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Goldstone bosons in Models with global symmetry, broken by Orbifolding

  1. Aug 23, 2011 #1

    I am interested in grand unification with extra dimensions. Especially the case when
    extra dimensions are broken by orbifolding.

    Now I am trying to understand how the Goldstonebosons appear in the spectrum of a
    theory with global (for example SU(N)) symmetry. From the Goldstonetheorem it follows
    that if this breaking is spontaneous then the particle spectrum has to include a set of
    massless bosons with the same quantum numbers as the broken generators.

    The model considered here consist of [itex]N_f[/itex] complex bosons [itex]\phi_i[/itex]. The lagrangian density is of the form
    [tex] L = \frac{1}{2}\partial_M \Phi (\partial^M \Phi)^{\dagger}= \frac{1}{2}\sum_{i=1}^{N_f}\partial_{M} \phi_i (\partial^{M} \phi_i)^{\dagger}.[/tex]
    Where [itex]M=0,1,2,3,4,5[/itex] and the fifth dimension is compactified to
    [itex]S^1[/itex] with Radius [itex]R[/itex]. Breaking via Orbifold is incorporated if one
    [tex] y \sim -y .[/tex]
    This yields that the Space-time is [itex]M\times S^1/Z_2[/itex] and the field can be
    written in as eigenstates of the Projection operator [itex]P[/itex] with the eigenvalues
    [itex]\pm1[/itex]. These eigenstates are given by
    [tex] \phi_+(x^{\mu},y) = \sum_n \phi^{(n)}(x^{\mu}) cos(\frac{ny}{R}),[/tex]
    [tex] \phi_-(x^{\mu},y) = \sum_n \phi^{(n)}(x^{\mu}) sin(\frac{ny}{R}). [/tex]
    We chose the parity of the fields to be
    [tex] \underbrace{(-1,...,-1}_{M_f ~\text{times}},1,...,1).[/tex]
    This leads to a Lagrangian density of the form
    [tex] L= \frac{1}{2}\sum_{i=1}^{M_f}\partial_{\mu} \phi_i^{(0)} (\partial^{\mu} \phi_i^{(0)})^{\dagger} + \frac{1}{2}\sum_{i=1}^{M_f}\sum_{n=-\infty,n\neq 0}^{\infty}\left(\partial_{\mu} \phi_i^{(n)} (\partial^{\mu} \phi_i^{(n)})^{\dagger} - \frac{n^2}{R^2} \phi_i^{(n)} \phi_i^{(n)\dagger} \right).[/tex]
    Most of the arguments given above a just summaries, and have to be discussed further.
    But for my purpose it is enough. Form the Lagrangian given above one finds that Goldstone
    bosons dose not appear in the spectrum.

    Maybe you can explain to me why they do not occur in the spectrum?

    I hope you understand my short review of the model, otherwise feel free to ask.

    best regards,

  2. jcsd
  3. Aug 23, 2011 #2
    This does not seem to be spontaneous breaking, which would be due to the VEV of a Higgs field. There is no energy level above which the unbroken symmetry is restored. Indeed in a spontanously broken theory the broken symmetries are still there, albeit non-linearly realized (via the Goldstone fields). This is not seem to be the case you describe here.
  4. Aug 23, 2011 #3
    Thx suprised for your answer.

    The model has a scale which is 1/R. An effective action in a regime of [itex]E << 1/R[/itex]
    would only consist of the zero modes of the fields with positive parity.

    I'm not sure about the symmetry above the breaking scale, but I think it could be
    restored either through the infinite series or through the fact that the zero Modes of
    the fields with negative parity can be included by adding a zero.

    The reason why this question comes up is that if this symmetry is gauged then the
    goldstone bosons appear as the components of the vector field of the extra dimensions.
    For example when the model above is gauged one would find that [itex]A^a_5(x)[/itex]
    are the goldstone bosons.

    Maybe the answer to the question is that this model is an effective description, in four
    dimensional field language, of an extra dimension model.

    What do you think?

    best regards,

  5. Aug 23, 2011 #4
    I thought the scale given by R is there even before symmetry breaking, and thus is independent from it. The breaking is achieved by imposing y ~ -y, which occurs at all scales, so in a sense the breaking scale is infinite and there is no unbrocken phase at any finite energy. This breaking does not involve the VEV of a scalar field, so I would not consider it as spontaneous.

    On the other hand, orbifolds are well known to occur in string theory and there the breaking occurs due to the background geometry, or vacuum state. In this sense, orbifolds do appear as spontaneous symmetry breakings. I am not sure, perhaps the Goldstone modes appear there as blow-up moduli that resolve or smoothen out the orbifold singularity.

    PS: yes, such models do occur in extra dimensional theories. But I have problems to understand orbifolds in field theory, anyway.
    Last edited: Aug 23, 2011
  6. Aug 24, 2011 #5
    I agree with you in this point, I think the problem is that no compactification mechanism is
    used, so that this model Lagrangian is just a effective description of theory which is
    spontaneously broken at the Planck scale.

    The question which bothers me is why do they occur when the symmetry is gauged? Maybe
    because the model consist of enough degree's of freedom (with the right quantum numbers)
    so that it is possible to identify the massless modes with the goldstone bosons.

    Maybe is this an explanation, but I do not know what "blow-up moduli" means.

    However thanks for your comments.
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