Do the virtual particles in quantum mechanics really exist?
FrediFizzx
American by J. Fleming of Madison, Wis. It is in the "Ask the Experts"
section and answered by Gordon Kane in the affirmative. He says;
"Virtual particles are indeed real -- they have observable effects that
physicists have devised ways of measuring. Quantum theory predicts that
every particle spends some time as a combination of other particles in
all possible ways."
He goes on to cite the Lamb shift as the first example of this. I
totally agree with Gordon Kane about this. For me, this would mean that
an electron viewed from a distance would even have a very tiny mix of
virtual top quarks to its total "make up". Discuss.
Best,
Fred Diether
Alf P. Steinbach
Good news for us non-physicists who tend to write incorrect statements
in public Usenet groups -- whatever we write, it will turn out to be
correct after all in someone's interpretation of quantum mechanics. To
wit, in a discussion of how black holes can possibly form at all in
finite outside time, I once referred to an electron (!) merrily worming
its way into the dark hole, but being split into individual quarks by
the infinite gravitational field gradient. Hah!
More seriously, define "really", "real" and "exist".
To me, something that has a discernible effect is real and really exists.
Is your question whether virtual particles have any effect?
They do. And according to the current consensus (is consensus
scientific?) they can turn non-virtual very quickly. Like Hawking
radiation.
But I think the interesting question here is what frame of reference
they choose, and why and how.
For example, if the vacuum is really seething with virtual particles, as
seems to be true, and if there's no preferred frame of reference, as
also seems to be true, shouldn't those virtual particles have
arbitrarily high kinetic energies as viewed from our frame of reference?
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
sr
FrediFizzx wrote:
IMHO, I don't think Kane understands advanced QFT.
You asked a similar thing on s.p.r. some time ago. I attempted an
answer then, but the thread didn't continue on very far. I append
a copy of that answer below, for convenience of s.p.f. readers:
---- Previous s.p.r. posting:
-----------------------------------------
I wrote:
>> The "bare electron + virtual cloud" picture
>> is a fictional artefact of tricky QFT math.
>> There is no such thing as a "bare electron"
>> in nature.
FrediFizzx asked:
> So are you saying that the "virtual cloud" is
> the electron?
No, of course I'm not saying that.
> If not, what is the electron then?
I'll probably just make it worse, but I'll try once...
A "bare" electron corresponds to an eigenstate of
a free Hamiltonian. Together, all these
eigenstates span a Hilbert space (aka Fock space)
representing arbitrary numbers of free (i.e:
non-interacting) electrons. This Hilbert space is
unphysical. To get a physically relevant Hilbert
space, one must include an interaction term in the
Hamiltonian. Physical electrons correspond to
eigenstates of this interacting Hamiltonian.
It used to be tacitly assumed that the basis of
free eigenstates could be used to expand the
interacting eigenstates. I.e: it was assumed that
the free Hilbert space coincided with the
interacting Hilbert space.
But people later realized this was an invalid
assumption: The interacting Hilbert space is
disjoint from the free one.
The notion of "bare electron + virtual cloud of
photons" is just a picture one gets when one tries
to express a mapping between free and interacting
Hilbert spaces in terms of operators from the free
Hilbert space.
The only states here that could legitimately be
considered physical are the eigenstates of the
interacting Hamiltonian. Unfortunately, no one
knows how to construct them non-perturbatively,
in general.
- strangerep.
FrediFizzx
news:1174272439.0...@e65g2000hsc.googlegroups.com...
>
> FrediFizzx wrote:
>
>> This was a question that appeared in the Mar. 2007 issue of
>> Scientific
>> American by J. Fleming of Madison, Wis. It is in the "Ask the
>> Experts"
>> section and answered by Gordon Kane in the affirmative. He says;
>>
>> "Virtual particles are indeed real -- they have observable effects
>> that
>> physicists have devised ways of measuring. Quantum theory predicts
>> that
>> every particle spends some time as a combination of other particles
>> in
>> all possible ways."
>>
>> He goes on to cite the Lamb shift as the first example of this. I
>> totally agree with Gordon Kane about this. For me, this would mean
>> that
>> an electron viewed from a distance would even have a very tiny mix of
>> virtual top quarks to its total "make up". Discuss.
>
> IMHO, I don't think Kane understands advanced QFT.
You must be thinking of a different Gordon Kane. ;-)
http://www.physics.lsa.umich.edu/department/directory/bio.asp?ID=164
I would suspect that he does understand advanced QFT.
> You asked a similar thing on s.p.r. some time ago. I attempted an
> answer then, but the thread didn't continue on very far. I append
> a copy of that answer below, for convenience of s.p.f. readers:
>
> ---- Previous s.p.r. posting:
> -----------------------------------------
>
> I wrote:
>
> >> The "bare electron + virtual cloud" picture
> >> is a fictional artefact of tricky QFT math.
> >> There is no such thing as a "bare electron"
> >> in nature.
>
> FrediFizzx asked:
>
> > So are you saying that the "virtual cloud" is
> > the electron?
>
> No, of course I'm not saying that.
>
> > If not, what is the electron then?
Sorry, I realized my question was a bit messed up from what you said
after posting it so I didn't pursue this on SPR. Even in my viewpoint
there is no such thing as a bare electron. As soon as you remove it
from the virtual pair field, it would no longer be an electron or any
other "particle" that we know of. Of course, this is just my current
humble viewpoint.
> I'll probably just make it worse, but I'll try once...
>
> A "bare" electron corresponds to an eigenstate of
> a free Hamiltonian. Together, all these
> eigenstates span a Hilbert space (aka Fock space)
> representing arbitrary numbers of free (i.e:
> non-interacting) electrons. This Hilbert space is
> unphysical. To get a physically relevant Hilbert
> space, one must include an interaction term in the
> Hamiltonian. Physical electrons correspond to
> eigenstates of this interacting Hamiltonian.
I think I and my model can agree with that.
> It used to be tacitly assumed that the basis of
> free eigenstates could be used to expand the
> interacting eigenstates. I.e: it was assumed that
> the free Hilbert space coincided with the
> interacting Hilbert space.
>
> But people later realized this was an invalid
> assumption: The interacting Hilbert space is
> disjoint from the free one.
I in my model, there is no such thing as the "free" one.
> The notion of "bare electron + virtual cloud of
> photons" is just a picture one gets when one tries
> to express a mapping between free and interacting
> Hilbert spaces in terms of operators from the free
> Hilbert space.
OK, but I don't see how this gets rid of virtual pairs even though we
can do without a bare electron. And it is not virtual photons I am
thinking of. It is virtual fermionic pairs.
> The only states here that could legitimately be
> considered physical are the eigenstates of the
> interacting Hamiltonian. Unfortunately, no one
> knows how to construct them non-perturbatively,
> in general.
Well, most HEP particle physicists don't seem to care. Most of them
like Gordon Kane, do take virtual particles to be real and make
successful calculations using them I would suppose.
How would you explain the Lamb Shift without using them? This is a real
measurable effect that I have never seen explained any other way yet.
Best,
Fred Diether
eugene_st...@usa.net
> > "Virtual particles are indeed real -- they have observable effects that
> > physicists have devised ways of measuring. Quantum theory predicts that
> > every particle spends some time as a combination of other particles in
> > all possible ways."
> > He goes on to cite the Lamb shift as the first example of this.
I think one needs to clearly understand the difference between
experimental facts and certain mathematical models invented to explain
these facts. The Lamb shift is a fact. It is real. On the other hand,
nobody has seen virtual particles directly. They are a part of an
elaborate mathematical model (QED) devised to explain the Lamb shift
(among other things). So, it would be premature to say that virtual
particles are "real", "exist", etc. Experimental facts stay forever,
but theories and mathematical models come and go. Are you sure that
QED is the final word and there cannot be another theory/model which
explains experiments just as well, or even better (e.g., without
annoying ultraviolet divergences)?
100 years ago everybody (except one guy) believed that aether is
"real". 2000 years ago everybody believed that epicycles are "real".
Don't fall in the same trap.
strangerep wrote:
> The only states here that could legitimately be
> considered physical are the eigenstates of the
> interacting Hamiltonian. Unfortunately, no one
> knows how to construct them non-perturbatively,
> in general.
This is correct, although I wanted to mention for the record that
there is a quite successful perturbative construction. It is called
the "dressed particle" approach. In this approach, already in the 2nd
perturbation order one obtains the Coulomb potential between
"physical" electrons (this accounts for 99% of the electron-electron
interactions). The 4th perturbation order is supposed to provide the
Lamb shift (by the way, without any participation of virtual particles
or "polarized vacuum"). However, explicit 4th order calculations have
not been performed yet.
A few relevant references:
O. W. Greenberg and S. S. Schweber, "Clothed particle operators in
simple models of quantum field theory", Nuovo Cim., 8 (1958), 378.
A. V. Shebeko and M. I. Shirokov, "Unitary transformations in quantum
field theory and bound states", http://www.arxiv.org/nucl-th/0102037.
E. V. Stefanovich, "Relativistic quantum dynamics" http://www.arxiv.org/physics/0504062
Eugene Stefanovich
FrediFizzx
news:5669q5F...@mid.individual.net...
>* FrediFizzx:
>> This was a question that appeared in the Mar. 2007 issue of
>> Scientific American by J. Fleming of Madison, Wis. It is in the "Ask
>> the Experts" section and answered by Gordon Kane in the affirmative.
>> He says;
>>
>> "Virtual particles are indeed real -- they have observable effects
>> that physicists have devised ways of measuring. Quantum theory
>> predicts that every particle spends some time as a combination of
>> other particles in all possible ways."
>>
>> He goes on to cite the Lamb shift as the first example of this. I
>> totally agree with Gordon Kane about this. For me, this would mean
>> that an electron viewed from a distance would even have a very tiny
>> mix of virtual top quarks to its total "make up". Discuss.
>
> Good news for us non-physicists who tend to write incorrect statements
> in public Usenet groups -- whatever we write, it will turn out to be
> correct after all in someone's interpretation of quantum mechanics.
> To wit, in a discussion of how black holes can possibly form at all in
> finite outside time, I once referred to an electron (!) merrily
> worming its way into the dark hole, but being split into individual
> quarks by the infinite gravitational field gradient. Hah!
;-) Perhaps "split" is not the right term to use.
> More seriously, define "really", "real" and "exist".
>
> To me, something that has a discernible effect is real and really
> exists.
>
> Is your question whether virtual particles have any effect?
>
> They do. And according to the current consensus (is consensus
> scientific?) they can turn non-virtual very quickly. Like Hawking
> radiation.
Yes, I also believe they do.
> But I think the interesting question here is what frame of reference
> they choose, and why and how.
>
> For example, if the vacuum is really seething with virtual particles,
> as seems to be true, and if there's no preferred frame of reference,
> as also seems to be true, shouldn't those virtual particles have
> arbitrarily high kinetic energies as viewed from our frame of
> reference?
Not necessarily. Virtual pairs basically only exist where real matter
or real radiation exist wrt our frame of reference. While they are off
mass shell and can have any energy, in normal energy level situations
they don't have high kinetic energies. It is possible in HEP
accelerators to produce virtual particles with very high kinetic
energies. So the theory goes. Now, I am wondering if there is another
"mass shell" that virtual particles do have to conform to? Can they
really have arbitrarily high energies? Or is there a limit that we just
don't know about yet? I suspect that there is. And it might have to do
with mini quantum black holes being formed.
Best,
Fred Diether
FrediFizzx
news:1174281693.2...@e1g2000hsg.googlegroups.com...
> FrediFizzx wrote:
>
>> > "Virtual particles are indeed real -- they have observable effects
>> > that
>> > physicists have devised ways of measuring. Quantum theory predicts
>> > that
>> > every particle spends some time as a combination of other particles
>> > in
>> > all possible ways."
>
>> > He goes on to cite the Lamb shift as the first example of this.
>
> I think one needs to clearly understand the difference between
> experimental facts and certain mathematical models invented to explain
> these facts. The Lamb shift is a fact. It is real. On the other hand,
> nobody has seen virtual particles directly. They are a part of an
> elaborate mathematical model (QED) devised to explain the Lamb shift
> (among other things). So, it would be premature to say that virtual
> particles are "real", "exist", etc. Experimental facts stay forever,
> but theories and mathematical models come and go. Are you sure that
> QED is the final word and there cannot be another theory/model which
> explains experiments just as well, or even better (e.g., without
> annoying ultraviolet divergences)?
No I am not sure but I have never seen the Lamb Shift explained any
other way yet. So I am just going to proceed with my model assuming the
obvious from what QED and QFT might be telling us unless something
better comes along. And sorry... so far I can't buy into your
instantaneous interactions if they are over an appreciable distance. No
one has actually seen a real electron either. Only what it does when
they interact with other particles. It is exactly the same thing with
virtual particles. We can only tell about them via interactions. Of
course it is more clear with real charged particles as they leave
tracks. But the track is not really the electron.
> 100 years ago everybody (except one guy) believed that aether is
> "real". 2000 years ago everybody believed that epicycles are "real".
> Don't fall in the same trap.
Well, I am not done yet with the quantum "vacuum" as a relativistic
medium which Volovik describes as the "new ether". My Quantum Vacuum
Charge concept depends highly on it. ;-) I am really just proceeding
down a path that is already partially laid out in front of us. Take a
chance; see where it goes. If it is a trap, hopefully not too hard of
one to get out of. ;-)
> strangerep wrote:
>
>> The only states here that could legitimately be
>> considered physical are the eigenstates of the
>> interacting Hamiltonian. Unfortunately, no one
>> knows how to construct them non-perturbatively,
>> in general.
>
> This is correct, although I wanted to mention for the record that
> there is a quite successful perturbative construction. It is called
> the "dressed particle" approach.
What is the difference between a "dressed particle" and one in the
virtual pairs viewpoint? What is it "dressed" with?
Best,
Fred Diether
Oh No
Quantum superpositions are better described as things which have a
probability of being, rather than things which actually are. There is in
fact a very small probability for the transient existence of virtual
hadrons in the vicinity of an electron.
Regards
--
Charles Francis
moderator sci.physics.foundations.
substitute charles for NotI to email
Oh No
>It used to be tacitly assumed that the basis of free eigenstates could
>be used to expand the interacting eigenstates. I.e: it was assumed that
>the free Hilbert space coincided with the
>interacting Hilbert space.
>
>But people later realized this was an invalid assumption: The
>interacting Hilbert space is disjoint from the free one.
>
>The notion of "bare electron + virtual cloud of photons" is just a
>picture one gets when one tries to express a mapping between free and
>interacting Hilbert spaces in terms of operators from the free Hilbert
>space.
>
>The only states here that could legitimately be considered physical are
>the eigenstates of the interacting Hamiltonian. Unfortunately, no one
>knows how to construct them non-perturbatively, in general.
>
between particles are permitted. An electron may emit/absorb a photon,
but it may not emit/absorb two photons within a very short time interval
of each other. This constraint is sufficient to remove both the
ultraviolet divergence and the Landau Pole. When this is done, Maxwell's
equations can be derived showing that the values of the "dressed" and
"bare" mass and charge are unchanged by virtual particles.
Oh No
>"sr" <stran...@yahoo.com.au> wrote in message news:1174272439.015030.
>145...@e65g2000hsc.googlegroups.com...
>>
>>
>> IMHO, I don't think Kane understands advanced QFT.
>
>You must be thinking of a different Gordon Kane. ;-)
>
>http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=FIND+A+KANE%2CG
>ORDON&FORMAT=www&SEQUENCE=ds%28d%29
>
>http://www.physics.lsa.umich.edu/department/directory/bio.asp?ID=164
>
>I would suspect that he does understand advanced QFT.
I would think that so long as no one understands QM, it is not possible
that anyone understands qft.
Robert
Right. On Mondays the proton pretends to be a neutrino, on Tuesdays
it's an electron, Wednesday is cross-dressing day so it can be a
neutral kaon or a pi meson, on Thursdays it's a quark but it can
change colors although no one is allowed to see it anyway, on Fridays
it's a high-flying photon, on Saturdays it's Higgs bozo and on Sundays
it takes a rest as a proton. Phew, talk about a hard-working particle!
My question is: are virtual physicists really real and, if so, do they
have the slightest clue?
With apologies,
Rob
www.amherst.edu/~rloldershaw
======================================= MODERATOR'S COMMENT:
Perhaps it shouldn't be approved, but I will class it as humour.
CarlBrannen
> FrediFizzx wrote:
> IMHO, I don't think Kane understands advanced QFT.
Gordon Kane has 28 of articles in Citebase.org,
the most cited with 146 citations. If he's ignorant,
then so are most of his peers.
> You asked a similar thing on s.p.r. some time ago. I attempted an
> answer then ...
There is a strong compulsion to spread the word when one is
in possession of a truth that other deny, but this whole subject
of the "reality of virtual particles" is a waste of time. QM
is a practical technique for calculation, it is not a description
of reality. The mathematical objects that appear in the
calculations may, or may not, be a fundamental part of
nature, no one really knows.
> >> The "bare electron + virtual cloud" picture
> >> is a fictional artefact of tricky QFT math.
> >> There is no such thing as a "bare electron"
> >> in nature.
The comparison that comes to mind is Mach's
insistence that atoms were only convenient mathematical
fictions convenient in chemical calculations. No one
had ever seen one in nature. You could say the
same thing about quarks, on the mass shell or off.
Sometimes mathematical fictions come to life,
sometimes they do not.
> The notion of "bare electron + virtual cloud of
> photons" is just a picture one gets when one tries
> to express a mapping between free and interacting
> Hilbert spaces in terms of operators from the free
> Hilbert space.
This is not a proof that virtual particles do not exist.
As Einstein(?) said, it would be enough if we could
understand just the electron. That we do not have
an understanding of the electron is clear from such
facts that its mass and mixing angles are arbitrary
parameters of the standard model.
> The only states here that could legitimately be
> considered physical are the eigenstates of the
> interacting Hamiltonian.
Everyone knows that the standard model is just
a low energy approximation of some unknown
higher energy theory. If you reject bare and
virtual particles because they are just approximations,
you need to apply this severity to the interaction
Hamiltonian stuff as well.
> Unfortunately, no one
> knows how to construct them non-perturbatively,
> in general.
I think that this is the central problem of QM,
in particular with respect to the color force.
I should write another post describing the
attributes of what the theory that does this
needs.
Carl
Oh No
>There is in fact a very small probability for the transient existence
>of virtual hadrons in the vicinity of an electron.
I should have been less precise, and talked about the possibility or
potential for this transient existence, or I should have said
probability amplitude rather than probability. Virtual particles do not
in general lead to meaningful probabilities. I do not myself think that
the lack of a meaningful probability is an argument that they are not
real.
Ken S. Tucker
> Thus spake sr <strange...@yahoo.com.au>
>
> >It used to be tacitly assumed that the basis of free eigenstates could
> >be used to expand the interacting eigenstates. I.e: it was assumed that
> >the free Hilbert space coincided with the
> >interacting Hilbert space.
>
> >But people later realized this was an invalid assumption: The
> >interacting Hilbert space is disjoint from the free one.
>
> >The notion of "bare electron + virtual cloud of photons" is just a
> >picture one gets when one tries to express a mapping between free and
> >interacting Hilbert spaces in terms of operators from the free Hilbert
> >space.
>
> >The only states here that could legitimately be considered physical are
> >the eigenstates of the interacting Hamiltonian. Unfortunately, no one
> >knows how to construct them non-perturbatively, in general.
>
> They can be constructed non-perturbatively if only discrete interactions
> between particles are permitted. An electron may emit/absorb a photon,
It's not the actual electron that emits and absorbs
energy, it's the atomic system. The so-called electron
orbital change involves the nucleus, in that relation.
The best I know is, an atom stores more or less
electrical potential energy in the relation of a e- to and p+,
basically a (-) and (+).
That energy is thought to be incremented and decremented
by photons.
I knee-jerk when I read an "electron emits a photon"
because in the rest frame of the electron the energy
of the e- is invariant. In that FoR the mass and
energy of the e- are presumed to be constant.
Regards
Ken
eugene_st...@usa.net
> No
> one has actually seen a real electron either. Only what it does when
> they interact with other particles. It is exactly the same thing with
> virtual particles. We can only tell about them via interactions. Of
> course it is more clear with real charged particles as they leave
> tracks. But the track is not really the electron.
This is a good point. Experimentalists never see "particles". They see
only movements of voltmeter arms, lines on the oscillograph screen,
and tracks in bubble chambers. However, it would be silly to think
that reality consists of such "movements", "lines", and "tracks". We
definitely need to assume the existence of some deeper level of
reality. The question is: how deep do we want to go? My idea is that
simply assuming the existence of "empty" vacuum, physical particles
and their interactions is sufficient for building a satisfactory
theory. "Polarizable" vacuum, "virtual" particles, strings,
superpartners, and other "deep" stuff is just an excessive baggage, in
my opinion.
> What is the difference between a "dressed particle" and one in the
> virtual pairs viewpoint? What is it "dressed" with?
I agree that "dressed particle" is not a good expression. It suggests
the existence of a virtual cloud around the particle, and I don't want
to make this suggestion. I use the words "clothed particles" or
"dressed particles" mainly for historical reasons: these names were
used for the last 50 years. I would prefer to call them "physical
particles".
Eugene.
eugene_st...@usa.net
> The comparison that comes to mind is Mach's
> insistence that atoms were only convenient mathematical
> fictions convenient in chemical calculations. No one
> had ever seen one in nature. You could say the
> same thing about quarks, on the mass shell or off.
> Sometimes mathematical fictions come to life,
> sometimes they do not.
You are right. For many centuries atoms were just "convenient
mathematical fictions". However, today we can isolate, see, and
manipulate individual atoms in experiments. So, it would be silly to
deny their "real" existence. I don't think we reached the same level
of confidence regarding quarks.
Eugene.
FrediFizzx
news:0GwaLaKLyl$FF...@charlesfrancis.wanadoo.co.uk...
> Thus spake FrediFizzx <fredi...@hotmail.com>
>>"sr" <stran...@yahoo.com.au> wrote in message
>>news:1174272439.015030.
>>145...@e65g2000hsc.googlegroups.com...
>>>
>>>
>>> IMHO, I don't think Kane understands advanced QFT.
>>
>>You must be thinking of a different Gordon Kane. ;-)
>>
>>http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=FIND+A+KANE%2CG
>>ORDON&FORMAT=www&SEQUENCE=ds%28d%29
>>
>>http://www.physics.lsa.umich.edu/department/directory/bio.asp?ID=164
>>
>>I would suspect that he does understand advanced QFT.
>
> I would think that so long as no one understands QM, it is not
> possible
> that anyone understands qft.
;-) Yeah, I said that wrong. I suspect he understands advanced QFT as
well as anyone else does. I was actually a bit surprised that he said
what he said about virtual particles being real as he never really
mentions them at all in his particle physics textbook. It seems the
assumption from the beginning is that they are real.
Best,
Fred Diether
FrediFizzx
In my viewpoint, I have to include all the virtual pairs that are in the
vicinity of a fermion as part of the description of that fermion. IOW,
the Higgs-like field environment is what really defines what a fermion
is. All fermions are confined in my viewpoint with neutrinos being the
least confined and quarks being the most confined. Neutrinos are the
closest we will ever get to the "bare" common entity that makes all
fermions. This scenario is just what seems to be a natural conclusion
of taking the quantum "vacuum" to be a relativistic medium of virtual
and "less than virtual" fermionic pairs. I could be wrong, but there
are clues that point in this direction so I will keep thinking about it.
Best,
Fred Diether
Oh No
>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message news:D2OXn
>BIwjl$FF...@charlesfrancis.wanadoo.co.uk...
>> probability of being, rather than things which actually are. There is
>>in
>> fact a very small probability for the transient existence of virtual
>> hadrons in the vicinity of an electron.
>
>In my viewpoint, I have to include all the virtual pairs that are in
>the vicinity of a fermion as part of the description of that fermion.
I work the other way, as is natural for Fock space. I construct the
model from isolated "bare" particles. When the particles are given the
possibility of interaction, the possibility of virtual particles is
automatic.
>IOW, the Higgs-like field environment is what really defines what a
>fermion is.
I don't like to see things in terms of Higgs type fields. IMV that can
only be a mathematical approximation to the true underlying situation.
> All fermions are confined in my viewpoint with neutrinos being the
>least confined and quarks being the most confined. Neutrinos are the
>closest we will ever get to the "bare" common entity that makes all
>fermions.
I don't see what is wrong with the particles themselves being
fundamental. That is what the equations say, imv.
>Fred Diether
Oh No
My feeling is we have. The resonances found in high energy experiments
seem to work perfectly in the quark model, as do transitions and jets. I
don't have any belief in gluons however.
Oh No
>On Mar 19, 1:23 am, "FrediFizzx" <fredifi...@hotmail.com> wrote:
>
>> No
>> one has actually seen a real electron either. Only what it does when
>> they interact with other particles. It is exactly the same thing with
>> virtual particles. We can only tell about them via interactions. Of
>> course it is more clear with real charged particles as they leave
>> tracks. But the track is not really the electron.
>
>
>This is a good point. Experimentalists never see "particles". They see
>only movements of voltmeter arms, lines on the oscillograph screen,
>and tracks in bubble chambers. However, it would be silly to think
>that reality consists of such "movements", "lines", and "tracks". We
>definitely need to assume the existence of some deeper level of
>reality. The question is: how deep do we want to go? My idea is that
>simply assuming the existence of "empty" vacuum, physical particles
>and their interactions is sufficient for building a satisfactory
>theory. "Polarizable" vacuum, "virtual" particles, strings,
>superpartners, and other "deep" stuff is just an excessive baggage, in
>my opinion.
I agree up to "virtual" particles. The problem here is with the word
"virtual". The theory makes complete sense to me if the "vacuum" is
really "void", a complete absence of properties and hence non-physical,
but particles are real, and have the same real properties whether we are
looking at them or not.
Oh No
>On Mar 19, 2:45 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake sr <strange...@yahoo.com.au>
>>
>>
>> >the eigenstates of the interacting Hamiltonian. Unfortunately, no one
>> >knows how to construct them non-perturbatively, in general.
>>
>> They can be constructed non-perturbatively if only discrete interactions
>> between particles are permitted. An electron may emit/absorb a photon,
>
>It's not the actual electron that emits and absorbs
>energy, it's the atomic system. The so-called electron
>orbital change involves the nucleus, in that relation.
>The best I know is, an atom stores more or less
>electrical potential energy in the relation of a e- to and p+,
>basically a (-) and (+).
In qed we only directly model electrons and photons. This is
particularly appropriate for isolated electrons and in scattering
experiments with electrons and positrons. Not an atom in sight. We
assume that similar behaviour is true of all charged particles, and that
an electron's behaviour when bound in an atom is fundamentally the same
in an atom as it is in isolation. While the energy for the emission of a
photon comes from the relationship between electron and nucleus, it is
the electron which emits the photon.
>That energy is thought to be incremented and decremented
>by photons.
>I knee-jerk when I read an "electron emits a photon"
>because in the rest frame of the electron the energy
>of the e- is invariant. In that FoR the mass and
>energy of the e- are presumed to be constant.
>Regards
>Ken
>
There isn't such a thing as the rest frame of the electron. If there
was, it wouldn't be an inertial frame since momentum is conserved when
the photon is emitted. The electron has to draw energy, e.g. from a
coulomb field, in order to emit a photon.
FrediFizzx
Hmm... I think I would have to disagree with that characterization
somewhat. The photon is created from the "vacuum" by the energy that
the dipole system gives up. A free electron can never absorb or emit a
photon. It is always at least a dipole situation for photon absorption
or emission. In QED, the nucleus is simply modeled as a point like
heavy elemental charge.
>>That energy is thought to be incremented and decremented
>>by photons.
>>I knee-jerk when I read an "electron emits a photon"
>>because in the rest frame of the electron the energy
>>of the e- is invariant. In that FoR the mass and
>>energy of the e- are presumed to be constant.
>>Regards
>>Ken
>>
> There isn't such a thing as the rest frame of the electron. If there
> was, it wouldn't be an inertial frame since momentum is conserved when
> the photon is emitted. The electron has to draw energy, e.g. from a
> coulomb field, in order to emit a photon.
I think Ken is right here. Electrons really don't emit or absorb
photons. In every case it takes at least a dipole situation. Photons
are created from and destroyed to the "vacuum" and the energy-momentum
is transferred. And in the case of an atom, most of the energy-momentum
gets transferred to or from the lightest particle. In electron -
electron scattering the energy-momentum is transferred equally to or
from both. Electrons emit or absorb the energy-momentum creating or
destroying photons from / to the "vacuum".
Best,
Fred Diether
FrediFizzx
news:1174324998.6...@l75g2000hse.googlegroups.com...
> On Mar 19, 1:23 am, "FrediFizzx" <fredifi...@hotmail.com> wrote:
>
>> No
>> one has actually seen a real electron either. Only what it does when
>> they interact with other particles. It is exactly the same thing
>> with
>> virtual particles. We can only tell about them via interactions. Of
>> course it is more clear with real charged particles as they leave
>> tracks. But the track is not really the electron.
>
>
> This is a good point. Experimentalists never see "particles". They see
> only movements of voltmeter arms, lines on the oscillograph screen,
> and tracks in bubble chambers. However, it would be silly to think
> that reality consists of such "movements", "lines", and "tracks". We
> definitely need to assume the existence of some deeper level of
> reality. The question is: how deep do we want to go? My idea is that
> simply assuming the existence of "empty" vacuum, physical particles
> and their interactions is sufficient for building a satisfactory
> theory. "Polarizable" vacuum, "virtual" particles, strings,
> superpartners, and other "deep" stuff is just an excessive baggage, in
> my opinion.
You could be right. However, you end up with instantaneous interactions
at a distance without some of the "excess baggage". ;-) Mainly
"polarizable vacuum" and "virtual particles". But virtual particles
being off mass shell does imply a kind of instantaneous interaction but
perhaps not at a distance. I would say more "superluminal" rather than
instantaneous. But could only be for very tiny "distances". Orthodox
superstring theory and SUSY do have some features I like but I do think
they are missing the mark. I like the atomism concept of string theory
and the "other world" concept that SUSY implies. I don't think fermions
transform to bosons or vise-versa. Of course in my relativistic medium
concept, there are only fermions and all gauge bosons are "wavicles" of
the fermionic medium. Fermions transform to fermions when going from
our world to the other. The other world just being the "Sea" that is
beyond our event horizon. Yeah, ya have to add a bunch more baggage to
make it all work. ;-) But it is fun to think about it.
>> What is the difference between a "dressed particle" and one in the
>> virtual pairs viewpoint? What is it "dressed" with?
>
>
> I agree that "dressed particle" is not a good expression. It suggests
> the existence of a virtual cloud around the particle, and I don't want
> to make this suggestion. I use the words "clothed particles" or
> "dressed particles" mainly for historical reasons: these names were
> used for the last 50 years. I would prefer to call them "physical
> particles".
Well, what I wanted to know is how are your "physical particles"
described compared to the "normal" viewpoint? Do they have extended
structure? Or what?
Best,
Fred Diether
sr
>> IMHO, I don't think Kane understands advanced QFT.
Fred Diether wrote:
> [...]
> I would suspect that he does understand advanced QFT.
I regretted writing that a few minutes after posting it.
(This is probably one of those instances where spf
moderators should probably reject the post for being
impolite to a 3rd party - that would have forced me to
express myself better, and less offensively.)
What I meant was that I suspected Kane did not have deep
familiarity with inequivalent representations, Haag's theorem,
and related matters that normally don't trouble practical
physicists in the slightest. My former QFT lecturer clearly
had a good understanding of practical QFT and calculations, etc,
and talked a lot about the vacuum seething with virtual particles.
But he was less able to talk at length about the things I just
mentioned, except to dismiss them (too glibly, I thought). Even
Weinberg does not mention them in his weighty tomes.
>> The notion of "bare electron + virtual cloud of
>> photons" is just a picture one gets when one tries
>> to express a mapping between free and interacting
>> Hilbert spaces in terms of operators from the free
>> Hilbert space.
> OK, but I don't see how this gets rid of virtual pairs
> even though we can do without a bare electron.
To express the mapping, one uses a formally unitary
transformation which involves an infinite number
of particles.
> And it is not virtual photons I am thinking of.
> It is virtual fermionic pairs.
These are just different variants of the same basic issue.
>> The only states here that could legitimately be
>> considered physical are the eigenstates of the
>> interacting Hamiltonian. Unfortunately, no one
>> knows how to construct them non-perturbatively,
>> in general.
> Well, most HEP particle physicists don't seem to care.
Agreed.
> Most of them like Gordon Kane, do take virtual particles
> to be real and make successful calculations using them
> I would suppose.
They don't "use" virtual particles to make the successful
calculations. They use perturbation series in QFT. Virtual
particles are just a way of picturing or interpreting the
various terms that arise in the series.
> How would you explain the Lamb Shift without using them?
> This is a real measurable effect that I have never seen
> explained any other way yet.
Same answer as in previous paragraph above, though I guess
it depends on what you mean by "explain". I'm guessing maybe
you meant "layman-accessible explanation"?
My (far less accessible) explanation is that accurate predictions
require full QED. Simply solving the Dirac eqn is not good enough.
- strangerep
eugene_st...@usa.net
> However, you end up with instantaneous interactions
> at a distance without some of the "excess baggage".
I know well how people dislike instantaneous action-at-a-distance.
Even Newton, who invented the first instantaneous potential, regarded
those who follow this idea too literally as ... mentally deprived, or
something of that sort.
On the other hand, we still don't have a clear experimental
measurement of the speed of propagation of Coulomb forces or
gravitational forces. The strongest "argument" against instantaneous
interactions usually comes from the special relativistic ban on
superluminal signal propagation. However, I am ready to prove that the
ban doesn't apply in this particular case. Though, this can be a topic
for a separate thread.
> Well, what I wanted to know is how are your "physical particles"
> described compared to the "normal" viewpoint? Do they have extended
> structure? Or what?
Stable "physical particles" are described by irreducible unitary
representations of the Poincare group. So, they have two intrinsic
parameters - mass and spin. If a particle participates in
interactions, then one can add to this list certain "charges".
Moreover, one can clearly define such measurable properties of a
particle as position, momentum, angular momentum, energy.
However, I wouldn't say that elementary particles, like electron, have
any structure, extended or otherwise. Even if they had a "structure"
we wouldn't be able to see it. "Never ask about something you cannot
observe".
Eugene.
FrediFizzx
> Thus spake FrediFizzx <fredi...@hotmail.com>
>>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
>>news:D2OXn
>>BIwjl$FF...@charlesfrancis.wanadoo.co.uk...
>>> Quantum superpositions are better described as things which have a
>>> probability of being, rather than things which actually are. There
>>> is
>>>in
>>> fact a very small probability for the transient existence of virtual
>>> hadrons in the vicinity of an electron.
>>
>>In my viewpoint, I have to include all the virtual pairs that are in
>>the vicinity of a fermion as part of the description of that fermion.
>
> I work the other way, as is natural for Fock space. I construct the
> model from isolated "bare" particles. When the particles are given the
> possibility of interaction, the possibility of virtual particles is
> automatic.
I can't do that in my model. What you might term an isolated bare
particle can't happen. Ever. Particles are always interacting with
"vacuum" virtual particles and they with "less than virtual" particles
in the "other world". And "real" particles of course interact with each
other via virtual particles.
>>IOW, the Higgs-like field environment is what really defines what a
>>fermion is.
>
> I don't like to see things in terms of Higgs type fields. IMV that can
> only be a mathematical approximation to the true underlying situation.
In my model, the geometry of interactions is so freakin' complicated I
will take a simplified approximation. ;-)
>> All fermions are confined in my viewpoint with neutrinos being the
>>least confined and quarks being the most confined. Neutrinos are the
>>closest we will ever get to the "bare" common entity that makes all
>>fermions.
>
> I don't see what is wrong with the particles themselves being
> fundamental. That is what the equations say, imv.
Well, nature could very well be that way, but I think it is too early to
tell for certain. I am just leaning more towards a string theory type
of atomism concept with a relativistic medium "twist" on it. IOW, ya
need the "particles" in the medium all interacting with each other to
get string loops by confinement. I actually imagine they are more like
branes than strings. But a "static" representation could be a double
looped string for fermions. Another twist is that there is a
circulation within the string that can go either direction. And string
tension is not really a consideration. So really quite a bit different
than orthodox superstring theory. And there is quite a bit more
"baggage" to really make this all work out. Massless point-like
entities constrained to "move" in a circular-like trajectory. It's a
string loop because its proper time is zero everywhere. In fact, time
is essentially meaningless at this level. Even stranger; space is
really also meaningless as this is where it is being defined in the
first place. A point-like entity is also "puffed-out". I think it
could be the source of what people regard as some of the quantum
mysteries. Well, some of my highly speculative thoughts here. ;-)
Your solution is really quite a bit more simple and elegant but I just
get the feeling that nature didn't give us all the different flavors of
particles as fundamental because there are clues in HEP experiments that
point to something more fundamental. Well, even the Dirac equation
itself seems to work for all fermions. We just have to plug in
different masses and other quantum numbers manually. It would be nice
if we found a natural generator for the masses and quantum numbers which
I think is only possible by going to a more fundamental picture of
atomism.
Best,
Fred Diether
Oh No
>Stable "physical particles" are described by irreducible unitary
>representations of the Poincare group. So, they have two intrinsic
>parameters - mass and spin. If a particle participates in interactions,
>then one can add to this list certain "charges". Moreover, one can
>clearly define such measurable properties of a particle as position,
>momentum, angular momentum, energy.
It is right to distinguish the former properties, mass, spin, and
charge, from the latter position, momentum, angular momentum, energy.
The former are properties which the particle has on its own, the latter
are properties which only exist in measurement. Strictly they are not
properties of a particle, but of the relationship a particle has, or can
have, with other matter.
Oh No
>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
>In my model, the geometry of interactions is so freakin' complicated I
>will take a simplified approximation. ;-)
I think the underlying structure should be simple. In qed the underlying
structure is extremely simple. Only two particles, electrons and
photons, only three properties, mass, spin and charge, and only one
interaction, an electron absorbs or emits a photon. The way in which
interactions build up to create all the structures we observe, such as
the structure of space time, is complex, but only because there are many
ways of building structures from the basic elements.
>
>>> All fermions are confined in my viewpoint with neutrinos being the
>>>least confined and quarks being the most confined. Neutrinos are the
>>>closest we will ever get to the "bare" common entity that makes all
>>>fermions.
>>
>> I don't see what is wrong with the particles themselves being
>> fundamental. That is what the equations say, imv.
>
>Well, nature could very well be that way, but I think it is too early to
>tell for certain.
The original argument for atomism, to avoid the paradoxes of Zeno, was
as true over two thousand years ago as it is today. Where we have
advanced is in creating formal mathematics and in having empirical
evidence so that we can avoid embellishing that basic idea with false
principles.
> I am just leaning more towards a string theory type
>of atomism concept with a relativistic medium "twist" on it. IOW, ya
>need the "particles" in the medium all interacting with each other to
>get string loops by confinement. I actually imagine they are more like
>branes than strings. But a "static" representation could be a double
>looped string for fermions. Another twist is that there is a
>circulation within the string that can go either direction. And string
>tension is not really a consideration. So really quite a bit different
>than orthodox superstring theory. And there is quite a bit more
>"baggage" to really make this all work out. Massless point-like
>entities constrained to "move" in a circular-like trajectory. It's a
>string loop because its proper time is zero everywhere. In fact, time
>is essentially meaningless at this level. Even stranger; space is
>really also meaningless as this is where it is being defined in the
>first place. A point-like entity is also "puffed-out". I think it
>could be the source of what people regard as some of the quantum
>mysteries. Well, some of my highly speculative thoughts here. ;-)
I find such speculations unnecessary if physics can be modelled without.
>Your solution is really quite a bit more simple and elegant but I just
>get the feeling that nature didn't give us all the different flavors of
>particles as fundamental because there are clues in HEP experiments
>that point to something more fundamental. Well, even the Dirac
>equation itself seems to work for all fermions. We just have to plug
>in different masses and other quantum numbers manually. It would be
>nice if we found a natural generator for the masses and quantum numbers
>which I think is only possible by going to a more fundamental picture
>of atomism.
It would indeed be nice not to have to plug in the numbers by hand. We
only have two numbers to plug in, mass and charge. If we are to be able
to show that actually these numbers are constrained and do not have to
be plugged in by hand, the first task must be to complete the details of
the mathematical structure which we already have. I don't put it past
possibility that one day it will be possible to show how the values we
observe for mass and charge arise, but to try and do so now, without a
complete mathematical structure, seems to me to be getting ahead of
ourselves. A problem which I think we must solve first, and which, from
things found about spin networks it appears is probably soluble, is to
see why we have three dimensions. I would not be at all surprised if the
existence of only charges of integral multiples of 1/3 is intrinsically
related to the fact of three dimensions. Perhaps when we have that
problem solved we will be ready to address the question of fundamental
masses. Not before, imv.
Oh No
>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message news:Y59hU
>Nc87v$FF...@charlesfrancis.wanadoo.co.uk...
>> Thus spake Ken S. Tucker <dyna...@vianet.on.ca>
>>>On Mar 19, 2:45 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> In qed we only directly model electrons and photons. This is
>> particularly appropriate for isolated electrons and in scattering
>> experiments with electrons and positrons. Not an atom in sight. We
>> assume that similar behaviour is true of all charged particles, and
>>that
>> an electron's behaviour when bound in an atom is fundamentally the
>>same
>> in an atom as it is in isolation. While the energy for the emission
>>of a
>> photon comes from the relationship between electron and nucleus, it is
>> the electron which emits the photon.
>
>Hmm... I think I would have to disagree with that characterization
>somewhat. The photon is created from the "vacuum" by the energy that
>the dipole system gives up. A free electron can never absorb or emit a
>photon. It is always at least a dipole situation for photon absorption
>or emission.
>
>I think Ken is right here. Electrons really don't emit or absorb
>photons. In every case it takes at least a dipole situation. Photons
>are created from and destroyed to the "vacuum" and the energy-momentum
>is transferred. And in the case of an atom, most of the energy-
>momentum gets transferred to or from the lightest particle. In
>electron - electron scattering the energy-momentum is transferred
>equally to or from both. Electrons emit or absorb the energy-momentum
>creating or destroying photons from / to the "vacuum".
The mathematical description is quite clear, imv. The photon field
operator creates or annihilates a photon on Fock space. The electron
field operator creates a positron or annihilates and electron, the
adjoint creates an electron or annihilates a positron. The interaction
operator is a simple statement that an electron emits or absorbs a
photon. The locality condition states that this happens at a point for
sizeless particles. I find this absolutely clear and unambiguous in the
mathematical language of qed, which is the language of quantum theory,
i.e. quantum logic.
When an isolated electron emits a photon, the photon must be reabsorbed.
Since the electron is isolated, the photon must be reabsorbed by the
emitting electron - that is a definitional truism, or the electron would
not be isolated. This is represented as a loop integral. Normally that
is thought to diverge. Actually it does not diverge. If one does not
piss all over basic mathematics which one should learn as a first year
undergraduate, and does the calculation properly this loop integral is
zero. One can describe that one of two ways. Either one can say that the
process is forbidden, and that there is no virtual cloud of photons
surrounding the electron, or one can say that the electron is surrounded
by virtual photons but that this has absolutely no effect on the theory,
a "dressed" electron is thus identical to a "bare" electron.
The interaction conserves momentum, and it conserves observed values of
energy. For an electron to emit a photon which is not reeabsorbed,
conservation of energy/momentum requires that the electron absorbs a
photon from somewhere else. That may be from the nucleus of an atom, or
it may be from other electrons. In either case the electron cannot be
said to be isolated. There is no interaction with a "vacuum". That idea
has as much validity as the ether in special relativity.
Mirror Space
> Thus spake eugene_stefanov...@usa.net
Hi Everybody!
It seems funny to me, but nobody mentioned the most important issue in
all this
"virtual particles" story.
If one to consider, say e-Nucleon --> e-Nucleon pion (just an example)
reaction
with virtual photon, or similar process with virtual W, Z - bozons -
there is no
principal difference for that matter.
"Virtual" just measure a level of off-massshellness for gamma, or W in
given
process. If kinematics will be chosen that given q^ will reach 0, the
particle
automatically will became a real. End of story.
Andy Inopin
Mirror Space
>
> - Показать цитируемый текст -
OOps, I'm sorry! It was meant of course q^2 and not q^ as I
misprint it.
Andy Inopin
maxwell
> On Mar 18, 9:45 pm, "FrediFizzx" <fredifi...@hotmail.com> wrote:
>
>
>
> > This was a question that appeared in the Mar. 2007 issue of Scientific
> > American by J. Fleming of Madison, Wis. It is in the "Ask the Experts"
> > section and answered by Gordon Kane in the affirmative. He says;
>
> > "Virtual particles are indeed real -- they have observable effects that
> > physicists have devised ways of measuring. Quantum theory predicts that
> > every particle spends some time as a combination of other particles in
> > all possible ways."
not one-to-one with reality - they are just intermediate steps in the
sequential logic flow. Do you think arbitrary decompositions of a
vector into 2 orthogonal components are both "real"? First define
"reality" & then work outwards.
======================================= MODERATOR'S COMMENT:
It is a major purpose of this group, and of this discussion, to disentangle from the mathematical theory what is an element of physical reality and what is not. If you can contribute to that discussion, please do. I will not generally accept posts which lecture to us on what we already know without making such a contribution.
Cl.Massé
5669q5F...@mid.individual.net
> To me, something that has a discernible effect is real and really exists.
>
> Is your question whether virtual particles have any effect?
>
> They do. And according to the current consensus (is consensus
> scientific?) they can turn non-virtual very quickly. Like Hawking
> radiation.
That's plain false. A particle turns real in first quantization as well,
but in first quantization there is no virtual particle.
> For example, if the vacuum is really seething with virtual particles, as
> seems to be true, and if there's no preferred frame of reference, as
> also seems to be true, shouldn't those virtual particles have
> arbitrarily high kinetic energies as viewed from our frame of reference?
Not alone can they have arbitrary high kinetic energies (up to perhaps the
Plank scale), but the total energy is infinite.
--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.
Cl.Massé
56689bF...@mid.individual.net
> This was a question that appeared in the Mar. 2007 issue of Scientific
> American by J. Fleming of Madison, Wis. It is in the "Ask the Experts"
> section and answered by Gordon Kane in the affirmative. He says;
>
> "Virtual particles are indeed real -- they have observable effects that
> physicists have devised ways of measuring. Quantum theory predicts that
> every particle spends some time as a combination of other particles in
> all possible ways."
>
> He goes on to cite the Lamb shift as the first example of this. I
> totally agree with Gordon Kane about this. For me, this would mean that
> an electron viewed from a distance would even have a very tiny mix of
> virtual top quarks to its total "make up". Discuss.
They are zero point energy that is unobservable. Of course, one may find a
plethora of scientific argument and endless discussions. But at the end of
the day, it is unobservable zero point energy. The Casimir effect is fully
explainable with the analogy of the classical Van der Waals force, without
even speaking about the vacuum energy. That's a case where even the
self-styled experts are wrong, since zero point energy is unobservable.
Every century has its fantasies, the last one was the existence of virtual
particle, the previous one was the possibility of the perpetual motion.
That one will perhaps be oranges that press themselves, or reduction of CO2
emission to cool down the planet.
eugene_st...@usa.net
> >Stable "physical particles" are described by irreducible unitary
> >representations of the Poincare group. So, they have two intrinsic
> >parameters - mass and spin. If a particle participates in interactions,
> >then one can add to this list certain "charges". Moreover, one can
> >clearly define such measurable properties of a particle as position,
> >momentum, angular momentum, energy.
>
> It is right to distinguish the former properties, mass, spin, and
> charge, from the latter position, momentum, angular momentum, energy.
> The former are properties which the particle has on its own, the latter
> are properties which only exist in measurement. Strictly they are not
> properties of a particle, but of the relationship a particle has, or can
> have, with other matter.
I agree about the distictinction of these two groups of properties.
Though I would formulate this distiction differently. I think that the
former properties do not depend on the particle's state: their
measurement yields the same value independent on the state of the
particle. The latter properties are state-dependent, i.e., there are
states with different position, momenta, etc.
Eugene.
Alf P. Steinbach
> "Alf P. Steinbach" <al...@start.no> a écrit dans le message de news:
> 5669q5F...@mid.individual.net
>
>> To me, something that has a discernible effect is real and really exists.
>>
>> Is your question whether virtual particles have any effect?
>>
>> They do. And according to the current consensus (is consensus
>> scientific?) they can turn non-virtual very quickly. Like Hawking
>> radiation.
>
> That's plain false.
Hawking radiation is well-known. It results from virtual particle pairs
forming right outside the event horizon of a black hole, one of each
pair falling into the hole, the other escaping as a real particle. The
mechanism is generally accepted but disputed[1], which is why I wrote
"consensus", but it's not plain false.
> A particle turns real in first quantization as well,
> but in first quantization there is no virtual particle.
Given that the previous assertion was just hogwash, this assertion is
probably also, but I'm not a physicist.
>> For example, if the vacuum is really seething with virtual particles, as
>> seems to be true, and if there's no preferred frame of reference, as
>> also seems to be true, shouldn't those virtual particles have
>> arbitrarily high kinetic energies as viewed from our frame of reference?
>
> Not alone can they have arbitrary high kinetic energies (up to perhaps the
> Plank scale), but the total energy is infinite.
Ditto for this assertion, sorry.
- Alf
Notes:
[1] <url: http://xxx.lanl.gov/abs/gr-qc/0304042/>.
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
eugene_st...@usa.net
> When an isolated electron emits a photon, the photon must be reabsorbed.
> Since the electron is isolated, the photon must be reabsorbed by the
> emitting electron - that is a definitional truism, or the electron would
> not be isolated. This is represented as a loop integral. Normally that
> is thought to diverge. Actually it does not diverge. If one does not
> piss all over basic mathematics which one should learn as a first year
> undergraduate, and does the calculation properly this loop integral is
> zero.
Why do you say so? The electron self-energy loop integral is infinite.
This is shown in every QFT textbook. Or I don't understand you
correctly?
Eugene.
Oh No
Every QFT textbook bar one, afaik. The one being Scharf, Finite QED.
> Or I don't understand you
>correctly?
You understand me correctly. It was fine for Dyson, Feynman, Schwinger
to do the calculations in a heuristic manner, which generated
divergences. It was even fine for them to deal with the divergences in
an ad hoc manner. It is not fine for succeeding generations to follow
the same arguments and to fail to straighten them out. As early as the
seventies it had been shown by Epstein and Glaser et al that the
ultraviolet divergences were to do with elementary errors in analysis,
the abuse of Wick's theorem, which effectively means the changing of
orders of integration where we do not have uniform convergence. I showed
the same thing in my own doctoral thesis, albeit a few years later.
The fact that this kind of abuse of mathematics will produce nonsensical
results has been known for about two hundred years and was well sorted
out by the latter half of the C19th. This is taught to first year maths
undergraduates. There was some excuse in the early days of qed, but for
modern text books like Peskin and Schroeder to produce the same stuff is
pure incompetence, imv. Based on this bad mathematics we have whole
reams of research into absolute nonsense, from the belief in the Higgs
boson, false arguments about renormalisability, the conceptual basis of
field theory, through to string theory.
Oh No
This is true. One might call the former absolute, fundamental or
elementary properties of the particle. The state describes what we know
of the particle, generally through measurement, and depends on the
particles relationship to other matter. Quantum weirdness as it applies
to states is a lot less weird when one recognises that quantum theory is
describing relationships and knowledge, rather than actual properties of
eugene_st...@usa.net
> You understand me correctly. It was fine for Dyson, Feynman, Schwinger
> to do the calculations in a heuristic manner, which generated
> divergences. It was even fine for them to deal with the divergences in
> an ad hoc manner. It is not fine for succeeding generations to follow
> the same arguments and to fail to straighten them out. As early as the
> seventies it had been shown by Epstein and Glaser et al that the
> ultraviolet divergences were to do with elementary errors in analysis,
> the abuse of Wick's theorem, which effectively means the changing of
> orders of integration where we do not have uniform convergence. I showed
> the same thing in my own doctoral thesis, albeit a few years later.
>
> The fact that this kind of abuse of mathematics will produce nonsensical
> results has been known for about two hundred years and was well sorted
> out by the latter half of the C19th. This is taught to first year maths
> undergraduates. There was some excuse in the early days of qed, but for
> modern text books like Peskin and Schroeder to produce the same stuff is
> pure incompetence, imv. Based on this bad mathematics we have whole
> reams of research into absolute nonsense, from the belief in the Higgs
> boson, false arguments about renormalisability, the conceptual basis of
> field theory, through to string theory.
I refuse to believe that Dyson, Feynman, Schwinger, Schweber, Bjorken,
Drell, Peskin, Schroeder, Weinberg, ... all made an elementary
mistake, and only Scharf got it right. Perhaps you can educate me (and
others) and show how the electron self-energy integral is calculated
by Scharf, so that zero is obtained rather than infinity? This is
rather simple integral, so, presumably, the idea can be explained
without complicated math.
Eugene.
FrediFizzx
>I wrote:
>
>>> IMHO, I don't think Kane understands advanced QFT.
>
> Fred Diether wrote:
>
>> [...]
>> I would suspect that he does understand advanced QFT.
>
> I regretted writing that a few minutes after posting it.
> (This is probably one of those instances where spf
> moderators should probably reject the post for being
> impolite to a 3rd party - that would have forced me to
> express myself better, and less offensively.)
We are new to doing moderation and we are letting some minor infractions
go through especially from quality posters such as yourself. We decided
that it might be better to discuss moderation policy on the group so as
to eventually form a set of rules that hopefully can be consistent and
that members of the group understand. We are more concerned with
harassing comments directed at the poster that one is replying to. For
sure, we will not tolerate that to the best of our abilities. But yes,
one should be careful to also be polite to third parties. I don't think
we will reject posts such as yours unless it was something more like
directly calling a third party stupid (or similar name calling). Now,
that is really impolite and offensive.
> What I meant was that I suspected Kane did not have deep
> familiarity with inequivalent representations, Haag's theorem,
> and related matters that normally don't trouble practical
> physicists in the slightest. My former QFT lecturer clearly
> had a good understanding of practical QFT and calculations, etc,
> and talked a lot about the vacuum seething with virtual particles.
> But he was less able to talk at length about the things I just
> mentioned, except to dismiss them (too glibly, I thought). Even
> Weinberg does not mention them in his weighty tomes.
That could be true. There are plenty of details to keep track of. Do
you mean that Weinberg doesn't mention virtual particles? Or the
issues? If virtual particles, G. Kane doesn't mention them in his
particle physics textbook either. I suppose there is an unmentioned
assumption that they are treated just like real particles.
[...]
>> Most of them like Gordon Kane, do take virtual particles
>> to be real and make successful calculations using them
>> I would suppose.
>
> They don't "use" virtual particles to make the successful
> calculations. They use perturbation series in QFT. Virtual
> particles are just a way of picturing or interpreting the
> various terms that arise in the series.
Don't you think that can be taken the other way around? If we make the
assumption that virtual particles are real, then they CAUSE the terms.
I suspect that is what most particle textbooks do.
>> How would you explain the Lamb Shift without using them?
>> This is a real measurable effect that I have never seen
>> explained any other way yet.
>
> Same answer as in previous paragraph above, though I guess
> it depends on what you mean by "explain". I'm guessing maybe
> you meant "layman-accessible explanation"?
It doesn't have to be "layman-accessible" but I would recommend being
able to describe it in words as best as possible. IOW, the math has to
connect to physics that should eventually result in a semi-classical
expression. It is physical processes that create the math in the first
place. I realize that it gets very difficult in many instances however.
> My (far less accessible) explanation is that accurate predictions
> require full QED. Simply solving the Dirac eqn is not good enough.
Of course. Solving the Dirac equation for this case does not reproduce
the Lamb shift. It was only possible to explain the Lamb shift in terms
of coupling to the "vacuum" field.
Best,
Fred Diether
Jay R. Yablon
> (This is probably one of those instances where spf
> moderators should probably reject the post for being
> impolite to a 3rd party - that would have forced me to
> express myself better, and less offensively.)
SR:
You have the distinction of being perhaps one of the only posters ever
on the Usenet who suggested that the moderators *should have* rejected
his own post. ;-)
Jay.
Oh No
>> My (far less accessible) explanation is that accurate predictions
>> require full QED. Simply solving the Dirac eqn is not good enough.
>
>Of course. Solving the Dirac equation for this case does not reproduce
>the Lamb shift. It was only possible to explain the Lamb shift in terms
>of coupling to the "vacuum" field.
I find this a very curious way of looking at it. The vacuum diagrams are
close loops with no external lines. They have no effect on physics. As
soon as you "couple" to them, they cease to be vacuum diagrams. What we
have is loops in propagators.
Oh No
>On Mar 20, 1:15 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
>> You understand me correctly. It was fine for Dyson, Feynman, Schwinger
>> to do the calculations in a heuristic manner, which generated
>> divergences. It was even fine for them to deal with the divergences in
>> an ad hoc manner. It is not fine for succeeding generations to follow
>> the same arguments and to fail to straighten them out. As early as the
>> seventies it had been shown by Epstein and Glaser et al that the
>> ultraviolet divergences were to do with elementary errors in analysis,
>> the abuse of Wick's theorem, which effectively means the changing of
>> orders of integration where we do not have uniform convergence. I showed
>> the same thing in my own doctoral thesis, albeit a few years later.
>>
>> The fact that this kind of abuse of mathematics will produce nonsensical
>> results has been known for about two hundred years and was well sorted
>> out by the latter half of the C19th. This is taught to first year maths
>> undergraduates. There was some excuse in the early days of qed, but for
>> modern text books like Peskin and Schroeder to produce the same stuff is
>> pure incompetence, imv. Based on this bad mathematics we have whole
>> reams of research into absolute nonsense, from the belief in the Higgs
>> boson, false arguments about renormalisability, the conceptual basis of
>> field theory, through to string theory.
>
>I refuse to believe that Dyson, Feynman, Schwinger, Schweber, Bjorken,
>Drell, Peskin, Schroeder, Weinberg, ... all made an elementary
>mistake,
An elementary mistake in a sense, but well buried in difficult
mathematics. The fact is that physicists had never had to pay attention
to such issues as uniform convergence when changing the order of
integration. They were always able to argue "this is physical so it must
be finite". That argument does not apply to the mathematical structure
of quantum theory. It should have been quickly recognised that it does
not apply, and the effect of changing the order of integration should
have been examined. It was one of the things I set out to examine in my
own thesis, and it has been examined by others before me. I don't think
there is any excuse for your attitude. For Dyson Feynman and Schwinger
to do a heuristic calculation to get an answer is fine. But to fail to
recognise even after the occurrence of divergence that the calculation
contains a mathematically illegal operation, and fail to analyse the
effect of that operation, that is not fine. That is plain sloppy and
mathematically incompetent. To say that "Feynman did it, so it must be
right" is not science and it is not the way mathematics is done. It is
necessary to construct the theory and the integrals carefully, and to
watch things like orders of integration. In the context of these
calculations, which are quite elaborate, that is much more difficult.
But there is still no excuse for not doing it. Nor is there any excuse
for writing down, as Peskin and Schroeder do, expressions equivalent to
1 + infinity + infinity^2 + ... = 1/(1-infinity) = 0
The effect of that, combined with the attitude "oh you get the right
answer so it must be right" together with hostility towards anyone who
tries to do the maths properly, is the reason why there are very few
mathematically competent theoretical physicists, and the reason why
theoretical physics has been stalled for fifty years.
>and only Scharf got it right.
Scharf merely wrote the text book, afaik. The original work on which it
was based was done by others.
>Perhaps you can educate me (and
>others) and show how the electron self-energy integral is calculated
>by Scharf, so that zero is obtained rather than infinity? This is
>rather simple integral, so, presumably, the idea can be explained
>without complicated math.
Doing it properly is complicated math.
sr
>>> As early as the seventies it had been shown by Epstein and
>>> Glaser et al that the ultraviolet divergences were to do with
>>> elementary errors in analysis, the abuse of Wick's theorem,
>>> which effectively means the changing of orders of integration
>>> where we do not have uniform convergence.
And later wrote:
> The fact is that physicists had never had to pay attention
> to such issues as uniform convergence when changing the order
> of integration.
I've read Scharf's book (albeit quite a while ago), and also some
of his more recent papers. I'm aware of his emphasis on improper
splitting of distributions as the root of all evil (i.e: if
one naively multiplies a tempered distribution by a theta fn,
the result is unlikely to remain a tempered distribution, and
so lots of familiar widely-used mathematical techniques become
inapplicable).
But could you please point me to exactly where the stuff about
uniform convergence and changing the order of integration is
discussed at length in this context?
Eugene Stefanovich wrote:
>> Perhaps you can educate me (and others) and show how the electron
>> self-energy integral is calculated by Scharf, so that zero is
>> obtained rather than infinity? This is rather simple integral, so,
>> presumably, the idea can be explained without complicated math.
> Doing it properly is complicated math.
I thought it all boiled down to Scharf/Epstein/Glaser using a
modified propagator, constructed perturbatively so as to retain
the nice properties of tempered distributions. (?)
- strangerep.
sr
I wrote:
>> [...] did not have deep
>> familiarity with inequivalent representations, Haag's theorem,
>> and related matters that normally don't trouble practical
>> physicists in the slightest. My former QFT lecturer clearly
>> had a good understanding of practical QFT and calculations, etc,
>> and talked a lot about the vacuum seething with virtual particles.
>> But he was less able to talk at length about the things I just
>> mentioned, except to dismiss them (too glibly, I thought). Even
>> Weinberg does not mention them in his weighty tomes.
> That could be true. There are plenty of details to keep track of. Do
> you mean that Weinberg doesn't mention virtual particles? Or the
> issues? [...]
I was referring to unitarily inequivalent representations, Haag's
theorem and related matters.
>>> Most of them like Gordon Kane, do take virtual particles
>>> to be real and make successful calculations using them
>>> I would suppose.
>> They don't "use" virtual particles to make the successful
>> calculations. They use perturbation series in QFT. Virtual
>> particles are just a way of picturing or interpreting the
>> various terms that arise in the series.
> Don't you think that can be taken the other way around?
No, I don't think that at all.
> If we make the assumption that virtual particles are real, then
> they CAUSE the terms. I suspect that is what most particle
> textbooks do.
Again, I don't think that's what most textbooks do at all.
Certainly they emphasize the pictorial Feynman diagram
approach, because those make the math easier to keep
track of. But the terms in the perturbation series are just
that: we took a Lagrangian, and applied a perturbation
approach in the interaction coupling constant.
>>> How would you explain the Lamb Shift without using them?
>>> This is a real measurable effect that I have never seen
>>> explained any other way yet.
>> Same answer as in previous paragraph above, though I guess
>> it depends on what you mean by "explain". I'm guessing maybe
>> you meant "layman-accessible explanation"?
> It doesn't have to be "layman-accessible" but I would recommend
> being able to describe it in words as best as possible. [...]
I don't know how to do this well for the more arcane aspects
of QFT. It's akin to explaining to a non-mathematician what
the real numbers are, and why there are more real numbers
between 0 and 1 than all of integers from 0 to infinity.
>> My (far less accessible) explanation is that accurate predictions
>> require full QED. Simply solving the Dirac eqn is not good enough.
> Of course. Solving the Dirac equation for this case does not reproduce
> the Lamb shift. It was only possible to explain the Lamb shift in terms
> of coupling to the "vacuum" field.
IMHO, that's a rather misleading explanation. But maybe it's the
only one that's useful until one can get more deeply into QFT math.
FrediFizzx
news:46000ada$0$4539$426a...@news.free.fr...
Well, we are not talking about the Casimir effect here but I do believe
the Van der Waals force also has a fluctuating vacuum field explanation.
We are talking about Lamb shift which is a very measurable effect that
seems to be due to "zero point energy". Has anyone been able to explain
Lamb shift without zpe? If not, I would have to argue that zpe is
indirectly observable.
Now, in the article G. Kane also goes on to say,
"...CERN produced millions of particles called Z bosons and measured
their mass very accurately. The measured value deviated a little from
the the mass apparently predicted by the Standard Model of particle
physics, but the difference could be explained by the time the Z spent
as a virtual top quark if such a quark had a certain mass. The mass of
the top quark, directly measured a few years later at Fermi National
Accelerator Laboratory in Batavia, Ill., agreed with that obtained from
the CERN analysis, providing another dramatic confirmation of our
understanding of virtual particles."
Seems to be more evidence to me.
Best,
Fred Diether
Oh No
>Earlier, Charles Francis wrote:
>
>>>> As early as the seventies it had been shown by Epstein and
>>>> Glaser et al that the ultraviolet divergences were to do with
>>>> elementary errors in analysis, the abuse of Wick's theorem,
>>>> which effectively means the changing of orders of integration
>>>> where we do not have uniform convergence.
>
>And later wrote:
>
>> The fact is that physicists had never had to pay attention
>> to such issues as uniform convergence when changing the order
>> of integration.
>
>I've read Scharf's book (albeit quite a while ago), and also some
>of his more recent papers. I'm aware of his emphasis on improper
>splitting of distributions as the root of all evil (i.e: if
>one naively multiplies a tempered distribution by a theta fn,
>the result is unlikely to remain a tempered distribution, and
>so lots of familiar widely-used mathematical techniques become
>inapplicable).
>
>But could you please point me to exactly where the stuff about
>uniform convergence and changing the order of integration is
>discussed at length in this context?
It appears to me that much of the book is a discussion of that at
length. One has the Dyson series using of a sum of time ordered
integrals with t_n < t_n-1 < ....< t_1 < t
S=Sum_n (-i)^n Int_-oo^oo dt_1 Int_-oo^t_1 dt_2 ...Int_-oo^t_n-1 dt_n
and one wants to rearrange the terms so that all the integrals run from
-oo to oo. After rearranging the terms the integrals are equivalent to a
sum of Feynman diagrams, and normally we find the familiar ultraviolet
divergence, but because the mathematical operations used are not
justified this does not prove that the Dyson series diverges. In fact
Scharf shows that it converges under a slightly dubious restriction of
the fields - but the restriction is at infinity, which is not the cause
of the ultraviolet divergence.
>Eugene Stefanovich wrote:
>
>>> Perhaps you can educate me (and others) and show how the electron
>>> self-energy integral is calculated by Scharf, so that zero is
>>> obtained rather than infinity? This is rather simple integral, so,
>>> presumably, the idea can be explained without complicated math.
>
>> Doing it properly is complicated math.
>
>I thought it all boiled down to Scharf/Epstein/Glaser using a
>modified propagator, constructed perturbatively so as to retain
>the nice properties of tempered distributions. (?)
Not really. One wants to replace an integral
Int_-oo^t_i dt_j + Int_t_i^oo dt_j (*)
with
Int _-oo^oo dt_j (**)
but the integrand diverges at t_i where it involves an undefinable equal
point multiplication between the fields
Scharf expresses the time ordering of the two parts using Heaviside step
functions.
Theta(t) = 0 t<0
Theta(t) = 1 t>=0
One normally writes
Theta(t) + Theta(-t) = 1
but this buries the limits which are implicit in an integral (*). Scharf
replaces the Heaviside step function with a continuous switching
function, which for convenience I will write
g(t,m) where g(0,m) =0
and m is a parameter such that g(t,m) -> Theta(t) as m -> infinity. The
switching function ensures convergence of the integrals, and with some
fairly ugly mathematics one ends up with finite results. The propagator
hasn't really been replaced, because it is restored normal in the limit,
at least for convergent diagrams, while the divergent diagrams have to
be treated in a more sophisticated manner.
I do things in a rather cruder manner in my own treatment. I replace the
integrals with discrete sums, and I exclude the equal point
multiplication on physical grounds, saying that an electron should not
be allowed to emit more than one photon at an instant of time. Then (**)
becomes
Sum_-oo^oo - delta_ti_tj
Then, when I let the discrete time interval go to zero, so that the sums
are replaced by integrals, it becomes explicit that (**) needs to be
replaced with
Int _-oo^oo dt_j - infinity (***)
in the case of a divergent diagram. The big limitation of my approach is
that, while it gives justification to the ad hoc procedure of
subtracting off the divergences, the infinity in (***) is a strictly
undefinable quantity. My analysis doesn't actually tell you how to
subtract it off to get right answers. One either has to go back to the
method of Epstein and Glaser, and use causal distributions or one has to
use general arguments to establish the properties of the finite part of
the integral - for example it turns out that vector properties of the
divergence in the vertex term are such that it makes no contribution to
the Schwinger correction to the gyromagnetic moment.
eugene_st...@usa.net
> >I thought it all boiled down to Scharf/Epstein/Glaser using a
> >modified propagator, constructed perturbatively so as to retain
> >the nice properties of tempered distributions. (?)
>
> Not really. One wants to replace an integral
>
> Int_-oo^t_i dt_j + Int_t_i^oo dt_j (*)
>
> with
>
> Int _-oo^oo dt_j (**)
>
> but the integrand diverges at t_i where it involves an undefinable equal
> point multiplication between the fields
What if instead of the time-ordered Feynman-Dyson theory one uses old-
fashioned perturbation theory? As far as I know, they are equivalent.
However, the latter does not require changes in time integration
orders. Calculations are more involved, but the same divergences
occur. So, I am not convinced that the order of integration has
anything to do with divergences.
Regards.
Eugene.
Cl.Massé
>> I think Ken is right here. [free] Electrons really don't emit or absorb
>> photons.
They do in the case of an interaction with another electron. It is possible
because the photon isn't on-shell, that is, E^2 - p^2 != 0.
Oh No
>Well, we are not talking about the Casimir effect here but I do believe
>the Van der Waals force also has a fluctuating vacuum field
>explanation. We are talking about Lamb shift which is a very measurable
>effect that seems to be due to "zero point energy". Has anyone been
>able to explain Lamb shift without zpe? If not, I would have to argue
>that zpe is indirectly observable.
I don't see where zero point energy comes in to the calculation of the
shift from Feynman rules. I think it is just one of those things people
talk about, but doesn't mean anything.
Cl.Massé
1174381524.3...@d57g2000hsg.googlegroups.com
> Hi Everybody!
>
> It seems funny to me, but nobody mentioned the most important issue in
> all this
> "virtual particles" story.
>
> If one to consider, say e-Nucleon --> e-Nucleon pion (just an example)
> reaction
> with virtual photon, or similar process with virtual W, Z - bozons -
> there is no
> principal difference for that matter.
>
> "Virtual" just measure a level of off-massshellness for gamma, or W in
> given
> process. If kinematics will be chosen that given q^ will reach 0, the
> particle
> automatically will became a real. End of story.
That's right. But there are two definitions for virtual. The one you talk
about corresponds to a particle number different from zero (see the thread
_Field quanta_ on sci.physics.research) And virtual particles of the
vacuum, that are the zero particle states in second quantization, that is,
zero point fluctuations.
FrediFizzx
> Thus spake FrediFizzx fredi...@hotmail.com
>
>>> I think Ken is right here. [free] Electrons really don't emit or
>>> absorb
>>> photons.
>
> They do in the case of an interaction with another electron. It is
> possible
> because the photon isn't on-shell, that is, E^2 - p^2 != 0.
Well, I was taught that photons are created from and destroyed to the
"vacuum". Photons are never "inside" an electron therefore electrons
don't emit them or absorb them. The energy-momentum is transfered thus
you can say that an electron releases the energy-momentum necessary to
create a photon from the "vacuum". But in every case it takes at least
a dipole situation. In your case above, there is a dipole situation at
the instant of interaction. At that instant, the electrons are no
longer "free". It doesn't matter whether the photon is virtual or not.
I would have to say that it is somewhat sloppy language to say that
electrons emit and absorb photons. It is the energy-momentum that is
emitted or absorbed by an electron. Not the photon particle itself.
Best,
Fred Diether
Oh No
perturbation expansion is not and contains divergences. To go from one
to the other requires that you interchange orders of integration. That
is when the divergences come in.
Oh No
>"Cl.Massé" <ret...@contactprospect.com> wrote in message news:460168f4$
>0$6778$426a...@news.free.fr...
>> Thus spake FrediFizzx fredi...@hotmail.com
>>
>>>> I think Ken is right here. [free] Electrons really don't emit or
>>>>absorb
>>>> photons.
>>
>> They do in the case of an interaction with another electron. It is
>>possible
>> because the photon isn't on-shell, that is, E^2 - p^2 != 0.
>
>Well, I was taught that photons are created from and destroyed to the
>"vacuum".
I don't know who taught you this but it is completely wrong.
>Photons are never "inside" an electron therefore electrons don't emit
>them or absorb them.
That is splitting hairs with language. The photon has no existence
before it is emitted and no existence after it is absorbed. Use the
words "create" and "annhilate" if you prefer.
> The energy-momentum is transfered thus you can say that an electron
>releases the energy-momentum necessary to create a photon from the
>"vacuum".
It has to get the energy-momentum from somewhere to maintain
conservation equations. It does that by absorbing at least one other
photon, which in turn comes from another charged particle, not from the
vacuum.
> But in every case it takes at least a dipole situation. In your case
>above, there is a dipole situation at the instant of interaction. At
>that instant, the electrons are no longer "free". It doesn't matter
>whether the photon is virtual or not. I would have to say that it is
>somewhat sloppy language to say that electrons emit and absorb photons.
I find the mathematical language of qed very precise. Saying an electron
emits or absorbs a photon is as precise a way of putting it into English
as I can imagine. I don't like to say the electron creates a photon,
because that is suggestive of volition.
>It is the energy-momentum that is emitted or absorbed by an electron.
>Not the photon particle itself.
The photon is the carrier of the energy-momentum
Oh No
>> If we make the assumption that virtual particles are real, then
>> they CAUSE the terms. I suspect that is what most particle
>> textbooks do.
>
>Again, I don't think that's what most textbooks do at all. Certainly
>they emphasize the pictorial Feynman diagram approach, because those
>make the math easier to keep track of. But the terms in the
>perturbation series are just that: we took a Lagrangian, and applied a
>perturbation approach in the interaction coupling constant.
This is true (except that I would call the interaction an operator or
more correctly an operator valued distribution, rather than a constant),
but I don't think it is the right way to do it. As I treat qed I start
with a Fock space of particles. Then I introduce an interaction between
particles. I never even mention a Lagrangian. In this way one sees that
virtual particles are just the same as real particles, and that
interactions between particles are described by Feynman diagrams. The
success of the approach may be judged by the fact that it is possible to
give a rigorous, divergence free, treatment of qed leading to Maxwell's
equations in the classical correspondence, none of which is true in the
standard approach, in that it is possible to explain gravity too, and
that it gives different predictions for cosmological redshift than
standard gtr, which as it turns out are supported empirically.
eugene_st...@usa.net
> The Dyson expansion is time ordered and converges in operator norm. The
> perturbation expansion is not and contains divergences. To go from one
> to the other requires that you interchange orders of integration. That
> is when the divergences come in.
I am not sure we are talking about the same thing here. I always
thought that Feynman-Dyson perturbation theory and old-fashioned
perturbation theory are just two equivalent solutions for the S-
matrix. In all examples that I worked out, these two approaches yield
the same results: finite low perturbation orders and infinite loop
integrals in higher orders. Yes, in order to change from O-F to F-D
one needs to rearrange the terms and switch the order of integrals,
however, I've never seen that these steps result in disappearance of
divergences.
One day I need to take a closer look at Epstein-Glaser and Scharf
works. However, so far I remain sceptical and prefer to remain in the
company of other more conventional authors.
Regards.
Eugene.
FrediFizzx
news:tUOeOiW1...@charlesfrancis.wanadoo.co.uk...
> Thus spake FrediFizzx <fredi...@hotmail.com>
>>"Cl.Massé" <ret...@contactprospect.com> wrote in message
>>news:460168f4$
>>0$6778$426a...@news.free.fr...
>>> Thus spake FrediFizzx fredi...@hotmail.com
>>>
>>>>> I think Ken is right here. [free] Electrons really don't emit or
>>>>>absorb
>>>>> photons.
>>>
>>> They do in the case of an interaction with another electron. It is
>>>possible
>>> because the photon isn't on-shell, that is, E^2 - p^2 != 0.
>>
>>Well, I was taught that photons are created from and destroyed to the
>>"vacuum".
>
> I don't know who taught you this but it is completely wrong.
It is not. Here is what Milonni says in "The Quantum Vacuum: An Intro.
to QED",
"In a process in which a photon is annihilated (absorbed), we can think
of the photon as making a transition into the vacuum state. Similarly,
when a photon is created (emitted), it is occasionally useful to imagine
that the photon has made a transition out of the vacuum state."
So not "completely" wrong. ;-) In my relativistic medium viewpoint, I
have to use the "from / to the vacuum" exclusively. And Milonni goes on
to quote Dirac that this was his viewpoint also.
>>Photons are never "inside" an electron therefore electrons don't emit
>>them or absorb them.
>
> That is splitting hairs with language. The photon has no existence
> before it is emitted and no existence after it is absorbed. Use the
> words "create" and "annhilate" if you prefer.
Certainly "create and annihilate" is the correct language to use. "Emit
and absorb" is simply shortcut language to use for the transfer of
energy-momentum which is really what is going on. Electrons emitting
and absorbing photons is really misleading to the actual physics, IMHO.
For the simple fact in the case of the hydrogen atom. It is the whole
atom system that absorbs or emits energy-momentum. Even though the
electron is mostly involved here, the nucleus does get / give up a tiny
bit of this transfer of energy-momentum. Therefore, the photon has to
be created from and destroyed to the "vacuum".
>> The energy-momentum is transfered thus you can say that an electron
>>releases the energy-momentum necessary to create a photon from the
>>"vacuum".
>
> It has to get the energy-momentum from somewhere to maintain
> conservation equations. It does that by absorbing at least one other
> photon, which in turn comes from another charged particle, not from
> the
> vacuum.
But not "absorbing" a photon; it only absorbs the energy-momentum and
the photon is gone. Energy-momentum released from the other charged
particle creates the photon from the vacuum then it goes to the electron
and is destroyed to the vacuum when the energy-momentum is transferred.
>> But in every case it takes at least a dipole situation. In your case
>>above, there is a dipole situation at the instant of interaction. At
>>that instant, the electrons are no longer "free". It doesn't matter
>>whether the photon is virtual or not. I would have to say that it is
>>somewhat sloppy language to say that electrons emit and absorb
>>photons.
>
> I find the mathematical language of qed very precise. Saying an
> electron
> emits or absorbs a photon is as precise a way of putting it into
> English
> as I can imagine. I don't like to say the electron creates a photon,
> because that is suggestive of volition.
I don't believe it is precise since in my viewpoint it is not
representative of the actual physics. The precise way to put it is that
energy-momentum is emitted or absorbed by an electron. But most people
substitute photon for "energy-momentum" as a shortcut. And as you say
below a photon is the carrier of the energy-momentum. But it is not THE
energy-momentum. It is simply the carrier.
>>It is the energy-momentum that is emitted or absorbed by an electron.
>>Not the photon particle itself.
>
> The photon is the carrier of the energy-momentum
Yes.
Best,
Fred Diether
FrediFizzx
> Thus spake FrediFizzx <fredi...@hotmail.com>
>>Well, we are not talking about the Casimir effect here but I do
>>believe
>>the Van der Waals force also has a fluctuating vacuum field
>>explanation. We are talking about Lamb shift which is a very
>>measurable
>>effect that seems to be due to "zero point energy". Has anyone been
>>able to explain Lamb shift without zpe? If not, I would have to argue
>>that zpe is indirectly observable.
>
> I don't see where zero point energy comes in to the calculation of the
> shift from Feynman rules. I think it is just one of those things
> people
> talk about, but doesn't mean anything.
Correction: Milonni says,
"As we shall see in Chapter 11, the 2p_{1/2} -- 2s_{1/2} splitting in
hydrogen due to vacuum polarization is about 27 Mhz. This is only about
1/40 of the total Lamb shift, but it is essential in the comparison of
theory with experiments. In other systems, such as muonic atoms, vacuum
polarization can be the _dominant_ contribution to the Lamb shift."
Via vacuum polarization, I do believe that effects of zpe are indirectly
observable.
Best,
Fred Diether
Oh No
Yes, but this is an example of what I mean when I say that it is just
something someone says without any meaning. The shift comes from a loop
in a propagator. That describes the creation and annihilation of a
particle antiparticle pair, and has nothing to do with an interaction
with a vacuum. The vacuum has no mathematical role in the structure of
qed, just as the ether has no role in special relativity.
Oh No
>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
>news:tUOeOiW1...@charlesfrancis.wanadoo.co.uk...
>> Thus spake FrediFizzx <fredi...@hotmail.com>
>>>"Cl.Massé" <ret...@contactprospect.com> wrote in message
>>>news:460168f4$
>>>0$6778$426a...@news.free.fr...
>>>> Thus spake FrediFizzx fredi...@hotmail.com
>>>>
>>>>>> I think Ken is right here. [free] Electrons really don't emit or
>>>>>>absorb
>>>>>> photons.
>>>>
>>>> They do in the case of an interaction with another electron. It is
>>>>possible
>>>> because the photon isn't on-shell, that is, E^2 - p^2 != 0.
>>>
>>>Well, I was taught that photons are created from and destroyed to the
>>>"vacuum".
>>
>> I don't know who taught you this but it is completely wrong.
>
>It is not. Here is what Milonni says in "The Quantum Vacuum: An Intro.
>to QED",
>
>"In a process in which a photon is annihilated (absorbed), we can think
>of the photon as making a transition into the vacuum state. Similarly,
>when a photon is created (emitted), it is occasionally useful to imagine
>that the photon has made a transition out of the vacuum state."
The only way I can parse this is to replace "vacuum state" with "empty
ket", but then it is just a convoluted way of saying that a photon
ceases to exist, or that it comes into being. The mathematical form of
the interaction, and the locality condition, is such that this only
happens in the presence of an electron. The presence of an electron is
not what I mean by "vacuum".
>
>So not "completely" wrong. ;-) In my relativistic medium viewpoint, I
>have to use the "from / to the vacuum" exclusively. And Milonni goes
>on to quote Dirac that this was his viewpoint also.
Dirac's work was prior to qed. He dealt with a semi-classical
approximation to qed in which the e.m. field was treated classically. If
one takes the theory at the deeper level of full qed, the field just
consists of the potential for photons to be exchanged with other charged
particles.
>
>>>Photons are never "inside" an electron therefore electrons don't emit
>>>them or absorb them.
>>
>> That is splitting hairs with language. The photon has no existence
>> before it is emitted and no existence after it is absorbed. Use the
>> words "create" and "annhilate" if you prefer.
>
>Certainly "create and annihilate" is the correct language to use.
>"Emit and absorb" is simply shortcut language to use for the transfer
>of energy-momentum which is really what is going on.
What is really going on is the exchange of photons.
> Electrons emitting and absorbing photons is really misleading to the
>actual physics, IMHO. For the simple fact in the case of the hydrogen
>atom. It is the whole atom system that absorbs or emits energy-
>momentum. Even though the electron is mostly involved here, the
>nucleus does get / give up a tiny bit of this transfer of energy-
>momentum. Therefore, the photon has to be created from and destroyed
>to the "vacuum".
That seems like a non-sequiter. There is an ongoing exchange of photons
between the nucleus and a bound electron. That is how the electron is
bound. Each photon originates and vanishes with a charged particle.
>
>>> The energy-momentum is transfered thus you can say that an electron
>>>releases the energy-momentum necessary to create a photon from the
>>>"vacuum".
>>
>> It has to get the energy-momentum from somewhere to maintain
>> conservation equations. It does that by absorbing at least one other
>> photon, which in turn comes from another charged particle, not from
>> the
>> vacuum.
>
>But not "absorbing" a photon; it only absorbs the energy-momentum and
>the photon is gone. Energy-momentum released from the other charged
>particle creates the photon from the vacuum then it goes to the
>electron and is destroyed to the vacuum when the energy-momentum is
>transferred.
Sounds backwards to me. Energy momentum does not even make sense without
a particle as carrier. It seems profoundly confusing to equate the empty
ket with vacuum. The creation and annihilation of photons is localised
at the electron, to whatever degree the electron may be said to be
localised. It is then more sensible to say the photon comes from an
electron and goes to an electron.
>
>>> But in every case it takes at least a dipole situation. In your case
>>>above, there is a dipole situation at the instant of interaction. At
>>>that instant, the electrons are no longer "free". It doesn't matter
>>>whether the photon is virtual or not. I would have to say that it is
>>>somewhat sloppy language to say that electrons emit and absorb
>>>photons.
>>
>> I find the mathematical language of qed very precise. Saying an
>> electron
>> emits or absorbs a photon is as precise a way of putting it into
>> English
>> as I can imagine. I don't like to say the electron creates a photon,
>> because that is suggestive of volition.
>
>I don't believe it is precise since in my viewpoint it is not
>representative of the actual physics.
It is representative of the mathematical structure of the physics. To my
way of thinking, the mathematical structure describes physics extremely
well.
>The precise way to put it is that energy-momentum is emitted or
>absorbed by an electron. But most people substitute photon for
>"energy-momentum" as a shortcut.
That is a much less accurate description of the mathematical structure.
I don't even think it parses.
> And as you say below a photon is the carrier of the energy-momentum.
>But it is not THE energy-momentum. It is simply the carrier.
Without which it makes no sense even to talk of this energy-momentum.
The photon is the fundamental entity, not energy-momentum.
>
>>>It is the energy-momentum that is emitted or absorbed by an electron.
>>>Not the photon particle itself.
>>
>> The photon is the carrier of the energy-momentum
>
>
FrediFizzx
> Thus spake FrediFizzx <fredi...@hotmail.com>
[...]
>>The precise way to put it is that energy-momentum is emitted or
>>absorbed by an electron. But most people substitute photon for
>>"energy-momentum" as a shortcut.
>
> That is a much less accurate description of the mathematical
> structure.
> I don't even think it parses.
It's not. Put your math aside if you can and actually think about the
mechanics. Pretend you are maybe like Faraday. Now, an electron has a
certain energy-momentum which it gives up part of it. How exactly does
that energy-momentum transform to create a photon so that it can be
carried by the photon? The electron certainly does not have the
structure to be able to do it. The photon doesn't have the structure
either and it wasn't even created yet anywise. This creation of the
photon does not just happen by magic. There is only one place left for
the structure to be to make the transformation work and that is the
quantum "vacuum". Q.E.D.
>> And as you say below a photon is the carrier of the energy-momentum.
>>But it is not THE energy-momentum. It is simply the carrier.
>
> Without which it makes no sense even to talk of this energy-momentum.
> The photon is the fundamental entity, not energy-momentum.
That is what I said. When an electron absorbs the energy-momentum from
the photon the photon is destroyed. If the photon is destroyed and no
longer exists, how can it possibly be absorbed by the electron? Please
answer me that one question.
Best,
Fred Diether
FrediFizzx
http://en.wikipedia.org/wiki/Vacuum_polarization
Of course, not the way you want to define the "vacuum" as the void. The
quantum "vacuum" does have a role in QED and QCD and it is not the same
thing as the void. All the virtual particles are part of the quantum
"vacuum".
Best,
Fred Diether
FrediFizzx
> Thus spake FrediFizzx <fredi...@hotmail.com>
>>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
>>Nc87v$FF...@charlesfrancis.wanadoo.co.uk...
>>> Thus spake Ken S. Tucker <dyna...@vianet.on.ca>
>>>>On Mar 19, 2:45 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
>>> In qed we only directly model electrons and photons. This is
>>> particularly appropriate for isolated electrons and in scattering
>>> experiments with electrons and positrons. Not an atom in sight. We
>>> assume that similar behaviour is true of all charged particles, and
>>>that
>>> an electron's behaviour when bound in an atom is fundamentally the
>>>same
>>> in an atom as it is in isolation. While the energy for the emission
>>>of a
>>> photon comes from the relationship between electron and nucleus, it
>>> is
>>> the electron which emits the photon.
>>
>>Hmm... I think I would have to disagree with that characterization
>>somewhat. The photon is created from the "vacuum" by the energy that
>>the dipole system gives up. A free electron can never absorb or emit
>>a
>>photon. It is always at least a dipole situation for photon
>>absorption
>>or emission.
>
>
>>
>>I think Ken is right here. Electrons really don't emit or absorb
>>photons. In every case it takes at least a dipole situation. Photons
>>are created from and destroyed to the "vacuum" and the energy-momentum
>>is transferred. And in the case of an atom, most of the energy-
>>momentum gets transferred to or from the lightest particle. In
>>electron - electron scattering the energy-momentum is transferred
>>equally to or from both. Electrons emit or absorb the energy-momentum
>>creating or destroying photons from / to the "vacuum".
>
>
> The mathematical description is quite clear, imv. The photon field
> operator creates or annihilates a photon on Fock space. The electron
> field operator creates a positron or annihilates an electron, the
> adjoint creates an electron or annihilates a positron. The interaction
> operator is a simple statement that an electron emits or absorbs a
> photon. The locality condition states that this happens at a point for
> sizeless particles. I find this absolutely clear and unambiguous in
> the
> mathematical language of qed, which is the language of quantum theory,
> i.e. quantum logic.
I think that is your particular interpretation of the math. An electron
can only absorb or emit energy-momentum. It cannot absorb or emit the
photon particle itself. The photon is created or destroyed at the point
of transfer. Energy-momentum is not a "thing" that can stand on its
own. Particles have a certain energy-momentum that they transfer
between each other. In the case of photons, they are created or
destroyed during the transfer. From where? There is only the "vacuum".
This is actually quite a fundamental issue that can only be explained
one way that I see.
> When an isolated electron emits a photon, the photon must be
> reabsorbed.
> Since the electron is isolated, the photon must be reabsorbed by the
> emitting electron - that is a definitional truism, or the electron
> would
> not be isolated.
Well, let's clarify what you mean by "isolated". By isolated do you
mean from other real fermions? Don't you think that if an electron
emits and reabsorbs a virtual photon that it is no longer "isolated" at
the instant the virtual photon exists?
> This is represented as a loop integral. Normally that
> is thought to diverge. Actually it does not diverge. If one does not
> piss all over basic mathematics which one should learn as a first year
> undergraduate, and does the calculation properly this loop integral is
> zero. One can describe that one of two ways. Either one can say that
> the
> process is forbidden, and that there is no virtual cloud of photons
> surrounding the electron, or one can say that the electron is
> surrounded
> by virtual photons but that this has absolutely no effect on the
> theory,
> a "dressed" electron is thus identical to a "bare" electron.
Of course in my relativistic medium viewpoint an electron is surrounded
by virtual fermionic pairs that are in direct physical contact with the
electron. So I would agree with what you are saying about the loop
integrals being zero as a sum. However in this viewpoint there is no
such animal as a bare electron since the medium defines what an electron
is in the first place. Electrons are always "dressed" to use that
language. In fact in this viewpoint, they are dressed out to infinity.
However, I am sure their effect of "tilting" the quantum "vacuum" is
lost into the noise of the medium pretty rapidly. IOW, I wouldn't
expect that we could detect the presence of a single electron at say a
meter in distance from it.
> The interaction conserves momentum, and it conserves observed values
> of
> energy. For an electron to emit a photon which is not reeabsorbed,
> conservation of energy/momentum requires that the electron absorbs a
> photon from somewhere else. That may be from the nucleus of an atom,
> or
> it may be from other electrons. In either case the electron cannot be
> said to be isolated. There is no interaction with a "vacuum". That
> idea
> has as much validity as the ether in special relativity.
Relativistic mediums are not ruled out by SR. The Standard Model says
you are wrong taking a Higgs-like field to be a relativistic medium. It
all depends on how you want to define the quantum "vacuum". I chose to
not equate it to the void. My viewpoint is consistent with the Standard
Model in that the mass of fermions is generated by the interaction with
a Higgs-like field. There is actually a fairly simple heuristic that
easily demonstrates this in the case of an electron. We will assume the
Higgs-like field is a relativistic medium of bound charges in which the
charge is +,- sqrt(hbar c) in CGS units. Start with the electron
compton wavelength expression and expand it.
m_e = 2pi hbar/(lambda_C c)
Expand using the relation w_C = 2pi c/lambda_C
m_e = 2pi hbar c/(lambda_C c^2)
m_e = (hbar c/w_C^2)(2pi/lambda_C)^3
We need to have the electronic charge in here so we use sqrt(alpha) =
e/sqrt(hbar c) and make the replacement,
m_e = ((e sqrt(hbar c))/w_C^2)[8pi^3/(lambda_C^3 sqrt(alpha))]
The expression in brackets is 1/volume so we can see that for the mass
of an electron we have an interaction between the electronic charge and
the bound charge of the Higgs-like field in a certain volume of space.
Of course the big assumption here is Quantum Vacuum Charge = sqrt(hbar
c). I am pretty confident that Higgs bosons will be found. Many of
them with different masses. It may have already started.
http://www.arxiv.org/abs/hep-ph/0610362
"Has HyperCP Observed a Light Higgs Boson?"
Best,
Fred Diether
Oh No
>> Thus spake FrediFizzx <fredi...@hotmail.com>
>
>[...]
>
>>>The precise way to put it is that energy-momentum is emitted or
>>>absorbed by an electron. But most people substitute photon for
>>>"energy-momentum" as a shortcut.
>>
>> That is a much less accurate description of the mathematical
>>structure.
>> I don't even think it parses.
>
>It's not. Put your math aside if you can and actually think about the
>mechanics.
No. I do not treat maths in the modern manner, that it is just formulae
to get an answer. The postulates of the maths should be abstracted from
the mechanics, and the results of the maths should also tell us about
about physics. This is why I reject quantisation and the Lagrangian
approach. That reduces maths to meaningless squiggles on paper.
Cannonical quantisation is not deductive reason and has no place either
in maths or in scientific theory.
>Pretend you are maybe like Faraday.
If I have to pretend, can I be Maxwell? :-)
>Now, an electron has a certain energy-momentum which it gives up part
>of it. How exactly does that energy-momentum transform to create a
>photon so that it can be carried by the photon? The electron certainly
>does not have the structure to be able to do it.
It has exactly the structure to do it. The structure is contained in the
manner of the coupling between the electron spin indices, the gamma
matrix and the photon creation/annihilation operator.
> The photon doesn't have the structure either and it wasn't even
>created yet anywise.
It has exactly the structure, vis the spin indices.
>This creation of the photon does not just happen by magic. There is
>only one place left for the structure to be to make the transformation
>work and that is the quantum "vacuum". Q.E.D.
Well, QED, NOT the quantum vacuum :-) The quantum vacuum has no role
whatsoever in QED. It is just words used in treatments which don't
understand QED. You cannot say the vacuum has structure to make the
transformation because the vacuum has no structure whereas the electron
and photon have precisely the required structure. That is QED.
>>> And as you say below a photon is the carrier of the energy-momentum.
>>>But it is not THE energy-momentum. It is simply the carrier.
>>
>> Without which it makes no sense even to talk of this energy-momentum.
>> The photon is the fundamental entity, not energy-momentum.
>
>That is what I said. When an electron absorbs the energy-momentum from
>the photon the photon is destroyed. If the photon is destroyed and no
>longer exists, how can it possibly be absorbed by the electron? Please
>answer me that one question.
It no longer exists because it has been absorbed by an electron. This
seems to be returning to a semantic quibble.
Oh No
>> Yes, but this is an example of what I mean when I say that it is just
>> something someone says without any meaning. The shift comes from a
>>loop
>> in a propagator. That describes the creation and annihilation of a
>> particle antiparticle pair, and has nothing to do with an interaction
>> with a vacuum. The vacuum has no mathematical role in the structure of
>> qed, just as the ether has no role in special relativity.
>
>http://en.wikipedia.org/wiki/Vacuum_polarization
This is a typical discussion such as field theorist produce. It is based
on muddy thinking. It is wrong because it is semantically inconsistent.
Essentially it is an oxymoron. The vacuum fluctuation diagrams have no
external lines. As the article says "the true vacuum contains short-
lived "virtual" particle-antiparticle pairs which are created in pairs
out of the Fock vacuum and then annihilate each other". The article then
confuses things. Having no external lines, these vacuum fluctuation
diagrams do not interact with matter, and they have no role in physics.
A loop in a propagator is quite a different diagram altogether.
>
>Of course, not the way you want to define the "vacuum" as the void.
A while back you had it defined as the empty ket, which is a reasonable
definition of void.
> The quantum "vacuum" does have a role in QED and QCD and it is not the
>same thing as the void. All the virtual particles are part of the
>quantum "vacuum".
If they are particles it is not a vacuum. QED.
Oh No
>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
>news:E5Xdo3GSm6$FF...@charlesfrancis.wanadoo.co.uk...
>> Thus spake FrediFizzx <fredi...@hotmail.com>
>>>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
>>>news:Y59hU
>>>Nc87v$FF...@charlesfrancis.wanadoo.co.uk...
>>>> Thus spake Ken S. Tucker <dyna...@vianet.on.ca>
>>>>>On Mar 19, 2:45 am, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>>
>
>I think that is your particular interpretation of the math.
Mathematics is a language which should be understood, not just a set of
mechanical rules which could be followed by a computer. If anyone says
otherwise, it is because it is a language which they do not understand.
> An electron
>can only absorb or emit energy-momentum. It cannot absorb or emit the
>photon particle itself. The photon is created or destroyed at the point
>of transfer. Energy-momentum is not a "thing" that can stand on its
>own. Particles have a certain energy-momentum that they transfer
>between each other. In the case of photons, they are created or
>destroyed during the transfer. From where? There is only the "vacuum".
>This is actually quite a fundamental issue that can only be explained
>one way that I see.
This is unduly repetitive. If you cannot say what this "vacuum" is or
how it appears in theory, it reduces to dogma. There is not only the
vacuum. There is the electron. The locality condition states that that
is where the photon comes from.
>> When an isolated electron emits a photon, the photon must be
>> reabsorbed.
>> Since the electron is isolated, the photon must be reabsorbed by the
>> emitting electron - that is a definitional truism, or the electron
>> would
>> not be isolated.
>
>Well, let's clarify what you mean by "isolated". By isolated do you
>mean from other real fermions? Don't you think that if an electron
>emits and reabsorbs a virtual photon that it is no longer "isolated" at
>the instant the virtual photon exists?
That issue is meaningless. This photon does not link to other matter,
and in any case contributes nothing to behaviour.
>
>> This is represented as a loop integral. Normally that
>> is thought to diverge. Actually it does not diverge. If one does not
>> piss all over basic mathematics which one should learn as a first year
>> undergraduate, and does the calculation properly this loop integral is
>> zero. One can describe that one of two ways. Either one can say that
>> the
>> process is forbidden, and that there is no virtual cloud of photons
>> surrounding the electron, or one can say that the electron is
>> surrounded
>> by virtual photons but that this has absolutely no effect on the
>> theory,
>> a "dressed" electron is thus identical to a "bare" electron.
>
>Of course in my relativistic medium viewpoint an electron is surrounded
>by virtual fermionic pairs that are in direct physical contact with the
>electron. So I would agree with what you are saying about the loop
>integrals being zero as a sum.
No. For an isolated electron, with mass-shell energy, the loop integrals
are individually zero. The sum which one finds in text books amounts to
1+infinity+infinity^2+...= 1/(1-infinity) = 0
This is absolute drivel, and shows only that the authors of text books
haven't got a clue.
>> The interaction conserves momentum, and it conserves observed values
>> of
>> energy. For an electron to emit a photon which is not reeabsorbed,
>> conservation of energy/momentum requires that the electron absorbs a
>> photon from somewhere else. That may be from the nucleus of an atom,
>> or
>> it may be from other electrons. In either case the electron cannot be
>> said to be isolated. There is no interaction with a "vacuum". That
>> idea
>> has as much validity as the ether in special relativity.
>
>Relativistic mediums are not ruled out by SR.
They are rendered meaningless, and hence non-physical.
> The Standard Model says
>you are wrong
Before one starts to understand qed, one should go back to first
principles and build it from scratch as a mathematically consistent
structure using only legitimate mathematics and deductive logic. I
started doing that for my doctoral thesis. It is what any decent
mathematician would have done. Most of what is said about qed in the
text books is enough to put any decent mathematician off. There are much
better jobs working for banks and oil companies.
Cl.Massé
56ajdmF...@mid.individual.net
> Hawking radiation is well-known. It results from virtual particle pairs
> forming right outside the event horizon of a black hole, one of each
> pair falling into the hole, the other escaping as a real particle. The
> mechanism is generally accepted but disputed[1], which is why I wrote
> "consensus", but it's not plain false.
There is exactly the same phenomenon with a very strong electric field. But
it can't be said: "the virtual pair turns real", only: "a high energy photon
decays into a particle-antiparticle pair." The reason is, the phenomena can
as well be described in first quantization, where there is no virtual
particle, up to perhaps very small radiative correction.
There is no known quantum of gravitational field. If there is one, the
mechanism is exactly the same. If not, the Dirac equation in first
quantization accounts very well for the pair creation.
There is no consensus about the virtual particles, even if almost all
subsidized physicists think they are "real", and more so when they have no
insight into second quantization.
Cl.Massé
1174423935.1...@d57g2000hsg.googlegroups.com
> I refuse to believe that Dyson, Feynman, Schwinger, Schweber, Bjorken,
> Drell, Peskin, Schroeder, Weinberg, ... all made an elementary
> mistake, and only Scharf got it right.
In the future, I'm certain that like in the Einstein controversy, the work
of Epstein and Glaser will be ascribed to Feynman. Feynman... Feynman...
that rings a bell... Wasn't he an actor in a movie about the Vietnam war?
No, I'm stupid, it's Rocky.
I thought that the physics input of Feynman was insignificant, and that he
only came up with mathematical formulations, but now I see he also botched
up the mathematical part.
More seriously, it has always been said, and I always believed, that an
electron can't interact with itself, i.e. can't absorb a photon it sent.
Wieso?
Cl.Massé
> Well, we are not talking about the Casimir effect here but I do believe
> the Van der Waals force also has a fluctuating vacuum field explanation.
That is one of the numerous equivalent mathematical formulation, but the
vacuum energy isn't necessary to describe them. (there is at least one
mathematical formulation that doesn't take it into account.)
> We are talking about Lamb shift which is a very measurable effect that
> seems to be due to "zero point energy". Has anyone been able to explain
> Lamb shift without zpe? If not, I would have to argue that zpe is
> indirectly observable.
The Lamb shift must be calculated from QED, but I wonder through which
infinite cancelling procedure. That said, even if there is an effect, it
isn't a *measurable* effect. Indeed, a measurable effect is a difference
between two measures, but it is impossible to make a measure without virtual
particles.
Besides, I don't know whether the Lamb shift is due to virtual particles, or
to the exchange of quanta, or still to the finite propagation speed of the
photons.
> Now, in the article G. Kane also goes on to say,
>
> "...CERN produced millions of particles called Z bosons and measured
> their mass very accurately. The measured value deviated a little from
> the the mass apparently predicted by the Standard Model of particle
> physics, but the difference could be explained by the time the Z spent
> as a virtual top quark if such a quark had a certain mass. The mass of
> the top quark, directly measured a few years later at Fermi National
> Accelerator Laboratory in Batavia, Ill., agreed with that obtained from
> the CERN analysis, providing another dramatic confirmation of our
> understanding of virtual particles."
>
> Seems to be more evidence to me.
Sorry, but not yet. He speaks about the top quark, but that is a spurious
statement. He should have spoken about a "top anti-top pair". And now, he
is no more speaking about virtual particles in the same sense. It was zero
point fluctuation, and now it is a particle allowed by the Heisenberg
uncertainty. Actually, here it is the decay of a Z boson into a top
anti-top pair, followed by a top anti-top annihilation. That phenomenon is
described by the electroweak theory in first quantization as well.
Saying what he said, Kane won't be contradicted neither by the working
physicists, who need to publish papers, neither by laypersons since it is so
badly explained that they don't get a chance to understand the first word.
What he call "understanding" means "agreement with experimental data."
FrediFizzx
> <eugene_st...@usa.net> a écrit dans le message de news:
> 1174423935.1...@d57g2000hsg.googlegroups.com
>
>> I refuse to believe that Dyson, Feynman, Schwinger, Schweber,
>> Bjorken,
>> Drell, Peskin, Schroeder, Weinberg, ... all made an elementary
>> mistake, and only Scharf got it right.
>
> In the future, I'm certain that like in the Einstein controversy, the
> work
> of Epstein and Glaser will be ascribed to Feynman. Feynman...
> Feynman...
> that rings a bell... Wasn't he an actor in a movie about the Vietnam
> war?
> No, I'm stupid, it's Rocky.
>
> I thought that the physics input of Feynman was insignificant, and
> that he
> only came up with mathematical formulations, but now I see he also
> botched
> up the mathematical part.
But it works for QED to extraordinary precision. Why?
> More seriously, it has always been said, and I always believed, that
> an
> electron can't interact with itself, i.e. can't absorb a photon it
> sent.
> Wieso?
What is "wieso"? Does that mean why? Even Feynman wondered about this
in 1964; "...we have allowed what is perhaps a silly thing, the
possibility of the 'point' electron acting on itself."
Best,
Fred Diether
Oh No
>On Mar 21, 2:05 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
>> The Dyson expansion is time ordered and converges in operator norm. The
>> perturbation expansion is not and contains divergences. To go from one
>> to the other requires that you interchange orders of integration. That
>> is when the divergences come in.
>
>I am not sure we are talking about the same thing here. I always
>thought that Feynman-Dyson perturbation theory and old-fashioned
>perturbation theory are just two equivalent solutions for the S-
>matrix. In all examples that I worked out, these two approaches yield
>the same results: finite low perturbation orders and infinite loop
>integrals in higher orders. Yes, in order to change from O-F to F-D
>one needs to rearrange the terms and switch the order of integrals,
>however, I've never seen that these steps result in disappearance of
>divergences.
I don't know what you mean by O-F perturbation theory, as but I have
been a bit imprecise. Scharf identifies the source of the divergence as
the abuse of Wick's theorem, where the ordering of operators are
switched, which I see as being tied to the order of integration.
Actually when I said order of integration, I should have said order of
taking limits. The time ordered integrals are improper integrals because
the integrand diverges at t_i=t_i-1. It is the limit implicit in the
bound of the integration which causes the trouble if it is not dealt
with carefully, not the one in the integration itself.
>
>One day I need to take a closer look at Epstein-Glaser and Scharf
>works. However, so far I remain sceptical and prefer to remain in the
>company of other more conventional authors.
>
>
Scharf doesn't answer all questions. For a start this is an S-matrix
approach. But he does take things forward a long way, and at leads he
deals in mathematics not wool pulling.
eugene_st...@usa.net
> > More seriously, it has always been said, and I always believed, that
> > an
> > electron can't interact with itself, i.e. can't absorb a photon it
> > sent.
> > Wieso?
>
> What is "wieso"? Does that mean why? Even Feynman wondered about this
> in 1964; "...we have allowed what is perhaps a silly thing, the
> possibility of the 'point' electron acting on itself."
Are we really interested in knowing how exactly the electron interacts
with itself? This cannot be observed anyway. Only interactions of this
"self-interacting electron" with other particles can be observed. So,
why not just fold this self-interaction into the definition of
"physical" or "dressed" electron? Then we can forget about strange
self-interacting "bare" electrons and express everything in terms of
real physical observable electrons with finite mass and charge. This
is the idea of the "dressed particle" approach.
Eugene.
Oh No
><eugene_st...@usa.net> a écrit dans le message de news:
>1174423935.1...@d57g2000hsg.googlegroups.com
>
>> I refuse to believe that Dyson, Feynman, Schwinger, Schweber, Bjorken,
>> Drell, Peskin, Schroeder, Weinberg, ... all made an elementary
>> mistake, and only Scharf got it right.
>
>In the future, I'm certain that like in the Einstein controversy, the work
>of Epstein and Glaser will be ascribed to Feynman. Feynman... Feynman...
>that rings a bell... Wasn't he an actor in a movie about the Vietnam war?
>No, I'm stupid, it's Rocky.
>
>I thought that the physics input of Feynman was insignificant, and that he
>only came up with mathematical formulations, but now I see he also botched
>up the mathematical part.
I think the physics input was inspired, but it fell on deaf ears and
Feynman learned not to argue the case. "Wouldn't it be funny" he said
"if these little diagrams actually had something to do with physics".
Actually they do, but we are taught, generation after generation, to
think of them as bookkeeping for calculations we are supposed to do but
not understand.
>
>More seriously, it has always been said, and I always believed, that an
>electron can't interact with itself, i.e. can't absorb a photon it sent.
>Wieso?
Usually it is said that this self interaction does take place, and then
one redefines what an electron is so as to get rid of it. I say that is
not necessary. For a free electron, when this interaction is calculated
correctly, it has zero amplitude and contributes nothing.
Cl.Massé
56dpm4F...@mid.individual.net
> It is not. Here is what Milonni says in "The Quantum Vacuum: An Intro.
> to QED",
>
> "In a process in which a photon is annihilated (absorbed), we can think
> of the photon as making a transition into the vacuum state. Similarly,
> when a photon is created (emitted), it is occasionally useful to imagine
> that the photon has made a transition out of the vacuum state."
>
> So not "completely" wrong. ;-) In my relativistic medium viewpoint, I
> have to use the "from / to the vacuum" exclusively. And Milonni goes on
> to quote Dirac that this was his viewpoint also.
It is certainly not "the photon" that makes a transition, it is the
electromagnetic field state. The em potential evolves as a function of
the local electronic current. After second quantization, that is exactly
the same, even if the em field state can now be written as a wave function
in the occupation number space, aka Fock space. The rest is complacent
gibberish, not worth unearthing Dirac.
> Certainly "create and annihilate" is the correct language to use. "Emit
> and absorb" is simply shortcut language to use for the transfer of
> energy-momentum which is really what is going on. Electrons emitting
> and absorbing photons is really misleading to the actual physics, IMHO.
> For the simple fact in the case of the hydrogen atom. It is the whole
> atom system that absorbs or emits energy-momentum. Even though the
> electron is mostly involved here, the nucleus does get / give up a tiny
> bit of this transfer of energy-momentum. Therefore, the photon has to
> be created from and destroyed to the "vacuum".
That implies faster than light information travel, since the proton and the
electron are spatially separated. It's blatantly inconsistent.
It is well known among rigorous thinkers that the authority and consensus
arguments are proof of nothing, the history of science give enough examples.
So one must be blunt in fighting against misconceptions stemming from them.
Cl.Massé
> Well, I was taught that photons are created from and destroyed to the
> "vacuum". Photons are never "inside" an electron therefore electrons
> don't emit them or absorb them. The energy-momentum is transfered thus
> you can say that an electron releases the energy-momentum necessary to
> create a photon from the "vacuum". But in every case it takes at least
> a dipole situation. In your case above, there is a dipole situation at
> the instant of interaction. At that instant, the electrons are no
> longer "free". It doesn't matter whether the photon is virtual or not.
What is the dipole? A *free* electron create a varying electromagnetic
field around itself, so it emits photons.
Ken S. Tucker
> Thus spake FrediFizzx fredifi...@hotmail.com
>
> >> I think Ken is right here. [free] Electrons really don't emit or absorb
> >> photons.
>
> They do in the case of an interaction with another electron. It is possible
> because the photon isn't on-shell, that is, E^2 - p^2 != 0.
>They do in the case of an interaction with another electron. It is possible
>because the photon isn't on-shell, that is, E2 - p2 != 0.
I do not see how 2 electrons can create a photon.
In an imaginary (e-) (e-) Coulomb reaction with
two electrons repelling, like,
<=(e-) (e-)=>
the current and change in current is zero.
Power = Volts x Current, since current is zero, no
power can be emitted, it's a Zero Watt light bulb.
So that imaginary repulsion cannot occur.
OTOH, a (+) (-) dipole reacts as,
(+)=> <=(-)
where we can get a real current by using a diagram
(reversing charge, reverses current) like,
<=(-) <=(-)
Now that has a increment in Voltage and Current
to yield a Power output, Non-Zero Watt bulb :-) .
The actual source of the origin of the photons is
another question, that probably involves the
mass-inertia of the relating charged particles.
Regards
Ken
Oh No
>
>It is certainly not "the photon" that makes a transition, it is the
>electromagnetic field state. The em potential evolves as a function of
>the local electronic current. After second quantization, that is exactly
>the same, even if the em field state can now be written as a wave function
>in the occupation number space, aka Fock space.
Second quantisation is normally presented as an irrational replacement,
but it can be justified by building the Fock space from photons. The
electromagnetic field is merely an expectation of observable quantities
to do with photons, so it certainly is photons which are transferred.
>
>That implies faster than light information travel, since the proton and the
>electron are spatially separated. It's blatantly inconsistent.
Spatial separation does not even make sense until after photons are
transferred.
Ken S. Tucker
I'd like to add, I physically studied electron (e-)
beams in CRT's and electron microscopes, and
I was quite surprised that the beam remained
collimated, inspite of the idea of inter-electron
repulsion, within the beam, due to Coulomb's
forces.
So for me, it's a bit of a mystery on how we're
able to sharply focus an electron beam, that
apparently ignores a Coulomb repulsion as
would be expected using classical theory
applied to the electron beam.
I'll speculate, two electrons can NOT form
a dipole, and therefore can NOT radiate,
to output a photon.
((Rather like two rabbits of the same sex
aren't making babies)).
What I understood as a TV antenna installer
was that the Xmitter, outputed Watts, (Power),
and it was my mission to install the dipoles
that would intercept those radiated Watts.
*Cummunications, is the exchange of power.*
((Tucker's formulating his 1st Law)).
Best regards
Ken S. Tucker
Cl.Massé
>> an electron can't interact with itself, i.e. can't absorb a photon it
>> sent. Wieso?
"FrediFizzx" <fredi...@hotmail.com> a écrit dans le message de news:
56g0cqF...@mid.individual.net
> What is "wieso"? Does that mean why?
That means wieso?, or perhaps how comes?
> Even Feynman wondered about this
> in 1964; "...we have allowed what is perhaps a silly thing, the
> possibility of the 'point' electron acting on itself."
Surely he meant: "we *calculated* what is perhaps *unreal*, ** the
*structureless* electron *interacting with* itself." The only relevance is
whether the theory agree with experiment. As usual, I fail to make sense
from his words. It seems as if he's speaking as The Creator.
When, after all that, they get infinities where the net effect is
inexistent, then with straight faces use dubious methods to put them under
the carpet, I can't help but laugh. Surely you are joking Mr. Feynman.
Cl.Massé
news: Av9u49cN...@charlesfrancis.wanadoo.co.uk
> Second quantisation is normally presented as an irrational replacement,
> but it can be justified by building the Fock space from photons. The
> electromagnetic field is merely an expectation of observable quantities
> to do with photons, so it certainly is photons which are transferred.
Well, it seems a sketch of the theory behind second quantization is worth
giving, so that we know what we speak about:
At the beginning is the em field, which obeys a classical wave equation.
The quantization of this motion is what is called second quantization, the
first being the quantization of a point particle into a wave function. The
em field is represented by a variable at each point, and the Maxwell
equation is a differential equation on these variables. It can be
diagonalized so that we get independent modes. There is a leeway of choice
for this modes, for example plane waves or spherical harmonics. The
differential equation of those modes is then the one of a harmonic
oscillator. By canonical quantization, the variables associated to the
modes are replaced by operators. It is as if a fictitious wave function of
the variables were introduced, on which the operator acts. The quantization
of the harmonic oscillators gives, for each mode, a set of eigenstates of
the Hamiltonian with equidistant energies, labelled with an integer. It
represents the number of "photons" in the mode. In the process, the em
field no longer has a well defined value. The state of the em field can now
be written as a complex linear combination of all those states, like:
Psi = (a|0>_i + b|1>_i + ... ; c|0>_j + d|1>_j + e|2>_j + ... ; ... )
Where 0, 1, 2 represents the number of photons in the mode i, j, ...
"A photon is created from the vacuum" means the transition:
|0> -> |1>
But the idea of a sudden transition comes from old QM, when there weren't
the projection postulate for a measurement. Actually, what happens is the
state evolves as:
a(t)|0> + b(t)|1>
where a and b are complex numbers whose evolution is derived from the
Maxwell equation.
BUT
We must be very cautious by saying that, since we are dealing with plane
waves or spherical harmonics, which have an infinite spatial extension.
>From causality, in every real process an infinity of modes is involved so
that the expectation value of the field cancels at infinity.
The vacuum is the state (|0>_i; |0>_j; ... ). In the vacuum state, the em
field is reduced to the zero point fluctuation. But as it is the ground
state, no energy can be extracted from it, although it has a formal infinite
energy, which means no power, therefore no effect.
> Spatial separation does not even make sense until after photons are
> transferred.
That is QED. Of course, one may want to replace it by a better theory.
Oh No
>"Oh No" <No...@charlesfrancis.wanadoo.co.uk> a écrit dans le message de
>news: Av9u49cN...@charlesfrancis.wanadoo.co.uk
>
>> Second quantisation is normally presented as an irrational replacement,
>> but it can be justified by building the Fock space from photons. The
>> electromagnetic field is merely an expectation of observable quantities
>> to do with photons, so it certainly is photons which are transferred.
>
>Well, it seems a sketch of the theory behind second quantization is worth
>giving, so that we know what we speak about:
Second quantisation is the replacement of a wave equation by an operator
equation. It is applied as much to the Dirac equation as to the em
field. First quantisation of the e.m. field would be replacing the field
with quantum wavefunctions for photons. Second quantisation is then
replacing the photon wave function with an operator. But really this is
a very silly way to approach qed imv. It is much better to start with a
Hilbert space of particle states, then you can define Fock space, and
then you can define the operators you need. That way you get to the same
operators, well motivated by legitimate mathematics, not through
canonical procedure. Much better too not to start with the e.m. field,
but to derive it as the expectation of observable operators.
>The vacuum is the state (|0>_i; |0>_j; ... ). In the vacuum state, the em
>field is reduced to the zero point fluctuation. But as it is the ground
>state, no energy can be extracted from it, although it has a formal infinite
>energy, which means no power, therefore no effect.
>
>> Spatial separation does not even make sense until after photons are
>> transferred.
>
>That is QED. Of course, one may want to replace it by a better theory.
>
It is special relativity also, if one understands it properly. The only
way I would replace it is to make interactions between electrons and
photons discrete. That way I can produce a consistent unified theory and
show the cause of gravity.
eugene_st...@usa.net
> Second quantisation is the replacement of a wave equation by an operator
> equation. It is applied as much to the Dirac equation as to the em
> field. First quantisation of the e.m. field would be replacing the field
> with quantum wavefunctions for photons. Second quantisation is then
> replacing the photon wave function with an operator. But really this is
> a very silly way to approach qed imv. It is much better to start with a
> Hilbert space of particle states, then you can define Fock space, and
> then you can define the operators you need. That way you get to the same
> operators, well motivated by legitimate mathematics, not through
> canonical procedure. Much better too not to start with the e.m. field,
> but to derive it as the expectation of observable operators.
I totally agree that starting from particle Hilbert spaces (rather
than from classical fields) is a much better way to introduce QFT.
This approach is taken in Weinberg's "The quantum theory of fields"
vol. 1, which I highly recommend.
Eugene.
Peter
> "Oh No" <No...@charlesfrancis.wanadoo.co.uk> a écrit dans le message de
> news: Av9u49cN...@charlesfrancis.wanadoo.co.uk
>
> > Second quantisation is normally presented as an irrational replacement,
> > but it can be justified by building the Fock space from photons. The
> > electromagnetic field is merely an expectation of observable quantities
> > to do with photons, so it certainly is photons which are transferred.
>
> Well, it seems a sketch of the theory behind second quantization is worth
> giving, so that we know what we speak about:
>
> At the beginning is the em field, which obeys a classical wave equation...
What exactly means "at the beginning", at the beginning of what?
> ...The quantization of this motion is what is called second quantization,
the
> first being the quantization of a point particle into a wave function...
Are you sure?
> ...The
> em field is represented by a variable at each point, and the Maxwell
> equation is a differential equation on these variables. It can be
> diagonalized so that we get independent modes...
The Maxwell eqs. diagonalized? ;-)
But more important, you discard the fact that the inner variable of a wave -
and thus analogue to the frequency of an oscillator - is its speed. Modes
occur through external influences (excitation, boundary conditions), hence,
frequency and wavelength are external parameters of a wave. If you don't take
this into account, you will perhaps run into troubles. For instance, in
normal modes, the spatial dependencies are treated classically, only the
temporal ones are quantized - a hardly consistent treatment (cf. Schleich,
'Quantization in Phase Space, Wiley, 2000, Ch. 10).
> ...
>
> "A photon is created from the vacuum" means the transition:
>
> |0> -> |1>
>
> But the idea of a sudden transition comes from old QM, when there weren't
> the projection postulate for a measurement. Actually, what happens is the
> state evolves as:
>
> a(t)|0> + b(t)|1>
>
> where a and b are complex numbers whose evolution is derived from the
> Maxwell equation.
>
> BUT
>
> We must be very cautious by saying that, since we are dealing with plane
> waves or spherical harmonics, which have an infinite spatial extension.
I told you above :-)
> >From causality, in every real process an infinity of modes is involved so
> that the expectation value of the field cancels at infinity.
Sounds crazy, doesn't it?
"an infinity of modes" is just unphysical.
> The vacuum is the state (|0>_i; |0>_j; ... ). In the vacuum state, the em
> field is reduced to the zero point fluctuation. But as it is the ground
> state, no energy can be extracted from it, although it has a formal
> infinite
> energy, which means no power, therefore no effect.
Sounds crazy, doesn't it?
Peter
FrediFizzx
Varying? And with respect to what? The dipole is formed by the other
electron the first one is interacting with. And come to think of it,
can any electron really be "free"? IOW, what is the definition of
"free"?
Best,
Fred Diether
maxwell
> "Cl.Massé" <ret...@contactprospect.com> wrote in message
>
> news:4603ff93$0$29035$426a...@news.free.fr...
>
>
>
> > 56dg3mF28h3v...@mid.individual.net
>
> >> Well, I was taught that photons are created from and destroyed to the
> >> "vacuum". Photons are never "inside" an electron therefore electrons
> >> don't emit them or absorb them. The energy-momentum is transfered
> >> thus
> >> you can say that an electron releases the energy-momentum necessary
> >> to
> >> create a photon from the "vacuum". But in every case it takes at
> >> least
> >> a dipole situation. In your case above, there is a dipole situation
> >> at
> >> the instant of interaction. At that instant, the electrons are no
> >> longer "free". It doesn't matter whether the photon is virtual or
> >> not.
>
> > What is the dipole? A *free* electron create a varying
> > electromagnetic
> > field around itself, so it emits photons.
>
> Varying? And with respect to what? The dipole is formed by the other
> electron the first one is interacting with. And come to think of it,
> can any electron really be "free"? IOW, what is the definition of
> "free"?
>
> Best,
>
> Fred Diether
Your assumption that electrons are never free follows from the
continuum assumption that underlies all physics; namely: "everything
interacts everywhere and always".
Peter
> Varying? And with respect to what? The dipole is formed by the other
> electron the first one is interacting with.
I agree.
The Green's functions theory of the polarizability of, say, a solid, is
basically a two-particle theory (within its many-body variant, the two
particles under consideration interact with the other particles as well). In
contrast, the density matrix theory of the polarizability of, say, a solid,
is basically a one-particle theory (within its many-body variant, the
particle under consideration interacts with the other particles as well).
Nevertheless, polarization is still the relative displacement of charges
against each other.
> And come to think of it,
> can any electron really be "free"? IOW, what is the definition of
> "free"?
no interaction (w.r.t. inertia, the gravitative interaction with the rest of
the universum is not counted)
Peter
Ken S. Tucker
On Mar 27, 9:59 am, Peter <end...@dekasges.de> wrote:
> "FrediFizzx" <fredifi...@hotmail.com> writes:
> > Varying? And with respect to what? The dipole is formed by the other
> > electron the first one is interacting with.
I went to look into the solar wind...
http://en.wikipedia.org/wiki/Solar_wind
Number's like 10^9 kg/s and 600 km/s
are quoted, (other ref's are available to
those figures) much of it consisting of p+
and e- particles.
Please correct me if I'm wrong. I searched
the literature on the subject, for some sort
of radiate EMR spectral output that I would
think could provide information about the
"dipoles" that would be expected to form
in the solar wind, well beyond the sun, and
wasn't able to find any yet.
I thought we might use that to emprically enter
data into the problem.
> I agree.
> The Green's functions theory of the polarizability of, say, a solid, is
> basically a two-particle theory (within its many-body variant, the two
> particles under consideration interact with the other particles as well). In
> contrast, the density matrix theory of the polarizability of, say, a solid,
> is basically a one-particle theory (within its many-body variant, the
> particle under consideration interacts with the other particles as well).
> Nevertheless, polarization is still the relative displacement of charges
> against each other.
>
> > And come to think of it,
> > can any electron really be "free"? IOW, what is the definition of
> > "free"?
Right, that's why I'm looking at the solar wind.
> no interaction (w.r.t. inertia, the gravitative interaction with the rest of
> the universum is not counted)
> Peter
I'm unsure about "inertia" for example, suppose
two electrons possess differing speeds like,
(-) ==========> electron "a"
(-)====> electron "b"
The Electro-potential energy (ab/r) varies, but
it may not necessarily emit radiation.
Regards
Ken
FrediFizzx
news:1174369436.1...@l75g2000hse.googlegroups.com...
> On Mar 19, 8:25 pm, "FrediFizzx" <fredifi...@hotmail.com> wrote:
>
>> However, you end up with instantaneous interactions
>> at a distance without some of the "excess baggage".
>
> I know well how people dislike instantaneous action-at-a-distance.
> Even Newton, who invented the first instantaneous potential, regarded
> those who follow this idea too literally as ... mentally deprived, or
> something of that sort.
I am not against superluminal interactions over sub-atomic "distances"
or smaller. I consider that to be the source of quantum uncertainty.
In the weirdness of it all, we really must take the stance that space
and time are being defined at this level. That is why I put distances
in quotes. For an example, we can't really localize an electron to a
"distance" smaller than on the order of the electron compton wavelength
divided by 2pi at low energies.
> On the other hand, we still don't have a clear experimental
> measurement of the speed of propagation of Coulomb forces or
> gravitational forces. The strongest "argument" against instantaneous
> interactions usually comes from the special relativistic ban on
> superluminal signal propagation. However, I am ready to prove that the
> ban doesn't apply in this particular case. Though, this can be a topic
> for a separate thread.
Are you making a new "force" other than the EM force? Coulomb forces
have to be described by the EM force which has the photon as the
carrier. Are you saying photons can propagate instantaneously in the
Coulomb case?
>> Well, what I wanted to know is how are your "physical particles"
>> described compared to the "normal" viewpoint? Do they have extended
>> structure? Or what?
>
>
> Stable "physical particles" are described by irreducible unitary
> representations of the Poincare group. So, they have two intrinsic
> parameters - mass and spin. If a particle participates in
> interactions, then one can add to this list certain "charges".
> Moreover, one can clearly define such measurable properties of a
> particle as position, momentum, angular momentum, energy.
>
> However, I wouldn't say that elementary particles, like electron, have
> any structure, extended or otherwise. Even if they had a "structure"
> we wouldn't be able to see it. "Never ask about something you cannot
> observe".
Oh sheesh, I wonder about stuff we can't observe all the time. ;-)
That will definitely not stop me from asking as I suspect asking can
lead to a more fundamental understanding. In my relativistic medium
viewpoint, all elementary particles do have *external* structure due to
the medium effect. And this is not really contrary to QFT nor the
Standard Model with a Higgs-like field. As soon as you say "mass", you
have to include the Higgs-like field since it is the generator of mass
for elementary fermions. I would have to say that the Higgs-like field
is also the source of intrinsic spin. I use the term "Higgs-like"
because in my model the Higgs bosons are not elementary. They are
always composites of virtual complementary fermions wrt our perspective.
Best,
Fred Diether
Cl.Massé
>> Second quantisation is the replacement of a wave equation by an operator
>> equation. It is applied as much to the Dirac equation as to the em
>> field. First quantisation of the e.m. field would be replacing the field
>> with quantum wavefunctions for photons. Second quantisation is then
>> replacing the photon wave function with an operator. But really this is
>> a very silly way to approach qed imv. It is much better to start with a
>> Hilbert space of particle states, then you can define Fock space, and
>> then you can define the operators you need.
But that's exactly the same. What is a Hilbert space if not a linear space
of wave functions? The only advantage I see is that it obfuscates physics
with abstract mathematical being, so that the inconsistencies and mysteries
are more difficult to find out. My opinion is that it is such gesticulation
that strayed most physicists from a sound understanding of the theory and
from real advances. "Create a photon from vacuum and transfer momentum to
it" is fiddle-faddle in my ear, and indeed it has no mathematical
formulation. As I pointed out, an infinity of photons must be created, and
I understand that it is very tricky to realize in a forest of bra and kets.
Even the very phrase "send a photon" is meaningless, since a photon has an
infinite spatial and temporal extension.
eugene_st...@usa.net
> For an example, we can't really localize an electron to a
> "distance" smaller than on the order of the electron compton wavelength
> divided by 2pi at low energies.
Then you should admit that there is no position operator in
relativistic quantum theory.
So, you deny the existence of the most basic and ubiquitous observable
- the position. Then you should explain how it happens that the
observable of position, which served us so well in non-relativistic
QM, suddenly disappears when we consider relativistic velocities.
My point of view is that the observable of position is equally well
defined in both non-relativistic and relativistic QM. In the latter
case, it is represented by the Newton-Wigner operator. Eigenfunctions
of this operator are delta-functions in the position space, and
electrons can be well localized in regions smaller than their Compton
wavelength.
> > On the other hand, we still don't have a clear experimental
> > measurement of the speed of propagation of Coulomb forces or
> > gravitational forces. The strongest "argument" against instantaneous
> > interactions usually comes from the special relativistic ban on
> > superluminal signal propagation. However, I am ready to prove that the
> > ban doesn't apply in this particular case. Though, this can be a topic
> > for a separate thread.
>
> Are you making a new "force" other than the EM force? Coulomb forces
> have to be described by the EM force which has the photon as the
> carrier. Are you saying photons can propagate instantaneously in the
> Coulomb case?
I am not accepting the idea that electromagnetic forces are produced
by exchanges of (virtual) photons. I think this is a wrong idea that
stems from too literal interpretation of Feynman diagrams. The
"dressed particle" approach tells me that all standard predictions of
QED (including the Lamb shift, anomalous magnetic moments, etc.) can
be obtained in a theory where physical electrons interact with each
other via instantaneous potentials.
Of course, only experiment can decide whether Coulomb and magnetic
interactions are retarded or instantaneous. I would love to see such
an experiment done.
> > "Never ask about something you cannot
> > observe".
>
> Oh sheesh, I wonder about stuff we can't observe all the time. ;-)
Please, stop doing that. Seriously.
Eugene.
Oh No
>On Mar 27, 3:14 pm, "FrediFizzx" <fredifi...@hotmail.com> wrote:
>
>> For an example, we can't really localize an electron to a
>> "distance" smaller than on the order of the electron compton wavelength
>> divided by 2pi at low energies.
>
>Then you should admit that there is no position operator in
>relativistic quantum theory.
>So, you deny the existence of the most basic and ubiquitous observable
>- the position. Then you should explain how it happens that the
>observable of position, which served us so well in non-relativistic
>QM, suddenly disappears when we consider relativistic velocities.
Fred makes a sound point. While the position operator exists, putting an
electron into a perfect eigenstate of position requires an infinite
energy. There is nothing new here. This problem exists in classical
electrodynamics also.
>> > On the other hand, we still don't have a clear experimental
>> > measurement of the speed of propagation of Coulomb forces or
>> > gravitational forces. The strongest "argument" against instantaneous
>> > interactions usually comes from the special relativistic ban on
>> > superluminal signal propagation. However, I am ready to prove that the
>> > ban doesn't apply in this particular case. Though, this can be a topic
>> > for a separate thread.
>>
>> Are you making a new "force" other than the EM force? Coulomb forces
>> have to be described by the EM force which has the photon as the
>> carrier. Are you saying photons can propagate instantaneously in the
>> Coulomb case?
Any relativistic wave function propagates instantly. Nonetheless
observable affects do not.
>I am not accepting the idea that electromagnetic forces are produced
>by exchanges of (virtual) photons. I think this is a wrong idea that
>stems from too literal interpretation of Feynman diagrams. The
>"dressed particle" approach tells me that all standard predictions of
>QED (including the Lamb shift, anomalous magnetic moments, etc.) can
>be obtained in a theory where physical electrons interact with each
>other via instantaneous potentials.
Potentials are merely expectations of the photon creation/annihilation
operator, more normally called a field operator. The standard dogma that
one should not take Feynman diagrams too literally has held science back
for 50 years imv. Far better to think about what Feynman diagrams really
mean as topological descriptions of physical processes.
>Of course, only experiment can decide whether Coulomb and magnetic
>interactions are retarded or instantaneous. I would love to see such
>an experiment done.
We know that observable effects are not instantaneous. We ought to think
more about whether EPR can be explained by retarded effects. That is
what the maths says happens.
>
>> > "Never ask about something you cannot
>> > observe".
>>
>> Oh sheesh, I wonder about stuff we can't observe all the time. ;-)
>
>
>Please, stop doing that. Seriously.
Again, this is a dogma which has held science back for fifty years. In
order to proceed to do research, one has to accept a paradigm in which
the best methodology for research, as used by Einstein and Feynman, is
prohibited. And then Smolin asks "why no new Einstein?"
eugene_st...@usa.net
> Fred makes a sound point. While the position operator exists, putting an
> electron into a perfect eigenstate of position requires an infinite
> energy. There is nothing new here.
You are right. Position eigenstates have infinite uncertainty of
momentum and energy. However, I don't see any contradiction here.
Momentum eigenstates also have infinite uncertainty of position.
However, this is not a reason to deny the existence of the momentum
operator, and the observability of momentum.
> We know that observable effects are not instantaneous.
How do you know that? Do you have a proof, either experimental or
theoretical?
Eugene.
FrediFizzx
news:guest.20070327164831$56...@news.killfile.org...
> "FrediFizzx" <fredi...@hotmail.com> writes:
>> And come to think of it,
>> can any electron really be "free"? IOW, what is the definition of
>> "free"?
>
> no interaction (w.r.t. inertia, the gravitative interaction with the
> rest of
> the universum is not counted)
But this "no interaction" can only be an abstract concept since if there
is no interaction then we don't even know that the electron is there.
If a "free" electron is "emitting or absorbing" (see, I even use the
shortcut myself) photons, they must be photons that we cannot detect.
;-)However in the Standard Model, electrons are always interacting with
a Higgs-like field so never really "free". In the case of a "free"
electron moving toward my rest postion at a constant velocity, could I
detect photons from it? I don't think so. At least not untill it is
within a certain very short range of the detector.
Best,
Fred Diether
Peter
> > > And come to think of it,
> > > can any electron really be "free"? IOW, what is the definition of
> > > "free"?
> > no interaction (w.r.t. inertia, the gravitative interaction with the rest
> of
> > the universum is not counted)
> > Peter
>
> I'm unsure about "inertia" for example, suppose
> two electrons possess differing speeds like,
>
> (-) ==========> electron "a"
>
> (-)====> electron "b"
>
> The Electro-potential energy (ab/r) varies, but
> it may not necessarily emit radiation.
If both electrons interact electrostatically, they are not free
Peter
Peter
> "Peter" <end...@dekasges.de> wrote in message
> news:guest.20070327164831$56...@news.killfile.org...
> > "FrediFizzx" <fredi...@hotmail.com> writes:
>
> >> And come to think of it,
> >> can any electron really be "free"? IOW, what is the definition of
> >> "free"?
> >
> > no interaction (w.r.t. inertia, the gravitative interaction with the
> > rest of
> > the universum is not counted)
>
> But this "no interaction" can only be an abstract concept since if there
> is no interaction then we don't even know that the electron is there.
Any subject of science is defined by abstraction.
Newton's Laws are won by abstraction as Maxwell's equations are (they discard
all other interactions in real matter).
> If a "free" electron is "emitting or absorbing" (see, I even use the
> shortcut myself) photons, they must be photons that we cannot detect.
A free electron moves along straight lines => no emission of photons (where
the energy of the photons should come from?)
> Best,
Peter
Peter
> On Mar 27, 6:56 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
> > Fred makes a sound point. While the position operator exists, putting an
> > electron into a perfect eigenstate of position requires an infinite
> > energy. There is nothing new here.
> You are right. Position eigenstates have infinite uncertainty of
> momentum and energy. However, I don't see any contradiction here.
> Momentum eigenstates also have infinite uncertainty of position.
> However, this is not a reason to deny the existence of the momentum
> operator, and the observability of momentum.
I would like to support this argument :-)
> > We know that observable effects are not instantaneous.
> How do you know that? Do you have a proof, either experimental or
> theoretical?
Perhaps not, because one cannot make predictictions about *all* future
physics and experiments. IMHO, the only sure what one can say is, that, (i),
ponderable matter cannot reach vacuum speed of light; (ii), electromagnetic
waves move with a final speed.
Peter
Oh No
>On Mar 27, 6:56 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>
>> Fred makes a sound point. While the position operator exists, putting an
>> electron into a perfect eigenstate of position requires an infinite
>> energy. There is nothing new here.
>
>You are right. Position eigenstates have infinite uncertainty of
>momentum and energy.
>However, I don't see any contradiction here.
>Momentum eigenstates also have infinite uncertainty of position.
>However, this is not a reason to deny the existence of the momentum
>operator, and the observability of momentum.
That is not the issue here. The issue is that the e.m. field associated
with a position eigenstate has infinite energy. This does not cause me
any problem, because I formulate the theory in terms of bare electrons
and I keep photon states mathematically distinct, but it is a problem
for your dressed particle approach in which the mass of your dressed
particle goes infinite in a position eigenstate. It appears to me that
if you do that then the position operator cannot be said to exist.
>> We know that observable effects are not instantaneous.
>
>
>How do you know that? Do you have a proof, either experimental or
>theoretical?
Yes, starting with Einstein's 1905 paper, brought up to date by
replacing speed of light with maximum speed of information, to the
locality condition in qed which shows directly that observable operators
have no amplitude outside the light cone, and going through to my own
demonstration that a short scale modification to qed which fixes the
mathematical problems will also modify the structure of space time
itself, yielding Einstein's field equation.
If you want to modify qed in some fundamental way such that you throw
out or change the locality condition, then you have a great deal of work
to do to show rigorously that your modification is consistent with
empirical evidence and that it is mathematically consistent. You can't
do that in an approach with attempts to deal with divergences by
including divergent quantities in the Hamiltonian. Waving your hands and
saying the resulting model works is not good enough. If it is based on
quantities with no mathematical existence, then it is not even a
mathematical model, and I would reject it too as a model of physics.
Oh No
Mathematically of course, the problem is that the standard model does
not exist.