Quark color charge

Apr22-04, 03:40 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','to olbar=no,location=no,scrollbars=yes,resizable=yes, status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usene t 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>I\'m trying to clarify my understanding of quark color charge.\nIt seems to me, that there is a linkage between quark color and electric\ncharge.\nFor example, a red up quark has a red color charge of 1 and an electric\ncharge of 2/3,\nwhile an antired up quark has a red color charge of -1 and an electric\ncharge of -2/3.\nIs it possible for an up quark to have a red color charge of 1 and an\nelectric charge of -2/3\n(or red color charge of -1 and electric charge of 2/3)?\nThanks in advance.\n\n\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>I'm trying to clarify my understanding of quark color charge.
It seems to me, that there is a linkage between quark color and electric
charge.
For example, a red up quark has a red color charge of 1 and an electric
charge of

while an antired up quark has a red color charge of -1 and an electric
charge of

.
Is it possible for an up quark to have a red color charge of 1 and an
electric charge of

(or red color charge of -1 and electric charge

Thanks in advance.

Apr24-04, 12:17 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','to olbar=no,location=no,scrollbars=yes,resizable=yes, status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usene t 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>Dave Snead wrote:\n\n> I\'m trying to clarify my understanding of quark color charge.\n> It seems to me, that there is a linkage between quark color and\n> electric charge.\n> For example, a red up quark has a red color charge of 1 and an\n> electric charge of 2/3,\n> while an antired up quark has a red color charge of -1 and an electric\n> charge of -2/3.\n> Is it possible for an up quark to have a red color charge of 1 and an\n> electric charge of -2/3\n> (or red color charge of -1 and electric charge of 2/3)?\n> Thanks in advance.\n\nThe colour charge is not linked in any way to the electric charge of the\nquarks, but is an additional independent gauge degree of freedom of the\nparticles of the standard model.\n\nTo understand it, one has to understand that the standard model bases on\nbasic symmetries of nature. According to Noether\'s theorem any\ncontinuous symmetry has a conservation law as consequence and vice\nversa.\n\nNow the standard model realises symmetries in a very elegant way, namely\nas local symmetries, i.e., you change the fields, describing in their\nquantised form the particles in a specific local way, and the equations\nof motion remain completely unchanged under those changes. The changes\nare called symmetry transformations. Taking all symmetry\ntransformations together, you obtain what mathematicians call a group.\n\nThe group, which concerns us here ist the socalled SU(3) group. This\ngroup can be defined as the set of all unitary complex 3x3 matrices\nwith determinant 1, and each such matrix can be generated by 8 linearly\nindependent "infinitesimal generators". We come back to this later.\n\nNow let\'s look on the quarks. The quarks are described by fields for a\nparticle of spin 1/2 (a socalled Dirac spinor). For each quark sort,\nthere are three such Dirac spinor fields. (Within the standard model we\nhave 6 sorts of quarks, also known as flavours of quarks, which are in\na certain sense also charges, but not related to colour but to the\nelectroweak interaction which we neglect in the following completely).\n\nThese three fields represent the three components of a complex vector\nthe SU(3) colour matrices operate on. That colour SU(3) is supposed to\nbe a symmetry transformation. According to the mathematical laws behind\nthe standard model, the quarks have thus to be described by equations\nwhich are invariant under such SU(3) transformations. As long as free\nquarks are concerned it is very easy to construct such equations. We\nhave just to write down the action in such a way that SU(3) doesn\'t\nchange it, and the end result is the free Dirac equation for each\nquark.\n\nSo far the transformation is only allowed to be global, i.e., the quark\nfields have to be multiplied by constant SU(3) matrices (i.e.,\nmatrices, which are not dependent on the space-time argument of the\nfield).\n\nNow the hard work is, to find the interactions between the quarks, and\nhere comes the idea of local field transformations into the game. Yang\nand Mills observed in 1954 that one obtains nice theories of\ninteracting particles by "gauging" a global symmetry to make it local.\n\nIf one transforms the quark fields with local colour SU(3) matrices the\ntrouble are the derivatives of these fields which appear in the action.\nNow this can be cured by introducing what mathematicians call a\nconnection, which describe how to parallel shift vectors (here the\nSU(3) vectors) such that the action also is invariant under local\ntransformations. The connections brings a new field into the game,\nnamely the socalled gauge bosons which have spin 1 (more precisely,\nthey have to be massless and thus are fields with helicity 1 rather\nthan spin 1).\n\nThe only other ingredient is the idea, that these fields shouls also\nrepresent some particles and thus they need also a kinetic term to\nbuild nice field equations. Also this is a more or less logical\nmathematical procedure. Had we done the same with the more easy group\nU(1) we had obtained QED. Thus the strong interactions are described as\na theory which is much alike QED but a little bit more complicated than\nthat due to the fact that the symmetry group SU(3) is a little bit more\ncomplicated.\n\nFirst of all, in the case of SU(3) we have 8 gauge bosons, since these\nare closely related to the infinitesimal generators of the gauge group\nSU(3). In our case of QCD the gauge bosons are called gluons, because\nthey are responsible for the strong force which glues together the\nquarks to hadrons (like the protons and neutrons which are part of the\nconstituents of the matter surrounding us).\n\nInstead of only one photon in the case of QED we thus have 8 gluons.\nAnother difference is that the gluons themselves carry charges. This is\ndue to the fact that the gauge group SU(3) in QCD is non-abelian. This\nmeans that two SU(3) transformations need not necessarily commute with\neach other, as it is familiar from rotations in our usual Euclidean 3-d\nspace. Thus we can look at the 8 gluons as one sort of particle\ncarrying 8 different charges, and thus gluons also strongly interact\nwith each other (while photons in QED do interact only very very weakly\ndue to quantum effects of higher order in the coupling).\n\nThe most important physical consequence of this self-interaction of the\ngluons is what is known as asymptotic freedom. This means that the\ntheory becomes a theory of weakly interacting particles only at very\nhigh bombarding energies of the particles, while the strong interaction\nbecomes really strong at low energies.\n\nOne mainly believes that this is the reason for the socalled quark\nconfinement, i.e., the notion that we never find free quarks or gluons\n(or any other coloured objects) in nature. All the strongly interacting\nparticles we study in accelerators are colour neutral objects. Either\nthey are bound states of a quark and an antiquark (which are known as\nmesons which all are bosons since they have integer valued spins) or\nstates of three quarks (where three the three SU(3) charges (colours)\nare combined such to build colour neutral particles, which are the\nbaryons like the proton and the neutron). Recently there was also\nevidence for the existence of particles consisting of five quarks (or\nmore precisely 4 quarks and one anti-quark), the socalled Pentaquarks.\n\nOn the other hand, asymptotic freedom of QCD tells us that if one\nsqueezes hadrons (i.e. mesons or baryons) together and heating this\nmatter up above a certain critical temperature one should find a hot\npiece of matter where quarks and gluons are whirling more or less\nfreely around. In analogy to the QED case, where at sufficiently high\ntemperatures the atoms of a gas "melt" to nuclei and electrons which\nare no longer bound together to atoms, this state of matter is called\nthe quark gluon plasma. There are strong hints that this state has been\nachieved at CERN (Geneva) and, even more likely, at the RHIC\n(Brookhaven, Long Island, NY), where heavy nuclei (heavy ions) are\nsmashed together at very high energies. This studies will be continued\nby even higher energies at the now being built new Large Hadron\nCollider at CERN. The knowledge about such dense states of matter is\nalso important for our understanding of the early universe, where there\nmust have been an epoche where quarks and gluons formed a quark gluon\nplasma before during the expansion and cooling of the universe these\ncombined to hadrons and these finally to nuclei (up to Lithium) which\ncould form stars and galaxies where all the heavier elements where\nformed.\n\n--\nHendrik van Hees Cyclotron Institute\nPhone: +1 979/845-1411 Texas A&M University\nFax: +1 979/845-1899 Cyclotron Institute, MS-3366\nhttp://theory.gsi.de/~vanhees/ College Station, TX 77843-3366\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>Dave Snead wrote:

> I'm trying to clarify my understanding of quark color charge.
> It seems to me, that there is a linkage between quark color and
> electric charge.
> For example, a red up quark has a red color charge of 1 and an
> electric charge of

> while an antired up quark has a red color charge of -1 and an electric
> charge of

.
> Is it possible for an up quark to have a red color charge of 1 and an
> electric charge of

> (or red color charge of -1 and electric charge

?
> Thanks in advance.

The colour charge is not linked in any way to the electric charge of the
quarks, but is an additional independent gauge degree of freedom of the
particles of the standard model.

To understand it, one has to understand that the standard model bases on
basic symmetries of nature. According to Noether's theorem any
continuous symmetry has a conservation law as consequence and vice
versa.

Now the standard model realises symmetries in a very elegant way, namely
as local symmetries, i.e., you change the fields, describing in their
quantised form the particles in a specific local way, and the equations
of motion remain completely unchanged under those changes. The changes
are called symmetry transformations. Taking all symmetry
transformations together, you obtain what mathematicians call a group.

The group, which concerns us here ist the socalled SU(3) group. This
group can be defined as the set of all unitary complex 3x3 matrices
with determinant 1, and each such matrix can be generated by 8 linearly
independent "infinitesimal generators". We come back to this later.

Now let's look on the quarks. The quarks are described by fields for a
particle of spin

socalled Dirac spinor). For each quark sort,
there are three such Dirac spinor fields. (Within the standard model we
have 6 sorts of quarks, also known as flavours of quarks, which are in
a certain sense also charges, but not related to colour but to the
electroweak interaction which we neglect in the following completely).

These three fields represent the three components of a complex vector
the SU(3) colour matrices operate on. That colour SU(3) is supposed to
be a symmetry transformation. According to the mathematical laws behind
the standard model, the quarks have thus to be described by equations
which are invariant under such SU(3) transformations. As long as free
quarks are concerned it is very easy to construct such equations. We
have just to write down the action in such a way that SU(3) doesn't
change it, and the end result is the free Dirac equation for each
quark.

So far the transformation is only allowed to be global, i.e., the quark
fields have to be multiplied by constant SU(3) matrices (i.e.,
matrices, which are not dependent on the space-time argument of the
field).

Now the hard work is, to find the interactions between the quarks, and
here comes the idea of local field transformations into the game. Yang
and Mills observed in 1954 that one obtains nice theories of
interacting particles by "gauging" a global symmetry to make it local.

If one transforms the quark fields with local colour SU(3) matrices the
trouble are the derivatives of these fields which appear in the action.
Now this can be cured by introducing what mathematicians call a
connection, which describe how to parallel shift vectors (here the
SU(3) vectors) such that the action also is invariant under local
transformations. The connections brings a new field into the game,
namely the socalled gauge bosons which have spin 1 (more precisely,
they have to be massless and thus are fields with helicity 1 rather
than spin 1).

The only other ingredient is the idea, that these fields shouls also
represent some particles and thus they need also a kinetic term to
build nice field equations. Also this is a more or less logical
mathematical procedure. Had we done the same with the more easy group
U(1) we had obtained QED. Thus the strong interactions are described as
a theory which is much alike QED but a little bit more complicated than
that due to the fact that the symmetry group SU(3) is a little bit more
complicated.

First of all, in the case of SU(3) we have 8 gauge bosons, since these
are closely related to the infinitesimal generators of the gauge group
SU(3). In our case of QCD the gauge bosons are called gluons, because
they are responsible for the strong force which glues together the
quarks to hadrons (like the protons and neutrons which are part of the
constituents of the matter surrounding us).

Instead of only one photon in the case of QED we thus have 8 gluons.
Another difference is that the gluons themselves carry charges. This is
due to the fact that the gauge group SU(3) in QCD is non-abelian. This
means that two SU(3) transformations need not necessarily commute with
each other, as it is familiar from rotations in our usual Euclidean 3-d
space. Thus we can look at the 8 gluons as one sort of particle
carrying 8 different charges, and thus gluons also strongly interact
with each other (while photons in QED do interact only very very weakly
due to quantum effects of higher order in the coupling).

The most important physical consequence of this self-interaction of the
gluons is what is known as asymptotic freedom. This means that the
theory becomes a theory of weakly interacting particles only at very
high bombarding energies of the particles, while the strong interaction
becomes really strong at low energies.

One mainly believes that this is the reason for the socalled quark
confinement, i.e., the notion that we never find free quarks or gluons
(or any other coloured objects) in nature. All the strongly interacting
particles we study in accelerators are colour neutral objects. Either
they are bound states of a quark and an antiquark (which are known as
mesons which all are bosons since they have integer valued spins) or
states of three quarks (where three the three SU(3) charges (colours)
are combined such to build colour neutral particles, which are the
baryons like the proton and the neutron). Recently there was also
evidence for the existence of particles consisting of five quarks (or
more precisely 4 quarks and one anti-quark), the socalled Pentaquarks.

On the other hand, asymptotic freedom of QCD tells us that if one
squeezes hadrons (i.e. mesons or baryons) together and heating this
matter up above a certain critical temperature one should find a hot
piece of matter where quarks and gluons are whirling more or less
freely around. In analogy to the QED case, where at sufficiently high
temperatures the atoms of a gas "melt" to nuclei and electrons which
are no longer bound together to atoms, this state of matter is called
the quark gluon plasma. There are strong hints that this state has been
achieved at CERN (Geneva) and, even more likely, at the RHIC
(Brookhaven, Long Island, NY), where heavy nuclei (heavy ions) are
smashed together at very high energies. This studies will be continued
by even higher energies at the now being built new Large Hadron
Collider at CERN. The knowledge about such dense states of matter is
also important for our understanding of the early universe, where there
must have been an epoche where quarks and gluons formed a quark gluon
plasma before during the expansion and cooling of the universe these
combined to hadrons and these finally to nuclei (up to Lithium) which
could form stars and galaxies where all the heavier elements where
formed.

--
Hendrik van Hees Cyclotron Institute
Phone:

Texas A&M University
Fax:

Cyclotron Institute,

http://theory.gsi.de/~vanhees/ College Station,

Apr24-04, 12:22 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','to olbar=no,location=no,scrollbars=yes,resizable=yes, status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usene t 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>Are you sure that quark color charge is related to\neletric charge ? Antiparticles are certainly linked to\ncolor charge, since an antiparticle has identical\nmass but opposite charge. An electron has charge +1,\nits antiparticle the anti-electron or positron has charge -1.\nAn up-quark u has electric charge +2/3, an anti-up-quark\nu-bar has electric charge -2/3.\n\nAll baryons and mesons are colorless, they contain an\neuqal mixture of red, green and blue. Baryons are three\nquark fermion states, for example the proton p=uud or\nneutron n = udd. They are characterized by RGB\nstates: u_r u_g d_b, u_g u_b d_r,..\nMesons are quark - antiquark boson states, they\nare characterized by R-(R-Bar)+B-(B-Bar)+G-(G-Bar)\ncolor states.\n\nAn up quark has always electric charge +2/3,\na down quark has always electric charge -1/3,\nregardless of the color which can be R,G or B.\nLike the eletric color, which can be changed\nby exchange of W+ and W- Bosons (the beta\ndecay), the color charge of a quark can be changed\nby gluon exchange: a R quark can emit a R-(B-Bar)\ngluon and become a B quark, and a R-Bar anti quark\ncan receive a R-(B-Bar) gluon and become a B-Bar\nanti-quark.\n\nR--->--- ------>------B\n\\ \\ R-(B-Bar)\nR---<----- -----<------B\n\nAnd the quarks are continuously exchanging\ngluons, so you can say which quark has which\ncolor. Color is a quantum number for quarks.\n\n\nSee for example\nQuarks and Leptons\nFrancis Halzen and Alan D. Martin\nJohn Wiley & Sons, 1984\n\n\n"Dave Snead" <dsnead6@charter.net> schrieb im Newsbeitrag\nnews:10896l3od6dv1f3@corp.supernews.c om...\n> I\'m trying to clarify my understanding of quark color charge.\n> It seems to me, that there is a linkage between quark color and electric\n> charge. For example, a red up quark has a red color charge of 1 and\n> an electric charge of 2/3, while an antired up quark has a red color\ncharge\n> of -1 and an electric charge of -2/3.\n> Is it possible for an up quark to have a red color charge of 1 and an\n> electric charge of -2/3 (or red color charge of -1 and electric\n> charge of 2/3)? Thanks in advance.\n>\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>Are you sure that quark color charge is related to
eletric charge ? Antiparticles are certainly linked to
color charge, since an antiparticle has identical
mass but opposite charge. An electron has charge +1,
its antiparticle the anti-electron or positron has charge -1.
An up-quark u has electric charge

an anti-up-quark
u-bar has electric charge

All baryons and mesons are colorless, they contain an
euqal mixture of red, green and blue. Baryons are three
quark fermion states, for example the proton

or
neutron n = udd. They are characterized by RGB
states:

..
Mesons are quark - antiquark boson states, they
are characterized by R-(R-Bar)+B-(B-Bar)+G-(G-Bar)
color states.

An up quark has always electric charge

a down quark has always electric charge

regardless of the color which can be R,G or B.
Like the eletric color, which can be changed
by exchange of W+ and W- Bosons (the $LaTeX Code: \\beta$
decay), the color charge of a quark can be changed
by gluon exchange: a R quark can emit a R-(B-Bar)
gluon and become a B quark, and a R-Bar anti quark
can receive a R-(B-Bar) gluon and become a B-Bar
anti-quark.

\ \ R-(B-Bar)
R---<----- -----<------B

And the quarks are continuously exchanging
gluons, so you can say which quark has which
color. Color is a quantum number for quarks.

See for example
Quarks and Leptons
Francis Halzen and Alan D. Martin
John Wiley & Sons, 1984

"Dave Snead" <dsnead6@charter.net> schrieb im Newsbeitrag
news:10896l3od6dv1f3@corp.supernews.com...
> I'm trying to clarify my understanding of quark color charge.
> It seems to me, that there is a linkage between quark color and electric
> charge. For example, a red up quark has a red color charge of 1 and
> an electric charge

while an antired up quark has a red color
charge
> of -1 and an electric charge of

.
> Is it possible for an up quark to have a red color charge of 1 and an
> electric charge of

red color charge of -1 and electric
> charge

? Thanks in advance.
>

Apr24-04, 09:18 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','to olbar=no,location=no,scrollbars=yes,resizable=yes, status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usene t 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>"Hendrik van Hees" <hees@comp.tamu.edu> wrote in message\nnews:c69e78\\$9gf58\\$1@ID-71437.news.uni-berlin.de...\n> Dave Snead wrote:\n>\n> > I\'m trying to clarify my understanding of quark color charge.\n> > It seems to me, that there is a linkage between quark color and\n> > electric charge.\n> > For example, a red up quark has a red color charge of 1 and an\n> > electric charge of 2/3,\n> > while an antired up quark has a red color charge of -1 and an electric\n> > charge of -2/3.\n> > Is it possible for an up quark to have a red color charge of 1 and an\n> > electric charge of -2/3\n> > (or red color charge of -1 and electric charge of 2/3)?\n> > Thanks in advance.\n>\n> The colour charge is not linked in any way to the electric charge of the\n> quarks, but is an additional independent gauge degree of freedom of the\n> particles of the standard model.\n>\n> To understand it, one has to understand that the standard model bases on\n> basic symmetries of nature. According to Noether\'s theorem any\n> continuous symmetry has a conservation law as consequence and vice\n> versa.\n>\n> Now the standard model realises symmetries in a very elegant way, namely\n> as local symmetries, i.e., you change the fields, describing in their\n> quantised form the particles in a specific local way, and the equations\n> of motion remain completely unchanged under those changes. The changes\n> are called symmetry transformations. Taking all symmetry\n> transformations together, you obtain what mathematicians call a group.\n>\n> The group, which concerns us here ist the socalled SU(3) group. This\n> group can be defined as the set of all unitary complex 3x3 matrices\n> with determinant 1, and each such matrix can be generated by 8 linearly\n> independent "infinitesimal generators". We come back to this later.\n>\n> Now let\'s look on the quarks. The quarks are described by fields for a\n> particle of spin 1/2 (a socalled Dirac spinor). For each quark sort,\n> there are three such Dirac spinor fields. (Within the standard model we\n> have 6 sorts of quarks, also known as flavours of quarks, which are in\n> a certain sense also charges, but not related to colour but to the\n> electroweak interaction which we neglect in the following completely).\n>\n> These three fields represent the three components of a complex vector\n> the SU(3) colour matrices operate on. That colour SU(3) is supposed to\n> be a symmetry transformation. According to the mathematical laws behind\n> the standard model, the quarks have thus to be described by equations\n> which are invariant under such SU(3) transformations. As long as free\n> quarks are concerned it is very easy to construct such equations. We\n> have just to write down the action in such a way that SU(3) doesn\'t\n> change it, and the end result is the free Dirac equation for each\n> quark.\n>\n> So far the transformation is only allowed to be global, i.e., the quark\n> fields have to be multiplied by constant SU(3) matrices (i.e.,\n> matrices, which are not dependent on the space-time argument of the\n> field).\n>\n> Now the hard work is, to find the interactions between the quarks, and\n> here comes the idea of local field transformations into the game. Yang\n> and Mills observed in 1954 that one obtains nice theories of\n> interacting particles by "gauging" a global symmetry to make it local.\n>\n> If one transforms the quark fields with local colour SU(3) matrices the\n> trouble are the derivatives of these fields which appear in the action.\n> Now this can be cured by introducing what mathematicians call a\n> connection, which describe how to parallel shift vectors (here the\n> SU(3) vectors) such that the action also is invariant under local\n> transformations. The connections brings a new field into the game,\n> namely the socalled gauge bosons which have spin 1 (more precisely,\n> they have to be massless and thus are fields with helicity 1 rather\n> than spin 1).\n>\n> The only other ingredient is the idea, that these fields shouls also\n> represent some particles and thus they need also a kinetic term to\n> build nice field equations. Also this is a more or less logical\n> mathematical procedure. Had we done the same with the more easy group\n> U(1) we had obtained QED. Thus the strong interactions are described as\n> a theory which is much alike QED but a little bit more complicated than\n> that due to the fact that the symmetry group SU(3) is a little bit more\n> complicated.\n>\n> First of all, in the case of SU(3) we have 8 gauge bosons, since these\n> are closely related to the infinitesimal generators of the gauge group\n> SU(3). In our case of QCD the gauge bosons are called gluons, because\n> they are responsible for the strong force which glues together the\n> quarks to hadrons (like the protons and neutrons which are part of the\n> constituents of the matter surrounding us).\n>\n> Instead of only one photon in the case of QED we thus have 8 gluons.\n> Another difference is that the gluons themselves carry charges. This is\n> due to the fact that the gauge group SU(3) in QCD is non-abelian. This\n> means that two SU(3) transformations need not necessarily commute with\n> each other, as it is familiar from rotations in our usual Euclidean 3-d\n> space. Thus we can look at the 8 gluons as one sort of particle\n> carrying 8 different charges, and thus gluons also strongly interact\n> with each other (while photons in QED do interact only very very weakly\n> due to quantum effects of higher order in the coupling).\n>\n> The most important physical consequence of this self-interaction of the\n> gluons is what is known as asymptotic freedom. This means that the\n> theory becomes a theory of weakly interacting particles only at very\n> high bombarding energies of the particles, while the strong interaction\n> becomes really strong at low energies.\n>\n> One mainly believes that this is the reason for the socalled quark\n> confinement, i.e., the notion that we never find free quarks or gluons\n> (or any other coloured objects) in nature. All the strongly interacting\n> particles we study in accelerators are colour neutral objects. Either\n> they are bound states of a quark and an antiquark (which are known as\n> mesons which all are bosons since they have integer valued spins) or\n> states of three quarks (where three the three SU(3) charges (colours)\n> are combined such to build colour neutral particles, which are the\n> baryons like the proton and the neutron). Recently there was also\n> evidence for the existence of particles consisting of five quarks (or\n> more precisely 4 quarks and one anti-quark), the socalled Pentaquarks.\n>\n> On the other hand, asymptotic freedom of QCD tells us that if one\n> squeezes hadrons (i.e. mesons or baryons) together and heating this\n> matter up above a certain critical temperature one should find a hot\n> piece of matter where quarks and gluons are whirling more or less\n> freely around. In analogy to the QED case, where at sufficiently high\n> temperatures the atoms of a gas "melt" to nuclei and electrons which\n> are no longer bound together to atoms, this state of matter is called\n> the quark gluon plasma. There are strong hints that this state has been\n> achieved at CERN (Geneva) and, even more likely, at the RHIC\n> (Brookhaven, Long Island, NY), where heavy nuclei (heavy ions) are\n> smashed together at very high energies. This studies will be continued\n> by even higher energies at the now being built new Large Hadron\n> Collider at CERN. The knowledge about such dense states of matter is\n> also important for our understanding of the early universe, where there\n> must have been an epoche where quarks and gluons formed a quark gluon\n> plasma before during the expansion and cooling of the universe these\n> combined to hadrons and these finally to nuclei (up to Lithium) which\n> could form stars and galaxies where all the heavier elements where\n> formed.\n>\n> --\n> Hendrik van Hees Cyclotron Institute\n> Phone: +1 979/845-1411 Texas A&M University\n> Fax: +1 979/845-1899 Cyclotron Institute, MS-3366\n> http://theory.gsi.de/~vanhees/ College Station, TX 77843-3366\n>\n\n\nHendrik,\n\nWhen you say\n\n> The colour charge is not linked in any way to the electric charge of the\n> quarks, but is an additional independent gauge degree of freedom of the\n> particles of the standard model.\n\nyou seem to imply that all 4 of these combinations are possible for a quark\n\nUp, red color charge = 1, electromagnetic charge = 2/3\nUp, red color charge = -1, electromagnetic charge = -2/3\nUp, red color charge = 1, electromagnetic charge = -2/3\nUp, red color charge = -1, electromagnetic charge = 2/3\n\nHowever I posted this query on sci.physics.particle and got this response\n\n>No, this not possible. "Color charge" is associated with\n>quarks and "anticolor charge" with antiquarks. A mixture\n>is not possible (only gluons consist of "color" and "anticolor").\n\n>Kind regards,\n>Dorian Credé\n\nWhich means the last two combinations are not possible.\nCan you resolve this discrepancy?\n\nThanks,\nDave Snead\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>"Hendrik van Hees" <hees@comp.tamu.edu> wrote in message
news:c69e78$9gf58$1@ID-71437.news.uni-berlin.de...
> Dave Snead wrote:
>
> > I'm trying to clarify my understanding of quark color charge.
> > It seems to me, that there is a linkage between quark color and
> > electric charge.
> > For example, a red up quark has a red color charge of 1 and an
> > electric charge of

> > while an antired up quark has a red color charge of -1 and an electric
> > charge of

.
> > Is it possible for an up quark to have a red color charge of 1 and an
> > electric charge of

> > (or red color charge of -1 and electric charge

?
> > Thanks in advance.
>
> The colour charge is not linked in any way to the electric charge of the
> quarks, but is an additional independent gauge degree of freedom of the
> particles of the standard model.
>
> To understand it, one has to understand that the standard model bases on
> basic symmetries of nature. According to Noether's theorem any
> continuous symmetry has a conservation law as consequence and vice
> versa.
>
> Now the standard model realises symmetries in a very elegant way, namely
> as local symmetries, i.e., you change the fields, describing in their
> quantised form the particles in a specific local way, and the equations
> of motion remain completely unchanged under those changes. The changes
> are called symmetry transformations. Taking all symmetry
> transformations together, you obtain what mathematicians call a group.
>
> The group, which concerns us here ist the socalled SU(3) group. This
> group can be defined as the set of all unitary complex 3x3 matrices
> with determinant 1, and each such matrix can be generated by 8 linearly
> independent "infinitesimal generators". We come back to this later.
>
> Now let's look on the quarks. The quarks are described by fields for a
> particle of spin

socalled Dirac spinor). For each quark sort,
> there are three such Dirac spinor fields. (Within the standard model we
> have 6 sorts of quarks, also known as flavours of quarks, which are in
> a certain sense also charges, but not related to colour but to the
> electroweak interaction which we neglect in the following completely).
>
> These three fields represent the three components of a complex vector
> the SU(3) colour matrices operate on. That colour SU(3) is supposed to
> be a symmetry transformation. According to the mathematical laws behind
> the standard model, the quarks have thus to be described by equations
> which are invariant under such SU(3) transformations. As long as free
> quarks are concerned it is very easy to construct such equations. We
> have just to write down the action in such a way that SU(3) doesn't
> change it, and the end result is the free Dirac equation for each
> quark.
>
> So far the transformation is only allowed to be global, i.e., the quark
> fields have to be multiplied by constant SU(3) matrices (i.e.,
> matrices, which are not dependent on the space-time argument of the
> field).
>
> Now the hard work is, to find the interactions between the quarks, and
> here comes the idea of local field transformations into the game. Yang
> and Mills observed in 1954 that one obtains nice theories of
> interacting particles by "gauging" a global symmetry to make it local.
>
> If one transforms the quark fields with local colour SU(3) matrices the
> trouble are the derivatives of these fields which appear in the action.
> Now this can be cured by introducing what mathematicians call a
> connection, which describe how to parallel shift vectors (here the
> SU(3) vectors) such that the action also is invariant under local
> transformations. The connections brings a new field into the game,
> namely the socalled gauge bosons which have spin 1 (more precisely,
> they have to be massless and thus are fields with helicity 1 rather
> than spin 1).
>
> The only other ingredient is the idea, that these fields shouls also
> represent some particles and thus they need also a kinetic term to
> build nice field equations. Also this is a more or less logical
> mathematical procedure. Had we done the same with the more easy group
> U(1) we had obtained QED. Thus the strong interactions are described as
> a theory which is much alike QED but a little bit more complicated than
> that due to the fact that the symmetry group SU(3) is a little bit more
> complicated.
>
> First of all, in the case of SU(3) we have 8 gauge bosons, since these
> are closely related to the infinitesimal generators of the gauge group
> SU(3). In our case of QCD the gauge bosons are called gluons, because
> they are responsible for the strong force which glues together the
> quarks to hadrons (like the protons and neutrons which are part of the
> constituents of the matter surrounding us).
>
> Instead of only one photon in the case of QED we thus have 8 gluons.
> Another difference is that the gluons themselves carry charges. This is
> due to the fact that the gauge group SU(3) in QCD is non-abelian. This
> means that two SU(3) transformations need not necessarily commute with
> each other, as it is familiar from rotations in our usual Euclidean 3-d
> space. Thus we can look at the 8 gluons as one sort of particle
> carrying 8 different charges, and thus gluons also strongly interact
> with each other (while photons in QED do interact only very very weakly
> due to quantum effects of higher order in the coupling).
>
> The most important physical consequence of this self-interaction of the
> gluons is what is known as asymptotic freedom. This means that the
> theory becomes a theory of weakly interacting particles only at very
> high bombarding energies of the particles, while the strong interaction
> becomes really strong at low energies.
>
> One mainly believes that this is the reason for the socalled quark
> confinement, i.e., the notion that we never find free quarks or gluons
> (or any other coloured objects) in nature. All the strongly interacting
> particles we study in accelerators are colour neutral objects. Either
> they are bound states of a quark and an antiquark (which are known as
> mesons which all are bosons since they have integer valued spins) or
> states of three quarks (where three the three SU(3) charges (colours)
> are combined such to build colour neutral particles, which are the
> baryons like the proton and the neutron). Recently there was also
> evidence for the existence of particles consisting of five quarks (or
> more precisely 4 quarks and one anti-quark), the socalled Pentaquarks.
>
> On the other hand, asymptotic freedom of QCD tells us that if one
> squeezes hadrons (i.e. mesons or baryons) together and heating this
> matter up above a certain critical temperature one should find a hot
> piece of matter where quarks and gluons are whirling more or less
> freely around. In analogy to the QED case, where at sufficiently high
> temperatures the atoms of a gas "melt" to nuclei and electrons which
> are no longer bound together to atoms, this state of matter is called
> the quark gluon plasma. There are strong hints that this state has been
> achieved at CERN (Geneva) and, even more likely, at the RHIC
> (Brookhaven, Long Island, NY), where heavy nuclei (heavy ions) are
> smashed together at very high energies. This studies will be continued
> by even higher energies at the now being built new Large Hadron
> Collider at CERN. The knowledge about such dense states of matter is
> also important for our understanding of the early universe, where there
> must have been an epoche where quarks and gluons formed a quark gluon
> plasma before during the expansion and cooling of the universe these
> combined to hadrons and these finally to nuclei (up to Lithium) which
> could form stars and galaxies where all the heavier elements where
> formed.
>
> --
> Hendrik van Hees Cyclotron Institute
> Phone:

Texas A&M University
> Fax:

Cyclotron Institute,

> http://theory.gsi.de/~vanhees/ College Station,

>

Hendrik,

When you say

> The colour charge is not linked in any way to the electric charge of the
> quarks, but is an additional independent gauge degree of freedom of the
> particles of the standard model.

you seem to imply that all 4 of these combinations are possible for a quark

Up, red color charge

electromagnetic charge

Up, red color charge

electromagnetic charge

Up, red color charge

electromagnetic charge

Up, red color charge

electromagnetic charge

However I posted this query on sci.physics.particle and got this response

>No, this not possible. "Color charge" is associated with
>quarks and "anticolor charge" with antiquarks. A mixture
>is not possible (only gluons consist of "color" and "anticolor").

>Kind regards,
>Dorian Credé

Which means the last two combinations are not possible.
Can you resolve this discrepancy?

Thanks,
Dave Snead

**mathman** · Apr27-04, 02:39 PM

The color charge and the electric charge are independent. Quarks have positive color charge, while anti-quarks have negative color charge (by convention). Electric charge for quarks are 2/3 (up, etc.) or -1/3 (down, etc.), while anit-quarks are reverse signed.

Apr27-04, 02:53 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','to olbar=no,location=no,scrollbars=yes,resizable=yes, status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usene t 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>Dave Snead wrote:\n\n> you seem to imply that all 4 of these combinations are possible for a\n> quark\n>\n> Up, red color charge = 1, electromagnetic charge = 2/3\n> Up, red color charge = -1, electromagnetic charge = -2/3\n> Up, red color charge = 1, electromagnetic charge = -2/3\n> Up, red color charge = -1, electromagnetic charge = 2/3\n>\n> However I posted this query on sci.physics.particle and got this\n> response\n>\n>>No, this not possible. "Color charge" is associated with\n>>quarks and "anticolor charge" with antiquarks. A mixture\n>>is not possible (only gluons consist of "color" and "anticolor").\n>\n>>Kind regards,\n>>Dorian Credé\n>\n> Which means the last two combinations are not possible.\n> Can you resolve this discrepancy?\n\nI hope so ;-). The point is that, as I wrote, colour charge (red, green,\nblue) labels the three components of quarks associated to the\nfundamental representation of the colour SU(3) group.\n\nNow, also the complex conjugated components of a vector in the\nfundamental-representation space of SU(3) form a(n irreducible)\nrepresentation of SU(3), which is not equivalent to the fundamental\nrepresentation. Thus, one has to distinguish between the colour SU(3)\nrepresentation of quarks (fundamental representation, often denoted\nsimply by 3) and this of anti-quarks (the conjugate complex of the\nfundamental representation, denoted by \\bar{3}).\n\nIf you neglect electromagnetic and weak interactions completely, all you\nneed to know is the QCD Lagrangian to get all this colour business\ncorrect. It\'s demanded by the general idea of non-abelian gauge\ninvariance (a la Yang and Mills):\n\nL=\\sum_{flavours} \\bar{q}_{flavour} (\\i \\fslash(\\partial)-m-g_s\n\\fslash{A}_{\\mu}) q_{flavour}+kinetic terms for the gauge field\n\nThe pattern of electric charges is a little bit more difficult to\nunderstand, since it comes from the electroweak theory, which is a\n"Higgsed gauge theory" of the gauge group SU(2) \\times U(1), where\nSU(2) is the weak-isospin group and U(1) the weak hypercharge group.\nAfter breaking this gauge symmetry spontaneously to U(1), where now\nU(1) is the electromagnetic charge gauge group, so that we get the\nright vector boson degrees of freedom (three heavy bosons which make\nthe W and Z bosons and one massless boson which is the photon).\n\nThen you have also to take into account that the mass eigenstates of the\nquarks, going into the QCD Lagrangian, are different from the flavour\neigenstates of the quarks, which thus have to mix through the CKM\nmatrix (Cabbibo-Kobayashi-Maskawa matrix).\n\nOne of the greatest miracles of the standard model, at least for me when\nI learned about it, is that the observed pattern of hypercharges and\nflavours, together with the three colour degrees of freedom of the\nquarks make the theory free of an anomaly of the local gauge group\nwhich would completely destroy the whole gauge picture behind it and\nmake it a meaningless conglomerate of stupid mathematics. But the\npotential anomaly (due to the fact that the weak interaction is a\nchiral theory, even with maximal violation of parity through the vector\nminus axial vector structure of the interaction) cancels exactly due of\nthe pattern of hypercharges or, if you like it better to express it in\nterms of the physical degrees of freedom, the electrical charges.\n\n--\nHendrik van Hees Cyclotron Institute\nPhone: +1 979/845-1411 Texas A&M University\nFax: +1 979/845-1899 Cyclotron Institute, MS-3366\nhttp://theory.gsi.de/~vanhees/ College Station, TX 77843-3366\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>Dave Snead wrote:

> you seem to imply that all 4 of these combinations are possible for a
> quark
>
> Up, red color charge

electromagnetic charge

> Up, red color charge

electromagnetic charge

> Up, red color charge

electromagnetic charge

> Up, red color charge

electromagnetic charge

>
> However I posted this query on sci.physics.particle and got this
> response
>
>>No, this not possible. "Color charge" is associated with
>>quarks and "anticolor charge" with antiquarks. A mixture
>>is not possible (only gluons consist of "color" and "anticolor").
>
>>Kind regards,
>>Dorian Credé
>
> Which means the last two combinations are not possible.
> Can you resolve this discrepancy?

I hope so ;-). The point is that, as I wrote, colour charge (red, green,
blue) labels the three components of quarks associated to the
fundamental representation of the colour SU(3) group.

Now, also the complex conjugated components of a vector in the
fundamental-representation space of SU(3) form a(n irreducible)
representation of SU(3), which is not equivalent to the fundamental
representation. Thus, one has to distinguish between the colour SU(3)
representation of quarks (fundamental representation, often denoted
simply by 3) and this of anti-quarks (the conjugate complex of the
fundamental representation, denoted $LaTeX Code: by \\bar{3})$ .

If you neglect electromagnetic and weak interactions completely, all you
need to know is the QCD Lagrangian to get all this colour business
correct. It's demanded by the general idea of non-abelian gauge
invariance (a la Yang and Mills):

$LaTeX Code: L=\\sum_{flavours} \\bar{q}_{flavour} (\\i \\fslash(\\partial)-m-g_s\\fslash{A}_{\\mu}) q_{flavour}+kinetic$ terms for the gauge field

The pattern of electric charges is a little bit more difficult to
understand, since it comes from the electroweak theory, which is a
"Higgsed gauge theory" of the gauge group SU(2) $LaTeX Code: \\times U(1),$ where
SU(2) is the weak-isospin group and U(1) the weak hypercharge group.
After breaking this gauge symmetry spontaneously to U(1), where now
U(1) is the electromagnetic charge gauge group, so that we get the
right vector boson degrees of freedom (three heavy bosons which make
the W and Z bosons and one massless boson which is the photon).

Then you have also to take into account that the mass eigenstates of the
quarks, going into the QCD Lagrangian, are different from the flavour
eigenstates of the quarks, which thus have to mix through the CKM
matrix (Cabbibo-Kobayashi-Maskawa matrix).

One of the greatest miracles of the standard model, at least for me when
I learned about it, is that the observed pattern of hypercharges and
flavours, together with the three colour degrees of freedom of the
quarks make the theory free of an anomaly of the local gauge group
which would completely destroy the whole gauge picture behind it and
make it a meaningless conglomerate of stupid mathematics. But the
potential anomaly (due to the fact that the weak interaction is a
chiral theory, even with maximal violation of parity through the vector
minus axial vector structure of the interaction) cancels exactly due of
the pattern of hypercharges or, if you like it better to express it in
terms of the physical degrees of freedom, the electrical charges.

--
Hendrik van Hees Cyclotron Institute
Phone:

Texas A&M University
Fax:

Cyclotron Institute,

http://theory.gsi.de/~vanhees/ College Station,

Apr27-04, 03:00 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','to olbar=no,location=no,scrollbars=yes,resizable=yes, status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usene t 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>"Dave Snead" <dsnead6@charter.net> wrote in message\nnews:10896l3od6dv1f3@corp.supernews.com.. .\n| I\'m trying to clarify my understanding of quark color charge.\n| It seems to me, that there is a linkage between quark color and electric\n| charge.\n\nOne linkage is that they have the same units. Another is that they describe\nattraction and repulsion. However, because of different symmetry\nprinciples, their sources are different and color charge is much more\ncomplex. But they do seem to have hbar*c in common for their coupling\nconstants. Hbar*c is charge^2 in cgs units, so that is probably the only\n"linkage" that they might have.\n\n| For example, a red up quark has a red color charge of 1 and an electric\n| charge of 2/3,\n| while an antired up quark has a red color charge of -1 and an electric\n| charge of -2/3.\n| Is it possible for an up quark to have a red color charge of 1 and an\n| electric charge of -2/3\n| (or red color charge of -1 and electric charge of 2/3)?\n| Thanks in advance.\n\nNo, you need to read a particle physics book. Most will have a good\nexplanation of color charge.\n\nFrediFizzx\n\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>"Dave Snead" <dsnead6@charter.net> wrote in message
news:10896l3od6dv1f3@corp.supernews.com...
| I'm trying to clarify my understanding of quark color charge.

seems to me, that there is a linkage between quark color and electric
| charge.

One linkage is that they have the same units. Another is that they describe
attraction and repulsion. However, because of different symmetry
principles, their sources are different and color charge is much more
complex. But they do seem to have $LaTeX Code: \\hbar*c$ in common for their coupling
constants. $LaTeX Code: \\Hbar*c$ is

in cgs units, so that is probably the only
"linkage" that they might have.

| For example, a red up quark has a red color charge of 1 and an electric
| charge of

| while an antired up quark has a red color charge of -1 and an electric
| charge of

| Is it possible for an up quark to have a red color charge of 1 and an
| electric charge of

red color charge of -1 and electric charge

| Thanks in advance.

No, you need to read a particle physics book. Most will have a good
explanation of color charge.

FrediFizzx

Apr28-04, 02:49 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','to olbar=no,location=no,scrollbars=yes,resizable=yes, status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usene t 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>"Dave Snead" <dsnead6@charter.net> wrote:\n> I\'m trying to clarify my understanding of quark color charge.\n\nIf you\'re familiar with color theory, the role of "chromaticity\ncoordinates" is played by the 2 Casimir operators corresponding to\nthe matrices\nlambda_3 = diag(1, 0, -1)\nand\nlambda_8 = diag(1, -2, 1) sqrt(1/3).\nIn the fundamental representation 3, the chromaticity coordinates\nare:\ncolor L3 L8\nred gs/2 gs/sqrt(12)\ngreen 0 -gs/sqrt(3)\nblue -gs/2 gs/sqrt(12).\n\nIn the conjugate representation 3*, they are:\n\ncolor L3 L8\ncyan -gs/2 -gs/sqrt(12)\nmagneta 0 gs/sqrt(3)\namber gs/2 -gs/sqrt(12),\n\nusing the names (cyan, magneta, amber) respectively for\n(anti-red, anti-green, anti-yellow); thus forming the 6 cardinal\npoints of the "Newton color wheel", in a hexagonal pattern.\n\ngs is the coupling constant for the SU(3) force.\n\nThe charge magnitudes for the quarks are given by:\nL1^2 + ... + L8^2 = 4/3 gs^2\nL3^2 + L8^2 = 1/3 gs^2;\n(and for the gluons, respectively: 3 gs^2 and gs^2).\n\nYou\'re right to see a relation between the quarks and charges; but\nnot quite the one you had in mind. It involves the quantum\nnumber\nG = 1/2 (baryon - lepton)\nwhich is\nG = 1/6 for quarks; -1/6 for anti-quarks\nG = -1/2 for leptons; 1/2 for leptons.\nThis plays the analogous role of "brightness" in color theory.\n\nIf, instead of electric charge Q, you consider Q-G, you get:\nQ-G = 1/2: up, neutrino, anti-down, positron\nQ-G = -1/2: down, electron, anti-up, anti-neutrino.\nIt\'s +/- 1/2 for all fermions and thus defines a q-bit. Call it "a".\n\nThere is also the distinction between left-handedness and\nright-handedness. Left-handed particles and right-handed particles\ncan interact by the weak force -- those with Q-G > 0 have an\nSU(2) charge of +g/2, those with Q-G < 0 have an SU(2) charge of\n-g/2. This is operator I3, which corresponds to the Pauli matrix\ntau_3 = diag(1 -1).\n\nSo, one gets:\nQ-G - 2 I3/g = -(Q-G) for SU(2) interacting particles\n= +(Q-G) for SU(2) non-interacting particles\nSo, this too forms a q-bit, which we\'ll call "b".\n\nBoth a and b are independent of the items relating to SU(3).\n\nNote also that the isospin magnitude\nI^2 = I1^2 + I2^2 + I3^2 = 3/4 g^2\nfor weakly interacting particles; 0 otherwise. Thus, one arrives\nat the invariant:\n3 (Q-G-I3)^3 + I^2 = 3/4\nwith\na = Q-G = +/- 1/2\nb = Q-G - 2 I3 = +/- 1/2.\n\nGoing back to SU(3), you\'ll notice that since G plays the\nrole of a "brightness" coordinate, it\'s natural to encapsulate\nall 6 quark colors, plus the 2 lepton "colors" (white = anti-lepton,\nblack = lepton) into one scheme.\n\ncolor G L3/gs L8/gs particle\nwhite 1/2 0 0 anti-lepton\nred 1/6 1/2 sqrt(1/12) red quark\ngreen 1/6 0 -sqrt(1/3) green quark\nblue 1/6 -1/2 sqrt(1/12) blue quark\ncyan -1/6 -1/2 -sqrt(1/12) anti-red anti-quark\nmagneta -1/6 0 sqrt(1/3) anti-green anti-quark\namber -1/6 1/2 -sqrt(1/12) anti-blue anti-quark\nblack -1/2 0 0 lepton.\n\nThe operators\nc = G - sqrt(1/3) L8/gs + L3/gs\nd = G + sqrt(4/3) L8/gs\ne = G - sqrt(1/3) L8/gs - L3/gs\nall take on only the values +/- 1/2 for all the fermions and so\neach define a q-bit. The combinations are:\ncolor c d e\nwhite + + +\nred + + -\ngreen + - +\nblue - + +\ncyan - - +\nmagneta - + -\namber + - -\nblack - - -.\n\nSince L^2 = 4/3 gs^2 for [anti-]quarks and 0 for [anti-]leptons,\nit follows that you also have the invariant:\n6G^2 + L^2/gs^2 = 3/2\nwith\nc = G - sqrt(1/3) L8/gs + L3/gs = +/- 1/2\nd = G + sqrt(4/3) L8/gs = +/- 1/2\ne = G - sqrt(1/3) L8/gs - L3/gs = +/- 1/2.\n\nSo, the 4 combinations of (a,b) bits and 8 combinations of (c,d,e)\nbits form two independent sets. The charge is just\nQ = a + (c+d+e)/3.\nThe other charges are:\nI3 = (a-b)/2\nY = (a+b)/2 + (c+d+e)/3 -- hypercharge\nG = (c+d+e)/3\nL3 = (c-d)/2\nL8 = (-c + 2d - e)/sqrt(12).\n\nAll of this bespeaks a symmetry structure given by:\nU(2)_{X,I} x U(3)_{G,L}\nwith\nU(2)_{X,I} = U(1)_X x SU(2)_I; X = Y-G = Q-G-I3\nU(3)_{G,L} = U(1)_G x SU(3)_L; G = (Baryon-Lepton)/2.\n\nIn the Standard Model, however, the right-neutrino and\nleft-antineutrino are not present. If you assume these exist and\nthat the neutrino is just given by an ordinary Dirac 4-spinor\n(instead of a 2-component Weyl spinor or Majorana spinor) then\nthe fermion space will factor into the representation:\n(1_{1/2} + 2_0 + 1_{-1/2})(1_{1/2} + 3_{1/6} + 3*_{-1/6} + 1_{-1/2})\n\nwhere the first set of numbers denotes m_x for SU(2) m-plet and\nY-G = x; the second set of numbers n_g denotes SU(3) n-plet and\nG = g. In detail, this given you:\n\n1_{1/2} 1_{1/2} = e*_L\n1_{1/2} 3_{1/6} = u_R (red,green,blue)\n1_{1/2} 3*_{-1/6} = d*_R (cyan,magneta,amber)\n1_{1/2} 1_{-1/2} = nu_R\n\n2_0 1_{1/2} = (e* nu*)_R\n2_0 3_{1/6} = (u d)_L (red,green,blue)\n2_0 3*_{-1/6} = (d* u*)_R (cyan,magneta,amber)\n2_0 1_{-1/2} = (nu e)_L\n\n1_{-1/2} 1_{1/2} = nu*_L\n1_{-1/2} 3_{1/6} = d_R (red,green,blue)\n1_{-1/2} 3*_{-1/6} = u*_L (cyan,magneta,amber)\n1_{-1/2} 1_{-1/2} = e_R\n\nwhere ()_L and ()_R respectively denote the left and right handed\nparticles. For instance, the 6-tuple 2_0 3_{1/6} consists of:\n\ncolor a b c d e a b c d e\nred u_L + - + + - d_L - + + + -\ngreen u_L + - + - + d_L - + + - +\nblue u_L + - - + + d_L - + - + +.\n\nThere is also the Higgs which has the structure\n\nphi = 2_{1/2} 1_0 + 2_{-1/2} 1_0.\n\nIt\'s of interest to note that the charges that would have to be\nassigned to the bits a,b,c,d,e would make them tuplets of the\nfollowing forms:\n(a,b) = 2_{1/2} 1_0; (c,d,e) = 1_0 3*_{-1/3}.\nSo, the Higgs components have the same characteristics as the\nfundamental a and b charges would have.\n\nFor each set, one computes the invariants:\n(a,b): 3X^2 + I^2 = 3/2; 6G^2 + L^2 = 0\n(c,d,e): 3X^2 + I^2 = 0; 6G^2 + L^2 = 2.\n\nThe gauge bosons form the structure:\n1_0 1_0 + 3_0 1_0 + 1_0 8_0;\nidentified respectively as\nTuplet Boson Force 3X^2 + I^2 6G^2 + L^2\n1_0 1_0 B Hypercharge 0 0\n3_0 1_0 W Weak force 2 0\n1_0 8_0 G Strong force 0 3;\nthe G\'s being the 8 gluons; W\'s the 3 weak bosons.\n\nThe photon and Z particle are linear combinations of B and W3 by\nphoton = B cos(T) + W3 sin(T)\nZ = W3 cos(T) - B sin(T)\nT = weak mixing angle;\nthe angle T is approximately equal to the smallest angle of the\n8-15-17 right triangle.\n\nInterestingly, if one took the product of the 6 tuples\n6 = 2_{1/2} 3*_{1/3} and 6* = 2_{-1/2} 3_{-1/3}\none would get:\n2_{1/2} x 2_{-1/2} = 1_0 + 3_0 for U(2)\n3*_{1/3} x 3_{-1/3} = 1_0 + 8_0 for U(3)\nthus resulting in the decomposition\n6 x 6* = 1_0 1_0 + 3_0 1_0 + 1_0 8_0 + 3_0 8_0.\n\nThe 24-tuple 3_0 8_0 would behave as W-G pairs, comprising the 24\npairings of W\'s with G\'s. They\'re not known to exist ... but\nwould be hard to disgintuish from W-G pairs.\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>"Dave Snead" <dsnead6@charter.net> wrote:
> I'm trying to clarify my understanding of quark color charge.

If you're familiar with color theory, the role of "chromaticity
coordinates" is played by the 2 Casimir operators corresponding to
the matrices
$LaTeX Code: \\lambda_3 =$ diag(1,

and
$LaTeX Code: \\lambda_8 =$ diag(1, $LaTeX Code: -2, 1) \\sqrt(1/3)$ .
In the fundamental representation 3, the chromaticity coordinates
are:
color L3 L8
red $LaTeX Code: gs/2 gs/\\sqrt(12)$
green $LaTeX Code: -gs/\\sqrt(3)$
blue $LaTeX Code: -gs/2 gs/\\sqrt(12)$ .

In the conjugate representation 3*, they are:

color L3 L8
cyan $LaTeX Code: -gs/2 -gs/\\sqrt(12)$
magneta $LaTeX Code: gs/\\sqrt(3)$
amber $LaTeX Code: gs/2 -gs/\\sqrt(12),$

using the names (cyan, magneta, amber) respectively for
(anti-red, anti-green, anti-yellow); thus forming the 6 cardinal
points of the "Newton color wheel", in a hexagonal pattern.

gs is the coupling constant for the SU(3) force.

The charge magnitudes for the quarks are given by:

.

LaTeX Code: . + L8^2 = 4/3 gs^2L3^2 + L8^2 = 1/3 gs^2;

(and for the gluons, respectively:

and

.

You're right to see a relation between the quarks and charges; but
not quite the one you had in mind. It involves the quantum
number

(baryon - lepton)
which is

for quarks;

for anti-quarks

for leptons; 1/2 for leptons.
This plays the analogous role of "brightness" in color theory.

If, instead of electric charge Q, you consider

you get:

up, neutrino, anti-down, positron

down, electron, anti-up, anti-neutrino.
It's

for all fermions and thus defines a q-bit. Call it "a".

There is also the distinction between left-handedness and
right-handedness. Left-handed particles and right-handed particles
can interact by the weak force -- those with

have an
SU(2) charge of

those with

have an SU(2) charge of

. This is operator I3, which corresponds to the Pauli matrix
$LaTeX Code: \\tau_3 =$ diag(1 -1).

So, one gets:

for SU(2) interacting particles

for SU(2) non-interacting particles
So, this too forms a q-bit, which we'll call "b".

Both a and b are independent of the items relating to SU(3).

Note also that the isospin magnitude

LaTeX Code: I^2 = I1^2 + I2^2 + I3^2 = 3/4 g^2

for weakly interacting particles; otherwise. Thus, one arrives
at the invariant:

with

LaTeX Code: a = Q-G = +/- 1/2b = Q-G - 2 I3 = +/- 1/2

.

Going back to SU(3), you'll notice that since G plays the
role of a "brightness" coordinate, it's natural to encapsulate
all 6 quark colors, plus the 2 lepton "colors" (white = anti-lepton,
black = lepton) into one scheme.

color

particle
white 1/2 anti-lepton
red $LaTeX Code: 1/6 1/2 \\sqrt(1/12)$ red quark
green 1/6 $LaTeX Code: -\\sqrt(1/3)$ green quark
blue $LaTeX Code: 1/6 -1/2 \\sqrt(1/12)$ blue quark
cyan $LaTeX Code: -1/6 -1/2 -\\sqrt(1/12)$ anti-red anti-quark
magneta $LaTeX Code: -1/6\\sqrt(1/3)$ anti-green anti-quark
amber $LaTeX Code: -1/6 1/2 -\\sqrt(1/12)$ anti-blue anti-quark
black

lepton.

The operators
$LaTeX Code: c = G - \\sqrt(1/3) L8/gs + L3/gsd = G + \\sqrt(4/3) L8/gse = G - \\sqrt(1/3) L8/gs - L3/gs$
all take on only the values

for all the fermions and so
each define a q-bit. The combinations are:
color c d e
white

red

green

blue

cyan

magneta

amber

black

.

Since

for [anti-]quarks and for [anti-]leptons,
it follows that you also have the invariant:

with
$LaTeX Code: c = G - \\sqrt(1/3) L8/gs + L3/gs = +/- 1/2d = G + \\sqrt(4/3) L8/gs = +/- 1/2e = G - \\sqrt(1/3) L8/gs - L3/gs = +/- 1/2$ .

So, the 4 combinations of (a,b) bits and 8 combinations of (c,d,e)
bits form two independent sets. The charge is just

.
The other charges are:

LaTeX Code: I3 = (a-b)/2Y = (a+b)/2 + (c+d+e)/3 --

hypercharge
$LaTeX Code: G = (c+d+e)/3L3 = (c-d)/2L8 = (-c + 2d - e)/\\sqrt(12).$

All of this bespeaks a symmetry structure given by:
$LaTeX Code: U(2)_{X,I}$ x $LaTeX Code: U(3)_{G,L}$
with
$LaTeX Code: U(2)_{X,I} = U(1)_X$ x $LaTeX Code: SU(2)_I; X = Y-G = Q-G-I3U(3)_{G,L} = U(1)_G$ x

(Baryon-Lepton)/2.

In the Standard Model, however, the right-neutrino and
left-antineutrino are not present. If you assume these exist and
that the neutrino is just given by an ordinary Dirac 4-spinor
(instead of a 2-component Weyl spinor or Majorana spinor) then
the fermion space will factor into the representation:
$LaTeX Code: (1_{1/2} + 2_0 + 1_{-1/2})(1_{1/2} + 3_{1/6} + 3*_{-1/6} + 1_{-1/2})$

where the first set of numbers denotes

for SU(2) m-plet and

the second set of numbers

denotes SU(3) n-plet and

. In detail, this given you:

$LaTeX Code: 1_{1/2} 1_{1/2} = e*_L1_{1/2} 3_{1/6} = u_R$ (red,green,blue)
$LaTeX Code: 1_{1/2} 3*_{-1/6} = d*_R$ (cyan,magneta,amber)
$LaTeX Code: 1_{1/2} 1_{-1/2} = \\nu_R2_0 1_{1/2} = (e* \\nu*)_R2_0 3_{1/6} = (u d)_L$ (red,green,blue)
$LaTeX Code: 2_0 3*_{-1/6} = (d* u*)_R$ (cyan,magneta,amber)
$LaTeX Code: 2_0 1_{-1/2} = (\\nu e)_L1_{-1/2} 1_{1/2} = \\nu*_L1_{-1/2} 3_{1/6} = d_R$ (red,green,blue)
$LaTeX Code: 1_{-1/2} 3*_{-1/6} = u*_L$ (cyan,magneta,amber)
$LaTeX Code: 1_{-1/2} 1_{-1/2} = e_R$

where

and

respectively denote the left and right handed
particles. For instance, the 6-tuple $LaTeX Code: 2_0 3_{1/6}$ consists of:

color a b c d e a b c d e
red

green

blue

.

There is also the Higgs which has the structure

$LaTeX Code: \\phi = 2_{1/2} 1_0 + 2_{-1/2} 1_0$ .

It's of interest to note that the charges that would have to be
assigned to the bits a,b,c,d,e would make them tuplets of the
following forms:
(a,b) $LaTeX Code: = 2_{1/2} 1_0;$ (c,d,e) $LaTeX Code: = 1_0 3*_{-1/3}$ .
So, the Higgs components have the same characteristics as the
fundamental a and b charges would have.

For each set, one computes the invariants:
(a,b):

LaTeX Code: 3X^2 + I^2 = 3/2; 6G^2 + L^2 =(c,d,e): 3X^2 + I^2 = 0; 6G^2 + L^2 = 2

.

The gauge bosons form the structure:

LaTeX Code: 1_0 1_0 + 3_0 1_0 + 1_0 8_0;

identified respectively as
Tuplet Boson Force

LaTeX Code: 3X^2 + I^2 6G^2 + L^21_0 1_0

B Hypercharge

W Weak force 2

G Strong force 3;
the G's being the 8 gluons; W's the 3 weak bosons.

The photon and Z particle are linear combinations of B and W3 by
photon

LaTeX Code: = B cos(T) + W3 sin(T)Z = W3 cos(T) - B sin(T)

T = weak mixing angle;
the angle T is approximately equal to the smallest angle of the

right triangle.

Interestingly, if one took the product of the 6 tuples
$LaTeX Code: 6 = 2_{1/2} 3*_{1/3}$ and $LaTeX Code: 6* = 2_{-1/2} 3_{-1/3}$
one would get:
$LaTeX Code: 2_{1/2} x 2_{-1/2} = 1_0 + 3_0$ for U(2)
$LaTeX Code: 3*_{1/3} x 3_{-1/3} = 1_0 + 8_0$ for U(3)
thus resulting in the decomposition
6

LaTeX Code: x 6* = 1_0 1_0 + 3_0 1_0 + 1_0 8_0 + 3_0 8_0

.

The 24-tuple

would behave as W-G pairs, comprising the 24
pairings of W's with G's. They're not known to exist ... but
would be hard to disgintuish from W-G pairs.

Apr30-04, 03:02 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','to olbar=no,location=no,scrollbars=yes,resizable=yes, status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usene t 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>"Dave Snead" <dsnead6@charter.net> wrote in message news:<10896l3od6dv1f3@corp.supernews.com>... \n> It seems to me, that there is a linkage between quark color and\n> electric charge.\n\nAs a followup to my previous reply, there is a 2nd way in which the\ntwo sets of charges are not independent.\n\nGiven the list of particles and their color charges (i.e., their\nbehavior under SU(3)) and isospins (their behavior under SU(2)),\nit turns out that there are severe constraints placed on whatever\nother charges may be included. The constraints are so as to make\nthe corresponding S-matrix expressions consistent so that you can\ndo perturbation theory; and the constraints are essentially a\nmodern-day analogue of Dirac\'s old charge quantization constraint;\nthough unrelated.\n\nThe constraints (almost) uniquely determine the assignment of\nelectric charges to the particles.\n\nAs mentioned before, "color" charge is actually a kind of\n2-dimensional charge which occupies a space closely analogous to\ncolor space; and is coordinatized by two quantum numbers that\ncorrespond to the SU(3) matrices lambda_3 = diag(1,0,-1) and\nlambda_8 = diag(1,-2,1)/sqrt(3).\n\nThe values for the quarks and anti-quarks are:\n\ncolor L3 L8\nred gs/2 gs/sqrt(12)\ngreen 0 -gs/sqrt(3)\nblue -gs/2 gs/sqrt(12)\ncyan -gs/2 -gs/sqrt(12)\nmagneta 0 gs/sqrt(3)\namber gs/2 -gs/sqrt(12),\n\nagain, using the names (cyan, magneta, amber) respectively for\n(anti-red, anti-green, anti-yellow).\n\nFor SU(2), you also have weak isospin I3, which takes on the values\n+/- g/2; where g is the coupling constant for SU(2), approximately\nequal to 17/8 e. The spectrum is as follows:\n\nI3 Particles\n0 right up; left anti-down, positron\ng/2 left up, neutrino; right anti-down, positron\n-g/2 left down, electron; right anti-up, anti-neutrino\n0 left up, neutrino; right anti-down, positron\n\nIf a right neutrino and left anti-neutrino were present they would\nbe included amongst the topmost entries.\n\nThe Standard Model also assigns a set of charges for hypercharge\nY, with values +/- g\'/6; +/- g\'/3; +/- 2g\'/3 for the [anti-]quarks;\n0; +/- g\'/2; +/- g\' for the [anti-]leptons where g\' is the U(1)\ncoupling constant, approximately equal to 17/15 e.\n\nThey take on the values -- which are listed alongside I3:\n\nY I3 Particles\ng\'(G + 1/2) 0 right up; left anti-down, positron\ng\' G g/2 left up, neutrino; right anti-down, positron\ng\' G -g/2 left down, electron; right anti-up, anti-neutrino\ng\'(G - 1/2) 0 left up, neutrino; right anti-down, positron\n\nwhere\nG = 1/2 (Baryon - Lepton);\n= +1/6 for quarks; -1/6 for anti-quarks;\n= +1/2 for anti-leptons; -1/2 for leptons.\n\nIt\'s the latter relation between Y and G that you picked up on, in\npart, in your question.\n\nIt\'s also the hypercharge Y which is deriveable; and out of this that\nyou get the expression for electric charge as:\nQ = e (I3/g + Y/g\')\n= (15 I3 + 8 Y)/17, approximately.\n(with (e/g)^2 + (e/g\')^2 = 1).\n\nThe consistency condition ultimately comes from the fact that the\nleft and right handed particles are assigned different values in\ntheir SU(2) spectrum. Ultimately this imposes a set of cubic\nrelations on the charges of the following forms:\n\nsum (ABC) for left = sum (ABC) for right\nfor all charges A, B, C\nsummed over all particles states in the spectrum.\n\nIn addition, there is also a requirement (coming out of the\nHiggs mechanism) that for all charge assignments, Q(...), one\nhas:\nQ(left down) - Q(right down)\n= Q(left electron) - Q(right electron)\n= Q(right up) - Q(left up)\n[ = Q(right neutrino) - Q(left neutrino); if you have a right neutrino].\n\nThe charge, by assumption is SU(2) and SU(3) symmetric so that the\nmembers of the same SU(2) and SU(3) multiplet must also have the\nsame assignments. In addition, anti-particles have opposite charges\nto one another.\n\nThe overall set of SU(2) and SU(3) assignments are listed in the\nfollowing table, along with the required assignments for any other\nputative charge Q; here I\'m only working with the whole number ratios\nof the respective charges, since that\'s all you need for writing down\nthe cubic constraints:\n\nQ I3 L3 L8 Q I3 L3 L8\nA+X 0 0 0 e*_L A 1 0 0 e*_R\nB+X 0 1 1 u_R red B 1 1 1 u_L red\nB+X 0 0 -2 u_R green B 1 0 -2 u_L green\nB+X 0 -1 1 u_R blue B 1 -1 1 u_L blue\n-B+X 0 -1 -1 d*_L cyan -B 1 -1 -1 d*_R cyan\n-B+X 0 0 2 d*_L magneta -B 1 0 2 d*_R magneta\n-B+X 0 1 -1 d*_L amber -B 1 1 -1 d*_R amber\n[-A+X 0 0 0 nu_R ] -A 1 0 0 nu_L\n\nQ I3 L3 L8 Q I3 L3 L8\nA -1 0 0 nu*_R [A-X 0 0 0 nu*_L]\nB -1 1 1 d_L red B-X 0 1 1 d_R red\nB -1 0 -2 d_L green B-X 0 0 -2 d_R green\nB -1 -1 1 d_L blue B-X 0 -1 1 d_R blue\n-B -1 -1 -1 u*_R cyan -B-X 0 -1 -1 u*_L cyan\n-B -1 0 2 u*_R magneta -B-X 0 0 2 u*_L magneta\n-B -1 1 -1 u*_R amber -B-X 0 1 -1 u*_L amber\n-A -1 0 0 e_L -A-X 0 0 0 e_R\n\nSince left particles are opposite to their corresponding right\nanti-particles; then the constraint\nsum (...) right = sum (...) left\n= - sum (...) right\nimplies\n2 sum (...) right = 0.\nSo, it\'s enough to only count the right-handed particles:\n\nQ I3 L3 L8\nA 1 0 0 e*_R\nB+X 0 1 1 u_R red\nB+X 0 0 -2 u_R green\nB+X 0 -1 1 u_R blue\n-B 1 -1 -1 d*_R cyan\n-B 1 0 2 d*_R magneta\n-B 1 1 -1 d*_R amber\n[-A+X 0 0 0 nu_R ]\nA -1 0 0 nu*_R\nB-X 0 1 1 d_R red\nB-X 0 0 -2 d_R green\nB-X 0 -1 1 d_R blue\n-B -1 -1 -1 u*_R cyan\n-B -1 0 2 u*_R magneta\n-B -1 1 -1 u*_R amber\n-A-X 0 0 0 e_R\n\nThe first thing to note is that the cubic combinations from I3, L3\nand L8 all add to 0 already; so the SU(2) and SU(3) parts are\nalready consistent with each other.\n\nThe sum of QQQ\'s is:\nA^3 - 3A^2 X - 3A X^2 - X^3 + 18 B X^2 = 0\n[18 B X^2 - 6 A X^2 = 0, if including nu_R]\n\nThe sums of terms of the forms QQZ, for the other charges Z is\nreadily seen to be 0 in all cases.\n\nThe sums of terms of the forms QYZ for the other charges\nY, Z with Y, Z different is also readily seen to be 0.\n\nFinally, the sums of terms of the form QYY for the other charges\nY are\nY = I3: 2A - 6B\nY = L3: 0\nY = L8: 0\n\nSo, the only other constraint equation is A = 3B. Substituting\nthis into the first equation yields:\nA^3 - 3 A^2 X + 3 A X^2 - X^3 = (A - X) = 0\n[or 6(3B - A) X^2 = 0; if including nu_R].\n\nIn the case where there\'s no nu_R, this implies X = A, which\nleads to the following assignments up to proportionality constant:\n\nQ\n6 e_L\n4 u_R\n3 e*_R, nu*_R\n2 d*_L\n1 d_L, u_L\n-1 d*_R, u*_R\n-2 d_R\n-3 e_L, nu_L\n-4 u*_L\n-6 e_R\n\nwhich is just the hypercharge, up to proportionality.\n\nIf, on the other hand, you include the right neutrino, then\nX remains undetermined and the assignments are, in general:\n\nQ\n3B+X e*_L\n3B e*_R, nu*_R\n3B-X nu*_L\nB+X u_R\nB d_L, u_L\nB-X d_R\n-B+X d*_L\n-B d*_R, u*_R\n-B-X u*_R\n-3B+X nu_R\n-3B e_L, nu_L\n-3B-X e_R\n\nwhich is just the linear combination:\nQ = 6B G + 2X (Y-G)\nwhere G = 1/2 (Baryon - Lepton). Requiring the right-neutrino to\nhave 0 charge will give you X = 3B,\nQ = 6B G + 6B (Y-G) = 6B Y\nthus recovering the hypercharge Y up to proportionality.\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>"Dave Snead" <dsnead6@charter.net> wrote in message news:<10896l3od6dv1f3@corp.supernews.com>...
> It seems to me, that there is a linkage between quark color and
> electric charge.

As a followup to my previous reply, there is a 2nd way in which the
two sets of charges are not independent.

Given the list of particles and their color charges (i.e., their
behavior under SU(3)) and isospins (their behavior under SU(2)),
it turns out that there are severe constraints placed on whatever
other charges may be included. The constraints are so as to make
the corresponding S-matrix expressions consistent so that you can
do perturbation theory; and the constraints are essentially a
modern-day analogue of Dirac's old charge quantization constraint;
though unrelated.

The constraints (almost) uniquely determine the assignment of
electric charges to the particles.

As mentioned before, "color" charge is actually a kind of
2-dimensional charge which occupies a space closely analogous to
color space; and is coordinatized by two quantum numbers that
correspond to the SU(3) matrices $LaTeX Code: \\lambda_3 = diag(1,0,-1)$ and
$LaTeX Code: \\lambda_8 = diag(1,-2,1)/\\sqrt(3)$ .

The values for the quarks and anti-quarks are:

color L3 L8
red $LaTeX Code: gs/2 gs/\\sqrt(12)$
green $LaTeX Code: -gs/\\sqrt(3)$
blue $LaTeX Code: -gs/2 gs/\\sqrt(12)$
cyan $LaTeX Code: -gs/2 -gs/\\sqrt(12)$
magneta $LaTeX Code: gs/\\sqrt(3)$
amber $LaTeX Code: gs/2 -gs/\\sqrt(12),$

again, using the names (cyan, magneta, amber) respectively for
(anti-red, anti-green, anti-yellow).

For SU(2), you also have weak isospin I3, which takes on the values

where g is the coupling constant for SU(2), approximately
equal to

. The spectrum is as follows:

I3 Particles
right up; left anti-down, positron
g/2 left up, neutrino; right anti-down, positron

left down, electron; right anti-up, anti-neutrino
left up, neutrino; right anti-down, positron

If a right neutrino and left anti-neutrino were present they would
be included amongst the topmost entries.

The Standard Model also assigns a set of charges for hypercharge
Y, with values

LaTeX Code: +/- gsingle-quote/6; +/- gsingle-quote/3; +/- 2gsingle-quote/3

for the [anti-]quarks;

LaTeX Code: 0; +/- gsingle-quote/2; +/- gsingle-quote

for the [anti-]leptons where g' is the U(1)
coupling constant, approximately equal to

.

They take on the values -- which are listed alongside I3:

Y I3 Particles

right up; left anti-down, positron
g' G g/2 left up, neutrino; right anti-down, positron
g' G

left down, electron; right anti-up, anti-neutrino

left up, neutrino; right anti-down, positron

where

(Baryon - Lepton);

for quarks;

for anti-quarks;

for anti-leptons;

for leptons.

It's the latter relation between Y and G that you picked up on, in
part, in your question.

It's also the hypercharge Y which is deriveable; and out of this that
you get the expression for electric charge as:

LaTeX Code: Q = e (I3/g + Y/gsingle-quote)= (15 I3 + 8 Y)/17,

approximately.
(with

LaTeX Code: (e/g)^2 + (e/gsingle-quote)^2 = 1)

.

The consistency condition ultimately comes from the fact that the
left and right handed particles are assigned different values in
their SU(2) spectrum. Ultimately this imposes a set of cubic
relations on the charges of the following forms:

sum (ABC) for left = sum (ABC) for right
for all charges A, B, C
summed over all particles states in the spectrum.

In addition, there is also a requirement (coming out of the
Higgs mechanism) that for all charge assignments, Q(...), one
has:
Q(left down) - Q(right down)
= Q(left electron) - Q(right electron)
= Q(right

Q(left up)
[ = Q(right neutrino) - Q(left neutrino); if you have a right neutrino].

The charge, by assumption is SU(2) and SU(3) symmetric so that the
members of the same SU(2) and SU(3) multiplet must also have the
same assignments. In addition, anti-particles have opposite charges
to one another.

The overall set of SU(2) and SU(3) assignments are listed in the
following table, along with the required assignments for any other
putative charge Q; here I'm only working with the whole number ratios
of the respective charges, since that's all you need for writing down
the cubic constraints:

A+X

A 1

B+X 1 1

red B 1 1 1

red
B+X

green B 1

green
B+X -1 1

blue B 1 -1 1

blue

cyan -B 1

cyan

2

magneta -B 1 2

magneta

1

amber -B 1 1

amber
$LaTeX Code: [-A+X\\nu_R ] -A$ 1 $LaTeX Code: \\nu_L$

Q I3 L3 L8 Q I3 L3 L8
A -1 $LaTeX Code: \\nu*_R [A-X\\nu*_L]B -1$ 1 1

red B-X 1 1

red
B -1

green B-X

green
B

1

blue B-X -1 1

blue

cyan

cyan

2

magneta

2

magneta

1

amber

1

amber

Since left particles are opposite to their corresponding right
anti-particles; then the constraint
sum (...) right = sum (...) left
= - sum (...) right
implies
2 sum (...) right = .
So, it's enough to only count the right-handed particles:

Q I3 L3 L8
A 1

B+X 1 1

red
B+X

green
B+X -1 1

blue
-B 1

cyan
-B 1 2

magneta
-B 1 1

amber
$LaTeX Code: [-A+X\\nu_R ]A -1\\nu*_R$
B-X 1 1

red
B-X

green
B-X -1 1

blue

cyan

2

magneta

1

amber

The first thing to note is that the cubic combinations from I3, L3
and L8 all add to already; so the SU(2) and SU(3) parts are
already consistent with each other.

The sum of QQQ's is:

LaTeX Code: A^3 - 3A^2 X - 3A X^2 - X^3 + 18 B X^2 =[18 B X^2 - 6 A X^2 = 0,

if including $LaTeX Code: \\nu_R]$

The sums of terms of the forms QQZ, for the other charges Z is
readily seen to be in all cases.

The sums of terms of the forms QYZ for the other charges
Y, Z with Y, Z different is also readily seen to be .

Finally, the sums of terms of the form QYY for the other charges
Y are

LaTeX Code: Y = I3: 2A - 6BY = L3:Y = L8:

So, the only other constraint equation is

. Substituting
this into the first equation yields:

LaTeX Code: A^3 - 3 A^2 X + 3 A X^2 - X^3 = (A - X) =

[or

if including $LaTeX Code: \\nu_R]$ .

In the case where there's $LaTeX Code: no \\nu_R,$ this implies

which
leads to the following assignments up to proportionality constant:

Q
6 $LaTeX Code: e_L4 u_R3 e*_R, \\nu*_R2 d*_L1 d_L, u_L-1 d*_R, u*_R-2 d_R-3 e_L, \\nu_L-4 u*_L-6 e_R$

which is just the hypercharge, up to proportionality.

If, on the other hand, you include the right neutrino, then
X remains undetermined and the assignments are, in general:

Q
$LaTeX Code: 3B+X e*_L3B e*_R, \\nu*_R3B-X \\nu*_LB+X u_RB d_L, u_LB-X d_R-B+X d*_L-B d*_R, u*_R-B-X u*_R-3B+X \\nu_R-3B e_L, \\nu_L-3B-X e_R$

which is just the linear combination:

where

(Baryon - Lepton). Requiring the right-neutrino to
have charge will give you

LaTeX Code: X = 3B,Q = 6B G + 6B (Y-G) = 6B Y

thus recovering the hypercharge Y up to proportionality.