2016
DOI: 10.1103/physrevlett.117.251601
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New Solutions for Non-Abelian Cosmic Strings

Abstract: We study the properties of classical vortex solutions in a non-Abelian gauge theory. A system of two adjoint Higgs fields breaks the SU(2) gauge symmetry to Z_{2}, producing 't Hooft-Polyakov monopoles trapped on cosmic strings, termed beads; there are two charges of monopole and two degenerate string solutions. The strings break an accidental discrete Z_{2} symmetry of the theory, explaining the degeneracy of the ground state. Further symmetries of the model, not previously appreciated, emerge when the masses… Show more

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Cited by 28 publications
(40 citation statements)
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“…The next simplest is an SU(2) gauge theory with two adjoint scalars, which has recently been studied numerically for the first time [33,34].…”
Section: Abelian Higgs Model In a Flrw Backgroundmentioning
confidence: 99%
“…The next simplest is an SU(2) gauge theory with two adjoint scalars, which has recently been studied numerically for the first time [33,34].…”
Section: Abelian Higgs Model In a Flrw Backgroundmentioning
confidence: 99%
“…We are also thinking of enlarging the model to accommodate other symmetries, such as the SU (2) × U (1) symmetry, to explore the presence of solutions within the non-Abelian context examined before in [35], and also the SU (3) × U (1) symmetry, to investigate color-magnetic structures in the dense quark matter scenario explored recently in [56]. The presence of non-Abelian symmetries makes the problem much harder, so the search for first order differential equations that solve the equations of motion is of current interest and would be welcome.…”
Section: Discussionmentioning
confidence: 99%
“…By integrating it, we get the value E = π, which matches with the value obtained through Eq. (12). We emphasize that the model engenders scale invariance, and this is the price we had to pay: we found stable finite energy global vortices, but we cannot decide on its size.…”
mentioning
confidence: 85%
“…Then, the energy is given by Eq. (12), which leads to E = π. Therefore, even though the permittivity is negative, the energy of the system is positive.…”
Section: P-3mentioning
confidence: 99%