[SOLVED] Renormalization

Mar2-05, 02:01 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>Hi all,\nI\'ve been learning about renormalization for the first time. It\nseems crazy --- we just take the divergences and put them into our mass\nand coupling constant! Why is this useful?\nOkay, I guess the "magic" of it is that, for a renormalizable\ntheory, /all/ of the divergences can be coped with in this way. But it\ndoesn\'t seem like we\'re /really/ coping with them, just sweeping them\nunder the carpet. Why doesn\'t it matter that our constants are now infinite?\nThanks in advance,\nJamie.\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>Hi all,
I've been learning about renormalization for the first time. It
seems crazy --- we just take the divergences and put them into our mass
and coupling constant! Why is this useful?
Okay, I guess the "magic" of it is that, for a renormalizable
theory,

of the divergences can be coped with in this way. But it
doesn't seem like we're /really/ coping with them, just sweeping them
under the carpet. Why doesn't it matter that our constants are now infinite?
Thanks in advance,
Jamie.

Mar3-05, 05:21 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>Perturbative renormalization a la particle physics seems a very mysterious\nthing. I think a clearer conceptual view of renormalization can be found by\nstarting with Wilson\'s formulation (Wilson and Kogut, Physics Reports,\n1974), including his description of the continuum limit, which is what\nperturbative renormalization is implicitly defining. The picture is\nsuperficially very different but much more comprehensible. Then you can move\non to understand the link between the two. This is not to say that the\nparticle physics method is not a more efficient algorithm.\n\nillywhacker;\n\n\n"Jamie Vicary" <jamievicary@gmail.com> a écrit dans le message de news:\nd02ld4\\$jfp\\$1@gemini.csx.cam.ac.uk...\n& gt; Hi all,\n> I\'ve been learning about renormalization for the first time. It\n> seems crazy --- we just take the divergences and put them into our mass\n> and coupling constant! Why is this useful?\n> Okay, I guess the "magic" of it is that, for a renormalizable\n> theory, /all/ of the divergences can be coped with in this way. But it\n> doesn\'t seem like we\'re /really/ coping with them, just sweeping them\n> under the carpet. Why doesn\'t it matter that our constants are now\n> infinite?\n> Thanks in advance,\n> Jamie.\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>Perturbative renormalization a la particle physics seems a very mysterious
thing. I think a clearer conceptual view of renormalization can be found by
starting with Wilson's formulation (Wilson and Kogut, Physics Reports,
1974), including his description of the continuum limit, which is what
perturbative renormalization is implicitly defining. The picture is
superficially very different but much more comprehensible. Then you can move
on to understand the link between the two. This is not to say that the
particle physics method is not a more efficient algorithm.

illywhacker;

"Jamie Vicary" <jamievicary@gmail.com> a écrit dans le message de news:
d02ld4$jfp$1@gemini.csx.cam.ac.uk...
> Hi all,
> I've been learning about renormalization for the first time. It
> seems crazy --- we just take the divergences and put them into our mass
> and coupling constant! Why is this useful?
> Okay, I guess the "magic" of it is that, for a renormalizable
> theory,

of the divergences can be coped with in this way. But it
> doesn't seem like we're /really/ coping with them, just sweeping them
> under the carpet. Why doesn't it matter that our constants are now
> infinite?
> Thanks in advance,
> Jamie.
>

Mar3-05, 05:22 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>Jamie Vicary wrote:\n> Hi all,\n> I\'ve been learning about renormalization for the first time. It\n> seems crazy --- we just take the divergences and put them into our mass\n> and coupling constant! Why is this useful?\n> Okay, I guess the "magic" of it is that, for a renormalizable\n> theory, /all/ of the divergences can be coped with in this way. But it\n> doesn\'t seem like we\'re /really/ coping with them, just sweeping them\n> under the carpet. Why doesn\'t it matter that our constants are now infinite?\n\nThe constants are infinite only if one takes a careless (and\nmeaningless) limit. it is a loose way of talking only. With\na little more care nothing needs to be swept under the carpet.\n\nThis is discussed at length in my theoretisches Physik FAQ at\nhttp://www.mat.univie.ac.at/~neum/physik-faq.txt\n\n\nArnold Neumaier\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>Jamie Vicary wrote:
> Hi all,
> I've been learning about renormalization for the first time. It
> seems crazy --- we just take the divergences and put them into our mass
> and coupling constant! Why is this useful?
> Okay, I guess the "magic" of it is that, for a renormalizable
> theory,

of the divergences can be coped with in this way. But it
> doesn't seem like we're /really/ coping with them, just sweeping them
> under the carpet. Why doesn't it matter that our constants are now infinite?

The constants are infinite only if one takes a careless (and
meaningless) limit. it is a loose way of talking only. With
a little more care nothing needs to be swept under the carpet.

This is discussed at length in my theoretisches Physik FAQ at
http://www.mat.univie.ac.at/~neum/physik-faq.txt

Arnold Neumaier

Mar3-05, 05:23 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>Hi Jamie,\n\nNot everyone accepts renormalization unquestioningly. Take a look at the\nrecent sci.physics.research discussion under the title "Connes & Marcolli\npaper on renormalization" if you are interested in some of the issues.\n\nChris Oakley.\n\n"Jamie Vicary" <jamievicary@gmail.com> wrote in message\nnews:d02ld4\\$jfp\\$1@gemini.csx.cam.ac.u k...\n> Hi all,\n> I\'ve been learning about renormalization for the first time. It\n> seems crazy --- we just take the divergences and put them into our mass\n> and coupling constant! Why is this useful?\n> Okay, I guess the "magic" of it is that, for a renormalizable\n> theory, /all/ of the divergences can be coped with in this way. But it\n> doesn\'t seem like we\'re /really/ coping with them, just sweeping them\n> under the carpet. Why doesn\'t it matter that our constants are now\ninfinite?\n> Thanks in advance,\n> Jamie.\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>Hi Jamie,

Not everyone accepts renormalization unquestioningly. Take a look at the
recent sci.physics.research discussion under the title "Connes & Marcolli
paper on renormalization" if you are interested in some of the issues.

Chris Oakley.

"Jamie Vicary" <jamievicary@gmail.com> wrote in message
news:d02ld4$jfp$1@gemini.csx.cam.ac.uk...
> Hi all,
> I've been learning about renormalization for the first time. It
> seems crazy --- we just take the divergences and put them into our mass
> and coupling constant! Why is this useful?
> Okay, I guess the "magic" of it is that, for a renormalizable
> theory,

of the divergences can be coped with in this way. But it
> doesn't seem like we're /really/ coping with them, just sweeping them
> under the carpet. Why doesn't it matter that our constants are now
infinite?
> Thanks in advance,
> Jamie.
>

Mar3-05, 05:24 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>Jamie Vicary wrote:\n> Hi all,\n> I\'ve been learning about renormalization for the first time. It\n> seems crazy --- we just take the divergences and put them into our\nmass\n> and coupling constant! Why is this useful?\n> Okay, I guess the "magic" of it is that, for a renormalizable\n> theory, /all/ of the divergences can be coped with in this way. But\nit\n> doesn\'t seem like we\'re /really/ coping with them, just sweeping them\n\n> under the carpet. Why doesn\'t it matter that our constants are now\ninfinite?\n> Thanks in advance,\n> Jamie.\n\n\nAn infinite solution to the group integral is the meaning of the\nmathematical theory problem. A proper abstract integral resolves the\nrenormalization dilemma. A simple way to cope is to accept\nrenormalization.\n\nAccept it as valid theory for the time being.\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>Jamie Vicary wrote:
> Hi all,
> I've been learning about renormalization for the first time. It
> seems crazy --- we just take the divergences and put them into our
mass
> and coupling constant! Why is this useful?
> Okay, I guess the "magic" of it is that, for a renormalizable
> theory,

of the divergences can be coped with in this way. But
it
> doesn't seem like we're /really/ coping with them, just sweeping them

> under the carpet. Why doesn't it matter that our constants are now
infinite?
> Thanks in advance,
> Jamie.

An infinite solution to the group integral is the meaning of the
mathematical theory problem. A proper abstract integral resolves the
renormalization dilemma. A simple way to cope is to accept
renormalization.

Accept it as valid theory for the time being.

Mar3-05, 05:24 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>Jamie Vicary wrote:\n> Hi all,\n> I\'ve been learning about renormalization for the first time. It\n> seems crazy --- we just take the divergences and put them into our mass\n> and coupling constant! Why is this useful?\n> Okay, I guess the "magic" of it is that, for a renormalizable\n> theory, /all/ of the divergences can be coped with in this way. But it\n> doesn\'t seem like we\'re /really/ coping with them, just sweeping them\n> under the carpet. Why doesn\'t it matter that our constants are now infinite?\n> Thanks in advance,\n> Jamie.\n>\n\nYou are not alone in your frustration. Dirac and Landau said the same\nthings. Welcome to the club! I have an answer to this puzzle which\nsatisfies, at least, myself: the problem is that in QED we use a\nwrong Hamiltonian.\nThere is a way to modify the Hamiltonian so that there are no more\ndivergences, virtual particles etc. but the phenomenally accurate\nQED results for the S-matrix are preserved.\n\nTake a look at the online book www.meopemuk.com/book.pdf\nHope you\'ll find some answers there.\n\nEugene Stefanovich.\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>Jamie Vicary wrote:
> Hi all,
> I've been learning about renormalization for the first time. It
> seems crazy --- we just take the divergences and put them into our mass
> and coupling constant! Why is this useful?
> Okay, I guess the "magic" of it is that, for a renormalizable
> theory,

of the divergences can be coped with in this way. But it
> doesn't seem like we're /really/ coping with them, just sweeping them
> under the carpet. Why doesn't it matter that our constants are now infinite?
> Thanks in advance,
> Jamie.
>

You are not alone in your frustration. Dirac and Landau said the same
things. Welcome to the club! I have an answer to this puzzle which
satisfies, at least, myself: the problem is that in QED we use a
wrong Hamiltonian.
There is a way to modify the Hamiltonian so that there are no more
divergences, virtual particles etc. but the phenomenally accurate
QED results for the S-matrix are preserved.

Take a look at the online book www.meopemuk.

.pdf
Hope you'll find some answers there.

Eugene Stefanovich.

Mar3-05, 05:26 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>Jamie Vicary wrote:\n> Hi all,\n> I\'ve been learning about renormalization for the first time. It\n> seems crazy --- we just take the divergences and put them into our\nmass\n> and coupling constant!\n> But it doesn\'t seem like we\'re /really/ coping with them, just\nsweeping\n> them under the carpet. Why doesn\'t it matter that our constants are\n> now infinite?\n\nHere\'s another way to look at it that goes against the grain.\n\nWhat scattering experiments to is, in effect, tomographically\nreconstruct the internal structure of a point source. The S-matrix is\nreally giving you a form factor for the source.\n\nWhen one refers to running the constants, masses, charges, etc., it\'s\nreally a case of inserting a *radial* dependence (or more generally, a\nradial and even angular dependence) of the parameters relative to the\nscattering center.\n\nCorrespondingly, the classical point source is endowed with internal\nstructure as a result.\n\nThe coupling actually has *2* factors in it: alpha = e^2/(2 h c\nepsilon_0): e and epsilon_0. Normally the latter is ignored since it\'s\ngenerally regarded as nothing more than an MKS fix which need not\nconcern a theoretical treatment.\n\nBIG MISTAKE!\n\nThe end result is that one is force to resort to the notion that it\'s\n"e", itself, that is varying with scale.\n\nInstead, it actually makes more sense to regard epsilon_0 as the one\nwho is doing the running, and not e. Then the picture suddenly becomes\na lot clearer on what\'s going on: you\'re treating the neighborhood of\nthe charge as a dielectric medium.\n\nCorresponding to alpha -> infinity as the momentum scale goes up, one\nhas epsilon -> 0, as r -> 0 for a point source. Given the relation (D\n= epsilon E), what this is asserting is that D is being smeared out,\nremaining at a finite value, as E goes to infinity as 1/r^2. One way\nto implement this idea, for instance is by the heuristic:\nepsilon = epsilon_0 (1 - P(kr) exp(-kr))\nfor some polynomial P(x). The requirement that the energy integral\nintegral (D.E/2 dV)\nconverge then places a constrain on what the polynomal P(x) may be:\nP(x) = 1 + x + x^2/2 + x^3 p(x).\nA heuristic for k is that it should be scaled to the natural length of\nthe source -- the compton wavelength. Then the energy integral will\nyield a result proportional to (m c^2).\n\nIf P(x) is chosen right, it will yield m c^2, exactly -- thus also\nsubsuming the \'running\' mass within this picture.\n\nFor anti-screening, epsilon_0 -> infinity and it\'s the E field that\ngoes to 0, as D goes to infinity as 1/r^2. Then a suitable starting\npoint would be:\nepsilon = epsilon_0/(1 - P(kr) exp(-kr)).\nAgain, the same conditions are required for the polynomial to give a\nfinite energy integral, with similar results. Assuming such a P(x) has\nbeen chosen, then short range, the potential V corresponding to E will\ngo like\nV ~~ r^2,\nas seen at short distances for quarks.\n\nThese are only heuristic ideas, but they get across what deeper\nmechanism may be lurking beneath the usual apparatus of renormalization\ntheory.\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>Jamie Vicary wrote:
> Hi all,
> I've been learning about renormalization for the first time. It
> seems crazy --- we just take the divergences and put them into our
mass
> and coupling constant!
> But it doesn't seem like we're /really/ coping with them, just
sweeping
> them under the carpet. Why doesn't it matter that our constants are
> now infinite?

Here's another way to look at it that goes against the grain.

What scattering experiments to is, in effect, tomographically
reconstruct the internal structure of a point source. The S-matrix is
really giving you a form factor for the source.

When one refers to running the constants, masses, charges, etc., it's
really a case of inserting a *radial* dependence (or more generally, a
radial and even angular dependence) of the parameters relative to the
scattering center.

Correspondingly, the classical point source is endowed with internal
structure as a result.

The coupling actually has *2* factors in it: $LaTeX Code: \\alpha = e^2/(2 h c\\epsilon_0): e$ and $LaTeX Code: \\epsilon_0$ . Normally the latter is ignored since it's
generally regarded as nothing more than an MKS fix which need not
concern a theoretical treatment.

BIG MISTAKE!

The end result is that one is force to resort to the notion that it's
"e", itself, that is varying with scale.

Instead, it actually makes more sense to regard $LaTeX Code: \\epsilon_0$ as the one
who is doing the running, and not e. Then the picture suddenly becomes
a lot clearer on what's going on: you're treating the neighborhood of
the charge as a dielectric medium.

Corresponding to $LaTeX Code: \\alpha ->$ infinity as the momentum scale goes up, one
has $LaTeX Code: \\epsilon -> 0,$ as

for a point source. Given the relation (D
$LaTeX Code: = \\epsilon E),$ what this is asserting is that D is being smeared out,
remaining at a finite value, as E goes to infinity as

. One way
to implement this idea, for instance is by the heuristic:
$LaTeX Code: \\epsilon = \\epsilon_0 (1 - P(kr) \\exp(-kr))$
for some polynomial P(x). The requirement that the energy integral
integral

converge then places a constrain on what the polynomal P(x) may be:
P(x)

A heuristic for k is that it should be scaled to the natural length of
the source -- the compton wavelength. Then the energy integral will
yield a result proportional to

.

If P(x) is chosen right, it will yield

exactly -- thus also
subsuming the 'running' mass within this picture.

For anti-screening, $LaTeX Code: \\epsilon_0 ->$ infinity and it's the E field that
goes to 0, as D goes to infinity as

. Then a suitable starting
point would be:
$LaTeX Code: \\epsilon = \\epsilon_0/(1 - P(kr) \\exp(-kr))$ .
Again, the same conditions are required for the polynomial to give a
finite energy integral, with similar results. Assuming such a P(x) has
been chosen, then short range, the potential V corresponding to E will
go like
V

as seen at short distances for quarks.

These are only heuristic ideas, but they get across what deeper
mechanism may be lurking beneath the usual apparatus of renormalization
theory.

Mar11-05, 03:31 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>Arnold Neumaier wrote:\n>\n> Don\'t ask for magic! Very primitively:\n>\n> In one dimension, integral_{-inf}^{+inf} dt/|t|^k converges for k>1,\n> but not for k=1. Thus one needs power counting to find out whether\n> certain integrals are likely to converge. in higher dimensions and in\n\n> field theory it is not really different.\n>\n> There is no shortcut to understanding the difference between\n> renormalizable theories and those which are not.\n\nI\'m not hoping to get a shortcut. I do have a rough understanding of\nhow to compute divergent parts of diagrams with the appropriate number\nof external lines so as to determine the counterterms necessary to\nrender the theory finite.\n\nI am trying to understand more concretely why it is said that\nultraviolet divergences is an indication of our ignorance of what\'s\ngoing on at short distances?\n\nParticularly for the phi^4 vs. phi^6 theory in 3+1 dimensions, I don\'t\nsee why we have less knowledge about what\'s going on at short distances\nthan we about phi^6. The only difference I see is the powers of the\nfield.\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>Arnold Neumaier wrote:
>
> Don't ask for magic! Very primitively:
>
> In one dimension, $LaTeX Code: integral_{-inf}^{+inf} dt/|t|^k$ converges for k>1,
> but not for k=1. Thus one needs power counting to find out whether
> certain integrals are likely to converge. in higher dimensions and in

> field theory it is not really different.
>
> There is no shortcut to understanding the difference between
> renormalizable theories and those which are not.

I'm not hoping to get a shortcut. I do have a rough understanding of
how to compute divergent parts of diagrams with the appropriate number
of external lines so as to determine the counterterms necessary to
render the theory finite.

I am trying to understand more concretely why it is said that
ultraviolet divergences is an indication of our ignorance of what's
going on at short distances?

Particularly for the $LaTeX Code: \\phi^4 vs. \\phi^6$ theory in 3+1 dimensions, I don't
see why we have less knowledge about what's going on at short distances
than we about $LaTeX Code: \\phi^6.$ The only difference I see is the powers of the
field.

Mar11-05, 03:31 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>What bothers me more than renormalization is the process of\nregularization. Renormalization is the process of removing the\ndivergences by absorbing them in physical constants such as mass and\ncharge. Renormalization, on the other hand, is a different process.\nRenormalization is required because, when standard perturbation theory\nis used, it is found that the photon has a divergent mass. However,\nunlike the electron, there is no mass term to absorb this divergence.\nRenormalizatoin is the mathematical process by which the divergence\nassociated with photon mass is removed. The problem I have with this\nis that the process is very "ad hoc" (in my opinion). The\nrenormalization of the electron mass has a physical bases, i.e, the\nbare mass is infinite which partially cancels the divergence that comes\nabout in perturbation theory. This, then, results in a finite\nexperimental mass. However the bare mass of the photon is presumed to\nbe zero so that there is no way to remove the divergent term.\n\nDan Solomon\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>What bothers me more than renormalization is the process of
regularization. Renormalization is the process of removing the
divergences by absorbing them in physical constants such as mass and
charge. Renormalization, on the other hand, is a different process.
Renormalization is required because, when standard perturbation theory
is used, it is found that the photon has a divergent mass. However,
unlike the electron, there is no mass term to absorb this divergence.
Renormalizatoin is the mathematical process by which the divergence
associated with photon mass is removed. The problem I have with this
is that the process is very "ad hoc" (in my opinion). The
renormalization of the electron mass has a physical bases, i.e, the
bare mass is infinite which partially cancels the divergence that comes
about in perturbation theory. This, then, results in a finite
experimental mass. However the bare mass of the photon is presumed to
be zero so that there is no way to remove the divergent term.

Dan Solomon

Mar12-05, 09:09 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>In my previous message I transposed "renormalization" with\n"regularizatoin" in several places. If anyone replied to that message\nI hope they reply to what I intended to say, not what I did say. To\nset the record straight I would like to repost with the correct message\nas follows:\n\nWhat bothers me more than renormalization is the process of\nregularization. Renormalization is the process of removing the\ndivergences by absorbing them in physical constants such as mass and\ncharge. Regularization, on the other hand, is a different process.\nRegularization is required because, when standard perturbation theory\nis used, it is found that the photon has a divergent mass. However,\nunlike the electron, there is no mass term to absorb this divergence.\nRegularization is the mathematical process by which the divergence\nassociated with photon mass is removed. The problem I have with this\nis that the process is very "ad hoc" (in my opinion). The\nrenormalization of the electron mass has a physical bases, i.e, the\nbare mass is infinite which partially cancels the divergence that\ncomes\nabout in perturbation theory. This, then, results in a finite\nexperimental mass. However the bare mass of the photon is presumed to\nbe zero so that there is no way to remove the divergent term.\n\nOK. I have carefully read this over and this time I am absolutely\ncertain that this is definitely-uh-probably-er-possibly-uh-maybe\ncorrect.\n\nDan Solomon\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 my previous message I transposed "renormalization" with
"regularizatoin" in several places. If anyone replied to that message
I hope they reply to what I intended to say, not what I did say. To
set the record straight I would like to repost with the correct message
as follows:

What bothers me more than renormalization is the process of
regularization. Renormalization is the process of removing the
divergences by absorbing them in physical constants such as mass and
charge. Regularization, on the other hand, is a different process.
Regularization is required because, when standard perturbation theory
is used, it is found that the photon has a divergent mass. However,
unlike the electron, there is no mass term to absorb this divergence.
Regularization is the mathematical process by which the divergence
associated with photon mass is removed. The problem I have with this
is that the process is very "ad hoc" (in my opinion). The
renormalization of the electron mass has a physical bases, i.e, the
bare mass is infinite which partially cancels the divergence that
comes
about in perturbation theory. This, then, results in a finite
experimental mass. However the bare mass of the photon is presumed to
be zero so that there is no way to remove the divergent term.

OK. I have carefully read this over and this time I am absolutely
certain that this is definitely-uh-probably-er-possibly-uh-maybe
correct.

Dan Solomon

Mar12-05, 09:09 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>> I am trying to understand more concretely why it is said that\n> ultraviolet divergences is an indication of our ignorance of what\'s\n> going on at short distances?\n\nFact: naive QFT in 3+1 dimensions leads to divergent scattering\namplitudes. Fact: all the ways of dealing with this problem are exactly\nthat. They "deal with" the problem rather than reformulating from\nscratch in such a way that the divergences never appear. If you cannot\nhandle mathematical ugliness, you should not be doing QFT.\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 am trying to understand more concretely why it is said that
> ultraviolet divergences is an indication of our ignorance of what's
> going on at short distances?

Fact: naive QFT in 3+1 dimensions leads to divergent scattering
amplitudes. Fact: all the ways of dealing with this problem are exactly
that. They "deal with" the problem rather than reformulating from
scratch in such a way that the divergences never appear. If you cannot
handle mathematical ugliness, you should not be doing QFT.

Mar12-05, 09:10 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>> What bothers me more than renormalization is the process of\n> regularization. Renormalization is the process of removing the\n> divergences by absorbing them in physical constants such as mass and\n> charge. Renormalization, on the other hand, is a different process.\n> Renormalization is required because, when standard perturbation theory\n> is used, it is found that the photon has a divergent mass. However,\n> unlike the electron, there is no mass term to absorb this divergence.\n> Renormalizatoin is the mathematical process by which the divergence\n> associated with photon mass is removed. The problem I have with this\n> is that the process is very "ad hoc" (in my opinion). The\n> renormalization of the electron mass has a physical bases, i.e, the\n> bare mass is infinite which partially cancels the divergence that comes\n> about in perturbation theory. This, then, results in a finite\n> experimental mass. However the bare mass of the photon is presumed to\n> be zero so that there is no way to remove the divergent term.\n\nUnderlying renormalization theory is the belief that one can selectively\nviolate basic laws of mathematics and still have enough left to make a\nmeaningful theory.\n\nI take the (minority) view that this is not possible. Even when one is\nusing approximations, one can use mathematics to determine the extent of\ntheir validity. But when one starts to deal in quantities that are,\nquite simply, meaningless, then the whole edifice collapses. Infinity\nminus infinity is indeterminate. That is all one needs to know.\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>> What bothers me more than renormalization is the process of
> regularization. Renormalization is the process of removing the
> divergences by absorbing them in physical constants such as mass and
> charge. Renormalization, on the other hand, is a different process.
> Renormalization is required because, when standard perturbation theory
> is used, it is found that the photon has a divergent mass. However,
> unlike the electron, there is no mass term to absorb this divergence.
> Renormalizatoin is the mathematical process by which the divergence
> associated with photon mass is removed. The problem I have with this
> is that the process is very "ad hoc" (in my opinion). The
> renormalization of the electron mass has a physical bases, i.e, the
> bare mass is infinite which partially cancels the divergence that comes
> about in perturbation theory. This, then, results in a finite
> experimental mass. However the bare mass of the photon is presumed to
> be zero so that there is no way to remove the divergent term.

Underlying renormalization theory is the belief that one can selectively
violate basic laws of mathematics and still have enough left to make a
meaningful theory.

I take the (minority) view that this is not possible. Even when one is
using approximations, one can use mathematics to determine the extent of
their validity. But when one starts to deal in quantities that are,
quite simply, meaningless, then the whole edifice collapses. Infinity
minus infinity is indeterminate. That is all one needs to know.

Mar14-05, 02:12 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>Chris Oakley wrote:\n>>What bothers me more than renormalization is the process of\n>>regularization. Renormalization is the process of removing the\n>>divergences by absorbing them in physical constants such as mass and\n>>charge. Renormalization, on the other hand, is a different process.\n>>Renormalization is required because, when standard perturbation theory\n>>is used, it is found that the photon has a divergent mass. However,\n>>unlike the electron, there is no mass term to absorb this divergence.\n>>Renormalizatoin is the mathematical process by which the divergence\n>>associated with photon mass is removed. The problem I have with this\n>>is that the process is very "ad hoc" (in my opinion). The\n>>renormalization of the electron mass has a physical bases, i.e, the\n>>bare mass is infinite which partially cancels the divergence that comes\n>>about in perturbation theory. This, then, results in a finite\n>>experimental mass. However the bare mass of the photon is presumed to\n>>be zero so that there is no way to remove the divergent term.\n>\n>\n> Underlying renormalization theory is the belief that one can selectively\n> violate basic laws of mathematics and still have enough left to make a\n> meaningful theory.\n>\n> I take the (minority) view that this is not possible. Even when one is\n> using approximations, one can use mathematics to determine the extent of\n> their validity. But when one starts to deal in quantities that are,\n> quite simply, meaningless, then the whole edifice collapses. Infinity\n> minus infinity is indeterminate. That is all one needs to know.\n>\n\nI tend to agree with you, and I offer a solution: forget about the\nquantization + renormalization + dressing nightmare and\n"infinity minus infinity" tricks involved there.\nJust take the RQD dressed particle Hamiltonian as a starting point\nfor doing physics. (I have a few low order terms of this Hamiltonian\npresented in the book; if you give me some time I\'ll derive higher\norder terms that should be more than enough to do practical\ncalculations). With this Hamiltonian you\'ll never meet divergent\nloop integrals, there is no need for regularization,\nrenormalization, and "infinity minus infinity" subtractions.\nAs long as you don\'t ask me where I got this Hamiltonian,\nyour sense of mathematical purity will not be offended.\n\n\nEugene Stefanovich.\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>Chris Oakley wrote:
>>What bothers me more than renormalization is the process of
>>regularization. Renormalization is the process of removing the
>>divergences by absorbing them in physical constants such as mass and
>>charge. Renormalization, on the other hand, is a different process.
>>Renormalization is required because, when standard perturbation theory
>>is used, it is found that the photon has a divergent mass. However,
>>unlike the electron, there is no mass term to absorb this divergence.
>>Renormalizatoin is the mathematical process by which the divergence
>>associated with photon mass is removed. The problem I have with this
>>is that the process is very "ad hoc" (in my opinion). The
>>renormalization of the electron mass has a physical bases, i.e, the
>>bare mass is infinite which partially cancels the divergence that comes
>>about in perturbation theory. This, then, results in a finite
>>experimental mass. However the bare mass of the photon is presumed to
>>be zero so that there is no way to remove the divergent term.
>
>
> Underlying renormalization theory is the belief that one can selectively
> violate basic laws of mathematics and still have enough left to make a
> meaningful theory.
>
> I take the (minority) view that this is not possible. Even when one is
> using approximations, one can use mathematics to determine the extent of
> their validity. But when one starts to deal in quantities that are,
> quite simply, meaningless, then the whole edifice collapses. Infinity
> minus infinity is indeterminate. That is all one needs to know.
>

I tend to agree with you, and I offer a solution: forget about the
quantization + renormalization + dressing nightmare and
"infinity minus infinity" tricks involved there.
Just take the RQD dressed particle Hamiltonian as a starting point
for doing physics. (I have a few low order terms of this Hamiltonian
presented in the book; if you give me some time I'll derive higher
order terms that should be more than enough to do practical
calculations). With this Hamiltonian you'll never meet divergent
loop integrals, there is no need for regularization,
renormalization, and "infinity minus infinity" subtractions.
As long as you don't ask me where I got this Hamiltonian,
your sense of mathematical purity will not be offended.

Eugene Stefanovich.

Mar15-05, 01:17 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>\njsolomon@mail.com wrote:\n\n> Renormalization is required because, when standard perturbation theory\n> is used, it is found that the photon has a divergent mass. However,\n> unlike the electron, there is no mass term to absorb this divergence.\n\nAre you speaking about the infrared divergence,\nor about the photon self-energy?\n\n\n> Renormalization is the mathematical process by which the divergence\n> associated with photon mass is removed. The problem I have with this\n> is that the process is very "ad hoc" (in my opinion).\n\nThe typical presentation might be ad hoc, but everything has a\nnatural explanation. Try the theoretical physics FAQ at\nhttp://www.mat.univie.ac.at/~neum/physics-faq.txt\n\n\nArnold Neumaier\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>jsolomon@mail.com wrote:

> Renormalization is required because, when standard perturbation theory
> is used, it is found that the photon has a divergent mass. However,
> unlike the electron, there is no mass term to absorb this divergence.

Are you speaking about the infrared divergence,
or about the photon self-energy?

> Renormalization is the mathematical process by which the divergence
> associated with photon mass is removed. The problem I have with this
> is that the process is very "ad hoc" (in my opinion).

The typical presentation might be ad hoc, but everything has a
natural explanation. Try the theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt

Arnold Neumaier

Mar15-05, 01:17 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>\nlost.and.lonely.physicist@ gmail.com wrote:\n\n> Arnold Neumaier wrote:\n>\n>>There is no shortcut to understanding the difference between\n>>renormalizable theories and those which are not.\n>\n> I\'m not hoping to get a shortcut. I do have a rough understanding of\n> how to compute divergent parts of diagrams with the appropriate number\n> of external lines so as to determine the counterterms necessary to\n> render the theory finite.\n>\n> I am trying to understand more concretely why it is said that\n> ultraviolet divergences is an indication of our ignorance of what\'s\n> going on at short distances?\n\nWhat is going on at short distances is what happens at large momenta.\nUV renormalization is the limit of arbitrarily large momenta.\nWe cannot measure anything here, hence have to substitute consistency\narguments for our ignorance.\n\n\n> Particularly for the phi^4 vs. phi^6 theory in 3+1 dimensions, I don\'t\n> see why we have less knowledge about what\'s going on at short distances\n> than we about phi^6. The only difference I see is the powers of the\n> field.\n\nIn Phi^4 theory, the power counting shows renormalizability.\nThus only few counterterms must be added to get a consistent\nfinite perturbative expansion at all orders. This means that a few\nparameters suffice to get a consistent theory which will be correct\nat the energies of interest (which should be essentially independent\nof what happens at the inaccessible large energies).\n\nIn phi^6 theory, the power counting shows nonrenormalizability.\nThus infinitely many counterterms must be added to get a consistent\nfinite perturbative expansion at all orders. This means that with a few\nparameters one can only get an effective low order theory, which may,\nhowever, still be good enough at the energies of interest.\nBut for better approximation, one needs to determine more and\nmore parameters...\n\n\nArnold Neumaier\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>lost.and.lonely.physicist@gmail.com wrote:

> Arnold Neumaier wrote:
>
>>There is no shortcut to understanding the difference between
>>renormalizable theories and those which are not.
>
> I'm not hoping to get a shortcut. I do have a rough understanding of
> how to compute divergent parts of diagrams with the appropriate number
> of external lines so as to determine the counterterms necessary to
> render the theory finite.
>
> I am trying to understand more concretely why it is said that
> ultraviolet divergences is an indication of our ignorance of what's
> going on at short distances?

What is going on at short distances is what happens at large momenta.
UV renormalization is the limit of arbitrarily large momenta.
We cannot measure anything here, hence have to substitute consistency
arguments for our ignorance.

> Particularly for the $LaTeX Code: \\phi^4 vs. \\phi^6$ theory in 3+1 dimensions, I don't
> see why we have less knowledge about what's going on at short distances
> than we about $LaTeX Code: \\phi^6$ . The only difference I see is the powers of the
> field.

In $LaTeX Code: \\Phi^4$ theory, the power counting shows renormalizability.
Thus only few counterterms must be added to get a consistent
finite perturbative expansion at all orders. This means that a few
parameters suffice to get a consistent theory which will be correct
at the energies of interest (which should be essentially independent
of what happens at the inaccessible large energies).

In $LaTeX Code: \\phi^6$ theory, the power counting shows nonrenormalizability.
Thus infinitely many counterterms must be added to get a consistent
finite perturbative expansion at all orders. This means that with a few
parameters one can only get an effective low order theory, which may,
however, still be good enough at the energies of interest.
But for better approximation, one needs to determine more and
more parameters...

Arnold Neumaier

Mar15-05, 01:17 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>\n> I tend to agree with you, and I offer a solution: forget about the\n> quantization + renormalization + dressing nightmare and\n> "infinity minus infinity" tricks involved there.\n> Just take the RQD dressed particle Hamiltonian as a starting point\n> for doing physics. (I have a few low order terms of this Hamiltonian\n> presented in the book; if you give me some time I\'ll derive higher\n> order terms that should be more than enough to do practical\n> calculations). With this Hamiltonian you\'ll never meet divergent\n> loop integrals, there is no need for regularization,\n> renormalization, and "infinity minus infinity" subtractions.\n> As long as you don\'t ask me where I got this Hamiltonian,\n> your sense of mathematical purity will not be offended.\n\nEugene,\n\nForgive me for pointing this out, but I could equally well say that if I\naccept the renormalization group equations as my starting point then I will\nnever have to do infinite subtractions. This is not wrong as the infinite\nsubtractions happen in the process of getting there from first principles.\nYour work also involves infinite subtractions (in the dressing\ntransformation) and cannot therefore have a long-term future. In the long\nterm mathematical chicanery cannot and will not be tolerated.\n\nChris\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 tend to agree with you, and I offer a solution: forget about the
> quantization + renormalization + dressing nightmare and
> "infinity minus infinity" tricks involved there.
> Just take the RQD dressed particle Hamiltonian as a starting point
> for doing physics. (I have a few low order terms of this Hamiltonian
> presented in the book; if you give me some time I'll derive higher
> order terms that should be more than enough to do practical
> calculations). With this Hamiltonian you'll never meet divergent
> loop integrals, there is no need for regularization,
> renormalization, and "infinity minus infinity" subtractions.
> As long as you don't ask me where I got this Hamiltonian,
> your sense of mathematical purity will not be offended.

Eugene,

Forgive me for pointing this out, but I could equally well say that if I
accept the renormalization group equations as my starting point then I will
never have to do infinite subtractions. This is not wrong as the infinite
subtractions happen in the process of getting there from first principles.
Your work also involves infinite subtractions (in the dressing
transformation) and cannot therefore have a long-term future. In the long
term mathematical chicanery cannot and will not be tolerated.

Chris