuTeV experiments, and even if not, look at (7.34) to
(7.44), which show the full and differential cross sections in the most
general form. This should help you with the NuTeV anomaly even if you don't
believe as I do that the magnetic monopole charge at least contributes to
this anomaly. Because these equations tell you how a vector boson (call it
the Z^u' if you wish) with mass > M_z would enhance the cross section
generally, whether the origin of that vector boson is from magnetic
monopoles or somewhere else. So, these give you a theoretical framework to
fit the data under a variety of assumptions that you may wish to make.

If you assume two or more massive bosons with mass > M_z, then there will be
further cross section terms for each new vector boson, as well as further
cross terms between pairs of vector bosons, the form of which can readily be
understood and deduced from (7.34) to (7.44). My own suspicion is that
there is also an electroweak-based Z^u' in the 1.3 TeV range in addition to
the M^u which mediates the magnetic monopole interaction here. This will
require extending the entire electroweak theory to consider weak and weak
hypercharge magnetic monopoles, and may well be the subject of my next
paper.

Once the cross section enhancement is known under whatever scenario one may
assume, the apparent impact on sin^2 theta_w can be deduced following the
steps shown in section 9. So, there is some good grist here for the NuTeV
folks. And for anyone who is interested in understanding magnetic monopoles
and chiral symmetries.

I also suggest a look at the conclusion.

From there, look at whatever you want.

Happy reading.

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com

 

[CarlB]

Jay, that is a cool paper. The mechanism you're suggesting for giving mass to the fermions seems to me to be similar to the old Zitterbewegung theory. My understanding is that this is in violation of Lorentz symmetry.

Another way of putting this: Your paper writes: "It may well be that here, the $$\psi_e = \psi_R$$ and $$\psi_m = \psi_L$$ are swallowing up one another to go from being two twocomponent fermions which are each massless, to being a single four-component Fermion which is massive."

This is a Higgsless mechanism for mass, a mechanism that I also believe in. The problem is Lorentz violation. To get an electron of spin +1/2 in the z direction, one must have a right handed electron moving +z and a left-handed electron moving -z. To move from one to the other is a violation of conservation of momentum. Any comments?

Carl

 

[Jay R. Yablon]

Hi Carl:

Thanks for your reply. I just saw your post or I would have replied sooner.

I guess I'd be interested in seeing mathematically how the Lorentz violation comes about in Zitterbewegung theory and how what I am doing matches up or not. It is a common practice to construct a four-component Fermion $$\psi$$ from the $$\psi_R$$ and $$\psi_L$$ using $$\psi = \psi_R + \psi_L$$, where Lorentz transformations essentially shift the relative magnitudes of the various components of the Dirac spinor, and in a sense, in addition of course to the (I believe, novel) connection with electric and magnetic charges, that is all we are suggesting here.

Beyond that, the comment about fermion mass is really just an aside from the main thrust of development. I'm curious about your overall view of this apporach to magnetic monopoles and whether you think we may be on to something for the NuTeV anomaly?

Thanks.

Jay.

Jay, that is a cool paper. The mechanism you're suggesting for giving mass to the fermions seems to me to be similar to the old Zitterbewegung theory. My understanding is that this is in violation of Lorentz symmetry.

Another way of putting this: Your paper writes: "It may well be that here, the $$\psi_e = \psi_R$$ and $$\psi_m = \psi_L$$ are swallowing up one another to go from being two twocomponent fermions which are each massless, to being a single four-component Fermion which is massive."

This is a Higgsless mechanism for mass, a mechanism that I also believe in. The problem is Lorentz violation. To get an electron of spin +1/2 in the z direction, one must have a right handed electron moving +z and a left-handed electron moving -z. To move from one to the other is a violation of conservation of momentum. Any comments?

Carl

 

[CarlB]

Dear Jay

I guess I'd be interested in seeing mathematically how the Lorentz violation comes about in Zitterbewegung theory and how what I am doing matches up or not.

I'm not sure if what you're doing matches up, but here's my understanding of the problem for my own version of fermion masses:

The problem with Lorentz violation when one attributes mass to left handed electrons turning into right handed electrons and vice versa appears when you write down the interactions in Feynman diagrams.

It's not called the "Zitterbewegung" problem in the literature. Since this method of giving mass to the electron violates Lorentz symmetry, it's not in the literature. The objection is what will happen when you talk to physicists about this sort of thing at a conference.

I believe that this method was known to Feynman, and must have been common knowledge back in the late 1940s. The reason for this is that Feynman implies that a version of this exists for adding mass to scalar particles (I seem to recall that they were massless Klein-Gordon). He wrote something like "Nobody knows what this means" with regard to the derivation. But the reference to this is not in the physics literature. Instead, it's a footnote for his excellent (high school math level) introduction to QED:

QED: The Strange Theory of Light and Matter
https://www.amazon.com/gp/product/0691083886/?tag=pfamazon01-20

As I said earlier, I believe that this is the correct way to add mass to massless particles (rather than Higgs). I partially wrote up a paper that included the Zitterbewegung creation of mass in QFT using (illegal) Feynman diagrams a year ago but it turns out that I've never released it. My hesitation to do so is entirely because of the problem with Lorentz invariance. I've talked about it over a napkin at a physics conference, to appreciative laughter (The idea is very simple in that it gives mass to the electron without any need for Higgs etc., but it is wildly heretical. The effect, to a physicist, is sort of like when you divide by zero in a proof you show to a mathematician.), but I've not written it up completely. Let me try and write it up here, but without the graphics.

The Zitterbewegung problem is that when you look for velocity eigenstates of the Dirac equation, one finds solutions only for speed c. For example:

"Even more surprising result [2] is obtained if we consider velocity $$\vec{\dot{x}}$$ in the Dirac formulation. Instantaneous group velocity of the electron has only values ±c in spite of the non-zero rest mass of electron. In addition, velocity of a free moving electron is not a constant of motion."
http://www.hait.ac.il/jse/vol0103/sep040706.pdf

Particles that travel at speed c are massless, so how does the electron get mass? Normally Zitterbewegung stuff is done at the QM level. It is when you take it to the QFT level and use it to explicitly create mass that you get into trouble with Lorentz invariance.

If the electron is to be made from massless subparticles, I suggest we take a bow to weak theory and assume that the electron is a composite made up of two particles that alternate, a left-handed electron and a right-handed electron. Let us work in the momentum representation. The propagators for a massless left/right-handed $$e_L, e_R$$ electron is:

$$i/p_L \; , i/p_R$$.

Notice that I'm leaving off the $$\gamma$$ stuff. You can add it back in, but the results is the same. That is, when you resum a simple set of Feynman diagrams you will turn a pair of massless spin-1/2 particles into a single massive spin-1/2 particle.

[Long post continued]

 

[CarlB]

I guess I'd be interested in seeing mathematically how the Lorentz violation comes about in Zitterbewegung theory and how what I am doing matches up or not.

[Long post continued]

Now consider the Feynman diagram that annihilates an $$e_L$$ and creates an $$e_R$$ with momentum $$p_R = p_L$$. This is a trivial type of Feynman diagram in that it has only one input and only one output. Add another diagram, one that does the reverse transformation. Together, these form a Zitterbewegung model for the electron.

Since we are considering the electron as a composite particle made up of a part e_L and a part e_R, we will have to have a two-component vector to model the electron. The two components of this vector will

correspond to the left and right handed parts of the electron. Since the two halves of the electron are scalar particles, we will require a vector of two complex numbers to model the electron. This is convenient because it is just the number of degrees of freedom we need.

To derive the propagator for the electron from the propagators for the e_L and e_R, we need to take into account the Zitterbewegung effect. That is, we need to take into account the interaction that converts a left-handed electron into a right-handed electron and vice-versa.

As is usual in QFT, we will need to resum the whole series of Feynman diagrams (that is, the series generated by the above two) in order to find the propagator for the electron. Fortunately, these two Feynman diagrams are particularly simple.

Let me write down the Feynman diagrams that we must resum, in order of their complexity:

$$e_L \to e_L$$
$$e_R \to e_R$$
$$e_L \to e_R$$
$$e_R \to e_L$$
$$e_R \to e_L \to e_R$$
$$e_L \to e_R \to e_L$$
$$e_R \to e_L \to e_R \to e_L$$
$$e_L \to e_R \to e_L \to e_R$$
...

To calculate these we need the usual Feynman rules. Note that since there are no loops, there is no need to do any integration. I need to specify a value to associate with the nodes, that is a probability, let us choose $im$.

Now I can resum the propagators. Note that the above series will resum not to a single propagator, but instead to 4. They will be the e_L -> e_L, the e_L -> e_R, the e_R -> e_L and the e_R -> e_R. Let's do

the e_L->e_L propagator first:

$$e_L \to e_L$$
$$+ e_L \to e_R \to e_L$$
$$+ e_L \to e_R \to e_L \to e_R \to e_L$$
$$+ ...$$

To sum these, remember the propagators defined above and that p = p_L = p_R.

$$L \to L =(i/p_L) + (i/p_L) I am (i/p_R) I am (i/p_L) + ...$$
$$L \to L =(i/p) + i (m/p)^2p + i (m/p)^4 p + ...$$
$$L \to L =(i/p)(1+(m/p)^2 + (m/p)^4 + ...$$
$$L \to L =\frac{i/p}{1-(m/p)^2}$$
$$L \to L =\frac{ip^2/p}{p^2-m^2}$$
$$L \to L =\frac{ip}{p^2-m^2}$$

I've left more stages in the algebra than necessary in order to hint where you will have to work harder to put the gamma matrices back in.

The p_R to p_R propagator will be similar:
$$L \to R =\frac{ip}{p^2-m^2}$$

The p_L to p_R propagator is as follows. The Feynman diagrams are:

$$e_L \to e_R$$
$$+ e_L \to e_R \to e_L \to e_R$$
$$+ e_L \to e_R \to e_L \to e_R \to e_L \to e_R$$
$$+ ...$$

These work out to be:

$$L \to R =(i/p_L) I am (i/p_R) + (i/p_L) I am (i/p_R) I am (i/p_L) I am (i/p_R) + ...)$$
$$L \to R = \frac{im}{p^2}(1 + (m/p)^2 + (m/p)^4 + ...)$$
$$L \to R = \frac{im}{p^2(1-(m/p)^2)}$$
$$L \to R = \frac{im}{p^2-m^2}$$

Similarly, R->L works out as:
$$R \to L = \frac{im}{p^2-m^2}$$

With all four propagators computed, we can put them together into a single equation using matrices. We assume our (spinors) consist of left-handed massless particles on the top part of a 2-vector, and right-handed massless particles on the bottom part. Then the desired resummation of the propagators is:

$$\left(\begin{array}{cc} L \to L \;&\; L \to R \\ R \to L \;&\; R \to R \end{array} \right)$$
$$=\left(\begin{array}{cc} \frac{ip}{p^2-m^2} \;&\; \frac{im}{p^2-m^2} \\ \frac{im}{p^2-m^2} \;&\; \frac{ip}{p^2-m^2} \end{array} \right)$$
$$=\frac{1}{p^2-m^2}\left(\begin{array}{cc} ip \; & \; I am \\ I am \; & ip \end{array}\right)$$

Uniting the left and right handed electron fields back to the same particle, one ends up with a propagator of:

$$\frac{i(p+m)}{p^2-m^2}$$

I tend to lose factors of i so you might check this. Also, note that the positron works out the same way and increases the size of the matrix from 2x2 to 4x4 as in the usual Dirac matrices.

Now in the above, there isn't any Lorentz violation. But if you consider an electron at rest with spin +1/2 in the z direction, for it to be composed of e_L and e_R portions, one must have that the velocities of these two subparticles be in opposite directions. This is incompatible with their momenta being identical. You can fix this by going back and making p_R = - p_L, but then you've lost conservation of momentum at the vertices of your Feynman diagrams.

Carl

 

[arivero]

This thread is now three citation steps away from a paper of Wilczek. It is cited by hep-ph/0505220 (Rivero-Gsponer), which is cited by hep-ph/0508301 (Koide), which is cited by hep-ph/0509295 (an article of Brian Patt, David Tucker-Smith and Frank Wilczek).

 

[Kea]

Koide formula

Perhaps someone has already commented on this, but it might be interesting to note the following:

The Koide mass formula

$$m_{e} + m_{\mu} + m_{\tau} = \frac{2}{3} (\sqrt{m_{e}} + \sqrt{m_{\mu}} + \sqrt{m_{\tau}})^{2}$$

can be rewritten in terms of the squareroots as

$$s_{e}^{2} + s_{\mu}^{2} + s_{\tau}^{2} = 4(s_{e}s_{\mu} + s_{e}s_{\tau} + s_{\tau}s_{\mu})$$

Via a simple change of coordinates

$$s_{e} + s_{\tau} = x_{1} \hspace{11mm} - s_{e} - \frac{s_{\mu}}{2} = x_{2} \hspace{11mm} \frac{s_{\mu}}{2} + s_{\tau} = x_{0}$$

the relation becomes

$$x_{1}^{2} = 4 x_{0} x_{2}$$

which is an instance (the $\rho = 0$ 'ground' case) of a coadjoint orbit for the obvious $SL(2,\mathbf{C})$ action on $\mathbf{C}^{3}$. In other words, the mass triplet is being described by a representation of (the double cover of) the Lorentz group, and the constraint on the vector is simply a quantization condition (a la Kirillov).

 

[arivero]

Via a simple change of coordinates

$$s_{e} + s_{\tau} = x_{1} \hspace{11mm} - s_{e} - \frac{s_{\mu}}{2} = x_{2} \hspace{11mm} \frac{s_{\mu}}{2} + s_{\tau} = x_{0}$$

Simple, but not uninteresting. One can rewrite as

$$2 s_\tau= x_0+x_1+x_2$$
$$2 s_e= -x_0+x_1-x_2$$
$$s_\mu= x_0 -x_1 -x_2$$

Or, for the masses
$$4 m_\tau= (x_0+x_1+x_2)^2=(x_0^2+x_1^2+x_2^2)+2(x_0x_1+x_1x_2+x_2x_0)$$
$$4 m_e= (-x_0+x_1-x_2)^2=(x_0^2+x_1^2+x_2^2)+2(-x_0x_1-x_1x_2+x_2x_0)$$
$$m_\mu= (x_0 -x_1 -x_2)^2=(x_0^2+x_1^2+x_2^2)+2(-x_0x_1+x_1x_2-x_2x_0)$$

The sums are,
$$m_\tau+m_e+m_\mu=\frac 32 (x_0^2+x_1^2+x_2^2) - 2 x_0x_1+2x_1x_2-x_2x_0$$
$$(s_\tau+s_e+s_\mu)^2=(x_0-x_2)^2=(x_0^2+x_2^2)-2x_0x_2$$

Hmm... I am tired. There is a mistake somwhere.

 

[Kea]

In other words

$$\frac{m_{\tau}}{m_{\mu}} = \frac{1}{4} \frac{(1 + y_{1} + y_{2})^{2}}{(1 - y_{1} - y_{2})^{2}}$$

$$\frac{m_{e}}{m_{\mu}} = \frac{1}{4} \frac{(- 1 + y_{1} - y_{2})^{2}}{(1 - y_{1} - y_{2})^{2}}$$

where $(y_{1},y_{2}) = (0.91,0.12)$

 

[CarlB]

Hmm... I am tired. There is a mistake somwhere.

Re:
$$s_{e} + s_{\tau} = x_{1} \hspace{11mm} - s_{e} - \frac{s_{\mu}}{2} = x_{2} \hspace{11mm} \frac{s_{\mu}}{2} + s_{\tau} = x_{0}$$

This makes the left hand side and right hand side of:

$$x_{1}^{2} = 4 x_{0} x_{2}$$

have differing signs.

Also, what is the "obvious SL(2,C) action on C^3".

[Edit]Oh, is it that you take the usual representation of SL(2,C) in 3x3 matrices and do matrix multiplication. As in

$$\left(\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{array}\right)$$

$$\left(\begin{array}{ccc}0 & 0 & 0 \\ 2 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right)$$

$$\left(\begin{array}{ccc}2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -2 \end{array}\right)$$

[/Edit]

Carl

 

[arivero]

Hmm, no. In the LHS
$$x_1^2=s_e^2+s_\tau^2+2 s_e s_\tau$$

And in the RHS
$$4 x_0 x_2=-(s_\mu^2+ 2s_\mu s_e + 2 s_\mu s_\tau + 4 s_e s_\tau)$$

Thus keeping the sign as it is, moving the first term to the LHS and the last of the LHS to the RHS we get
$$s_e^2+s_\tau^2+s_\mu^2=-2 s_\mu s_e -2 s_\mu s_\tau -6 s_e s_\tau$$
Or changing the sign, we get
$$s_e^2+s_\tau^2-s_\mu^2=2 s_\mu s_e +2 s_\mu s_\tau +2 s_e s_\tau$$

I can not see how Kea's is equivalent to Koide's.

 

[Kea]

Oops - I'd better recheck sums!

 

[CarlB]

An argument for $$1/\alpha^3 = 2573380$$, based on requiring that the ratios of two actions be an exact integer is here:

http://www.fervor.demon.co.uk/noteto.htm

From:
http://www.fervor.demon.co.uk/

Meanwhile, I'm finding a suspicious relationship between Fibonacci series and the equation for the phase angle $$\delta$$ in my version of Koide's lepton mass relationship. After a bunch of theory, you get:

$$\tan(2 \delta)/\sqrt{3} = \kappa$$

then an important cosine is

$$\cos(\theta_B) = \frac{1-\kappa}{1+3\kappa} = 29/73$$

$$29 = F_3^2 + F_5^2$$
$$73 = F_4^2 + F_6^2$$

where $$F_N$$ is the Nth Fibonacci number, 1, 1, 2, 3, 5, 8 ...

I don't think that this is of any physical significance. It gives a value for $$\delta = .22226$$ that is too high. But I'm including it here for entertainment only.

Carl

 

[CarlB]

A Very Simple Empirical Neutrino Mass Formula
Wojciech Krolikowski
http://www.arxiv.org/abs/hep-ph/0510355

This references Koide's lepton mass formula, but the neutrinos don't follow the 1.5 factor, at least according to my calculator, even when you include possible negative square roots.

Note that this is another application of that interesting number, 29.

Also, my idea for using a Lorentz violation in a hidden dimension to arrange for the breaking of electroweak symmetry is no longer unique as a group in Europe is submitted a similar idea:
http://arxiv.org/abs/hep-ph/0510373

The difference is that my paper (so far it's at http://www.brannenworks.com/PPANIC05.pdf but it needs corrections) is written from the assumption that the wave equation is the fundamental physical object while their's is written with the usual Lagrangian / Hamiltonian formalism.

Carl

 

[arivero]

I happenned to give a look today to hep-ex/0511027, from the CERN. There in page 126 the new measure estimate to W mass is given as
$$m_W=80.392\pm0.039$$. It is still to be seen how this new fit enters in the next edition of the particle data group evaluation, but any influence will affect possitively the estimates in messages #41 to #88 of this thread (and leaves out the need for corrections such as the one of message #63). Let's wait.

Note also hep-ex/0509008v2 pg 224, where the prediction of $$m_W$$ from the rest of parameters of electroweak model is said to be calculated as 80.363$$\pm$$0.032

 

[arivero]

Funny, tonight I have found the first approximation of the equations starting this thread,
$$\ln {m_\mu \over m_e}= 2\pi - 3 {1\over \pi}$$
$$\ln {m_\tau \over m_\mu}= \pi - {1\over \pi}$$
in
Andreas Blumhofer, Marcus Hutter Nucl.Phys. B484 (1997) 80-96
http://arxiv.org/abs/hep-ph/9605393

 

[CarlB]

Arivero,

I guess I should mention that I'm soon to release a paper that describes an alternative method of calculating probabilities than the spinor method.

Basically, the idea is to go back to the density matrix and write probabilities using geometric principles directly from the density matrix, ignoring the fact that it was derived from a spinor. I can show that when one does this, if one chooses a restricted type of representation for a Clifford algebra, then one obtains the same probabilities that are seen in the usual spinor representation. But it turns out that ALL the representations that one sees in the literature are of these restricted type.

The usual spinor probabilities, when translated into density matrix form, become traces. The way I'm calculating probabilites is with the geometric squared magnitude instead.

As an example, consider some 2x2 matrix over the complexes. Such a matrix can always be written as a sum over complex multiples of unity and the Pauli spin matrices. That is:

$$M = \alpha_1\hat{1} + \alpha_x\sigma_x + \alpha_y\sigma_y + \alpha_z\sigma_z$$

where $$\alpha_\chi$$ are complex constants. Then define

$$|M|^2_G = \sum_\chi |\alpha_\chi|^2.$$

If you happen to have two pure density matrices, for example $$\rho$$ and $$\rho'$$, then the probability of transition between the two states described by the two matrices is

$$P_{\rho,\rho'} = \tr(\rho\rho')$$

It turns out that the above probability is proportional to $$|\rho\rho'|_G^2$$. Furthermore, the same thing applies to all the usual representations of the Dirac algebra. The constant of proportionality is just $$tr(\hat{1})$$ or 2 for the Pauli algebra and 4 for the Dirac algebra.

However, there do exist representations where the geometric probability is not equivalent to the spinor probability (or the trace), and these apply to the masses of the fermions.

As an example of the "general" representations where the geometric probability is not equivalent to the standard ones, one must understand more about representations than I can put into this short note. I'll put it into a successor.

The reason one ends up with square roots in the mass matrix is because the fundamental probability relation, when defined in terms of spinors, is 4th order. That is, P = <A|B><B|A> has four contributions from spinors. The natural fields are the density matrices, and in that representation, probabilities are proportional to squares (or products) of the objects, as we would expect. The usual requirement that a wave function be normalized, <A|A>=1 would better be written as <A|A><A|A> = 1.

Carl

 

[CarlB]

A geometric characterization of the representation of a Clifford algebra in matrices using primitive idempotents (ideals):

We can fully characterize a representation of a Clifford algebra in complex matrices NxN by knowing the Clifford algebra equivalents of N+1 simple matrices that are all primitive idempotents. There are N "diagonal primitive idempotents". The first N are just the diagonal matrices that have a single one somewhere along the diagonal and are everywhere else zero. We will label these by $$\iota_j$$ where "j" gives the position on the diagonal, 0 to N-1, where the one is. But this is not quite enough to specify a representation. To complete the specification, we can add the "democratic primitive idempotent". This is the matrix that has 1/N in every location, which we will call $$\iota_D$$.

To get any specific position on the matrix, for example the (j,k) spot, we can simply multiply:

$$M_{jk} = \iota_j \; \iota_D \; \iota_k$$

Thus it is clear that if you know these N+1 elements of the Clifford algebra, then you know the representation.

Now the odd thing about ALL the usual representations of Clifford algebras used in physics is that the degrees of freedom contained in $$\iota_D$$ are completely independent of the degrees of freedom contained in the set of $$\iota_j$$. For example, in the Pauli algebra, the diagonal primitive idempotents have z as a degree of freedom, while the counter diagonal elements have x or y. These are completely independent, in terms of degrees of freedom. The same thing is true for the various representations of the Dirac matrices that appear in the literature.

Now it turns out that it is possible to make representations where there are degrees of freedom shared between the diagonal and the democratic primitive idempotents. And it is these representations where the geometric probability comes out different from the usual trace.

To explain more about this, I would have to go into more detail on how one obtains the primitive idempotents (or primitive ideals) of a Clifford algebra. But it's very beautiful and I can't wait to release it. Right now, I'm working on designing a biodiesel plant design instead.

Perhaps I should remind the reader that the "democratic mixing matrices" also appear naturally in the Koide mass formula.

Carl

 

[arivero]

Right now, I'm working on designing a biodiesel plant design instead.

You mean you are being paid? Interesting, even if offtopic. Lot of deforestation will follow (as it is said of soja beans and of palm oil).

Perhaps I should remind the reader that the "democratic mixing matrices" also appear naturally in the Koide mass formula.

Yeah, you should . It has been a long time ago since this thread was alive, and without this reference it seems a bit off.

 

[CarlB]

Now it turns out that it is possible to make representations where there are degrees of freedom shared between the diagonal and the democratic primitive idempotents. And it is these representations where the geometric probability comes out different from the usual trace.

I should exhibit a representation of SU(2) which does not satisfy the requirement that the geometrically defined squared magnitude is equivalent to the usual spinor squared magnitude (and therefore to the traces). So here it is:

$$\sigma_x = \left(\begin{array}{cc}0.0&0.2\\5.0&0.0\end{array}\right)$$

$$\sigma_y = \left(\begin{array}{cc}0.0&-0.2i\\5.0i&0.0\end{array}\right)$$

$$\sigma_z = \left(\begin{array}{cc}1.0&0.0\\0.0&-1.0\end{array}\right)$$

This corresponds to a democratic primitive idempotent of

$$\iota_D = (1.0 + 2.6\sigma_x + 2.4\sigma_x\sigma_z)/2.0$$

and the diagonal primitive idempotents are the usual:

$$\iota_\pm = (1.0 \pm \sigma_z)/2.0$$

From the above, you can see that the degrees of freedom of both the diagonal primitive idempotents and the democratic primitive idempotent include the sigma_z. That is, (sigma_x)(sigma_x sigma_z) = sigma_z so that the democratic primitive idempotent encroaches on the turf of the diagonal primitive idempotents.

To see that the above violates the relationship between squared magnitude and spinor squared magnitude one may consider the element |x><x| where |x> is the spin-1/2 eigenvector for spin in the x direction. Of course <x|x> = 1, but ||x><x||x><x|| in the geometric language is very large.

Oh, and I get paid for the biodiesel design if the design gets accepted by the customer. As far as deforestation, this shouldn't contribute much. It wil use US oil crops and US surplus oils (i.e. yellow grease, etc.).

In the tropics, its pretty clear that the oil palm is the way to go. Those things are prolific. But I really don't think science has found the optimum oil producing plant to grow in the US type climate.

Carl