Re: Renormalization in QED

From: kenseto (kenseto_at_erinet.com)
Date: 07/21/04 

Date: Wed, 21 Jul 2004 16:35:36 -0400

"Bjoern Feuerbacher" <feuerbac@thphys.uni-heidelberg.de> wrote in message
news:cdlv2q$h5u$1@news.urz.uni-heidelberg.de...
> kenseto wrote:
> > "Bjoern Feuerbacher" <feuerbac@thphys.uni-heidelberg.de> wrote in
message
> > news:cdlbe5$djp$1@news.urz.uni-heidelberg.de...
> >
> >>kenseto wrote:
> >>
>
> [snip]
>
> >>>>>>We have examples where NM failed. There are *no* examples where the
SM
> >>>>>>failed.
> >>>>>
> >>>>>
> >>>>>It failed to include a theory of quantum gravity.
> >>>>
> >>>>It never intended to do that, so that is not a failure.
> >>>
> >>>
> >>>Ah....but it is a failure of the basic model of SM
> >>
> >>Err, we were talking about failures of the SM, not about
> >>failures of "the basic model of SM" (that does not even
> >>make sense - the basic model of the Standard Model???).
> >>Movement of goalposts noted.
> >
> >
> > Not moving goalposts....
>
> Yes, they moved.
>
>
> >I was merely pointing out the deficiencies of SM.
>
> That the SM does not include quantum gravity is not a failure
> or deficiency of it, since it never was intended to do that.

Ah...but how do they know that SM does not include a theory
of quantum gravity? It is obvious that they tried to use SM to
describe gravity but failed. The reason for the failure is that they
can't re-normalize the infinities away.
>
>
>
> >>>which says
> >>>that all the forces of nature are mediated by the exchanging
> >>>of messenger particles
> >>
> >>Err, that is a mathematical consequence of the basic postulates
> >>of QFT. You don't know the basic postulates of QFT, right? If
> >>you think you do, please list them.
>
> I notice that you ignored my questions about the basic
> postulates of QFT. Big surprise.

So what are the SM postulates??
>
>
> > Since SM includes QFT and the basic postulates of QFT failed
> > to give a viable theory of quantum gravity then this is a
> > failure of the SM.
>
> No, that's a non sequitur (that does not follow).

It follows.
>
> Additionally, it is still not clear if QFT can give a viable
> theory of quantum gravity or not.

Not so far even though tremendous efforts were expanded on it.

>People are still working
> on that question, and in recent years it turned out that
> if one uses non-perturbative methods, one can make quite big
> advances. That's a quite vivid area of research and by no
> means a settled question so far.

Yeah...working on the question. But to date it shows a failure of QFT.
>
>
> [snip]
>
>
> >>Before the Higgs was invented, the SM did not exist. So saying
> >>that the Higgs was invented to give the massless particles in
> >>the SM mass makes no sense.
> >
> >
> > So are you saying that those equations that contains the
> > massless particles are not yet the SM but after the Higgs
> > is introduced into those equations then they become the SM???
>
> With the caveat that the equations *are* not the SM, but
> they *describe* the SM, I am saying *exactly* that, yes.

So what is SM??
>
>
> [snip]
>
> >>You did not address my point that the SM predicts a certain
> >>energy range for the Higgs, and we have searched only less than
> >>half of that energy range so far.
> >
> >
> > As I said before if the Higgs is not found in the full energy
> > range they will
> > simply invent another epicycle to fix the problem.
> > And this process will go on and on.
>
> 1) I already addressed this.
> 2) This does not change my point here: that we have not found
> the Higgs so far is *not* a failure of the SM, since we have
> not searched in the whole range where it should exist so far!

It is a failure of SM.
>
>
> [snip]
>
>
> >>>SM has no explanation for any of the problems outlined.
> >>
> >>Err, hint: that's why people try to find possible extensions
> >>of the SM.
> >
> >
> > Err, hint: this course of action is not scientific.
>
> It is not scientific to search for alternative theories
> when the current theory seems to have some potential
> problems???

It is not too scientific and imaginative if the same failed model
is use to develop the alternate theory.
>
>
> > One needs to realize that
> > the basic assumption (that all forces of nature are mediated
> > by messenger particles) of SM have failed.
>
> What's your evidence that this assumption has indeed failed?

The graviton failed to mediate the force of gravity.
>
>
> > It's time to explore new ideas.
>
> Yes. String theory, for example.

It can't be tested experimentally.
>
>
> > MM represent such a new idea.
>
> MM is not physics, but a bunch of rhetoric and unsupported
> assertions.

Then why are you wasting my time??
>
>
>
>
> [snip]
>
> >>You failed to address my actual point above that apparently
> >>MM could live with and without a Higgs, and that this is quite
> >>flexible.
> >
> >
> > MM says that all sorts unstable particles can exist and they are
> > all make from orbiting S-Particles. If you want to call this flexibility
> > that's OK with me.
>
> What I am calling flexible here is the simple fact that MM
> makes no definite prediction if the Higgs exists or not.

So??
>
> [snip
>
>
> >>Fine. SM says that the Higgs *does* exist. So, if the Higgs is
> >>indeed found, that is evidence for the SM, but not for MM,
> >>since MM did not predict that the Higgs *should* exist.
> >
> >
> > <sigh> MM says that the Higgs can exist but is no big deal. Why?
> > Because it is just another unstable particle.
>
> Err, the SM predicts not only the existence of the Higgs, but
> also some of its properties (spin etc.). So do you admit
> that the SM makes more and more specific predictions as MM here?

NO...what ever properties the Higgs has is just the properties of that
unstable particle.
>
>
> >>>However even if the Higgs exists MM does not say that the Higgs
> >>>impart mass to real particles such as the electron or the quarks.
> >>
> >>Since that is the *definition* of the Higgs boson, if a particle
> >>is discovered which does not do that, one could not call it the
> >>Higgs.
> >
> >
> > That is the SM definition for the Higgs. The MM definition for
> > the Higgs is that it is just another unstable particle.
>
> "just another unstable particle" is not a definition.
> It's perhaps a description, but in no way a definition!

Saying that the Higgs give mass to a massless particle is just an assertion.
>
>
> [snip]
>
>
> >>>>Err, no, the basic model of SM was *not* that "all particles are
> >>>>massless". Apparently you yet again misunderstood (or willfully
> >>>>misinterpreted) a pop science source which you read about the SM.
> >>>
> >>>
> >>>No your runt brothers in this NG say so..... specifcally Bilge.
> >>
> >>As I said above: you misunderstood or willfully misinterpreted.
> >
> >
> > ROTFLOL....I didn't know that all particles are massless until
> > Bilge mentioned it. So how could I misinterpreted that?
>
> *sigh* And you go on, in this case misinterpreting what *I*
> actually said above.

????????????????
>
>
> >>>>>Because orbiting one E-String will eliminate the effect of the CRE
> >>>>>force.
> >>>
> >>>>1) How does this eliminate the effect of the CRE force?
> >>>
> >>>
> >>>When an S-Particle orbits one E-String there is no CXRE force. Why?
> >>>Because CRE force is derived from the divergent geometry between
> >>>two or more E-Strings
> >>
> >>So what??? If there are two divergent E-Strings, and one of them
> >>is orbited by an S-particle, there is a CRE force, isn't there???
> >
> >
> > No...the orbiting S-Particle experiences no CRE force.
>
> Why not???

How can one E-String cause a CRE force on an S-Particle orbiting it??
. Perhaps you don't know what is the CRE force? I suggest that you
read my paper again.
>
>
> > It will experience
> > a CRE force if it is orbiting around the two or more E-Strings.
>
> Why?

Because those E-Strings are diverging from each other. This means that
the orbit of the S-Paticles in incresing with time. This is what the CRE f
orce is all about.

>
>
> [snip]
>
>
> >>Hence my original question still stands:
> >>why is only the *smallest* orbital diameter around one E-String
> >>stable?
> >
> >
> > There is only one orbital diameter around one E-String and that
> > orbit has the smallest diameter compared to orbits around two
> > or more E-Strings
>
> Interesting. In earlier times, it sounded as if there are
> several possible orbital diameters even for an orbit around
> only one E-String.

NO...

Did I misunderstood you there, or have you
> withdrawn that idea now simply?

You misunderstood. I was comparing orbit around one E-String with
those orbit around two or more E-Strings.
>
>
> [snip]
>
>
> >>>I think that they have limited stability.
> >>
> >>What's your evidence for that?
> >
> >
> > What is your evidence for their stability?
>
> <http://pdg.lbl.gov/2004/tables/lxxx.pdf>
>
> I notice that you use one of your usual tactics: instead
> of backing up your own assertions with evidence, simply
> act as if the other one had to bring up contrary evidence.
> How convenient.

I am not going to look it up. You give us a summary of it.
>
> [snip]
>
>
> >>>No...the muon neutrino and tau neutrino decay into electron
> >>>neutrinos.
> >>
> >>Evidence for that assertion, please.
>
> I notice that yet again, you did not bother to provide
> evidence for your assertion. How convenient. It's really
> easy to make assertions if one does not have to back them
> up, right?
>
>
>
> > So are you saying that the muon neutrino is stable and that
> > it does not decay into electron neutrino?
>
> It does not decay into an electron neutrino, but there are
> oscillations, i.e. the muon neutrino can periodically turn
> into a electron or tau neutrino and back to a muon neutrino.
> There is a heap of evidence for that. Start with SuperKamiokande
> and SNO.

Are you sayinbg that a muon neurtino in the lab will oscillate between
an electron or tau neutrino and back to a muon neutrino? Is this really
observed in the lab? Or is this just a math prediction because the
neutrino from the sun show less electron neutrinos than expected?
>
>
> [snip]
>
>
> >>>An unbounded up quarks is less stable than an electron.
> >>
> >>Why?

Because it has a larger orbital diameter which, in turn, is cause by
the S-Particle orbiting around two or more E-String. BTW this
causes the Up quark to have less than one unit of electric change
and cause the up quark to have a larger mass than the electron.
> >
> >
> > Because its S-Particle is orbiting around more than one E-String.
> >
> >>
> >>>Up quarks are stable in a bounded state.
> >>
> >>Bounded to other quarks, or what do you mean? How does this
> >>increase stability?
> >
> >
> > Stacked interaction with other up quarks. This increase the
> > stability because
> > their electric charges add up to more than 1 unit of charge.
>
> How does the increases electric charge increase the stability?

It is too complicated to explain here. Read the paper more carefully.
>
> And what about two stacked down quarks, where the electric
> charges do *not* add up to more than 1 unit of charge?

A neutron has two stacked down quarks for a combined electric
change of 2/3. That's why the neutron has limited stability.
>
>
> > BTW any particle
> > has electric charge one or greater is stable.
>
> Why?
>
> And: do you mean the absolute value of the charge here?
> I.e. are particles with charge greater or equal to one
> and with charge less or equal to minus one stable?
> Or only one of these groups?

All particles with +/- unit charge are stable.
>
>
> [snip]
>
>
> >>>NO...An S-Particle will feel the repulsive force from all surrounding
> >>>E-Strings.
> >>>So it will maintain its motion in the E-Matrix
> >>
> >>Please show mathematically that the repulsive force from
> >>the surrounding E-Strings will keep the S-particle on an orbit
> >>around one E-String.
> >
> >
> > No you do it. It's about time you contribute something.
>
> Fine. If you tell me how the repulsive force from the surrounding
> E-Strings depends on distance (and the angles), what distances
> they have to each other, and how big the orbital diameter is
> (and the force law - see below), I will do the calculation.

I don't have these data.
>
> You don't expect me to be able to do a calculation without
> having any data to start with, do you?

I don't know what you are trying to calculate.
>
>
> [snip]
>
>
> >>You did not answer my question above. Is F=m*a valid for
> >>S-particles or not?
> >
> >
> > F=m*a is not valid for S-Particle.
>
> Thanks.
>
> Then how are the acceleration of an S-particle and the
> force acting on it related?

The S-Particle is not accelerating. It has a constant state of
orbiting absolute motion (Motion in the E-Matrix).

Ken Seto

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