Alfred Einstead
Apr21-04, 03:25 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"FrediFizzx" <fredifizzx@hotmail.com> wrote:\n> What would the "extra U(1) sector for the baryon-lepton number" represent\n> physically? I don\'t get this.\n\nAdding to my previous reply: there are some other consequences, all\nvery interesting, of having the extra force corresponding to this\nU(1) gauge symmetry.\n\nAtoms would *not* be neutral, for one. The quantum number\nG = 1/2 (Baryon - Lepton) is +1/2 for protons and neutrons,\n-1/2 for electrons (and -1/2 for neutrinos). So, there would be\na positive charge corresponding to the number of neutrons in\nthe atom.\n\nThis means that the force would be repulsive for ordinary matter.\nThis limits the size of the coupling constant, if the force is\nlong-range, to being much smaller than the coupling corresponding\nto gravity ... unless what you\'re seeing in the way of "gravity"\nis actually the residual left over from a much stronger gravity\ncancelled out partly by this force.\n\nIn that case, there would be a dependence on the type of matter,\nleading (for instance) to a differential effect on different isotopes\nof the same element.\n\nOtherwise, the gravity you see is the gravity that\'s actually\nthere and the coupling constant for this extra force is incredibly\nsmall (which includes the case posited by the Standard Model\nof a coupling constant of 0).\n\nIf the force is short range, then this adds at least one extra\ndegree of freedom that needs to be accounted for by the\nsymmetry breaking mechanism; so that there\'d be 5 Higgs modes or\nmore, instead of 4.\n\nIn all cases, you\'d have the following effects:\n(a) a slight difference in the behavior of matter and\nanti-matter near the Earth (anti-matter with a\nslightly stronger attraction)\n(b) the above-mentioned dependence on isotopes\n(c) a small attraction between electrons and neutrons\nWhether they\'re measureable or not depends on how small the\ncoupling is (if not 0) or how large the boson mass is (if the\nforce is short range).\n\nOther consequences are also entailed in the large scale; especially\nif the force is long-range. All the celestial matter; planets,\nstars, etc.; would have a net positive G charge, making the visible\npart of the galaxy a huge cosmic G charge. The otherwise-sterile\nright neutrinos (as well as the left neutrinos) would all be negative\nG charges; and the resulting situation would be analogous to that of\nan atom -- a positvely charged galactic core composed of ordinary\nluminous matter, surrounded by a huge cloud of neutrinos.\n\nIf the remained a net positive charge, even with the neutrino\ncloud, then there\'d also be net repulsive force between galaxies\nleading to an expansion faster than that which is accounted for\nby gravitational effects alone.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"FrediFizzx" <fredifizzx@hotmail.com> wrote:
> What would the "extra U(1) sector for the baryon-lepton number" represent
> physically? I don't get this.
Adding to my previous reply: there are some other consequences, all
very interesting, of having the extra force corresponding to this
U(1) gauge symmetry.
Atoms would *not* be neutral, for one. The quantum number
G = 1/2 (Baryon - Lepton) is +1/2 for protons and neutrons,
-1/2 for electrons (and -1/2 for neutrinos). So, there would be
a positive charge corresponding to the number of neutrons in
the atom.
This means that the force would be repulsive for ordinary matter.
This limits the size of the coupling constant, if the force is
long-range, to being much smaller than the coupling corresponding
to gravity ... unless what you're seeing in the way of "gravity"
is actually the residual left over from a much stronger gravity
cancelled out partly by this force.
In that case, there would be a dependence on the type of matter,
leading (for instance) to a differential effect on different isotopes
of the same element.
Otherwise, the gravity you see is the gravity that's actually
there and the coupling constant for this extra force is incredibly
small (which includes the case posited by the Standard Model
of a coupling constant of 0).
If the force is short range, then this adds at least one extra
degree of freedom that needs to be accounted for by the
symmetry breaking mechanism; so that there'd be 5 Higgs modes or
more, instead of 4.
In all cases, you'd have the following effects:
(a) a slight difference in the behavior of matter and
anti-matter near the Earth (anti-matter with a
slightly stronger attraction)
(b) the above-mentioned dependence on isotopes
(c) a small attraction between electrons and neutrons
Whether they're measureable or not depends on how small the
coupling is (if not 0) or how large the boson mass is (if the
force is short range).
Other consequences are also entailed in the large scale; especially
if the force is long-range. All the celestial matter; planets,
stars, etc.; would have a net positive G charge, making the visible
part of the galaxy a huge cosmic G charge. The otherwise-sterile
right neutrinos (as well as the left neutrinos) would all be negative
G charges; and the resulting situation would be analogous to that of
an atom -- a positvely charged galactic core composed of ordinary
luminous matter, surrounded by a huge cloud of neutrinos.
If the remained a net positive charge, even with the neutrino
cloud, then there'd also be net repulsive force between galaxies
leading to an expansion faster than that which is accounted for
by gravitational effects alone.
> What would the "extra U(1) sector for the baryon-lepton number" represent
> physically? I don't get this.
Adding to my previous reply: there are some other consequences, all
very interesting, of having the extra force corresponding to this
U(1) gauge symmetry.
Atoms would *not* be neutral, for one. The quantum number
G = 1/2 (Baryon - Lepton) is +1/2 for protons and neutrons,
-1/2 for electrons (and -1/2 for neutrinos). So, there would be
a positive charge corresponding to the number of neutrons in
the atom.
This means that the force would be repulsive for ordinary matter.
This limits the size of the coupling constant, if the force is
long-range, to being much smaller than the coupling corresponding
to gravity ... unless what you're seeing in the way of "gravity"
is actually the residual left over from a much stronger gravity
cancelled out partly by this force.
In that case, there would be a dependence on the type of matter,
leading (for instance) to a differential effect on different isotopes
of the same element.
Otherwise, the gravity you see is the gravity that's actually
there and the coupling constant for this extra force is incredibly
small (which includes the case posited by the Standard Model
of a coupling constant of 0).
If the force is short range, then this adds at least one extra
degree of freedom that needs to be accounted for by the
symmetry breaking mechanism; so that there'd be 5 Higgs modes or
more, instead of 4.
In all cases, you'd have the following effects:
(a) a slight difference in the behavior of matter and
anti-matter near the Earth (anti-matter with a
slightly stronger attraction)
(b) the above-mentioned dependence on isotopes
(c) a small attraction between electrons and neutrons
Whether they're measureable or not depends on how small the
coupling is (if not 0) or how large the boson mass is (if the
force is short range).
Other consequences are also entailed in the large scale; especially
if the force is long-range. All the celestial matter; planets,
stars, etc.; would have a net positive G charge, making the visible
part of the galaxy a huge cosmic G charge. The otherwise-sterile
right neutrinos (as well as the left neutrinos) would all be negative
G charges; and the resulting situation would be analogous to that of
an atom -- a positvely charged galactic core composed of ordinary
luminous matter, surrounded by a huge cloud of neutrinos.
If the remained a net positive charge, even with the neutrino
cloud, then there'd also be net repulsive force between galaxies
leading to an expansion faster than that which is accounted for
by gravitational effects alone.