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Could it be that Higgs's boson does not exist, but Higgs mechanism does?

  1. Apr 9, 2008 #1
    Non-Higgs boson process of the generating of elementary particle masses in QFT

    As is known in the experiments on LEP and Bevatron, Higgs's bosons were not discovered. It is supposed that an energy amount bigger than 160 GeV is here needed. But some scientists doubt the detection of Higgs in LHC under higher energies. E.g., already after the first unsuccessful results prof. Peter Woit had discussed on his blog with a post-doc the following
    (see http://www.math.columbia.edu/~woit/wordpress/archives/000002.html ):

    “Fred Says:
    March 24th, 2004 at 6:58 pm
    Hi Peter..
    If they don’t find any SUSY at LHC, I doubt it will be the death knell of String Theory. Alas, there are enough degrees of freedom there to just argue that it breaks at a higher scale, no big deal!
    The nasty thing of course is that it just adds more fine tuning to minimal SUSY, and much of the original point would be lost.
    I’m more interested though, in the (seemingly absurd) case that the LHC doesn’t discover the Higgs! AFAICS, all reasonable models put it firmly in reach of the LHC.. If we don’t find it, well, something drastic has got to give.

    Peter Says:
    March 24th, 2004 at 7:11 pm
    Hi Fred!
    The interesting thing about Gross’s talk was that he was kind of going out on a limb on the issue of supersymmetry at the LHC. I’m more and more convinced that he and a lot of others are getting discouraged about string theory and no supersymmetry at the LHC will be the final straw.
    The worst possible thing for particle theory would be if the standard model Higgs shows up, behaving like a standard elementary scalar field with a certain mass. Then we would still be in the situation of having no idea where the Higgs potential or couplings come from, and no prospects for doing experiments in our lifetime to find out.
    I’m also hoping the LHC doesn’t find a standard Higgs field, but evidence for some more interesting way of breaking electroweak gauge symmetry, one that we haven’t thought of yet”
    .(citation end)

    Further I want to discuss such “some more interesting way of breaking gauge symmetry, one that we haven’t thought of yet”, but which already exists within the framework of the quantum field theory.
  2. jcsd
  3. Apr 9, 2008 #2


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    I would think that it would be a normal thing to do: Consider ALL alternatives.
    I have been looking at the following paper on Gamma-Ray Emission.
    I assume that this is the highest energy that has been detected.
    Still no Higgs.
    Discovery of TeV Gamma-Ray Emission from the Cygnus Region of the Galaxy
    (Submitted on 21 Nov 2006)

    frequencies as high as 2.9e+27 Hz have been detected from astrophysical sources

    2.99792458e+27 = 1e-17 cm
    I started my blog with the question, “What if they don’t find the Higgs?”
  4. Apr 9, 2008 #3
    Isn't Technicolor an example of a model where the Higgs Mechanism is used but the Higgs Boson is not present? (I.E. because instead of the Higgs as we know it they have multiple Higgs-like particles which jointly invoke the Higgs mechanism.)
  5. Apr 10, 2008 #4
    I will consider more simple case in the framework of QED, but I think that it could be extended on QCD.
  6. Apr 10, 2008 #5
    I will first remind briefly Higgs’s mechanism (in more detail see the thread https://www.physicsforums.com/showthread.php?t=224611 )

    On the Higgs mechanism of the acquisition of particles’ masses in the Standard Model

    Standard Model (SM) describes all fundamental interactions of elementary particles (with exception of gravitation). These interactions are called gauge, since here the interaction fields are produced by gauge transformations. Interaction in SM is achieved by means of the exchange of the gauge bosons: mass-free photons and gluons, and also by three massive W and Z vector bosons.
    In the initial stage of SM theory building all fundamental particles are mass-free. They acquire masses as a result of symmetry breaking. It is assumed that symmetry breaking is caused by a certain scalar field (Higgs's field), which contains the massive scalar bosons of Higgs. Interaction of this field with the particles generates its masses. The procedure of the description of the appearance of particles’ masses is called Higgs's mechanism. Its description is well known (see e.g. the above thread).
  7. Apr 10, 2008 #6
    Now let us examine the above-mentioned process of particles’ mass generation in regard to one simple case in the framework of QFT, in parallel to Higgs's mechanism.

    Photon as gauge field
    Thus, the theory of Standard Model (or theory of Young-Mills gauge fields) is the theory of gauge mass-free vector fields, in which the particles acquire mass due to the spontaneous breaking of the gauge symmetry of vacuum.
    One of the examples of such generation of massive fields (particles) can be
    considered the case of the electron-positron pair photoproduction:

    (A) [tex]\gamma +N\to e^++e^-+N\quad ,[/tex]

    Actually, the photon [tex]\gamma [/tex] is the mass-free gauge vector boson. EM
    nuclear field [tex]N[/tex] initiates its transformation into two massive particles:
    electron and positron, but it does not disappear, similarly to Higgs's
    field. The fields of electron and positron [tex]e^+,e^-[/tex] are the spinors, which
    are converted differently in comparison with the vector field. Thus, it is
    possible to say that the reaction (A) describes the process of the symmetry
    breaking of initial mass-free vector field to generate the massive spinor
    Using Feynman's diagrams, in the framework of SM we can very accurately calculate all characteristics of particles with the exception of charge and mass. Nevertheless, the reaction (A) remains mysterious: we do not know, how the process of the transformation of mass-free boson field into massive fields takes place. Let us attempt to imagine it. Let us examine the Feynman diagram of the above reaction of pair production (see Fig.):

    In the framework of SM in the vertex of diagram the interaction of the particles’ currents takes place. But it is obvious that namely here, before the electron-positron production, the transformation of photon into the electron-positron pair begin. This means that here the interaction of photon with nuclear field occurs. As is known the existing theory does not examine interaction of photon with the nucleus field. Then, what transformation could this be?
    Here there are several evidences to answer this question.
    1) Photon is particle of EM wave. As is known, this particle is space-nonlocal (see thread
    https://www.physicsforums.com/showthread.php?t=32102&page=4 ). Moreover, this particle can be considered as the simplest boson string, similar to boson string of the string theory (see the thread
    https://www.physicsforums.com/showthread.php?t=225278 ).
    Then, the above question can be formulated differently: what does occur with this EM string in the strong nuclear field? As is known in this case the trajectory of EM wave must be bent. In other words, photon must perform some transformation, during which its trajectory is bent.
    2) As is shown in the above thread, this twirling transformation is identical to gauge transformation and this fully corresponds to ideas of SM. Actually, (Rayder, L. QFT) gauge transformations is the transformation of rotation in the space of the internal symmetry of particles. As is known, Pauli matrix and three first (photons’) Gell-Mann matrices are the generators of the gauge transformation of groups SU(2) and SU(3) respectively, which describe the rotation in the plane and in the 3D space, correspondingly.
    3) Due to the transformation, which takes place in the vertex of diagram, the mass-free photon must become the massive particles. To occur this, Higgs's theory prompts that the intermediate (massive) bosons must participate here. In other words, it is possible to assume that the acquisition of electron and positron mass occurs through the intermediate massive boson.

    Basing on this evidences, let us suppose that in the vertex of Feynman's diagram is achieved the gauge transformation of “linear” photon in the closed (“non-linear”) photon string. It is understandable that in this case the photon must acquire the rest mass (i.e. it must become massive intermediate boson), since its center of rotation can remain at one place. (Imagine the model: the electromagnetic wave, which moves in the closed optical fiber or wave-guide, seemingly acquires, as whole, the rest mass).
    Thus, we have two stages, parallel to Higgs's mechanism: during the first stage in the nuclear field occurs the gauge transformation of mass-free photon into “massive” photon (i.e. into special intermediate boson). On the second - the transformation of massive boson into two massive fermions takes place.
    These ideas can be translated to the mathematical language. In this case it can be shown that the matrix form of the equations of such curvilinear closed EM strings mathematically completely coincides with the quantum equations of vector and spinor particles. In other words, such non-Higgs interpretation of the process of mass generating lies completely within the framework of Standard Model.

    Attached Files:

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  8. Apr 10, 2008 #7


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    There are more than one way to skin a cat.

    Maybe this is what you are looking for.
    Cosmological Implications of a Scale Invariant Standard Model
    Authors: Pankaj Jain, Subhadip Mitra, Naveen K. Singh
    (Submitted on 14 Jan 2008)
    We generalize the standard model of particle physics such it displays global scale invariance. The gravitational action is also suitably modified such that it respects this symmetry. This model is interesting since the cosmological constant term is absent in the action. We find that the scale symmetry is broken by the recently introduced cosmological symmetry breaking mechanism. This simultaneously generates all the dimensionful parameters such as the Newton's gravitational constant, the particle masses and the vacuum or dark energy. We find that in its simplest version the model predicts the Higgs mass to be very small, which is ruled out experimentally. We further generalize the model such that it displays local scale invariance. In this case the Higgs particle disappears from the particle spectrum and instead we find a very massive vector boson. Hence the model gives a consistent description of particle physics phenomenology as well as fits the cosmological dark energy.
    Some might be interested in reading what else I found in "Looking into the proton" blog.
  9. Apr 11, 2008 #8
    Thanks. It is interesting and similar. But in my analysis (see below) Higgs boson is absolutely absent, but the mathematics is similar to Higgs mechanism.
  10. Apr 11, 2008 #9
    Now, following the Feynman diagram, we attempt to describe mathematically the process of the pair electron-positron photoproduction by means of the intermediate stage with massive vector boson.
    We will use here only final formulas, the detail proof of which it can be found on the thread https://www.physicsforums.com/showthread.php?t=225278

    Quantum equation of the photon
    According to QED (see Akhiezer and Berestetskii) “it is natural to assume Maxwell's equation as the equations of the field, which describes the quantum-mechanical states of the photon. This, together with the relationship , is sufficient for constructing of the theory of photon”. Thus, let us accept, as is done in QED, that the quantized equation of EM wave describes photon.
    For the certainty we will examine the circularly polarized photon moving
    along the axis [tex]y[/tex]. The Feynman diagram lines of photon (``linear'' quantized electromagnetic string) [tex]\gamma [/tex] corresponds to the linear equation:
    (eq1) [tex]
    \left[ {\left( {\hat {\alpha }_o \hat {\varepsilon }} \right)^2-c^2\left(
    {\hat {\vec {\alpha }}\hat {\vec {p}}} \right)^2} \right]\Phi =0\quad ,
    where [tex]\hat {\varepsilon }=i\hbar \frac{\partial }{\partial t},\hat {\vec
    {p}}=-i\hbar \vec {\nabla }[/tex] are the operators of energy and momentum,
    correspondingly, and [tex]\vec {\Phi }(y)[/tex] is certain matrix, which contains the components of wave function:
    (eq2) [tex]
    \Phi =\left( {{\begin{array}{*{20}c}
    {{\rm E}_x } \hfill \\
    {{\rm E}_z } \hfill \\
    {i{\rm H}_x } \hfill \\
    {i{\rm H}_z } \hfill \\
    \end{array} }} \right)=\left( {{\begin{array}{*{20}c}
    {\Phi _1 } \hfill \\
    {\Phi _2 } \hfill \\
    {\Phi _3 } \hfill \\
    {\Phi _4 } \hfill \\
    \end{array} }} \right),
    [tex]\left\{ {\hat {\alpha }_0 ,\hat {\vec {\alpha }}} \right\}[/tex] are Dirac's
    matrixes; moreover, energy and momentum of photon are quantized:
    [tex]\varepsilon =\hbar \omega [/tex] and [tex]p=\hbar k[/tex] .

    Equation of “massive” intermediate boson (photon)
    Due to the gauge transformation (i.e. because of the passage from the
    straight path to the curvilinear) [tex]\hat {R}\Phi \to \Psi [/tex], instead of the
    linear quantized electromagnetic field [tex]\vec {\Phi }(y)[/tex] the non-classical
    quantum wave field (particle) [tex]\Psi \left( y \right)[/tex] is produced. The
    obtained equation takes the form:
    (eq3) [tex]
    \left( {\hat {\varepsilon }^2-c^2\hat {\vec {p}}^2-m_p^2 c^4} \right)\Psi
    (eq4) [tex]
    \Psi =\left( {{\begin{array}{*{20}c}
    {{\rm E}'_x } \hfill \\
    {{\rm E}'_z } \hfill \\
    {i{\rm H}'_x } \hfill \\
    {i{\rm H}'_z } \hfill \\
    \end{array} }} \right)=\left( {{\begin{array}{*{20}c}
    {\Psi _1 } \hfill \\
    {\Psi _2 } \hfill \\
    {\Psi _3 } \hfill \\
    {\Psi _4 } \hfill \\
    \end{array} }} \right)\quad ,
    This equation is similar to the scalar equation of Klein-Gordon, but it is equation of vector particle. This fact is not difficult to show, using its electromagnetic record. (Let us recall also that the account of Higgs's mechanism usually begins from the Klein-Gordon-like equation (see thread https://www.physicsforums.com/showthread.php?t=224611 ).
    Because of the gauge transformation of the initial photon, the transport of its fields along the curvilinear trajectory takes place. This forms two tangential electric currents with opposite direction for each half-period of photon. Moreover, the characteristics of current are unambiguously connected with the Ricci connection coefficients (or in the general case, with Christoffel connection coefficients) and with the mass of the convoluted photon (by the way: the above connection among charge and mass explains, why the charge and mass renormalizations are always connected). The intermediate photon mass is equal obviously to the photon energy, divided into the square of light speed.
    It is not difficult to understand that the symmetry breaking of the initial field here actually occurs: from the linear photon it become nonlinear and then is broken into two spinor paticles. But the role of Higgs's field in this case plays not the Higgs’s boson, but nuclear field.
    It is remain to obtain the spinor particles.
  11. Apr 11, 2008 #10
    Equations of the fermions
    Making the factorization of the equation (3) and multiplying it from left to
    the Hermitian- conjugated wave function we will obtain the equation:
    (eq5) [tex]
    \Psi ^+\left( {\hat {\alpha }_o \hat {\varepsilon }-c\hat {\vec {\alpha
    }}\cdot \hat {\vec {p}}-K} \right)\left( {\hat {\alpha }_o \hat {\varepsilon
    }+c\hat {\vec {\alpha }}\cdot \hat {\vec {p}}+K} \right)\Psi =0\quad ,
    where (according to calculations; see the mentioned thread) additive term
    has the form [tex]K=\hat {\beta }m_p c^2[/tex], and constant [tex]m_p =\frac{\hbar \omega _p }{c^2}[/tex] is mass of the ``massive'' photon.
    From the equations (5) follow two equations for the advanced and regarded
    waves, which are the particle equations with external field: of electron in
    the field of positron, and vice versa:
    (eq6') [tex]\left[ {\hat {\alpha }_0 \left( {\hat {\varepsilon }+\varepsilon _{ex} }
    \right)+c\hat {\vec {\alpha }}\cdot \left( {\hat {\vec {p}}+\vec {p}_{ex} }
    \right)+\hat {\beta }m_e c^2} \right]\psi =0\quad ,
    (eq6’’) [tex]
    \psi ^+\left[ {\hat {\alpha }_0 \left( {\hat {\varepsilon }-\varepsilon
    _{ex} } \right)+c\hat {\vec {\alpha }}\cdot \left( {\hat {\vec {p}}-\vec
    {p}_{ex} } \right)+\hat {\beta }m_e c^2} \right]=0,[/tex]
    [tex]\psi =\left( {{\begin{array}{*{20}c}
    {E_x } \hfill \\
    {E_z } \hfill \\
    {iH_x } \hfill \\
    {iH_z } \hfill \\
    \end{array} }} \right)=\left( {{\begin{array}{*{20}c}
    {\psi _1 } \hfill \\
    {\psi _2 } \hfill \\
    {\psi _3 } \hfill \\
    {\psi _4 } \hfill \\
    \end{array} }} \right)[/tex]
    is some quantum wave function, which corresponds to
    electromagnetic field after division of the twirled photon and it is spinor
    contrary to [tex]\Psi [/tex], which is vector wave function.

    Let us note that the appearance of the retarded and advanced spinor waves explains, why on Feynman's diagram the electron moves in the reverse time, than positron.

    A question about the reason for photon fission into two particles can arise. Apparently, it is possible to assume that fission of the convoluted massive photon (intermediate boson) occurs due to the repulsion of the currents of each half-period of photon.
    Many consequences of this photon fission into two charged particles are given on the thread https://www.physicsforums.com/showthread.php?t=225278

    Thus, it occurs that within the framework of SM there is non-Higgs mechanism of the mass generation of elementary particles, which is actually connected with the breaking of gauge symmetry and with many other formal components of Higgs's mechanism.
    Last edited: Apr 11, 2008
  12. Apr 12, 2008 #11


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    I’m going to paraphrase … you help with the corrections.

    In order to “look inside” the photon we must make some asumption about its structure/composition.
    We must obey the speed of light, (299,792,458 m/s.)

    The photon moves forward 4 Plank length and meanwhile the amplitude goes from zero, up one Planck length, down one Planck length to zero, then it goes down one Planck length and back up to zero.

    I imagine a hoop that has a circumference of 4 Planck length. (A string)
    In that hoop there is a Squezed energy that represent the amplitude (a 2 brane).
    I put a mark on the hoop and the brane so that I’ll know when they are back into allignment.
    Now I make the brane go around the inside of the hoop once, U(1) while the hoop moves forward 4 Planck length.

    If I make the hoop spin as it is moving forward then, if I identify the front and the back as different, the hoop will make one full rotation with the front being back to where it started as it moves forward 4 Planck length. The hoop would then look like a sphere, a 2d brane surface.

    If I make the “squeezed energy”, (2 brane) rotate, “twirl”, as it goes around the inside of the hoop, it will also have 4 positions as it goes around.

    Something funny occurs from the point of view of an outside observer when the spin and the “twirled” are included, motions can appear stationary, too slow or too fast.

    Of course, this is scalable.
    If the photon amplitude is represented by a string vibrating up and down, (travelling in a sine pattern) then the biggest amplitude would be c/4.

    Since the photon is also moving forward at the speed of light, then, if we had a particle/string/scalar/brane to represent the amplitude, we would have to combine the two motions and we would end up with c/6 for the maximum amplitude, (299,792,458 /6 = 49,965,408.
    This would be two tetra with sides of 49,965,408 m)
    Is this falsifiable?
  13. Apr 12, 2008 #12
  14. Apr 12, 2008 #13
    As addition, let us examine briefly some mathematical correspondences with mathematical description of Higgs's mechanism (see https://www.physicsforums.com/showthread.php?t=224611 )

    Let's write down the interaction Lagrangian of photon with atom nucleus:
    (eq7) [tex]
    L=\Phi ^+\left( {\hat {\alpha }_o \hat {\varepsilon }-c\hat {\vec {\alpha
    }}\cdot \hat {\vec {p}}} \right)\left( {\hat {\alpha }_o \hat {\varepsilon
    }+c\hat {\vec {\alpha }}\cdot \hat {\vec {p}}} \right)\Phi +L\left( {\Phi
    ,N(\vec {r},t)} \right)+L\left( {N(\vec {r},t)} \right),
    where [tex]\Phi [/tex]is photon wave function (2); [tex]N=N\left( {\vec {r},t} \right)[/tex]
    is the nuclear field, which description we don't know in the points, closed
    to nucleus center; the term [tex]L\left( {N(\vec {r},t)} \right)[/tex] is
    conditionally the nucleus Lagrangian and [tex]L\left( {\Phi ,N(\vec {r},t)}
    \right)[/tex] is the photon-nucleus interaction Lagrangian.
    After a photon twirls to the massive boson, the Lagrangian of the vector
    massive boson of equation (5) will be the following:
    (eq8) [tex]
    L=\Psi ^+\left( {\hat {\alpha }_o \hat {\varepsilon }-c\hat {\vec {\alpha
    }}\cdot \hat {\vec {p}}+\hat {\beta }m_p c^2} \right)\left( {\hat {\alpha
    }_o \hat {\varepsilon }+c\hat {\vec {\alpha }}\cdot \hat {\vec {p}}-\hat
    {\beta }m_p c^2} \right)\Psi + \\
    +L\left( {N(\vec {r},t)} \right) \\
    Without the nucleus Lagrangian (which doesn’t work now), equation (8) can be recorded in the form:
    (eq9) [tex]
    L=\partial _\mu \Psi ^+\partial ^\mu \Psi -\left( {m_p c^2} \right)^2\Psi
    ^+\Psi \quad ,
    Obviously, the energy-momentum conservation law is valid for the massive
    particle. In this case, (taking into account that [tex]\Psi ^+\Psi =E^2+H^2[/tex],
    [tex]\Psi ^+\hat {\vec {\alpha }}\Psi =\frac{1}{2}\left[ {\vec {E}\times \vec
    {H}} \right][/tex] and so forth) we have:
    (eq10) [tex]
    m_p^2 c^4=\left( {\varepsilon ^2-c^2\vec {p}^2} \right)=\left( {\Delta \tau
    } \right)^2\left( {U^2-c^2\vec {g}^2} \right)= \\
    =\frac{(\Delta \tau )^2}{\left( {8\pi } \right)^2}\left[ {\left( {\vec
    {E}^2+\vec {H}^2} \right)^2-4\left( {\vec {E}\times \vec {H}} \right)^2}
    \right] \\
    \end{array}\quad ,
    where [tex]U[/tex]and [tex]g[/tex] is energy and momentum densities of electromagnetic (non-Maxwellian) field, and [tex]\Delta \tau [/tex] is volume of integration].
    (eq11) [tex]
    m_p^2 c^4=M^2\left[ {\left( {\psi ^+\psi } \right)^2-4\left( {\psi ^+\hat
    {\vec {\alpha }}\psi } \right)^2} \right]\quad ,
    where [tex]M={\Delta \tau } \mathord{\left/ {\vphantom {{\Delta \tau } {8\pi }}} \right. \kern-\nulldelimiterspace} {8\pi }[/tex]. Obviously, this nonlinear term corresponds to the self-action of the photon field parts.
    It is not difficult to see that now the interaction Lagrangian of equation (9) has a similarity with Higgs's potential, by means of which is in SM the spontaneous symmetry breaking produced. As this known, spontaneous symmetry breaking consists in the fact that from the mass-free vector field, which has two spin states (analogs of the state “linear” photon) and the mass-free scalar field [tex]\phi [/tex] (analog of nuclear field) appears massive vector particle. It is remarkable, that this breaking appears due to nonlinear interaction of Higgs's field [tex]\phi [/tex] with itself. The self-interaction energy can be written down in the form of the potential [tex]V(\phi )=\lambda ^2\left( {\left| \phi \right|^2-\eta ^2} \right)^2[/tex], where [tex]\left| \phi \right|^2[/tex]is isoscalar, [tex]\lambda [/tex] is the dimensionless parameter, [tex]\eta [/tex] is the parameter, which has the dimensionality of mass.
    We did not investigate in detail this parallelism. But the analysis, carried out above, indicates the possibility of describing of the particle mass production without the presence of Higgs's boson, but with help of mechanism, similar to the Higgs mechanism.

    Akhiezer, A.I. and Berestetskiy, V.B. Quantum electrodynamics. 1965
    Rayder, L. The quantum field theory. 1985.
  15. May 13, 2011 #14


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    Science Advisor

    Do you mean the scenario in which ALL the degrees of freedom of the Higgs field(s) are "eaten" by the gauge bosons, such that no degrees of freedom are left to give a new particle?
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