Re: Cluster Decomposition [Archive]

Oct17-04, 01:46 PM

Arnold Neumaier

Oct19-04, 03:29 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\nMM wrote:\n> Arnold Neumaier wrote:\n>\n> > Cluster decomposition means essentally that you cannot\n> > scatter a particle at a very distant particle. So what\n> > happens on the moon is irrelevant to experiments on\n> > the earth (except those depending on moonshine).\n> > Cluster decomposition is the basis of all physics.\n>\n> One thing perplexes me about the cluster decomposition\n> principle. It\'s self-evidently true for interactions\n> that become weaker with distance, but what about QCD\n> which gets stronger with distance?\n\nFor clusters consisting of single quarks, it\'s wrong there,\nexcept at high energies, where the perturbation expansion\nmakes sense because of asymptotic freedom. The \'correct\'\ncluster decomposition is always one for bound states,\nas can be seen from a more detailed nonrelativistic analysis.\n\nThis is not apparent from Weinberg\'s argument,\nsince perturbation theory breaks down in the presence of\nbound states.\n\n\nArnold Neumaier\n\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>MM wrote:
> Arnold Neumaier wrote:
>
> > Cluster decomposition means essentally that you cannot
> > scatter a particle at a very distant particle. So what
> > happens on the moon is irrelevant to experiments on
> > the earth (except those depending on moonshine).
> > Cluster decomposition is the basis of all physics.
>
> One thing perplexes me about the cluster decomposition
> principle. It's self-evidently true for interactions
> that become weaker with distance, but what about QCD
> which gets stronger with distance?

For clusters consisting of single quarks, it's wrong there,
except at high energies, where the perturbation expansion
makes sense because of asymptotic freedom. The 'correct'
cluster decomposition is always one for bound states,
as can be seen from a more detailed nonrelativistic analysis.

This is not apparent from Weinberg's argument,
since perturbation theory breaks down in the presence of
bound states.

Arnold Neumaier