Quark colour charge [Archive]

alistair

Jun7-04, 04:54 AM

Alfred Einstead

Jun8-04, 04:03 AM

<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>\nalistair@goforit64.fsnet.co.uk (alistair) wrote:\n> Why does a gluon have colour charge but a photon does not have\n> electric charge?\n\nBecause the corresponding classical Yang-Mills fields have\ncharge non-conservation. Part of the charge & current can\nbe added to or taken from a test body by the field; and\nactually resides in the field.\n\nClassically, the equations for the field are:\nF^a_{mn} = d_m A^a_n - d_n A^a_m + f^a_{bc} A^b_m A^c_n\n[summation convention used]\n\nwith the field F being *non-linear* in the potential A. The\nf\'s are the structure constants for a Lie algebra:\n[Y_a, Y_b] = f^c_{ab} Y_c.\nFor SU(2) these would be (with suitable scaling)\n[T+,T-] = 2 T0\n[T0,T+] = T+\n[T0,T-] = -T-\nso\nf^0_{+-} = -f^0_{-+} = 2\nf^+_{0+} = -f^+_{+0} = 1\nf^-_{-0} = -f^-_{0-} = 1\nf^a_{bc} = 0 else\n\nThe generalization of the inhomogeneous part of the Maxwell\nequations would be:\nd_m F_a^{nm} = J_a^n + f^b_{ac} A^c_m F_b^{mn}\n(I might have the sign wrong on the last term).\n\nThe field, itself, has a non-zero charge and current proportional\nto the product of the potential A and the field F; to a degree\nproportional to the structure constants f.\n\n[And the equations corresponding to the homogeneous part of\nMaxwell\'s equations have non-zero magnetic charge and currrent\nterms on the right, also proportional to A, F and f.]\n\nThe matter current J_a^n is NOT conserved by itself. Only the\ncombination of it and the part contained in the field above\nis.\n\nWhen quantized, the field therefore is represented by bosons\nthat carry part of the said charge and current.\n\nAnd it also means that a test body\'s charge (e_a), as a\nfunction e_a(s) of proper time s, would also vary as follows:\ne_a\'(s) = -f^b_{ac} A^c_m e_b(s),\nreflecting the gain or loss to or from the field.\n\nTechnically, all the above is also true of the electromagnetic\nfield; since it has a non-trivial intersection with the SU(2)\npart of the electroweak field.\n\nThe photon is electrically neutral since its corresponding\nLie algebra unit Q has 0 commutator with itself:\n[Q,Q] = 0\nbut its T+ and T- charges are not 0; since [Q,T+] and [Q,T-] are\nNOT 0! The photon is not neutral. Only the boson, Y, corresponding\nto the U(1) part of the electroweak force is neutral. And the Y\nis NOT the photon. It\'s a mixture of the photon and the Z. And\nat the same time, the photon and Z are mixtures of Y and the\nboson W0 corresponding to T0; so both the photon and Z have\nnon-zero T+ and T- charges.\n\n> And when physicists say that the colour force and the electric force\n> are united at a certain temperature, does this mean that they have the\n> same strength or that the colour charge has become electric charge?\n\nIt means electric charge is screened, color charge is anti-screened;\nthat the charges you actually see from macroscopic distance are\ndrastically different, on account of this screening, from the\ncharges seen from close up; and that when seen from the right\ndistance, both sets of charges are multiples of the same unit\ncharge (more precisely: the 2 electroweak coupling constants\nand the color force coupling constant all meet at a common\nvalue, when seen at a specific length scale).\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>alistair@goforit64.fsnet.co.uk (alistair) wrote:
> Why does a gluon have colour charge but a photon does not have
> electric charge?

Because the corresponding classical Yang-Mills fields have
charge non-conservation. Part of the charge & current can
be added to or taken from a test body by the field; and
actually resides in the field.

Classically, the equations for the field are:
F^{a_}{mn} = d_m A^{a_n} - d_n A^{a_m} + f^{a_}{bc} A^{b_m} A^{c_n}
[summation convention used]

with the field F being *non-linear* in the potential A. The
f's are the structure constants for a Lie algebra:
[Y_a, Y_b] = f^{c_}{ab} Y_c.
For SU(2) these would be (with suitable scaling)
[T+,T-] = 2 T0[T0,T+] = T+[T0,T-] = -T-
so
f^{0_}{+-} = -f^{0_}{-+} = 2f^+_{0+} = -f^+_{+0} = 1f^-_{-0} = -f^-_{0-} = 1f^{a_}{bc} = else

The generalization of the inhomogeneous part of the Maxwell
equations would be:
d_m F_a^{nm} = J_a^n + f^{b_}{ac} A^{c_m} F_b^{mn}
(I might have the sign wrong on the last term).

The field, itself, has a non-zero charge and current proportional
to the product of the potential A and the field F; to a degree
proportional to the structure constants f.

[And the equations corresponding to the homogeneous part of
Maxwell's equations have non-zero magnetic charge and currrent
terms on the right, also proportional to A, F and f.]

The matter current J_a^n is NOT conserved by itself. Only the
combination of it and the part contained in the field above
is.

When quantized, the field therefore is represented by bosons
that carry part of the said charge and current.

And it also means that a test body's charge (e_a), as a
function e_a(s) of proper time s, would also vary as follows:
e_a'(s) = -f^{b_}{ac} A^{c_m} e_b(s),
reflecting the gain or loss to or from the field.

Technically, all the above is also true of the electromagnetic
field; since it has a non-trivial intersection with the SU(2)
part of the electroweak field.

The photon is electrically neutral since its corresponding
Lie algebra unit Q has commutator with itself:
[Q,Q] =
but its T+ and T- charges are not 0; since [Q,T+] and [Q,T-] are
NOT 0! The photon is not neutral. Only the boson, Y, corresponding
to the U(1) part of the electroweak force is neutral. And the Y
is NOT the photon. It's a mixture of the photon and the Z. And
at the same time, the photon and Z are mixtures of Y and the
boson W0 corresponding to T0; so both the photon and Z have
non-zero T+ and T- charges.

> And when physicists say that the colour force and the electric force
> are united at a certain temperature, does this mean that they have the
> same strength or that the colour charge has become electric charge?

It means electric charge is screened, color charge is anti-screened;
that the charges you actually see from macroscopic distance are
drastically different, on account of this screening, from the
charges seen from close up; and that when seen from the right
distance, both sets of charges are multiples of the same unit
charge (more precisely: the 2 electroweak coupling constants
and the color force coupling constant all meet at a common
value, when seen at a specific length scale).