lost.and.lonely.physicist@gmail.com
Jul7-05, 10:07 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>In deriving the beta function of, say, QED using dimensional\nregularization we get the relation (up to 1 loop)\n\nbeta[e] = - (epsilon/2) e - e dlog[Z_e]/dln[mu] (1)\n\nand\n\nZ_e = 1 + e^2 A / epsilon\n\nwhere e is the coupling, Z_e is the renormalization of the coupling, A\nis some constant that does not depend on mu or epsilon and mu is the\narbitrary scale introduced to make the dimension of the coupling the\nsame as if epsilon was zero.\n\nNow if I do the following math\n\ndlog[Z_e]/dln[mu]\n= (1/(1 + e^2 A / epsilon))(d/dln[mu])( 1 + e^2 A / epsilon )\n= (2 e A/(epsilon + e^2 A)) beta[e]\n\nIs this correct? It seems extremely elementary, but if I trust my\nresults\n\nbeta[e](1 + 2 e^2 A/(epsilon + e^2 A) ) = - (epsilon/2)e\n\nTaking the limit epsilon going to zero I get that the beta function for\nQED is zero up to one loop!\n\nWhat seems to be usually done is we taylor expand the denominator of\n1/(1 + e^2 A / epsilon) \\approx 1 - e^2 A / epsilon and so\n\ndlog[Z_e]/dln[mu]\n= 2 A e beta[e] / epsilon + Order[e^4] (2)\n\nand next we iterate it, putting (1) for beta into (2)\n\nbeta[e]\n= - (epsilon/2) e - 2 A e^2 beta[e] / epsilon + Order[e^4]\n= - (epsilon/2) e - 2 A e^2 (- (epsilon/2) e - 2 A e^2 beta[e] /\nepsilon + Order[e^4]) / epsilon + Order[e^4]\n= - (epsilon/2) e + A e^3 + (2 A e^2)^2 beta[e] / epsilon^2 + . . .\n\nand we say it\'s approximately equal to\n\n= A e^3\n\n(For instance A = 1/12pi^3 for pure QED.)\n\nHow can we ignore the epsilon? How can we taylor expand in powers of e\nwhen epsilon is supposed to be tiny in dimensional regularization? Even\nif we do taylor expand and iterate what about the 1/epsilon^2 and\npossibly even more infinite quantities?\n\nThis issue really bothers me a lot because I\'m sure there\'s something\nI\'m not understanding here, since the QED coupling does indeed run as\ncalculated, right? Any clarification would be deeply appreciated.\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>In deriving the \beta function of, say, QED using dimensional
regularization we get the relation (up to 1 loop)
\beta[e] = - (\epsilon/2) e - e dlog[Z_e]/dln[\mu][/itex] (1)
and
Z_e = 1 + e^2 A / \epsilon
where e is the coupling, Z_e is the renormalization of the coupling, A
is some constant that does not depend on \mu or \epsilon and \mu is the
arbitrary scale introduced to make the dimension of the coupling the
same as if \epsilon was zero.
Now if I do the following math
dlog[Z_e]/dln[\mu]= (1/(1 + e^2 A / \epsilon))(d/dln[\mu])( 1 + e^2 A / \epsilon )= (2 e A/(\epsilon + e^2 A)) \beta[e]
Is this correct? It seems extremely elementary, but if I trust my
results
\beta[e](1 + 2 e^2 A/(\epsilon + e^2 A) ) = - (\epsilon/2)e
Taking the limit \epsilon going to zero I get that the \beta function for
QED is zero up to one loop!
What seems to be usually done is we taylor expand the denominator of
1/(1 + e^2 A / \epsilon) \approx 1 - e^2 A / \epsilon and so
dlog[Z_e]/dln[\mu]= 2 A e \beta[e] / \epsilon + Order[e^4] (2)
and next we iterate it, putting (1) for \beta into (2)
[itex]\beta[e]= - (\epsilon/2) e - 2 A e^2 \beta[e] / \epsilon + Order[e^4]= - (\epsilon/2) e - 2 A e^2 (- (\epsilon/2) e - 2 A e^2 \beta[e] /\epsilon + Order[e^4]) / \epsilon + Order[e^4]= - (\epsilon/2) e + A e^3 + (2 A e^2)^2 \beta[e] / \epsilon^2 + . . .
and we say it's approximately equal to
= A e^3
(For instance A = 1/12pi^3 for pure QED.)
How can we ignore the \epsilon? How can we taylor expand in powers of e
when \epsilon is supposed to be tiny in dimensional regularization? Even
if we do taylor expand and iterate what about the 1/\epsilon^2 and
possibly even more infinite quantities?
This issue really bothers me a lot because I'm sure there's something
I'm not understanding here, since the QED coupling does indeed run as
calculated, right? Any clarification would be deeply appreciated.
regularization we get the relation (up to 1 loop)
\beta[e] = - (\epsilon/2) e - e dlog[Z_e]/dln[\mu][/itex] (1)
and
Z_e = 1 + e^2 A / \epsilon
where e is the coupling, Z_e is the renormalization of the coupling, A
is some constant that does not depend on \mu or \epsilon and \mu is the
arbitrary scale introduced to make the dimension of the coupling the
same as if \epsilon was zero.
Now if I do the following math
dlog[Z_e]/dln[\mu]= (1/(1 + e^2 A / \epsilon))(d/dln[\mu])( 1 + e^2 A / \epsilon )= (2 e A/(\epsilon + e^2 A)) \beta[e]
Is this correct? It seems extremely elementary, but if I trust my
results
\beta[e](1 + 2 e^2 A/(\epsilon + e^2 A) ) = - (\epsilon/2)e
Taking the limit \epsilon going to zero I get that the \beta function for
QED is zero up to one loop!
What seems to be usually done is we taylor expand the denominator of
1/(1 + e^2 A / \epsilon) \approx 1 - e^2 A / \epsilon and so
dlog[Z_e]/dln[\mu]= 2 A e \beta[e] / \epsilon + Order[e^4] (2)
and next we iterate it, putting (1) for \beta into (2)
[itex]\beta[e]= - (\epsilon/2) e - 2 A e^2 \beta[e] / \epsilon + Order[e^4]= - (\epsilon/2) e - 2 A e^2 (- (\epsilon/2) e - 2 A e^2 \beta[e] /\epsilon + Order[e^4]) / \epsilon + Order[e^4]= - (\epsilon/2) e + A e^3 + (2 A e^2)^2 \beta[e] / \epsilon^2 + . . .
and we say it's approximately equal to
= A e^3
(For instance A = 1/12pi^3 for pure QED.)
How can we ignore the \epsilon? How can we taylor expand in powers of e
when \epsilon is supposed to be tiny in dimensional regularization? Even
if we do taylor expand and iterate what about the 1/\epsilon^2 and
possibly even more infinite quantities?
This issue really bothers me a lot because I'm sure there's something
I'm not understanding here, since the QED coupling does indeed run as
calculated, right? Any clarification would be deeply appreciated.