Weinberg on the anomaly

Al.Rivero@gmail.com

<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>Does anybody have more details about the argument developed in\nWeinberg, v II p 381-382 (around formula 22.3.39) where it is argued\nthat "the anomalies in a given set of symmetries therefore are\nunaffected by the possible presence of fermions with a mass allowed by\nthese symmetries"?\n\nWeinberg does not give bibliography for this part of the topic; any\npointer?\n\nIf I understand it well, it means that if for instance the top mass is\nallowed by the fundamental symmetry group, then the top could not\ncontribute to anomalies in the fundamental theory and, by \'t hooft\nprinciple, it could not contribute neither to anomalies in effective\ntheories built upon the fundamental.\n\nI wonder if I could then to conclude that mesons of the third family\nare unaffected by the chiral anomaly.\n\nYours,\n\nAlejandro\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>Does anybody have more details about the argument developed in
Weinberg, v II [itex]p 381-382[/itex] (around formula 22.3.39) where it is argued
that "the anomalies in a given set of symmetries therefore are
unaffected by the possible presence of fermions with a mass allowed by
these symmetries"?

Weinberg does not give bibliography for this part of the topic; any
pointer?

If I understand it well, it means that if for instance the top mass is
allowed by the fundamental symmetry group, then the top could not
contribute to anomalies in the fundamental theory and, by 't hooft
principle, it could not contribute neither to anomalies in effective
theories built upon the fundamental.

I wonder if I could then to conclude that mesons of the third family
are unaffected by the chiral anomaly.

Yours,

Alejandro

markwh04@yahoo.com

<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>Al.Rivero@gmail.com wrote:\n> I wonder if I could then to conclude that mesons of the third family\n> are unaffected by the chiral anomaly.\n\nThe generations are the degeneracy of the charge spectrum. Therefore,\neach one is a clone in charge space and exactly the same cancellations\napplies to each. The only essential restriction requires is that a\ncomplete copy of the first generation appear in the 2nd and 3rd. So,\nfor instance, if there\'s a bottom, there\'d have to be a top.\n\nThe constraint to remove the anomaly corresponds to Tr(gamma^5 Y_a\n{Y_b,Y_c}) = 0, where the Y\'s are the gauge generators.\n\nThe SU(2) and SU(3) generators automatically fall through correctly; as\ndoes anything else that is parity-symmetric. The only culprit is the\nhypercharge Y.\n\nIn fact, if you start out with just SU(2) and SU(3) and ask what other\nc(generation-independent) charges can fit with these; you get one of\ntwo results:\n(1) If neutrinos are represented as Weyl particles, with a missing\nright-neutrino; left-antineutrino sector; then you get the\nhypercharge\n(2) If neutrinos are represented as Dirac particles, including an\n(as-of-yet unseen) right-neutrino; left-antineutrino; then\nyou get the hypercharge AND the (baryon - lepton) quantum\nnumber.\n\nCalling G = 1/2 (baryon - lepton), and Y = hypercharge (scaled to -1\nfor the right-electron), you find the culprit behind the result:\nY - G behaves as the right-handed analogue of isospin\nand particles naturally arrange into 4 classes given by\n(Y - G) = +/- 1/2; I_3 = +/- 1/2.\n\nSo, the anomaly removal has a tendency to drive SU(2) into a\n"protective envelope" algebra U(2) which contains the extra generator Y\n- G; or into the larger parity-symmetric algebra SU(2)_L x SU(2)_R,\nwith Y - G = I3_R.\n\nIn other words, the anomaly is basically forcing a parity symmetric\nalgebra to re-emerge.\n\nIt\'s of interest to note that the charge generator Q (electric charge)\nwhich remains after symmetry breaking will then take on the revealing\nform\nQ = I3 + Y = I3 + I3_R + G;\ni.e., a parity symmetric residue of a possible SU(2)_R x SU(2)_L.\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>Al.Rivero@gmail.com wrote:
> I wonder if I could then to conclude that mesons of the third family
> are unaffected by the chiral anomaly.

The generations are the degeneracy of the charge spectrum. Therefore,
each one is a clone in charge space and exactly the same cancellations
applies to each. The only essential restriction requires is that a
complete copy of the first generation appear in the 2nd and 3rd. So,
for instance, if there's a bottom, there'd have to be a top.

The constraint to remove the anomaly corresponds to [itex]Tr(\gamma^5 Y_a{Y_b,Y_c}) = 0,[/itex] where the Y's are the gauge generators.

The SU(2) and SU(3) generators automatically fall through correctly; as
does anything else that is parity-symmetric. The only culprit is the
hypercharge Y.

In fact, if you start out with just SU(2) and SU(3) and ask what other
c(generation-independent) charges can fit with these; you get one of
two results:
(1) If neutrinos are represented as Weyl particles, with a missing
right-neutrino; left-antineutrino sector; then you get the
hypercharge
(2) If neutrinos are represented as Dirac particles, including an
(as-of-yet unseen) right-neutrino; left-antineutrino; then
you get the hypercharge AND the (baryon - lepton) quantum
number.

Calling [itex]G = 1/2[/itex] (baryon - lepton), and Y = hypercharge (scaled to -1
for the right-electron), you find the culprit behind the result:
[itex]Y - G[/itex] behaves as the right-handed analogue of isospin
and particles naturally arrange into 4 classes given by
[itex](Y - G) = +/- 1/2; I_3 = +/- 1/2[/itex].

So, the anomaly removal has a tendency to drive SU(2) into a
"protective envelope" algebra U(2) which contains the extra generator Y
[itex]- G;[/itex] or into the larger parity-symmetric algebra [itex]SU(2)_L x SU(2)_R,[/itex]
with [itex]Y - G = I3_R[/itex].

In other words, the anomaly is basically forcing a parity symmetric
algebra to re-emerge.

It's of interest to note that the charge generator Q (electric charge)
which remains after symmetry breaking will then take on the revealing
form
[itex]Q = I3 + Y = I3 + I3_R + G;[/itex]
i.e., a parity symmetric residue of a possible [itex]SU(2)_R x SU(2)_L[/itex].

Al.Rivero@gmail.com

<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>\n\nmarkwh04@yahoo.com ha escrito:\n> Al.Rivero@gmail.com wrote:\n> > I wonder if I could then to conclude that mesons of the third family\n> > are unaffected by the chiral anomaly.\n>\n> The generations are the degeneracy of the charge spectrum. Therefore,\n> each one is a clone in charge space and exactly the same cancellations\n> applies to each. The only essential restriction requires is that a\n> complete copy of the first generation appear in the 2nd and 3rd. So,\n> for instance, if there\'s a bottom, there\'d have to be a top.\n\nYep, and on another side, the mass of the top is very differen from the\nother two generations, and I wonder if it indicates a different\nunderlying symmetry and thus a different anomaly.\n\n> The constraint to remove the anomaly corresponds to Tr(gamma^5 Y_a\n> {Y_b,Y_c}) = 0, where the Y\'s are the gauge generators.\n>\n> The SU(2) and SU(3) generators automatically fall through correctly; as\n> does anything else that is parity-symmetric. The only culprit is the\n> hypercharge Y.\n>\n\n> (...)\n>\n> In other words, the anomaly is basically forcing a parity symmetric\n> algebra to re-emerge.\n\n...\n\n> It\'s of interest to note that the charge generator Q (electric charge)\n> which remains after symmetry breaking will then take on the revealing\n> form\n> Q = I3 + Y = I3 + I3_R + G;\n> i.e., a parity symmetric residue of a possible SU(2)_R x SU(2)_L.\n\n...\n\n\nBut, can your arguments be used to show that the chiral anomaly\nreponsible of pion decay emerges from the gauge anomaly of the stardard\nmodel?\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>markwh04@yahoo.com ha escrito:
> Al.Rivero@gmail.com wrote:
> > I wonder if I could then to conclude that mesons of the third family
> > are unaffected by the chiral anomaly.
>
> The generations are the degeneracy of the charge spectrum. Therefore,
> each one is a clone in charge space and exactly the same cancellations
> applies to each. The only essential restriction requires is that a
> complete copy of the first generation appear in the 2nd and 3rd. So,
> for instance, if there's a bottom, there'd have to be a top.

Yep, and on another side, the mass of the top is very differen from the
other two generations, and I wonder if it indicates a different
underlying symmetry and thus a different anomaly.

> The constraint to remove the anomaly corresponds to [itex]Tr(\gamma^5 Y_a[/itex]
> [itex]{Y_b,Y_c}) = 0,[/itex] where the Y's are the gauge generators.
>
> The SU(2) and SU(3) generators automatically fall through correctly; as
> does anything else that is parity-symmetric. The only culprit is the
> hypercharge Y.
>

> (...)
>
> In other words, the anomaly is basically forcing a parity symmetric
> algebra to re-emerge.

...

> It's of interest to note that the charge generator Q (electric charge)
> which remains after symmetry breaking will then take on the revealing
> form
> [itex]Q = I3 + Y = I3 + I3_R + G;[/itex]
> i.e., a parity symmetric residue of a possible [itex]SU(2)_R x SU(2)_L[/itex].

...

But, can your arguments be used to show that the chiral anomaly
reponsible of pion decay emerges from the gauge anomaly of the stardard
model?

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