All the lepton masses from G, pi, e

[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

$$s_{e}^{2} + s_{\mu}^{2} + s_{\tau}^{2} = 4(s_{e}s_{\mu} +<br /> s_{e}s_{\tau} + s_{\tau}s_{\mu})$$

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

 

[arivero]

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:

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

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

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

It is more a sort of deformation of SU(2) than a representation of [the infinitesimal generators of] SU(2), isn't it? People expects selfadjointness to get unitary generators, thus unitary group, which is the thing that the U stands for, after all. Just terminology issue, but it can be a communication problem.

 

[CarlB]

I'm sure I've been sloppy with my terminology. I tend to be a very calculation oriented person. Let me try again. SU(2) is the group, su(2) is the algebra.

SU(2) refers to the fact that one can represent the elements of the group by special unitary 2x2 matrices. But that is not the only representation of the group, nor are the Pauli matrices the only representation of the algebra su(2).

The set of deformed matrices are a representation of su(2) in that they satisfy the usual commutation relation:

$$\sigma_j\sigma_k-\sigma_k\sigma_j = 2i \epsilon_{jkm}\sigma_m$$.

If we wanted to describe the SU(2) that this defines, then we just take exponentials of the above. But since the above is identical to the usual su(2), we'll just get a deformation of the usual SU(2). Uh, what's the correct word for "deformation"? In this case it's something like isomorphism or homeomorphism or...

By the way, the better (i.e. geometric) way of writing the above relation is to replace i with $$\sigma_x \sigma_y \sigma_z$$.

Carl

 

[arivero]

But since the above is identical to the usual su(2), we'll just get a deformation of the usual SU(2). Uh, what's the correct word for "deformation"? In this case it's something like isomorphism or homeomorphism or...

Hmm I think it is not going to be an isomorphism; supposse that in the original representation of SU(2) I build unitary elements $$A_i=\exp ( a_i^\mu \sigma_\mu)$$ from the sigmas and triplets of real numbers; this is the standard business. Now you should have an invertible map from this real triplet in in the standard representation to the real triplet components in your representation, ie a map $$(a^\mu) \leftrightarrow (b^\mu)$$. And then in order to check that you representation it really an iso-, and not a deformation, you should be sure of the details, for instance that a composition of any three elements giving the identity, $$A_1 A_2 A_3 = I$$ in the original representation should still give the identity in your mapped representation. I doubt it, because your sigmas are not idempotent anymore. So let it to be called "deformation".

 

[arivero]

By the way, the better (i.e. geometric) way of writing the above relation is to replace i with $$\sigma_x \sigma_y \sigma_z$$.

Yeah, it is a very depth trick because somehow you want the product of the three sigma to have a meaning as volume element (or dual to it), the product of two as surface elements (or dual to it) and so on. But all these topics should be theme for a separate, well indexed, thread somewhere. Clifford things.

 

[CarlB]

I doubt it, because your sigmas are not idempotent anymore.

I'm not sure where you're going here. $$\sigma_x$$ is never idempotent. They square to unity, in both cases. And the idempotents you get from them, for example,

$$(1+\sigma_x)/2.0$$

is idempotent in both representations. In the usual, one obtains:

$$\left(\begin{array}{cc}0.5&0.5\\0.5&0.5\end{array}\right)$$

In the deformed representation, one obtains:

$$\left(\begin{array}{cc}0.5&2.5\\0.1&0.5\end{array}\right)$$

One can easily verify that both these representations of $$(1+\sigma_x)/2.0$$ are, in fact, idempotent.

One can define a Clifford algebra as a vector space. On doing this, one obtains that addition is automatically identical between representations, and in this case the multiplication is identical too. So I'm pretty sure that they're both equally valid representations of su(2), and after exponentiation, SU(2).

Anyway, all this gets back to quarks and leptons in that I've discovered a beautiful representation of both the quarks and leptons that requires this as a basis for the theory.

What is going on here, in the making of the representations non unitary, is that we are preserving the usual quantum mechanics for the individual particles (i.e. the ideal generated by (1+\sigma_x) is isomorphic to the usual spin-1/2 no matter the representation), but are changing the relationship between different particles.

This shows up in the lepton mass formula because the different particles are different types (in terms of color). So you need a little more freedom than is available when you assume that the spinors (different particles) are separated from each other in the usual method of unitary representations. But they're still good su(2) reps.

It's kind of a long and involved thing and I'm splitting it up into two papers both of which are long and involved. The one having to do with representations also gets into the meaning of gauges and why Hestenes' factorization of the Dirac equation has an orientation built into it. For more on that subject, see the excellent short comment by Baylis:
http://www.arxiv.org/abs/quant-ph/0202060

Yet another way of describing this is that I am rejecting the splitting of the Banach space (or density matrix) into a Hilbert space (or spinors) because there are always multiple ways (i.e. a gauge) to do this. The Banach space is more physical because it does not depend on gauge. Along this line, but restricted to a single particle and the U(1) complex phase gauge, those who are fans of Bohmian mechanics need to read Hiley's paper which generalizes Bohmian mechanics to density matrices:
http://www.arxiv.org/abs/quant-ph/0005026

In terms of Schwinger's elegant measurement algebra, this amounts to writing the theory strictly in terms of simple measurements M(a), and rejecting the use of general measurements M(a,b). My paper shows that when one uses M(a,b), one automatically implies an orientation gauge similar to the one that Hestenes uses. And this gets back into the assumption that the vacuum is a physical state.

Carl