# Lorentz Violation of the Standard Model

1. Jan 12, 2006

### jcyoon

Recently I have posted a preprint on the web (http://xxx.lanl.gov/abs/hep-ph/0502142)
that contradicts the conclusion of the unification of electroweak interactions,
which is the very reason the Standard Model is awarded Nobel prize.

Before putting this preprint, I have asked as many leading researchers as
possible to find out any logical failure of my argument. And I have been
trying to publish it to a journal, but I am having trouble in communication
because of my low profile and the magnitude of its claim. However, I have
not faced anyone who could refute this preprint as most of their
arguments are inconsistent with the fundamentallity of the Dirac equations.
I am deeply ashamed of myself to ask this issue in public, but if I am mistaken,
it would be better to put an end to this now than never. Please be open-minded
to my challenging question and kindly reply me if you or someone you know have
a reasonalbe and rigorous argument against it.

Here is the outline of preprint.

1. Massive particle cannot be represented by pure L,R chirality.
2. Standard Model started with L,R chirality with massless fermion assumption.
Even after the fermion gets its mass term by Higgs mechanism, the fermions are
decribed by L,R chirality as if massless.
3. One may justify this, but those justifications are not free of massless
limit approximation
that violates Lorentz invariance or denial of unification.
4. Therefore, the unification of electroweak is only approximate, which is
inconsistent with Lorentz invariance. That means it looks like an unification,
but it is not true.

Thanks,
J.C. Yoon

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2. Mar 10, 2006

### CarlB

Dr Yoon,

I think that your papers here are unappreciated. If they are ignored, it's because people really have no answer to your observations. I'll reread them this weekend and see what kind of questions I can come up with. Thanks for being willing to defend them here in public.

Carl

3. Mar 16, 2006

### Tom Mattson

Staff Emeritus
Yes, in the reviewing forum it was remarked that this issue is "tricky". I've been waiting for Spring Break to read both this paper and Blackforest's in detail.

4. Mar 16, 2006

### CarlB

I eventually found that it has been discussed at some length on sci.physics.research:

https://www.physicsforums.com/forumdisplay.php?f=123

Having downloaded the papers a few days ago, I have two comments. First, I think that it should be obvious that it is possible for an underlying theory to be not Lorentz symmetric, while the effective theory could be so. By "Lorentz symmetric" I mean that the effective theory could not only provide results that satisfy Lorentz symmetry, but could be composed of elements that are also Lorentz symmetric.

From the comments at sci.physics.research, it appears that one must grant JC Yoon's claim that the standard model is, in fact, composed of components that are not Lorentz symmetric. On the other hand, the standard model does provide Lorentz symmetric results. My conclusion from this is that the "theory of everything" is unlikely to be Lorentz symmetric.

My second comment is on an earlier paper by JC Yoon. In it, he mentioned, almost in passing, that the CKM mixing angle matrix for the quarks, when interpreted as a matrix of coupling constants, implies that charge is not conserved. This was said to be due to the fact that the CKM matrix has complex phases involved. I would love to see a reference from the literature for this. Ah, here's the quote:

http://www.arxiv.org/abs/hep-ph/0211005

Carl

5. Mar 17, 2006

### jcyoon

Dear CarlB,

The paper you mentioned "The Origin of CP Violation" is meant
to be brief and supported by other following papers. That is why
you may find some statements rather obscure.

What I tried to point out is that the complex coupling constant
from the CP violating phase in CKM implies non-hermitian Hamiltonian
between individual quark interactions and therefore the observable
such as charge is not constant in time evolution so that it violates
the conservation of charge.

However, now I myself find this argument rather inaccurate
as it involves more complicated aspects, which I would like to reserve
until the Lorentz violation issue resolves.

Thanks,
J.C. Yoon

6. Mar 22, 2006

### CarlB

Okay, let me put this another way.

Like you, I concluded that the standard model is not Lorentz compatible. But my thinking was different. Several years ago I discovered the "Euclidean relativity" being discussed on another forum in the "independent research" here on PhysicsForums, and realized that it would form a natural way of getting a Wick rotation into the foundations of physics. Thus, the connection between Wick rotated QFT and statistical mechanics would become reversed, and the natural description of QM would be the Wick rotation of a Euclidean world.

But it is very difficult to talk to physicists about violations of Lorentz symmetry. From reading the comments on sci.physics.research I see that you have discovered this as well. So I decided to work on the problem by violating Lorentz symmetry in a more subtle fashion.

There are a series of theorems that were begun by Coleman and Mandula that tell us (more or less) that because of Lorentz symmetry, it is impossible to mix the internal and external symmetries of particles. For someone who would like to explain the elementary particles geometrically, this is a devastating result. However, if the standard model is not truly Lorentz symmetric, or more generally, if the underlying theory below the standard model is not Lorentz symmetric, then the Coleman Mandula no-go theorems do not apply.

But no one really wants to listen to these sorts of ideas.

On the other hand, there is a particularly elegant foundation for QFT devised by Julian Schwinger called the Measurement Algebra, or SMA. The SMA uses projection operators such as "M(e)" to indicate a measurement, or filter, that allows a beam of particles to enter, but only allows an "e" or electron, to exit. Thus the algebra is naturally a model of a generalization of Stern-Gerlach filters.

The SMA is compatible with the density matrix formulation of QM, in terms of the rules of the algebra, and you can write the usual Pauli projection operators to model actual Stern-Gerlach filters.

I believe that Coleman-Mandula puts restrictions on how you can mix the Pauli projection operators (which are external) with the projection operators that pick out particular particles. However, if you geometrize the SMA using a Clifford algebra, you soon find that the elementary fermions do not quite follow the natural structure of the primitive idempotents (i.e. elementary projection operators) of a Clifford algebra. You can, however, make the elementary particles match such a structure by assuming subparticles.

And when you examine the structure of those subparticles, you end up with cross-generation operators that give you relationships between the leptons of different generations. This is something that the standard model does not provide.

An example of the formulas that drop out is the Koide mass formula. Here's a reference which gives the eigenvalue form that I found that follows from the above ideas:
http://www.arxiv.org/abs/hep-ph/0505220

Note that the above formula is exact to experimental error (6 places). I recently extended the above formula for the charged lepton masses to compute the masses of the neutral leptons and put it up on this website as post #187 here:

This gets back to your comment in the following way: The above formulas were generated by assuming that the generations of leptons are made up of mixtures of the same subparticles, but with different phases. In order to apply the requirement that their associated projection operators be idempotent, i.e. Y^2 = Y, one must conclude that color charge is not conserved.

Now the formulas given above are mass formulas. I'm busily working on generalizing these formulas to get formulas for the mixing angles.

I really don't think that you will make much progress convincing the physics establishment that the standard model is not Lorentz invariant. People believe what they want to believe. However, a formula for the mixing angles of the neutrinos relative to the charged leptons, or that of the up and down quarks (CKM or MNS if I recall), would surely be a beautiful result, and since the underlying model already violates Lorentz symmetry through a violation of the Coleman Mandula prescription, it will then follow that the standard model violates Lorentz symmetry. And look at how much attention Koide's formula has in the literature. An extension to the neutrinos and mixing angles is exactly what physics needs.

So please do not hesitate to explain. I've given here as good a reason as I can produce as to why these things are connected, and why I believe that solving the problem of understanding the mixing angles will, eventually, also solve the problem of understanding the Lorentz violations inherent to the standard model.

Carl

7. Apr 25, 2006

### CarlB

Okay, the not completely obvious answer to this question came to me.

If a force is conservative with respect to an observable, then any closed path causes no change in the observable. For example, energy is conserved in Newton's gravity, so when a planet returns to the same position in an orbit it has the same energy.

For the case of a table of coupling constants, one finds out whether or not it is conservative by tracing different paths through the coupling constants. In the case of the CKM or MNS matrices, one does this by tracking a state through the matrix, through multiple transitions.

When you trace through a series of transitions in a table of coupling constants, you multiply the consecutive coupling constants. Each coupling constant will contribute a probability and a phase. Now the probabilities have nothing to do with conservation per se. They simply indicate the probability that this particular phase is applicable.

But the product of the phases may or may not be conserved. As an example, suppose that you had a table of coupling constants between a red, green, and blue state and they were of the form:

$$\left(\begin{array}{ccc}1&e^{+i\delta}&e^{-i\delta}\\ e^{-i\delta}&1&e^{+i\delta}\\ e^{+i\delta}&e^{+i\delta}&1\end{array}\right)$$

The above is to be interpreted as a table of (R,G,B) by (R,G,B) coupling constants.

If you made the RGR sequence of coupling transitions, then you would pick up a + delta in the first step from R to G (i.e. position 1,2 in the matrix), and then a - delta in the step from G to R. The net change in phase is zero, the path is conservative.

But if you made the RGBR sequence, you would return to R with a phase of 3 delta, and this is the negative of what the RBGR phase would give. Therefore, phase is not conserved by the above coupling constants unless delta is of form $$2n \pi/3$$.

Now the above case was simplified in that I assumed that the same matrix could be used for repeated application. That is, I assumed that the outgoing state was of the same type as the incoming state. This is the case explicitly with my model of the lepton masses, but the case with the CKM and MNS matrices is not so clear.

The CKM and MNS matrices tell you how to move from the flavor eigenstates to the mass eigenstates. In order to presume that this can be repeated, and therefore that one could fail to conserve phase, one must imagine that one is performing a sequence of measurements that alternate between weak force and mass measurement.

Now the presumed "tribimaximal" form of the MNS matrix is real, so there are no phases to not conserve. It is the CKM matrix that includes phases, so the question is this: Is there a non conserved phase in the CKM matrix?

Now getting back to JC Yoon's original comment, the fact that phase is not conserved through the mixing matrices does not imply that charge is not conserved. Phase is still a global symmetry of the system and so is conserved.

As an example of a much simpler system that changes phase (but clearly conserves charge), consider the sequence of Stern-Gerlach filters that keep particles with spin-1/2 oriented in the z, y, x and z directions. The overall picked up phase is pi/4. Ignoring amplitudes (i.e. real factors) and keeping only complex phases:

$$(1+\sigma_z)(1+\sigma_x)(1+\sigma_y)(1+\sigma_z) = e^{i\pi/4}(1+\sigma_z)$$

We can see this explicitly by building up a table of coupling constants. Here I'll keep the amplitudes, and use matrix positions 1, 2 and 3 to stand for x, y and z:

$$\frac{1}{2}\left(\begin{array}{ccc}2&1+i&\sqrt{2}\\ 1-i&2&\sqrt{2}\\ \sqrt{2}&\sqrt{2}&2\end{array}\right)$$

The individual entries in the above are of the form <a|b>, where a and b take x, y and z.

And certainly a Stern-Gerlach experiment conserves charge.

Carl

Last edited: Apr 25, 2006
8. Oct 30, 2006

### charris

i agree carlB

9. Nov 5, 2006

### CarlB

You know, I'm now no longer convinced that a generalized Stern-Gerlach experiment conserves charge. I think it does if the particle coming in arrives diagonalized relative to the S-G apparatus, but otherwise, an S-G experiment can change charge.

For example, suppose your S-G sent positive charged particles up and negative charged particles down. So long as you only put particles that were eigenstates of charge into such an experiment it will conserve charge. But if you could arrange for a mixed state particle, then it will be turned into a pure state by that S-G experiment. To make a mixed particle one would naturally use an S-G apparatus that split particles according to a projection operator that mixed charge. Then you can arrange to not conserve charge with S-G experiments, just the same way you can arrange for spin up to turn into spin down in a series of S-G experiments with different orientations.

Carl

10. Jun 25, 2007

### Kay zum Felde

Hi,

I've red your paper and I've a question about your Hamiltonian H_{weak}. Aren't there missed the generators of the SU(2) group, usually denoted by t ? I am not an expert of the electroweak theory, but aren't there different generators for the left-handed and the right-handed electron (S. Weinberg: "Quantum Theory of Fields, page 305 ff.). The generator for the left-handed electron is two dimensional unity matrix and the generator for the right-handed electron is just a single number. Therefore the Hamiltonian should look different ?

Greetings Kay

11. Jun 25, 2007

### jcyoon

Dear Kay zum Felde,

Thanks for your interest in my paper.

The Hamiltonian H_{weak} in my paper is consist of massive fermion fields($\psi$), whereas the electroweak theory in S. Weinberg: "Quantum Theory of Fields II, page 305 is based on massless fermion, though it is rather ambiguous.

In the middle of page 305, it denotes $e_{L,R}$ as the left- and right-handed "parts"(chirality) of the electron field, that is, $e_{L,R}$ represent only parts of full solution of massive fermion, while they could represent for whole fermion only when massless. In experimental measurements, there is no way to detect only the part of massive fermion (chirality). The best we can do is to approximate experimentally measured helicity to chilarity, but a left-handed helicity electron can be observed as right-handed helicity depending on the reference frame and therefore we cannot distinguish left- and right-handed massive fermions to introduce different generators.

You may find my discussion with Prof. S. Weinberg and others helpful in this links, http://www.jcyoon.com/phpBB/

Sincerely yours,
J.C. Yoon

12. Jun 26, 2007

### Kay zum Felde

Hi,

I've red the comment of Steven Weinberg, which you've posted on your site. I think, that you should carefully read it again. The massless fermions Weinberg implies, your're are talking about, are the neutrinos, even if nobody knows, if they are really massless (my actual knowledge, may be this is now known). And there are yet no equations, that describe neutrinos. Electrons aren't massless in the electroweak theory as of my knowledge. The electroweak theory is composed of representations of SU(2)xU(1), where U(1) describes the electromagnetical and neutrino part and SU(2) the part of the the W-particles involved in the weak interaction. U(1) is not broken by the Higgs mechanism. SU(2) is broken by Higgs effect and this is leading to mass for the W-particles. The W-particles and the photon mediate the interaction. So, what is left is the question, whethere the neutrino is massless or not, and this is yet not decided. The noble price is usually given to theories and to experimental research, when they are verified to common belief. I'm sure people who decides which theory or which experiment or in general which work is worth a noble price know also what are the weaknesses of the theories and experiments. Nobody is perfect and at least theories are only given the noble prices, when they agree with experiments to some certain degree. Noone knows yet a theory, that explains everything. Every theory has its boundaries. The electroweak theory works in its boundaries and if you research about the neutrinos, then you'll find discussions whether they've mass or not.

Greetings

Kay

13. Jun 27, 2007

### jcyoon

Dear Kay zum Felde,

I agree with you that Professor Weinberg's reply itself was regarding neutrino, which was quite accurate. As you may have already read my response to his, I have suggested to put aside the nuetrino issue, since the issue I have raised is regarding massive fermions such as electrons and muons, especially the parity-violating asymmetry in SLAC E158, which was critical to accept the Stadard Model. In this sense, Professor Gerard 't Hooft said that Professor Weinberg's statement was incorrect, suspecting that it has been badly tanken out of context, which is not (from the debates with Professor Gerad 't Hooft).

Yes, electrons are massive but their definition and properties in the electronweak theory are not massive. Electrons were massless before the Higgs mechanism, represented by definite chirality $\psi_{L,R}$, but after its mass term has been obtained the definition of fields $\psi_{L,R}$ remains the same which does not satiesfy the equations of motion, the Dirac equations. Therefore, electrons are massive, but still have the property of massless fermion with definite chirality, contradicting the Dirac equations the solution of which is always mixed of two chiralities.

Yes, every theory has its boundaries, but the problem is that the boundaries were mistaken. The Standard Model stepped out its boundaries and its claim of electroweak unification and experimental verification have been accepted as more accurate than it should be.

Sincerely yours,
J.C. Yoon

Last edited: Jun 27, 2007
14. Jun 27, 2007

### Kay zum Felde

Hi,

I think, your're right, when you say there're a lot questions open. On the other hand, my experience with articles that are simply proposing this and that is fundamentally not allright have indeed not a lot chances to be published. What you should do get some paper published is: find a better theory. This will be difficult.

You start from the idea, that electrons have no mass before a Higgs-particle assigns mass to them. I've red in the book by Warren Siegel: 'Fields' (free on the web at www.arxiv.org, a good one covering a lot things), that chirality theory offers a lot different possibilities about how a symmetry can be broken. There's (according how I interprete this) no agreement, how this is going to be happen.

May be Weinberg was expressing with his comment about the Dirac-equation, that there's a lot to do. For example the "classical" Dirac-equation shows only antiunitary and antilinearly considering time-reversal symmetry. I found a couple of days ago an article from Wei Min Jin in the archive, who argues about this and presents a new equation (by the way nonlinear in the gauge-fields), which offers unitary time-reversal symmetry. So I think, a good idea is to read more about chirality-theories. They are as I believe very popular right now.

greetings Kay

15. Oct 19, 2007

### Hans de Vries

Dear JC Yoon.

The Electroweak theory based on the Dirac spinor formulation is fully Lorentz Invariant. Your issue with this theory stems from a very often encountered miss understanding of the physical meaning of helicity which I would like to clarify. Now, the electroweak interactions are based on two independent contravariant four-vectors, V and A, the vector and axial current.

$$\begin{array}{l c l c l c l} \bar{\psi}\gamma^\mu\psi & = & \psi_L^\dag\tilde{\sigma}^\mu \psi_L & + & \psi_R^\dag \sigma^\mu \psi_R & \quad & \mbox{vector} \\ \\ \bar{\psi}\gamma^5\gamma^\mu\psi & = & \psi_L^\dag\tilde{\sigma}^\mu \psi_L & - & \psi_R^\dag \sigma^\mu \psi_R & \quad & \mbox{axial vector} \end{array}$$

1) Vector current

The first is simply the charge/current density $J_V$. Its space components $J_V^1, J_V^2, J_V^3$ are zero for the (massive) fermion at rest while its time component $J_V^0$ represents the charge density. This vector transforms like a Lorentz vector in the same way as the energy momentum vector.

2) Axial current

The second (axial) vector $J_A$ represents the spin-density of the Dirac field. It points in the direction of the spin of the field which can be any direction, in principle independent of the direction of the motion the electron at non-relativistic speeds. This vector reduces to the classical 3-component spin vector in the rest frame of the fermion.

In other frames it transforms just as any other relativistic spin 4-vector, classically or quantum mechanically: If the spin is transversal to the direction of motion then the space components do not change. If the spin is parallel to the motion then it increases with a vector gamma.

3) Helicity

Now, the Electroweak interactions depend exclusively on the Lorentz invariant vectors described above. To correctly understand the role of the term helicity we reorder the relativistic transformation formula for spin to explicitly show how it grows from the rest-frame spin $\vec{J}_{A}^{\ rest}$ to the spin $\vec{J}_{A}$ at velocity $\beta$. See Jackson (11.159)

$$\vec{J}_A\ =\ \vec{J}_{A}^{\ rest}\ +\ \frac{\gamma^2}{\gamma+1}\ (\ \vec{\beta}\cdot \vec{J}_{A}^{\ rest}\ )\ \vec{\beta}$$

As, you see, the vector grows always in the direction parallel to $\beta$ and the magnitude (and sign) depends on the inner product of $\vec{J}_{A}^{\ rest}$ and $\beta$. Now at the ultra-relativistic speed of collider experiments the spin will grow and grow and always in the direction parallel to the motion. In the limit there are only two possibilities, parallel or anti-parallel to the motion determined by the sign of the inner product.

Thus Helicity is a relativistic effect and is identical for classical and quantum mechanical objects. The same happens for a rotating rock moving at a relativistic speed. What goes into the electro-weak equations is not a value for the helicity but the values of the two Lorentz invariant four-vectors, which, at very high speeds, happen to be almost aligned to each other. Therefor the theory itself is Lorentz invariant.

Regards, Hans

16. Oct 19, 2007

### jcyoon

Dear Hans de Vries,

First, L or R from $$\psi_{L,R}$$ denotes chirality, not helicity mentioned in 3) Helicity.
In the Standard Model, the definition of Left-handed field $$\psi_{L}$$ is given by
$$\psi_{L} \equiv (1 - \gamma^{5})\psi$$
which indicates only part of the exact massive Dirac solutions of $$\psi$$,
while helicity should be defined for a whole field $$\psi$$ representing massive fermions.
See Peskin and Schroeder, p44.

Second, the best place to examine Lorentz violation is in the explicit calculation of matrix elements
that is supposed to be Lorentz invariant, where the combination of vector and axial
currents is considered, not separately. For example, the calculation in E. Derman and W.
Marciano, Ann. Phys. 121, 147 (1979) for SLAC E158 shows that the matrix elements for
incoming electrons with Left- and Right-handed helicities are different, while the Left-handed
helicity state of electron can be observed as Right-handed helicity
under a proper Lorentz transformations rigorously according to the massive Dirac equations.

Sincerely yours,
J.C. Yoon

Last edited: Oct 19, 2007
17. Oct 19, 2007

### Hans de Vries

Dear J.C.

Indeed, the above is what I was taking for granted in my post.

What is important is the physical interpretation. Chiralities lead to sums of
contravariant vectors which behavior can be analyzed by studying the
mathematics of the spinor bi-linears. The behavior of these vectors under
Lorentz transform is what I am trying to convey in my post.

At the other hand, helicity dependent physics occurs in our daily lives,
even in such ordinary things like drilling a hole in the wall:

Left handed helicity (as observed by the wall)

$$\vec{J}_V \quad \mbox{move drill into wall}$$
$$\vec{J}_A \quad \mbox{rotate drill left wise}$$

Right handed helicity (as observed by the wall)

$$\vec{J}_V \quad \mbox{move drill into wall}$$
$$\vec{J}_A \quad \mbox{rotate drill right wise}$$

Only the right handed helicity will leave a hole in the wall. It doesn't
matter that the helicity is different in other reference frames,
(Euclidian or Lorentzian) It's the helicity in the frame of the wall
which determines if you get a hole or not.

A fully Lorentz invariant definition of Helicity can be constructed by
combining three vectors: The speed of the particle (drill) and the
speed of the target (wall) plus the spin vector of the particle (drill).

Regards, Hans

Last edited: Oct 19, 2007
18. Oct 19, 2007

### jcyoon

Dear Hans de Vries,

Thanks for an interesting anology of wall.
Though I am not fond of anology, let me try this time.

Helicity is defined by whether spin direction is parallel or antiparallel
to momentum. For the example of wall, we can observe the Left-handed
helicity drill as Right-handed helicity drill in the reference frame faster than
the speed of drilling where the direction of drilling is opposite to the spin
orientation. Then, the claim that only Left-handed helicity drilling is valid is
inconsistent with this Lorentz transformed observation. Also, in real world
we can drill a wall whether it rotates right- or left-wise.

$$\vec{J}_V \quad \mbox{move drill into wall}$$
$$\vec{J}_A \quad \mbox{rotate drill left wise}$$

Also, the definition of helicity in Quantum Field Theory and its Lorentz-variant property
can be found in Peskin and Schroeder, p47 and for your anology helicty corresponding to
Quantum Field Theory should be defined by whether drilling rotation is parallel or
antiparallel to the drill velocity (Note that speed is Lorentz-scalar).

Sincerely yours,
J.C. Yoon

Last edited: Oct 19, 2007
19. Oct 22, 2007

### BDOA

Found an interesting paper on restoring lorentz symmetry to the SM

Hi, hope you'll be interested this paper i found,

Like J.C. Yoon, He claims that weak force in the standard model, thus not
respect the full, orthochronous lorenz symmetry (boosts, rotations and CPT),
but he also claims that even adding a right handed weak force, would
not restored the symmetry. Instead one most add a right handed weak
force, and an axial force, which has opposite charges for left and right
spinning particles.

I found the above paper, because of my own work (which now needs paper reviewing),
upon adding an axial force to majorana neutrinos.

http://www.geocities.com/ch1rality/

BDOA

20. Dec 13, 2007

### reilly

Hans -- Bravo.
Regards,
Reilly