Lorentz Violation of the Standard Model

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J.C. Yoon - 18 Jan 2006 09:56 GMT

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.
You may find more information and FAQ on my homepage http://www.jcyoon.com

In this paper, I am asking whether the unification of the Standard Model is
rigorous enough to claim as a fundamental nature of nature.

As I am fully aware how bold my claim is, I have asked many leading
researchers as possible to find out any logical failure of my argument.
But I have not faced any reasonable refutation, rather their abrupt
discontinuances after some discussions have convinced me that I am at least
on the right track.

I don't know why, but I am still ashamed of myself to ask this issue in
public, but if I am mistaken, it would be better put an end to this now than
never.

Please be open-minded to my challenging questions and kindly reply me if you
or someone you know have a reasonable and rigorous argument against it.

Here is the outline of preprint.

1. No massive particle can be described by only one of left- and
right-handed "chirality" fermion field according to the Dirac equations.

2. The Standard Model starts with the assumption of massless fermion
labeling them as purely left- or right-handed chiral fermions. After the
massless fermion obtained its mass term, they are still labeled as left- and
right-handed fermion. If it is left- and right-handed chirality, then it
contradicts with the Dirac equation as there is no massive fermion with
purely one chirality. If it is left- and right-handed helicity, then it is
only an approximation that violates Lorentz invariance.

3. Therefore, the unification of electroweak interactions is at best
Lorentz-violating approximation, not a fundamental property of nature.

Thanks,
J.C. Yoon

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Chris Dams - 19 Jan 2006 15:47 GMT

Dear J.C Yoon,

>1. No massive particle can be described by only one of left- and
>right-handed "chirality" fermion field according to the Dirac equations.

Yes, this is true.

>2. The Standard Model starts with the assumption of massless fermion
>labeling them as purely left- or right-handed chiral fermions. After the
[quoted text clipped - 3 lines]
>purely one chirality. If it is left- and right-handed helicity, then it is
>only an approximation that violates Lorentz invariance.

No, it does not work this way. While it is true that the fermion field is
excited only through what one might call left-handed component, it is not
true that this results in only one possible spin-state. If one tries to
consider left-handed and right-handed particles separately, one will soon
find that because the field has become massive, a left-handed particle can
turn into a right-handed one and vice versa. This indicates that the
particles no longer correspond to left/right-handed states in the field and
should be diagonalized anew. If the mass is small the new particles will be
close to the previous massless left/right-handed eigenstates but they will
not be identical. The consquence of this is that both spin states can be
produced (four states if one includes anti-particles). What is still true is
that the "wrong" states have a much smaller probability to be produced.

I suggest that you further study the relation between (on-shell) particle
eigenstates and the fields from which they arise. This relation is a little
more complicated then you seem to be aware of.

Best wishes,
Chris

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J.C. Yoon - 20 Jan 2006 07:14 GMT

Dear Chris Dams,

I am posting this reply again, for I think my response to you is lost.

I would like to appreciate your kind comment and suggestion for further
study.
But I am afraid that I still do not see why my argument fails.

> I suggest that you further study the relation between (on-shell) particle
> eigenstates and the fields from which they arise. This relation is a little
> more complicated then you seem to be aware of.

Yes, I agree the relation between particle eigenstates and the field is
complicated, but I am not sure what papers I should consult with for
the better understanding. In my opinion, the interpretation of particle and
field
in my paper is most accepted and most rigorous one we can find. One may
bring an argument of interacting field but QFT is based on the locality of
interaction
and therefore asymptotic completeness and most experiments we know is also
based on the same interpretation of the relation between particle
eigenstates
and the fields. If you are aware of such paper that clarify the relation of
particle
and field other than what I mentioned, please let me know its references.

> If the mass is small the new particles will be
> close to the previous massless left/right-handed eigenstates but they will
> not be identical. The consquence of this is that both spin states can be
> produced (four states if one includes anti-particles). What is still true is
> that the "wrong" states have a much smaller probability to be produced.

Here "a much smaller probability" is shown as "the Lorentz-violation
approximation"
in my article. No matter how small its probability is, it still violates
Lorentz
invariance.

Thanks,
J.C. Yoon

Reply to this Message

Chris Dams - 20 Jan 2006 09:54 GMT

Dear J.C. Yoon,

>> I suggest that you further study the relation between (on-shell) particle
>> eigenstates and the fields from which they arise. This relation is a
>little
>> more complicated then you seem to be aware of.

>Yes, I agree the relation between particle eigenstates and the field is
>complicated, but I am not sure what papers I should consult with for
>the better understanding.

In the introductory chapter of my PhD-thesis I explore this issue a bit
more. My thesis is not yet on the web but will be in some time. Usually
this is done in the operator formalism, but I prefer to use the
path-integral formalism. In that case one can define particles by giving
their creation and annihilation configurations. In the case of the massive
fermions these would be

\bar a(k,e,t) = int d^3x \bar psi(x) gamma^0 e^{-ikx} u_e(k)

for the cration of a particle with momentum k (k^2=m^2), spin e and the
condition k.e=0 at time t. The anti-particle would be created by

\bar b(k,e,t) = int d^3x \bar v_e(k) gamma^0 e^{-ikx} psi(x)

The particle annihilation configurations are the Dirac-conjugates of
this. These configurations are chosen such that

<a(k',e,t') \bar a(k,e,t)> = (2*pi)^3 2*omega)k delta^3(k'-l) theta(t'-t).

This can be shown to lead to wave-packets with the expected behaviour.
Where <...> is the expectation value to be calculated using the path
integral.

>> If the mass is small the new particles will be
>> close to the previous massless left/right-handed eigenstates but they will
>> not be identical. The consquence of this is that both spin states can be
>> produced (four states if one includes anti-particles). What is still true
>is
>> that the "wrong" states have a much smaller probability to be produced.

>Here "a much smaller probability" is shown as "the Lorentz-violation
>approximation"
>in my article. No matter how small its probability is, it still violates
>Lorentz
>invariance.

I am afraid that I do not really understand your argument then. As far as
I understood is goes as: because of the (1-gamma^5)/2 in the interaction
vertices only the left-handed fermions can be produced (2 different particle
states); however a massive Dirac fields supports four particle eigenstates;
because 2!=4 something is wrong. Is this a right way to present your
argument?

What one should consider to see what kind of particle can be produced
from a particular interaction vertex is the correlation of the annihilation
configuration with that vertex. E.g., the amplitude for producing a massive
fermion contains factors

<a(k,e,t) \bar\psi(x)> (1-gamma^5)/2,

where x is the position of the interaction vertex and also the chiral
projector that is contained in the interaction vertex is written down.
Both spins +e and -e can be produced in this case.

Best,
Chris Dams

Reply to this Message

J.C. Yoon - 29 Jan 2006 23:27 GMT

Dear Chris Dams,

I am afraid that I still fully understand your notations.

The way you interpret (1 -gamma^5)/2 is from weak interaction
and the unification of electroweak from the Standard Model suggest
a different interpretation from your argument.

In the Standard Model, the electroweak interaction is unified since it
interprets
(1 - gamma^5)/2 not as a part of interaction structure as you did, but a
part
of particle so that QED and weak interaction has only gamma^{\mu}.

Thanks,
J.C. Yoon

> I am afraid that I do not really understand your argument then. As far as
> I understood is goes as: because of the (1-gamma^5)/2 in the interaction
[quoted text clipped - 16 lines]
> Best,
> Chris Dams

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Natural Science Forum / Physics / Particle Physics / January 2006

Lorentz Violation of the Standard Model