<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\nThere are two families of unstable quarks (top, bottom,strange and\ncharm)\nand one family of stable quarks (up and down).Are there any reasons to\nthink that there should be another family of stable quarks so that\nthere are two unstable and two stable families?\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>There are two families of unstable quarks (top, bottom,strange and
charm)
and one family of stable quarks (up and down).Are there any reasons to
think that there should be another family of stable quarks so that
there are two unstable and two stable families?
Ulmo
Jul16-04, 03:02 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nalistair@goforit64.fsnet.co.uk (alistair) wrote in message news:<861c1b21.0407151425.1234b9f6@posting.google.com>...\n> There are two families of unstable quarks (top, bottom,strange and\n> charm)\n> and one family of stable quarks (up and down).Are there any reasons to\n> think that there should be another family of stable quarks so that\n> there are two unstable and two stable families?\n\n\nThe reason they are stable is because they are lightest.\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 in message news:<861c1b21.0407151425.1234b9f6@posting.google.com>...
> There are two families of unstable quarks (top, bottom,strange and
> charm)
> and one family of stable quarks (up and down).Are there any reasons to
> think that there should be another family of stable quarks so that
> there are two unstable and two stable families?
The reason they are stable is because they are lightest.
alistair
Jul19-04, 04:08 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>\n\n\n\nUlmo said:\n> the stable quarks are the lightest quarks\n\nAlistair says:\n\nThe stable quarks are the lightest known quarks.\nOther known quarks can yield the stable quarks.\nBut if a new stable quark existed with a mass just above the\nmass of a down quark, would it be stable?\nIt would not experience the weak force because if\nit did then a new neutrino would be associated with it\nand this would increase the rate of decay of the z boson,\nand would be at odds with observational evidence.\nBut this absence of the weak interaction would explain\nwhy other known quarks don\'t become the new stable quark.\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>Ulmo said:
> the stable quarks are the lightest quarks
Alistair says:
The stable quarks are the lightest known quarks.
Other known quarks can yield the stable quarks.
But if a new stable quark existed with a mass just above the
mass of a down quark, would it be stable?
It would not experience the weak force because if
it did then a new neutrino would be associated with it
and this would increase the rate of decay of the z boson,
and would be at odds with observational evidence.
But this absence of the weak interaction would explain
why other known quarks don't become the new stable quark.
Igor
Jul19-04, 04:09 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>\n\n\n\nalistair@goforit64.fsnet.co.uk (alistair) wrote in message news:<861c1b21.0407151425.1234b9f6@posting.google.com>...\n> There are two families of unstable quarks (top, bottom,strange and\n> charm)\n> and one family of stable quarks (up and down).Are there any reasons to\n> think that there should be another family of stable quarks so that\n> there are two unstable and two stable families?\n\nThere are also two generations of unstable lepton pairs. One of the\nbest ways to think of these are as "excited states" which always\neventually decay back into the stable ground states, although the\nprocess by which this occurs is much more complicated than what\nhappens with atomic states. Decay occurs in two steps by way of the\nweak interaction, rather than by emission of photons, but the general\nidea is still basically the same.\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 in message news:<861c1b21.0407151425.1234b9f6@posting.google.com>...
> There are two families of unstable quarks (top, bottom,strange and
> charm)
> and one family of stable quarks (up and down).Are there any reasons to
> think that there should be another family of stable quarks so that
> there are two unstable and two stable families?
There are also two generations of unstable lepton pairs. One of the
best ways to think of these are as "excited states" which always
eventually decay back into the stable ground states, although the
process by which this occurs is much more complicated than what
happens with atomic states. Decay occurs in two steps by way of the
weak interaction, rather than by emission of photons, but the general
idea is still basically the same.
alistair
Jul20-04, 04:25 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>\nthoovler@excite.com (Igor) wrote in message news:<d434b6c6.0407182318.530dc24e@posting.google.com>...\n> alistair@goforit64.fsnet.co.uk (alistair) wrote in message news:<861c1b21.0407151425.1234b9f6@posting.google.com>...\n> > There are two families of unstable quarks (top, bottom,strange and\n> > charm)\n> > and one family of stable quarks (up and down).Are there any reasons to\n> > think that there should be another family of stable quarks so that\n> > there are two unstable and two stable families?\n>\n> There are also two generations of unstable lepton pairs. One of the\n> best ways to think of these are as "excited states" which always\n> eventually decay back into the stable ground states, although the\n> process by which this occurs is much more complicated than what\n> happens with atomic states. Decay occurs in two steps by way of the\n> weak interaction, rather than by emission of photons, but the general\n> idea is still basically the same.\n\nAlistair writes:\n\nDoesn\'t the fact that there are two unstable quark generations and two\nunstable lepton generations suggest that quarks and leptons are the\nsame in some way? Quarks and leptons both have charge, rest mass and\nspin but differ in that leptons are not supposed to experience the\ncolour force.But how do we know that leptons don\'t experience the\ncolour force and how do we know that leptons are not made from quarks\nof some kind.Wouldn\'t it make more sense if they were?\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>thoovler@excite.com (Igor) wrote in message news:<d434b6c6.0407182318.530dc24e@posting.google.com>...
> alistair@goforit64.fsnet.co.uk (alistair) wrote in message news:<861c1b21.0407151425.1234b9f6@posting.google.com>...
> > There are two families of unstable quarks (top, bottom,strange and
> > charm)
> > and one family of stable quarks (up and down).Are there any reasons to
> > think that there should be another family of stable quarks so that
> > there are two unstable and two stable families?
>
> There are also two generations of unstable lepton pairs. One of the
> best ways to think of these are as "excited states" which always
> eventually decay back into the stable ground states, although the
> process by which this occurs is much more complicated than what
> happens with atomic states. Decay occurs in two steps by way of the
> weak interaction, rather than by emission of photons, but the general
> idea is still basically the same.
Alistair writes:
Doesn't the fact that there are two unstable quark generations and two
unstable lepton generations suggest that quarks and leptons are the
same in some way? Quarks and leptons both have charge, rest mass and
spin but differ in that leptons are not supposed to experience the
colour force.But how do we know that leptons don't experience the
colour force and how do we know that leptons are not made from quarks
of some kind.Wouldn't it make more sense if they were?
mathman
Jul20-04, 04:51 AM
Both theory and observation have the six quarks mentioned as all there is. There is some evidence that strange quarks may be stable in certain (astronomical, not on earth) situations.
Ulmo
Jul21-04, 05:03 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>\n\n> Alistair writes:\n>\n> Doesn\'t the fact that there are two unstable quark generations and two\n> unstable lepton generations suggest that quarks and leptons are the\n> same in some way? Quarks and leptons both have charge, rest mass and\n> spin but differ in that leptons are not supposed to experience the\n> colour force.But how do we know that leptons don\'t experience the\n> colour force and how do we know that leptons are not made from quarks\n> of some kind.Wouldn\'t it make more sense if they were?\n\n\nThere is overwhelming evidence from Rutherford type experiments that\nhadrons such as potons have internal structure and that leptons such\nas electrons do not.\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 writes:
>
> Doesn't the fact that there are two unstable quark generations and two
> unstable lepton generations suggest that quarks and leptons are the
> same in some way? Quarks and leptons both have charge, rest mass and
> spin but differ in that leptons are not supposed to experience the
> colour force.But how do we know that leptons don't experience the
> colour force and how do we know that leptons are not made from quarks
> of some kind.Wouldn't it make more sense if they were?
There is overwhelming evidence from Rutherford type experiments that
hadrons such as potons have internal structure and that leptons such
as electrons do not.
alistair
Jul21-04, 05:03 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>\nmathman <mathnucl@optonline.net> wrote in message news:<mathman.19opdk@physicsforums.com>...\n> Both theory and observation have the six quarks mentioned as all there\n> is. There is some evidence that strange quarks may be stable in\n> certain (astronomical, not on earth) situations.\n\n\nThe standard model can\'t predict the number of quark families.\nThe reason usually given for believing there are only three\nis that the decay rate of the z boson would be greater if there were\nmore families.Strange quarks are believed to be stable in strange\nstars ( nobody knows if they really exist yet)under high pressure\nconditions.But would all quarks be stable under a high enough\npressure? Why would high pressure cause quarks to become stable?\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>mathman <mathnucl@optonline.net> wrote in message news:<mathman.19opdk@physicsforums.com>...
> Both theory and observation have the six quarks mentioned as all there
> is. There is some evidence that strange quarks may be stable in
> certain (astronomical, not on earth) situations.
The standard model can't predict the number of quark families.
The reason usually given for believing there are only three
is that the decay rate of the z boson would be greater if there were
more families.Strange quarks are believed to be stable in strange
stars ( nobody knows if they really exist yet)under high pressure
conditions.But would all quarks be stable under a high enough
pressure? Why would high pressure cause quarks to become stable?
Thomas Dent
Jul21-04, 05:47 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>\n\n\nalistair@goforit64.fsnet.co.uk (alistair) wrote\n\n> > > There are two families of unstable quarks (top, bottom,strange and\n> > > charm)\n> > > and one family of stable quarks (up and down).Are there any reasons to\n> > > think that there should be another family of stable quarks so that\n> > > there are two unstable and two stable families?\n\nNo. The number of families is an experimentally determined number, and\nexperiments have, so far as I know, not indicated the need for any\nother quarks. The ratio of the hadronic cross-section in high-energy\ne+e- scattering to the leptonic cross-section is a measure of the\nnumber of quarks at a given energy, and the number comes out as 5\nabove the charm threshold. Plus the top quark, which is too massive to\nbe made at e+e- colliders, makes six.\n\n\n> > There are also two generations of unstable lepton pairs. One of the\n> > best ways to think of these are as "excited states" which always\n> > eventually decay back into the stable ground states, although the\n> > process by which this occurs is much more complicated than what\n> > happens with atomic states.\n\nActually, as far as we know, there are only heavy unstable leptons (mu\nand tau), not lepton pairs. The leptons are in weak doublets with\nneutrinos which have very small masses and (so far as we know) do not\ndecay. Also, the mu and tau do not seem to be excited states of the\nelectron, since the electron (unlike the atom) does not have\ndetectable internal structure which could become excited. They really\nseem to be different particles. Although, in various speculative\ntheories, there might be unification of the different families, there\nis no direct experimental evidence for this.\n\n\n> Alistair writes:\n>\n> Doesn\'t the fact that there are two unstable quark generations and two\n> unstable lepton generations suggest that quarks and leptons are the\n> same in some way?\n\nThis idea was put forward in the 1970\'s by Pati and Salam who wrote\nthe famous paper "Lepton number as the fourth colour". However, the\nextended colour symmetry has to be broken at a very high energy so\nthat leptons behave differently at low energies. The gluons which\ntransform quarks to leptons and vice versa are very massive and their\neffects are very much suppressed. Otherwise the proton would decay\nquickly.\n\n> Quarks and leptons both have charge, rest mass and\n> spin but differ in that leptons are not supposed to experience the\n> colour force.\n\nNot just that they are "not supposed to", but it is an experimental\nfact that they don\'t.\n\n> But how do we know that leptons don\'t experience the\n> colour force\n\nExperimental data on lepton scattering. For example e+e- scattering is\ndescribed by the electromagnetic and weak forces perfectly well. Also,\nif you imagine an atom where the electron experiences a colour force,\nits behaviour would be quite different. And the proton might be\nunstable.\n\nIt\'s an experimental fact that some particles are weakly-interacting\nand some are strongly-interacting, and by definition the leptons are\nthe former.\n\n> and how do we know that leptons are not made from quarks\n> of some kind. Wouldn\'t it make more sense if they were?\n\nThat\'s a matter of taste. So-called "preonic" models attempt to\ndescribe the known particles as composites made up of "preons" which\nare similar to quarks by being confined. But it\'s very difficult to\nimagine a model which can fit the experimental data, which are\nconsistent with the leptons not having any internal structure.\n\nDear Alistair, why don\'t you go and take an Open University course on\nparticle physics instead of trying to do it on usenet?\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
> > > There are two families of unstable quarks (top, bottom,strange and
> > > charm)
> > > and one family of stable quarks (up and down).Are there any reasons to
> > > think that there should be another family of stable quarks so that
> > > there are two unstable and two stable families?
No. The number of families is an experimentally determined number, and
experiments have, so far as I know, not indicated the need for any
other quarks. The ratio of the hadronic cross-section in high-energy
e+e- scattering to the leptonic cross-section is a measure of the
number of quarks at a given energy, and the number comes out as 5
above the charm threshold. Plus the top quark, which is too massive to
be made at e+e- colliders, makes six.
> > There are also two generations of unstable lepton pairs. One of the
> > best ways to think of these are as "excited states" which always
> > eventually decay back into the stable ground states, although the
> > process by which this occurs is much more complicated than what
> > happens with atomic states.
Actually, as far as we know, there are only heavy unstable leptons (\mu
and \tau), not lepton pairs. The leptons are in weak doublets with
neutrinos which have very small masses and (so far as we know) do not
decay. Also, the \mu and \tau do not seem to be excited states of the
electron, since the electron (unlike the atom) does not have
detectable internal structure which could become excited. They really
seem to be different particles. Although, in various speculative
theories, there might be unification of the different families, there
is no direct experimental evidence for this.
> Alistair writes:
>
> Doesn't the fact that there are two unstable quark generations and two
> unstable lepton generations suggest that quarks and leptons are the
> same in some way?
This idea was put forward in the 1970's by Pati and Salam who wrote
the famous paper "Lepton number as the fourth colour". However, the
extended colour symmetry has to be broken at a very high energy so
that leptons behave differently at low energies. The gluons which
transform quarks to leptons and vice versa are very massive and their
effects are very much suppressed. Otherwise the proton would decay
quickly.
> Quarks and leptons both have charge, rest mass and
> spin but differ in that leptons are not supposed to experience the
> colour force.
Not just that they are "not supposed to", but it is an experimental
fact that they don't.
> But how do we know that leptons don't experience the
> colour force
Experimental data on lepton scattering. For example e+e- scattering is
described by the electromagnetic and weak forces perfectly well. Also,
if you imagine an atom where the electron experiences a colour force,
its behaviour would be quite different. And the proton might be
unstable.
It's an experimental fact that some particles are weakly-interacting
and some are strongly-interacting, and by definition the leptons are
the former.
> and how do we know that leptons are not made from quarks
> of some kind. Wouldn't it make more sense if they were?
That's a matter of taste. So-called "preonic" models attempt to
describe the known particles as composites made up of "preons" which
are similar to quarks by being confined. But it's very difficult to
imagine a model which can fit the experimental data, which are
consistent with the leptons not having any internal structure.
Dear Alistair, why don't you go and take an Open University course on
particle physics instead of trying to do it on usenet?
Igor
Jul22-04, 05:16 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>\ntdent@auth.gr (Thomas Dent) wrote in message news:<cb504c2c.0407210124.596a0b6e@posting.google.com>...\n> alistair@goforit64.fsnet.co.uk (alistair) wrote\n>\n> > > > There are two families of unstable quarks (top, bottom,strange and\n> > > > charm)\n> > > > and one family of stable quarks (up and down).Are there any reasons to\n> > > > think that there should be another family of stable quarks so that\n> > > > there are two unstable and two stable families?\n>\n> No. The number of families is an experimentally determined number, and\n> experiments have, so far as I know, not indicated the need for any\n> other quarks. The ratio of the hadronic cross-section in high-energy\n> e+e- scattering to the leptonic cross-section is a measure of the\n> number of quarks at a given energy, and the number comes out as 5\n> above the charm threshold. Plus the top quark, which is too massive to\n> be made at e+e- colliders, makes six.\n>\n>\n> > > There are also two generations of unstable lepton pairs. One of the\n> > > best ways to think of these are as "excited states" which always\n> > > eventually decay back into the stable ground states, although the\n> > > process by which this occurs is much more complicated than what\n> > > happens with atomic states.\n>\n> Actually, as far as we know, there are only heavy unstable leptons (mu\n> and tau), not lepton pairs. The leptons are in weak doublets with\n> neutrinos which have very small masses and (so far as we know) do not\n> decay.\n\n\nYou\'re essentially correct, but Feynman diagrams can be interpreted in\na few different ways, and there is plenty of evidence that neutrinos\noscillate between different flavors, just as quarks do, albeit in a\nmuch simpler fashion since they only appear to interact weakly. But\nyou do make a very good point.\n\n\n\n>Also, the mu and tau do not seem to be excited states of the\n> electron, since the electron (unlike the atom) does not have\n> detectable internal structure which could become excited. They really\n> seem to be different particles. Although, in various speculative\n> theories, there might be unification of the different families, there\n> is no direct experimental evidence for this.\n\n\nI never really meant to imply that heavier leptons were actually\nexcited states of the more stable ones. I was just using that as an\nanalogy, which is why I put it in quotes. Since I was trying to keep\nthings as simple as possible, I omitted a few of the details, but did\nexplain that it was somewhat analogous to photon emission in an atom,\nbut a two step process involving not only decay, but pair production.\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>tdent@auth.gr (Thomas Dent) wrote in message news:<cb504c2c.0407210124.596a0b6e@posting.google.com>...
> alistair@goforit64.fsnet.co.uk (alistair) wrote
>
> > > > There are two families of unstable quarks (top, bottom,strange and
> > > > charm)
> > > > and one family of stable quarks (up and down).Are there any reasons to
> > > > think that there should be another family of stable quarks so that
> > > > there are two unstable and two stable families?
>
> No. The number of families is an experimentally determined number, and
> experiments have, so far as I know, not indicated the need for any
> other quarks. The ratio of the hadronic cross-section in high-energy
> e+e- scattering to the leptonic cross-section is a measure of the
> number of quarks at a given energy, and the number comes out as 5
> above the charm threshold. Plus the top quark, which is too massive to
> be made at e+e- colliders, makes six.
>
>
> > > There are also two generations of unstable lepton pairs. One of the
> > > best ways to think of these are as "excited states" which always
> > > eventually decay back into the stable ground states, although the
> > > process by which this occurs is much more complicated than what
> > > happens with atomic states.
>
> Actually, as far as we know, there are only heavy unstable leptons (\mu
> and \tau), not lepton pairs. The leptons are in weak doublets with
> neutrinos which have very small masses and (so far as we know) do not
> decay.
You're essentially correct, but Feynman diagrams can be interpreted in
a few different ways, and there is plenty of evidence that neutrinos
oscillate between different flavors, just as quarks do, albeit in a
much simpler fashion since they only appear to interact weakly. But
you do make a very good point.
>Also, the \mu and \tau do not seem to be excited states of the
> electron, since the electron (unlike the atom) does not have
> detectable internal structure which could become excited. They really
> seem to be different particles. Although, in various speculative
> theories, there might be unification of the different families, there
> is no direct experimental evidence for this.
I never really meant to imply that heavier leptons were actually
excited states of the more stable ones. I was just using that as an
analogy, which is why I put it in quotes. Since I was trying to keep
things as simple as possible, I omitted a few of the details, but did
explain that it was somewhat analogous to photon emission in an atom,
but a two step process involving not only decay, but pair production.
Alfred Einstead
Jul22-04, 05:16 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>\ntdent@auth.gr (Thomas Dent) wrote in message\n> > and how do we know that leptons are not made from quarks\n> > of some kind. Wouldn\'t it make more sense if they were?\n>\n> That\'s a matter of taste. So-called "preonic" models attempt to\n> describe the known particles as composites made up of "preons" which\n> are similar to quarks by being confined. But it\'s very difficult to\n> imagine a model which can fit the experimental data, which are\n> consistent with the leptons not having any internal structure.\n\n"preon" means internal structure. It\'s not difficult to\nconstruct a model for the particle spectrum that fits what\'s\nactually seen.\n\nAssign the following:\nPre A: Hypercharge: g\'/4\nWeak Isospin: -g/4\nColor neutral, 0 (baryon-lepton) number\nPre B: Hypercharge: g\'/4\nWeak Isospin: g/4\nColor neutral, 0 (baryon-lepton) number\nPre C: Hypercharge: -g\'/6\nColor (SU(3)) casimir charges: (gs/4, gs/(4 sqrt(3)))\n(baryon-lepton): -1/3\nPre D: Hypercharge: -g\'/6\nColor SU(3) casimir numbers: (0, -gs/(2 sqrt(3)))\n(baryon-lepton): -1/3\nPre E: Hypercharge: -g\'/6\nColor SU(3) casimir numbers: (-gs/4, gs/(4 sqrt(3)))\n(baryon-lepton): -1/3\n\nwhere g\', g and gs are respectively the U(1), SU(2) and SU(3)\ncoupling constants of the Standard Model.\n\nThe 32 combinations containing one each of (Pre A or Anti-Pre A),\netc. gives you the 32 particle & anti-particle states of each\ngeneration. Right neutrinos and left anti-neutrinos are the\nstates corresponding to (A,B,C,D,E) and\n(anti-A,anti-B,anti-C,anti-D,anti-E). Excluding these combinations\ngives you the 30 states of the Standard Model.\n\n(C,D,E) and (anti-C,anti-D,anti-E) have SU(3) charges that are\n1/2 those of the 3 and 3* representations, and so are identified\nessentially as fundamental charges for the colors (red, green, blue)\nand the anti-colors (cyan, magneta, amber).\n\n(A,B) and (anti-A, anti-B) have 1/2 the charges of the 2 and 2*\nrepresentations of SU(2).\n\nThe fermion spinor would comprise the 32 (or 30) complex components\nand would be parametrized by the 5 quantum numbers (a,b,c,d,e)\n= +/- each. Including a 6th (redundant) index, s, for spin; the\ncomponents of the Dirac spinor psi_{abcdes} would satisfy conjugacy\nrelations of the form:\nC[psi]_{abcdes} = psi_{-a -b -c -d -e -s}\npsi_{abcdes}* = sign(abcdes) C[psi]_{abcdes}\n\nThe representation of the gamma matrices is somewhat involved.\nParity flips the b index, which leads to a relatively simple\nrepresentation for gamma^0. gamma^5, if I recall correctly,\ninvolves the a, b and s indices. The pauli matrices\nsigma^1, sigma^2 and sigma^3 involve just the s index. Once\nyou have these, you have the Dirac matrices, as well.\n\nSU(2) flavor-changing interactions swap (A,Anti-B) <-> (Anti-A,B).\nThe 4 Higgs modes flip A<->Anti-A, or B<->Anti-B. The SU(3)\ncolor-changing interactions effect the swaps\n(C,Anti-D) <-> (Anti-C,D); (D,Anti-E) <-> (Anti-D,E) and\n(E,Anti-C) <-> (Anti-E,C).\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>tdent@auth.gr (Thomas Dent) wrote in message
> > and how do we know that leptons are not made from quarks
> > of some kind. Wouldn't it make more sense if they were?
>
> That's a matter of taste. So-called "preonic" models attempt to
> describe the known particles as composites made up of "preons" which
> are similar to quarks by being confined. But it's very difficult to
> imagine a model which can fit the experimental data, which are
> consistent with the leptons not having any internal structure.
"preon" means internal structure. It's not difficult to
construct a model for the particle spectrum that fits what's
actually seen.
Assign the following:
Pre A: Hypercharge: g'/4
Weak Isospin: -g/4
Color neutral, (baryon-lepton) number
Pre B: Hypercharge: g'/4
Weak Isospin: g/4
Color neutral, (baryon-lepton) number
Pre C: Hypercharge: -g'/6
Color (SU(3)) casimir charges: (gs/4, gs/(4 \sqrt(3)))
(baryon-lepton): -1/3
Pre D: Hypercharge: -g'/6
Color SU(3) casimir numbers: (0, -gs/(2 \sqrt(3)))
(baryon-lepton): -1/3
Pre E: Hypercharge: -g'/6
Color SU(3) casimir numbers: (-gs/4, gs/(4 \sqrt(3)))
(baryon-lepton): -1/3
where g', g and gs are respectively the U(1), SU(2) and SU(3)
coupling constants of the Standard Model.
The 32 combinations containing one each of (Pre A or Anti-Pre A),
etc. gives you the 32 particle & anti-particle states of each
generation. Right neutrinos and left anti-neutrinos are the
states corresponding to (A,B,C,D,E) and
(anti-A,anti-B,anti-C,anti-D,anti-E). Excluding these combinations
gives you the 30 states of the Standard Model.
(C,D,E) and (anti-C,anti-D,anti-E) have SU(3) charges that are
1/2 those of the 3 and 3* representations, and so are identified
essentially as fundamental charges for the colors (red, green, blue)
and the anti-colors (cyan, magneta, amber).
(A,B) and (anti-A, anti-B) have 1/2 the charges of the 2 and 2*
representations of SU(2).
The fermion spinor would comprise the 32 (or 30) complex components
and would be parametrized by the 5 quantum numbers (a,b,c,d,e)
= +/- each. Including a 6th (redundant) index, s, for spin; the
components of the Dirac spinor \psi_{abcdes} would satisfy conjugacy
relations of the form:
C[\psi]_{abcdes} = \psi_{-a -b -c -d -e -s}\psi_{abcdes}* = sign(abcdes) C[\psi]_{abcdes}
The representation of the \gamma matrices is somewhat involved.
Parity flips the b index, which leads to a relatively simple
representation for \gamma^0. \gamma^5, if I recall correctly,
involves the a, b and s indices. The pauli matrices
\sigma^1, \sigma^2 and \sigma^3 involve just the s index. Once
you have these, you have the Dirac matrices, as well.
SU(2) flavor-changing interactions swap (A,Anti-B) <-> (Anti-A,B).
The 4 Higgs modes flip A<->Anti-A, or B<->Anti-B. The SU(3)
color-changing interactions effect the swaps
(C,Anti-D) <-> (Anti-C,D); (D,Anti-E) <-> (Anti-D,E) and
(E,Anti-C) <-> (Anti-E,C).
Creighton Hogg
Jul22-04, 11:22 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>\n\nOn 22 Jul 2004, Alfred Einstead wrote:\n\n>\n> tdent@auth.gr (Thomas Dent) wrote in message\n> > > and how do we know that leptons are not made from quarks\n> > > of some kind. Wouldn\'t it make more sense if they were?\n> >\n> > That\'s a matter of taste. So-called "preonic" models attempt to\n> > describe the known particles as composites made up of "preons" which\n> > are similar to quarks by being confined. But it\'s very difficult to\n> > imagine a model which can fit the experimental data, which are\n> > consistent with the leptons not having any internal structure.\n>\n> "preon" means internal structure. It\'s not difficult to\n> construct a model for the particle spectrum that fits what\'s\n> actually seen.\n>\n> Assign the following:\n> Pre A: Hypercharge: g\'/4\n> Weak Isospin: -g/4\n> Color neutral, 0 (baryon-lepton) number\n> Pre B: Hypercharge: g\'/4\n> Weak Isospin: g/4\n> Color neutral, 0 (baryon-lepton) number\n> Pre C: Hypercharge: -g\'/6\n> Color (SU(3)) casimir charges: (gs/4, gs/(4 sqrt(3)))\n> (baryon-lepton): -1/3\n> Pre D: Hypercharge: -g\'/6\n> Color SU(3) casimir numbers: (0, -gs/(2 sqrt(3)))\n> (baryon-lepton): -1/3\n> Pre E: Hypercharge: -g\'/6\n> Color SU(3) casimir numbers: (-gs/4, gs/(4 sqrt(3)))\n> (baryon-lepton): -1/3\n>\n> where g\', g and gs are respectively the U(1), SU(2) and SU(3)\n> coupling constants of the Standard Model.\n>\n> The 32 combinations containing one each of (Pre A or Anti-Pre A),\n> etc. gives you the 32 particle & anti-particle states of each\n> generation. Right neutrinos and left anti-neutrinos are the\n> states corresponding to (A,B,C,D,E) and\n> (anti-A,anti-B,anti-C,anti-D,anti-E). Excluding these combinations\n> gives you the 30 states of the Standard Model.\n\nOkay, well I think I understand what you did here, but I\'m still not quite\nsure how well it would fit the real experimental data. Do the masses of\nthe particles work out correctly this way, and what about sizes? Why are\nthey bound so tightly that the electron appears structureless if the only\ncharges involved are the ones from the standard model? If the electron\nis has a substructure with color in it, then why didn\'t we see evidence of\npomeron exchange between electron and positron in LEP?\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>On 22 Jul 2004, Alfred Einstead wrote:
>
> tdent@auth.gr (Thomas Dent) wrote in message
> > > and how do we know that leptons are not made from quarks
> > > of some kind. Wouldn't it make more sense if they were?
> >
> > That's a matter of taste. So-called "preonic" models attempt to
> > describe the known particles as composites made up of "preons" which
> > are similar to quarks by being confined. But it's very difficult to
> > imagine a model which can fit the experimental data, which are
> > consistent with the leptons not having any internal structure.
>
> "preon" means internal structure. It's not difficult to
> construct a model for the particle spectrum that fits what's
> actually seen.
>
> Assign the following:
> Pre A: Hypercharge: g'/4
> Weak Isospin: -g/4
> Color neutral, (baryon-lepton) number
> Pre B: Hypercharge: g'/4
> Weak Isospin: g/4
> Color neutral, (baryon-lepton) number
> Pre C: Hypercharge: -g'/6
> Color (SU(3)) casimir charges: (gs/4, gs/(4 \sqrt(3)))
> (baryon-lepton): -1/3
> Pre D: Hypercharge: -g'/6
> Color SU(3) casimir numbers: (0, -gs/(2 \sqrt(3)))
> (baryon-lepton): -1/3
> Pre E: Hypercharge: -g'/6
> Color SU(3) casimir numbers: (-gs/4, gs/(4 \sqrt(3)))
> (baryon-lepton): -1/3
>
> where g', g and gs are respectively the U(1), SU(2) and SU(3)
> coupling constants of the Standard Model.
>
> The 32 combinations containing one each of (Pre A or Anti-Pre A),
> etc. gives you the 32 particle & anti-particle states of each
> generation. Right neutrinos and left anti-neutrinos are the
> states corresponding to (A,B,C,D,E) and
> (anti-A,anti-B,anti-C,anti-D,anti-E). Excluding these combinations
> gives you the 30 states of the Standard Model.
Okay, well I think I understand what you did here, but I'm still not quite
sure how well it would fit the real experimental data. Do the masses of
the particles work out correctly this way, and what about sizes? Why are
they bound so tightly that the electron appears structureless if the only
charges involved are the ones from the standard model? If the electron
is has a substructure with color in it, then why didn't we see evidence of
pomeron exchange between electron and positron in LEP?
Thomas Dent
Jul22-04, 01:07 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\nwhopkins@csd.uwm.edu (Alfred Einstead) wrote\n\n> tdent@auth.gr (Thomas Dent) wrote in message\n> > > and how do we know that leptons are not made from quarks\n> > > of some kind. Wouldn\'t it make more sense if they were?\n> >\n> > That\'s a matter of taste. So-called "preonic" models attempt to\n> > describe the known particles as composites made up of "preons" which\n> > are similar to quarks by being confined. But it\'s very difficult to\n> > imagine a model which can fit the experimental data, which are\n> > consistent with the leptons not having any internal structure.\n>\n> "preon" means internal structure. It\'s not difficult to\n> construct a model for the particle spectrum that fits what\'s\n> actually seen.\n>\n> Assign the following:\n> Pre A: Hypercharge: g\'/4\n> Weak Isospin: -g/4\n> Color neutral, 0 (baryon-lepton) number\n> Pre B: Hypercharge: g\'/4\n> Weak Isospin: g/4\n> Color neutral, 0 (baryon-lepton) number\n> Pre C: Hypercharge: -g\'/6\n> Color (SU(3)) casimir charges: (gs/4, gs/(4 sqrt(3)))\n> (baryon-lepton): -1/3\n> Pre D: Hypercharge: -g\'/6\n> Color SU(3) casimir numbers: (0, -gs/(2 sqrt(3)))\n> (baryon-lepton): -1/3\n> Pre E: Hypercharge: -g\'/6\n> Color SU(3) casimir numbers: (-gs/4, gs/(4 sqrt(3)))\n> (baryon-lepton): -1/3\n>\n> where g\', g and gs are respectively the U(1), SU(2) and SU(3)\n> coupling constants of the Standard Model.\n>\n> The 32 combinations containing one each of (Pre A or Anti-Pre A),\n> etc. gives you the 32 particle & anti-particle states of each\n> generation. Right neutrinos and left anti-neutrinos are the\n> states corresponding to (A,B,C,D,E) and\n> (anti-A,anti-B,anti-C,anti-D,anti-E). Excluding these combinations\n> gives you the 30 states of the Standard Model.\n>\n> (C,D,E) and (anti-C,anti-D,anti-E) have SU(3) charges that are\n> 1/2 those of the 3 and 3* representations, and so are identified\n> essentially as fundamental charges for the colors (red, green, blue)\n> and the anti-colors (cyan, magneta, amber).\n>\n> (A,B) and (anti-A, anti-B) have 1/2 the charges of the 2 and 2*\n> representations of SU(2).\n>\n> The fermion spinor would comprise the 32 (or 30) complex components\n> and would be parametrized by the 5 quantum numbers (a,b,c,d,e)\n> = +/- each. Including a 6th (redundant) index, s, for spin; the\n> components of the Dirac spinor psi_{abcdes} would satisfy conjugacy\n> relations of the form:\n> C[psi]_{abcdes} = psi_{-a -b -c -d -e -s}\n> psi_{abcdes}* = sign(abcdes) C[psi]_{abcdes}\n>\n> The representation of the gamma matrices is somewhat involved.\n> Parity flips the b index, which leads to a relatively simple\n> representation for gamma^0. gamma^5, if I recall correctly,\n> involves the a, b and s indices. The pauli matrices\n> sigma^1, sigma^2 and sigma^3 involve just the s index. Once\n> you have these, you have the Dirac matrices, as well.\n>\n> SU(2) flavor-changing interactions swap (A,Anti-B) <-> (Anti-A,B).\n> The 4 Higgs modes flip A<->Anti-A, or B<->Anti-B. The SU(3)\n> color-changing interactions effect the swaps\n> (C,Anti-D) <-> (Anti-C,D); (D,Anti-E) <-> (Anti-D,E) and\n> (E,Anti-C) <-> (Anti-E,C).\n\nFine, fine, but why don\'t we see individual preons? Why don\'t we\napparently see any signature of the preonic substructure in e+e-\nscattering? What is the strong force that confines them together two\nby two? What energy would we have to go to to see some signal of the\nsubstructure? Where do the different generations and CKM mixing come\nin?\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>whopkins@csd.uwm.edu (Alfred Einstead) wrote
> tdent@auth.gr (Thomas Dent) wrote in message
> > > and how do we know that leptons are not made from quarks
> > > of some kind. Wouldn't it make more sense if they were?
> >
> > That's a matter of taste. So-called "preonic" models attempt to
> > describe the known particles as composites made up of "preons" which
> > are similar to quarks by being confined. But it's very difficult to
> > imagine a model which can fit the experimental data, which are
> > consistent with the leptons not having any internal structure.
>
> "preon" means internal structure. It's not difficult to
> construct a model for the particle spectrum that fits what's
> actually seen.
>
> Assign the following:
> Pre A: Hypercharge: g'/4
> Weak Isospin: -g/4
> Color neutral, (baryon-lepton) number
> Pre B: Hypercharge: g'/4
> Weak Isospin: g/4
> Color neutral, (baryon-lepton) number
> Pre C: Hypercharge: -g'/6
> Color (SU(3)) casimir charges: (gs/4, gs/(4 \sqrt(3)))
> (baryon-lepton): -1/3
> Pre D: Hypercharge: -g'/6
> Color SU(3) casimir numbers: (0, -gs/(2 \sqrt(3)))
> (baryon-lepton): -1/3
> Pre E: Hypercharge: -g'/6
> Color SU(3) casimir numbers: (-gs/4, gs/(4 \sqrt(3)))
> (baryon-lepton): -1/3
>
> where g', g and gs are respectively the U(1), SU(2) and SU(3)
> coupling constants of the Standard Model.
>
> The 32 combinations containing one each of (Pre A or Anti-Pre A),
> etc. gives you the 32 particle & anti-particle states of each
> generation. Right neutrinos and left anti-neutrinos are the
> states corresponding to (A,B,C,D,E) and
> (anti-A,anti-B,anti-C,anti-D,anti-E). Excluding these combinations
> gives you the 30 states of the Standard Model.
>
> (C,D,E) and (anti-C,anti-D,anti-E) have SU(3) charges that are
> 1/2 those of the 3 and 3* representations, and so are identified
> essentially as fundamental charges for the colors (red, green, blue)
> and the anti-colors (cyan, magneta, amber).
>
> (A,B) and (anti-A, anti-B) have 1/2 the charges of the 2 and 2*
> representations of SU(2).
>
> The fermion spinor would comprise the 32 (or 30) complex components
> and would be parametrized by the 5 quantum numbers (a,b,c,d,e)
> = +/- each. Including a 6th (redundant) index, s, for spin; the
> components of the Dirac spinor \psi_{abcdes} would satisfy conjugacy
> relations of the form:
> C[\psi]_{abcdes} = \psi_{-a -b -c -d -e -s}
> \psi_{abcdes}* = sign(abcdes) C[\psi]_{abcdes}
>
> The representation of the \gamma matrices is somewhat involved.
> Parity flips the b index, which leads to a relatively simple
> representation for \gamma^0. \gamma^5, if I recall correctly,
> involves the a, b and s indices. The pauli matrices
> \sigma^1, \sigma^2 and \sigma^3 involve just the s index. Once
> you have these, you have the Dirac matrices, as well.
>
> SU(2) flavor-changing interactions swap (A,Anti-B) <-> (Anti-A,B).
> The 4 Higgs modes flip A<->Anti-A, or B<->Anti-B. The SU(3)
> color-changing interactions effect the swaps
> (C,Anti-D) <-> (Anti-C,D); (D,Anti-E) <-> (Anti-D,E) and
> (E,Anti-C) <-> (Anti-E,C).
Fine, fine, but why don't we see individual preons? Why don't we
apparently see any signature of the preonic substructure in e+e-
scattering? What is the strong force that confines them together two
by two? What energy would we have to go to to see some signal of the
substructure? Where do the different generations and CKM mixing come
in?
alistair
Jul22-04, 02:02 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nThomas Dent said:\n>if you imagine an atom where the electron experiences a colour force,\n>its behaviour would be quite different\n\n\nAn electron in a hydrogen atom would not experience the colour force\nat\n10^ - 10 metres which is its average radius.\n\nThomas Dent:\n\n>Pati and Salam who wrote\n>the famous paper "Lepton number as the fourth colour". However, the\n>extended colour symmetry has to be broken at a very high energy so\n>that leptons behave differently at low energies. The gluons which\n>transform quarks to leptons and vice versa are very massive and their\n>effects are very much suppressed. Otherwise the proton would decay\n>quickly.\n\n\nPerhaps the gluons could be very massive in the early universe\nbut are not so massive now.\n\nCreighton Hogg wrote:\n\n> If the electron has a substructure with color in it, then why didn\'t\n> we see evidence of pomeron exchange between electron and positron in\n> LEP?\n\n\nPerhaps pomerons of the right energy (some theorists say pomerons are\nmade from glueballs which are made from gluons)can only exist in the\nconditions of the early universe.\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>Thomas Dent said:
>if you imagine an atom where the electron experiences a colour force,
>its behaviour would be quite different
An electron in a hydrogen atom would not experience the colour force
at
10^ - 10 metres which is its average radius.
Thomas Dent:
>Pati and Salam who wrote
>the famous paper "Lepton number as the fourth colour". However, the
>extended colour symmetry has to be broken at a very high energy so
>that leptons behave differently at low energies. The gluons which
>transform quarks to leptons and vice versa are very massive and their
>effects are very much suppressed. Otherwise the proton would decay
>quickly.
Perhaps the gluons could be very massive in the early universe
but are not so massive now.
Creighton Hogg wrote:
> If the electron has a substructure with color in it, then why didn't
> we see evidence of pomeron exchange between electron and positron in
> LEP?
Perhaps pomerons of the right energy (some theorists say pomerons are
made from glueballs which are made from gluons)can only exist in the
conditions of the early universe.
FrediFizzx
Jul23-04, 06:34 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>\n\n"Thomas Dent" <tdent@auth.gr> wrote in message\nnews:cb504c2c.0407220756.e4092fa@posting.google.com...\n\n| whopkins@csd.uwm.edu (Alfred Einstead) wrote\n\n\n| > SU(2) flavor-changing interactions swap (A,Anti-B) <-> (Anti-A,B).\n| > The 4 Higgs modes flip A<->Anti-A, or B<->Anti-B. The SU(3)\n| > color-changing interactions effect the swaps\n| > (C,Anti-D) <-> (Anti-C,D); (D,Anti-E) <-> (Anti-D,E) and\n| > (E,Anti-C) <-> (Anti-E,C).\n|\n| Fine, fine, but why don\'t we see individual preons? Why don\'t we\n| apparently see any signature of the preonic substructure in e+e-\n| scattering? What is the strong force that confines them together two\n| by two? What energy would we have to go to to see some signal of the\n| substructure? Where do the different generations and CKM mixing come\n| in?\n\nIMHO, it is not "substructure". It is external structure due to the quantum\nvacuum. You have to think somewhat inside-out. I don\'t think many particle\nphysicists now-a-days would argue that the quantum vacuum (QV) is not a\ndielectric medium that screens charge. It is more than just a dielectric\nmedium. We are pretty sure it has all the conjugate fermion pairs in it.\nThis effectively makes all fermions a mix of all other fermions to varying\ndegrees when you look at the complete picture because they all couple to\ncharge. All "bare" fermions are identical if it were possible to isolate\nthem from the quantum vacuum. But it is not possible to isolate them from\nthe quantum vacuum so you *always* have to take the QV into consideration.\nNeutrinos are somewhat the "odd man out". Do they form conjugate pairs?\n\nFrediFizzx\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>"Thomas Dent" <tdent@auth.gr> wrote in message
news:cb504c2c.0407220756.e4092fa@posting.google.com...
| whopkins@csd.uwm.edu (Alfred Einstead) wrote
| > SU(2) flavor-changing interactions swap (A,Anti-B) <-> (Anti-A,B).
| > The 4 Higgs modes flip A<->Anti-A, or B<->Anti-B. The SU(3)
| > color-changing interactions effect the swaps
| > (C,Anti-D) <-> (Anti-C,D); (D,Anti-E) <-> (Anti-D,E) and
| > (E,Anti-C) <-> (Anti-E,C).
|
| Fine, fine, but why don't we see individual preons? Why don't we
| apparently see any signature of the preonic substructure in e+e-
| scattering? What is the strong force that confines them together two
| by two? What energy would we have to go to to see some signal of the
| substructure? Where do the different generations and CKM mixing come
| in?
IMHO, it is not "substructure". It is external structure due to the quantum
vacuum. You have to think somewhat inside-out. I don't think many particle
physicists now-a-days would argue that the quantum vacuum (QV) is not a
dielectric medium that screens charge. It is more than just a dielectric
medium. We are pretty sure it has all the conjugate fermion pairs in it.
This effectively makes all fermions a mix of all other fermions to varying
degrees when you look at the complete picture because they all couple to
charge. All "bare" fermions are identical if it were possible to isolate
them from the quantum vacuum. But it is not possible to isolate them from
the quantum vacuum so you *always* have to take the QV into consideration.
Neutrinos are somewhat the "odd man out". Do they form conjugate pairs?