Why does a strong magnetic field disrupt the nucleus? The quarks cannot be magnetic monopoles because they carry electric charge. The quarks are confined in a duel superconductive vacuum formed by 'magnetic charges"
http://arxiv.org/pdf/1008.1055v2.pdf Superconductivity of QCD vacuum in strong magnetic field *We show that in a sufficiently strong magnetic field the QCD vacuum may undergo a transition to a new phase where charged mesons are condensed. In this phase the vacuum behaves as an anisotropic inhomogeneous superconductor which supports superconductivity along the axis of the magnetic field. In the directions transverse to the magnetic field the superconductivity is absent. The magnetic field-induced anisotropic superconductivity { which is realized in the cold vacuum, i.e. at zero temperature and density { is a consequence of a non-minimal coupling of the p mesons to the electromagnetic field. The onset of the superconductivity of the charged p mesons should also induce an inhomogeneous superuidity of the neutral p0 mesons. We also argue that due to simple kinematical reasons a strong enough magnetic field makes the lifetime of the p mesons longer by closing the main channels of the strong decays of the p mesons into charged pions. * In other words, in a strong enough magnetic field, pions will condence out of the vacuum and disrupt the nucleus. Also, this effect generates an electric current of quarks along the magnetic field axis provided the densities of left- and right-handed quarks are not equal. In the cold matter the external magnetic field may create spatially inhomogeneous structures which are made of quark condensates. --------------------------------------------------------------------------------------------------------------- http://homepages.uni-regensburg.de/~eng14891/qcdB_workshop/pdf/QCDB_Mueller.pdf When QED meets QCD Is slide 21 what happens in LENR? On Sun, Mar 22, 2015 at 3:15 PM, Axil Axil <[email protected]> wrote: > From an old post as follows: > > Strong magnetic fields can break down the vacuum, to result in the > creation of p mesons. > > > http://homepages.uni-regensburg.de/~eng14891/qcdB_workshop/pdf/QCDB_Callebaut.pdf > > Holographic study of magnetically induced p meson condensation > > QCD in strong magnetic Fields > > In the last slide > > Conclusion: back to objectives > > Studied e_ect: *r *meson condensation > > phenomenological models: Bc = m2 > > *r *= 0.6 GeV2 > > lattice simulation: slightly higher value of Bc _ 0.9 GeV2 > > holographic approach: > > can the *r *meson condensation be modeled? Yes > > can this approach deliver new insights? e.g. taking into account > > constituents, e_ect on Bc > > Up and down quark constituents of the *p *meson can be modeled as > separate branes, each responding to the magnetic Field by changing their > embedding. This is a modeling of the chiral magnetic catalysis e_ect. We > take this into account and _nd also a string e_ect on the mass, leading to > a Bcrit _ 0.78 GeV2 > > In simple terms, if the magnetic field is strong enough, the vacuum > breaks down an p meson start to condense into existence. > > On Sun, Mar 22, 2015 at 3:05 PM, Axil Axil <[email protected]> wrote: > >> http://arxiv.org/pdf/1203.5699.pdf >> >> *The p and A mesons in strong abelian magnetic field in SU(2) lattice >> gauge theory.* >> >> >> The paper explains how a P mason can be created from condensation in the >> vacuum by a magnetic field of the proper strength. >> >> On Sun, Mar 22, 2015 at 2:21 PM, Jones Beene <[email protected]> wrote: >> >>> Yes – positrons are not seen – and this is especially true since a >>> specialized meter which is tuned to detect positrons was used in some of >>> Rossi’s testing - with a sensitivity of picowatts. There was nothing above >>> background. >>> >>> >>> >>> Mesons are also not possible – since they are not produced by fusion nor >>> by radioactive decay - only as products of very high-energy interactions >>> such as cosmic rays or accelerators. >>> >>> >>> >>> *From:* David Roberson >>> >>> >>> >>> If electrons were the only particle generated the system would assume a >>> negative charge that continues to increase over time. Most reactions with >>> electrically neutral particles generate an equal number of positive charged >>> particles as negative ones. >>> >>> The fact that there are no 511 keV gammas suggests to me that no >>> positrons are being generated during the process. >>> >>> >>> >> >> >
