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Solar neutrinos and Gravitational differences of a chemical nature
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Eugene Stefanovich
science forum Guru


Joined: 24 Mar 2005
Posts: 519

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

Arnold Neumaier wrote:

Quote:

All renormalization until today follows the same pattern,
including yours. One does certain formal computations at
finite cutoff and at some point where it no longer harms
moves the cutoff to infinity, being left with approximate
formulas at some (fixed or variable) loop order which no
longer contain a cutoff and have finite values.


In Tomonaga-Schwinger-Feynman theory the cutoff-dependent
counterterms are
added to the Hamiltonian. When the cutoff momentum is finite, everything
is finite: the Hamiltonian, the counterterms, the S-matrix. However,
when the cutoff goes to infinity
the conterterms do not have a finite limit. That's what I call
"the Hamiltonian is infinite". The S-matrix does have a definite limit,
but the Hamiltonian does not.


This is only because you ignore wave function renormalization,
which is of course necessary to get finite matrix elements
for the Hamiltonian.

The limiting Hamiltonian is perturbatively well-defined in the
physical Hilbert space obtained as limit of renormalized Hilbert
spaces at finite cutoff, as the cutoff goes to infinity.
For this, it is best to work in the functional Schroedinger picture.


Maybe I should be even more precise: The Hamiltonian in T-S-F theory
is a polynomial function of creation and annihilation operators of
(bare) particles. In the limit of infinite cutoff the coefficients
of the terms in this polynom do not tend to finite values.
(I simply say that they are infinite). RQD approach suggests to
apply a unitary transformation to this Hamiltonian (you can also
consider this as a unitary redefinition of particles, i.e., the
transition from bare to dressed particles) which makes the coefficients
in the Hamiltonian finite (in the limit of infinite cutoff).

This is discussed at length in my recent article hep-th/0503076
where I use a simple model of QFT as an example. I would appreciate
any comments/opinions/ideas about this paper.

Quote:

In the Glazek-Wilson theory and in RQD both the Hamiltonian and
the S-matrix have finite limits at infinite cutoff.

I probably haven't stressed enough that I am always talking about
theories in the limit of infinite cutoff.


It is less prone to misunderstanding if you avoid the sloppy use
of 'infinite' and talk about the limiting process.

I'll try to be less sloppy in the future. You can change

the word 'infinite' in my previous posts to the
'does not have a finite value in the limit of infinite cutoff'.

Eugene Stefanovich.
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I.Vecchi
science forum Guru Wannabe


Joined: 05 May 2005
Posts: 124

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: non-GR theories of gravity Reply with quote

Joseph Lazio wrote:
Quote:
... we know of one kind of dark matter already,
neutrinos. They appear to have some mass and they do not interact
via
the electromagnetic force (i.e., they are dark). If there's one kind
of dark matter, why not another?


Yeah, why not?
Because physics is about making testable predictions, not about waving
problems away. MOND has an impressive series of accurate predictions
to its credit (and it has apparently failed some tests too ), while
dark matter is virtually unfalsifiable ([1]).

Well, since I am at it, I'll make a prediction too. It's a follow up
to my other post in this thread.

Consider MOND's basic equation
mu(a/a0)*a=-NABLA(GPN) (1)
GPN being the Newtonian gravitational potential and mu being a function
such that m(x)=1 for x>>1 and m(x)=x for x<<1. Milgrom regards a0~10
exp(-Cool sec cm exp(2) as a fundamental constant.

Having noticed that missing mass is mostly needed in dim objects, one
may speculate that a0 and its fellow free parameter M/L in MOND are
actually phenomenological parameters adjusting for the capacity of the
information channel (CoIC) relaying the observer and the observed
object (or more accurately, the corresponding quantum fluctuation) , in
the galaxies' case the hydrogen atoms from which information is
extracted.
For a standard candle CoIC decreases as 1/Dist exp(2), where Dist
denotes the distance between the observer and the candle. In general
CoIC is something like InfC/Dist exp(2), InfC being the the amount of
information the object churns out, which scales with (and can be
replaced by) its luminosity L.
One can then replace (1) with
mu(a/k(CoiC))*a=-NABLA(GPN) (2)
where k is a decreasing function, rapidly approaching 0 as CoIC
increases and such that k(CoIC)~a0 when CoIC has a value corresponding
to systems where MOND has been verified. Equation (2) encodes the
conjecture that the observed dynamics deviates from the Newtonian
regime as the capacity of the information channel sinks.

In the surveys I have inspected, distant galaxies are not considered,
since they yield data of inferior quality (precisely because they have
a small CoIC). On the basis of (2) I conjecture that, once properly
measured, they will exhibit large discrepancies between visible and
classical Newtonian dynamical mass at accelerations larger than a0.

I take bets on the outcome.

IV

[1] http://www.arxiv.org/abs/astro-ph/0204521
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Eugene Stefanovich
science forum Guru


Joined: 24 Mar 2005
Posts: 519

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

Arnold Neumaier wrote:
Quote:
Eugene Stefanovich wrote:

Igor Khavkine wrote:


There is absolutely no relation between bare particles and what are
called "virtual particles". The latter are simply not particles, they
are squiggles on paper.

There is a relation. When you draw a real electron in QED you first draw
a line corresponding to a "bare" electron. Then you add lines of virtual
photons that begin and end on the bare line. Then you add small
virtual electron-positron loops to the virtual photon lines, etc.
So, loosely speeking, in QED real electron = bare electron + coat of
virtual particles.


Not 'in QED' but only 'in standard perturbative QED'.
One can do QED in many ways, and depending on how it is done,
what is virtual is very different. In NRQED, which is the version
of QED used for high agguracy calculations of the Lamb shift
(and hence responsible for the supreme trust in QED),
there are no virtual particles at all.


How can it be so? From what I read about NRQED, its Hamiltonian
contains trilinear terms, like a^+c^+a. This means that if you
prepare a state of one electron a^+|0> at time=0, then after a
short time, this state will evolve into exp(iHt)a^+|0> which
is an infinite linear combination of multiparticle states.
That's what I call dressing by a coat of virtual particles.
That's what I find unphysical in all QFT theories which are
not presented in the dressed particle form.
Of course, this problem can be remedied by a unitary transform
to dressed particles or by a (equivalent) unitary transformation of
the Hamiltonian. But I haven't seen if anybody have done that for
NRQED.

Eugene Stefanovich.
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Eugene Stefanovich
science forum Guru


Joined: 24 Mar 2005
Posts: 519

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

Igor Khavkine wrote:

Quote:
Can you give a reference to your approach besides the PDF file
from your website?

In addition to the book www.meopemuk.com/book.pdf
there are two papers: Ann. Phys. 292 (2001), 139 and hep-th/0503076.
A couple of related papers can be also found on
www.geocities.com/meopemuk

Quote:
When you draw a real electron in QED you first draw
a line corresponding to a "bare" electron. Then you add lines of virtual
photons that begin and end on the bare line. Then you add small
virtual electron-positron loops to the virtual photon lines, etc.
So, loosely speeking, in QED real electron = bare electron + coat of
virtual particles.


That is one way to look at it, but is not the only one. I need not draw
bare particle lines at all. If I wish, I can only draw dressed lines.
In this formulation scattering amplitudes can still be expanded in
terms of Feynman diagrams, but only dressed propagators will be used.
Wherever Feynman diagrams will have internal lines and loops,
integration over the momenta of these internal lines is still required.
These internal lines whose momenta must be integrated over are commonly
referred to as "virtual particles". In this formulation, there are no
bare particles, but still there are virtual particles. Conclusion: bare
particles have nothing to do with virtual particles and vice versa.
And, I'm afraid I'm repeating myself, "virtual particles" are not
particles at all!

I think we are moving closer to agreement here.
If you are interested only in the S-matrix, you have a great freedom
in choosing the Hamiltonian, particle representation, and Feynman rules.
you can use the bare particle representation + renormalization (as it
was done more that 50 years ago) and get exact results. However, if you
are interested in the time evolution of wave functions, you MUST choose
the Hamiltonian in the dressed particle representation. Otherwise,
you may easily get a wrong physical interpretation. Otherwise, in your
theory a single electron will dissociate into a bunch of multiparticle
states, and you may get a wrong impression that two electrons interact
with each other by throwing virtual particles.

This never happens in the dressed particle representation. A single
electron remains single at all times. Two electrons interact via
instantaneous potentials like V = a^+a^+aa. Of course, when the
S-matrix is calculated with these potentials you'll meet products
like V^2, V^3, etc. which can be represented by loop diagrams in the
Feynman's technique.

The point we were discussing was how real particles interact.
The answer is not obvious unless you express the interaction operator
in terms of creation and annihilation operators of real (dressed)
particles. If you do that, then the instantaneous (not retarded) form
of interaction V = a^+a^+aa becomes obvious.

Quote:


According to classical electromagnetic theory, electromagnetic waves
(= real photons in my language) can be created only by charges in
accelerated motion. Now, let us take a van-der-Graaf generator with
two
highly charged metallic balls. There is no movement of charges
accelerated or otherwise. So, there are no electromagnetic waves in
this
case. If you put the generator in a dark room, place photographic
plates everywhere around it, and wait for 100 years, you'll not find
a
single blackened spot on any of the plates. So, there was no single
real photon in the room. However, there was a huge electrostatic
attraction between the two
balls. How are you going to apply your statement "Electromagnetic
waves
== electromagnetic interaction" in this case?


Yes, I will apply that statement. And if you will indeed perform the
experminet with a van der Graaf generator, you will see spots on the
plates. Well, perhaps the photographic paper is not sensitive for low
enough fequencies, in that case I suggest a simple radio reciever. You
say that electromagnetic radiation is associated with accelerating
charges. But tell me, how do you charge up a van der Graaf generator
without accelerating any charges? Even if you assume that the electric
fields have settled into a static configuration, the statement that
this interaction can still be explained with electromagnetic waves is
nothing but the statement that the electric field can be written in a
Fourier decomposition in terms of different frequency components.

I am sure that after all moving parts in the generator have stopped
and static field has settled down, you will not be able to detect
neither visible nor radio waves (photons) around it. There could be
some radiation due to tunneling of electrons between the balls,
I am not sure about that. But let us disregard this exotic possibility.

Quote:

If you've ever witnessed interference in radio or TV signals, have you
ever wondered why it is called "static"?

That's wrong usage of the word. There could be no EM waves emitted
by a static set of charges in the ground state.

Quote:

And if you think of the thought experiment that I propose to you (take
to relatively stationary particles, perturb one of them at event (x,t)
and measure the event (x',t') where the second particle will start
feeling the effects) you'll see that particle one cannot be perturbed
without suffering acceleration and thus emitting radiation. There is
nothing mysterious that the space-time separation between events (x,t)
and (x',t') can only be light-like.

I disagree. According to my interpretation of dressed particle QED,
the second particle at (x',t) will start to feel the perturbation of
the first particle (x,t) immediately (t'=t) due to the direct Coulomb
and magnetic a^+a^+aa terms in the Hamiltonian. There will be another
kick felt by the second particle at time t' = t + |x' - x|/c
due to REAL photons emitted by the accelerated 1st particle.
This second kick is what is usually described by the transverse
electromagnetic
wave. The first kick (at t'=t) is due to action-at-a-distance.

So far there were no experiments unambiguously confirming/rejecting
my proposal. The closest thing is the observation of superluminal
"photon tunneling" in FTIR. However, I agree that this experiment
allows different interpretations. Let's wait and see when
experimentalists will come up with something better.


Quote:


As I understand, you think that the Coulomb interaction was
transmitted by some material carrier (field) which cannot be detected
by itself (it doesn't blacken photographic plates, it doesn't leave
traces in bubble chambers, etc). In other words, the field is a form
of matter quite different from real particles (photons, electrons,
neutrinos, etc). I am not talking about the transverse
electromagnetic
field which freely propagates in space and, in my language, is
equivalent to the flow of real particles photons. I am talking about
the static Coulomb field. The only way to detect such a field is by
its
action on charged particles.


Blackening photographic plates is not the only way to detect a field.
Your eyes, your radio, and charged particles in the ionosphere do a
good job as well. The electromagnetic field is detectable and changes
in the electromagnetic field and the field itself *are* electromagnetic
waves, that is can they can be decomposed in oscillating Fourier modes.

Eyes and radio detect transverse EM fields, that I prefer to call
particles (photons). In RQD they are equivalent to freely propagating
photons. Static (Coulomb and magnetic) fields can only
be detected by forces exerted on test charges. In RQD, these forces are
described by the terms like a^+a^+aa in the Hamiltonian.
These forces are instantaneous (the force acting on particle A depends
on the position and velocity of particle B at the same time instant).
From this I conclude that EM interaction is a combination of
action-at-a-distance and "exchange" of real photons moving at the
speed of light.

Quote:


My interpretation is that there is no such field. The interactions
between charged particles can be described more economically by
using direct instantaneous interparticle potentials
(Coulomb and magnetic). I understand that currently there is no
convincing experiment that allows us to distinguish between our two
interpretations. So, I admit that your point of view has a right to
exist. All I am asking from you is to admit that my point of view
does not
violate any sacred physical principle, does not contradict
experiment,
and also has a right for existence. Let future experiments decide who
was right.


Your interpretation is at odds with not only with the conventional one,
but from the above I can also see that you do not recognize well
established experimental evidence. So far I have not seen you propose
an experiment that would be able to falsify your hypothesis. Care to
speculate?

The idea of the experiment is simple (its practical implementation is not):
as you said, take 2 charges at some distance, wiggle the first charge,
and record the time when the second charge starts to wiggle in response.

As I discuss in subsection 12.3.5 of the book, the FTIR experiment
is very close to this ideal: when the light wave or microwave falls on
the glass-air interface it reflects and induces vibration of dipoles
on the interface. This creates time-varying Coulomb and magnetic fields
in the gap. They are usually called "evanescent waves". It is important
that the transverse field is not present in the gap (it totally
reflects), so there is no admixture of the retarded interactions to
the forces acting between dipoles on two sides of the gap. According
to my views, these forces are purely instantaneous. The induced
oscillation of dipoles on the other side of the gap creates EM wave
(photons) propagating in the other piece of glass. This makes an
impression that photons have tunnelled across the gap faster than
light.

I thought about changing this experiment by eliminating the transverse
fields entirely. For example, one can use ultrasound waves instead
of microwaves. If one can generate a wave of optical phonons
(polarization) in the material, this wave will also reflect from the
interface and create time-varying dipole moments there. These dipoles
will induce dipoles on the other side of the gap, and a corresponding
ultrasound wave there. If I am right, then the dipoles on both
sides of the gap interact instantaneously, and the whole process
will be seen as superluminal tunneling of ultrasound across the gap.
Unfortunately the period of ultrasound vibrations is many orders
of magnitude larger that the time required for light to cross the
gap. So, it is nearly impossible to distinguish superluminal and
subluminal situations in such experiment.

I invite you to give some thought to possible experimental
tests of "retarded vs. instantaneous" interactions. In my opinion,
such an experiment is long overdue.

Eugene Stefanovich.
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Arnold Neumaier
science forum Guru


Joined: 24 Mar 2005
Posts: 379

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

Eugene Stefanovich wrote:
Quote:
Arnold Neumaier wrote:

Could you please give me references to works where "closed path"
dynamical predictions were compared with experiment. It would be
especially useful if some quantum field effects (radiative corrections)
were involved and if the studied system was simple (not plasma
or solid state, etc.) and at zero temperature.

One important application of the CTP formalism is the derivation of
quantum Boltzmann equations; these are used to model the motion of
plasmas, where plenty of experimental material is available.
But you have to look it up yourself; I don't have details and don't
want to do your research.


Arnold Neumaier
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Arnold Neumaier
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Posts: 379

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

Franz Heymann wrote:
Quote:
"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
news:422EB74C.5020806@univie.ac.at...

Franz Heymann wrote:

"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
news:422AC49F.3030502@univie.ac.at...

Franz Heymann wrote

Unstable particles are modelled exactly like stable particles,
namely as external lines in a Feynman diagram.
Virtual particles in Feynman diagrams are exactly those
which are not given by external lines.

Hence what is real and what is virtual is not affected by a
diagram rotation - this only affects what is input oand what is

output.

I am not happy with that.
Let me illustrate my unhappiness with a specific example.
Take the resonant elastic scattering case:

pi + p --> Z --> pi + p

I hope we both agree that the Z is produced as a real but unstable
particle in the s-channel.

No. Z is a virtual particle since it is an intermediate line.

You are doing nothing to alleviate my unhappiness.
If the Z were a virtual prticle, it would not be on its mass shell and
a mass measurement would not be possible by studying the reaction.
It is not on its (complex) mass shell when it is in an internal line as
in your diagram.

When measured, one has instead a diagram where the Z is detected,
and hence on an external line of Feynman diagrams associated to
the coresponding cross section!

In ordinary scattering experiments the existence of the Z shows in
resonances of the cross section. This is evidence for the presence
of the Z in the action, and can be used to calculate properties of Z,

But it is observed, and its (complex) mass is actually measured.

It is _not_ observed in an experiment where the Feynman diagram
you drew is part of the calculation.

It _can_ be observed, but only in a process analyzed with
Feynman diagrams where the Z is an external line.

I maintain that it is an external line which has its far end
terminated in the decay of the resonance.

Fine; then we agree. For an external line remains external under
rotating the diagram.

Initially you argued that rotation changes virtual particles
to real ones; it was only that which I objected to.


Quote:
In atomic physics, a photon may be absorbed by an atom in its ground
state. The excited state very quickly decays back to the ground state
by the emission of a photon.

Are you now saying that the excited state is not a real, but a virtual
state?

No. There is a big difference between the two.

real = external line = on-shell = observable
virtual = internal line = off-shell = ficiticous

excited states are bound states and usually appear _only_
as external lines.


Arnold Neumaier
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Igor Khavkine
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Joined: 01 May 2005
Posts: 607

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

Eugene Stefanovich wrote:
Quote:
Igor Khavkine wrote:


Guessing the form of the quantum Hamiltonian is called
quantization.
Afterward, you keep the Hamiltonian *operator* the same while its
*matrix elements* may change because the basis in which they are
written
is may change. This is true regardless of how many times you claim
that
your unitary transformation changes the Hamiltonian.

There are two equivalent approaches to dressing. One approach
promoted
by Shirokov and Shebeko is, as you said, to keep the Hamilton
operator
of QED the same while change the definition of particles (from bare
particles
to dressed particles) and rewrite the Hamiltonian in terms of dressed
particle operators. Another approach is to keep the original particle
operators (just call them dressed) and to apply the unitary dressing
transformation directly to the Hamiltonian. Both approaches are
completely
equivalent (just as the Schroedinger and Heisenberg pictures of
quantum
mechanics are equivalent). This is discussed in subsection 12.1.10 of
my book. If you prefer to use the language of the first approach,
this
is OK with me. The confusion arose because I used the language of the
2nd approach in my book. I say that the unitary dressing
transformation
changes the Hamiltonian because the functional dependence of the
Hamiltonian on the dressed particle operators is different from
the functional dependence of the original QED Hamiltonian on the bare
particle operators.

There to "approaches" as you call them are exactly the same. What is
confusing that the meaning of your words does not coincide with what
you actually want to say. You give Shirokov and Shebeko as a reference
for the first approach. I can support that with any textbook on linear
algebra. Can you give a reference to your approach besides the PDF file
from your website? Part of good communication is the use of language
that is appropriate and unambiguous for both parties.

Quote:
You are right, I do need bare particles BEFORE the correct
Hamiltonian is
written. I need them to go through the 3 steps mentioned above. I
hate to
do that, but I do not have other choice at the moment. My point was
that I
do not need bare or virtual particles AFTER the RQD Hamiltonian is
written.


There is absolutely no relation between bare particles and what are
called "virtual particles". The latter are simply not particles,
they
are squiggles on paper.

There is a relation. When you draw a real electron in QED you first
draw
a line corresponding to a "bare" electron. Then you add lines of
virtual
photons that begin and end on the bare line. Then you add small
virtual electron-positron loops to the virtual photon lines, etc.
So, loosely speeking, in QED real electron = bare electron + coat of
virtual particles.

That is one way to look at it, but is not the only one. I need not draw
bare particle lines at all. If I wish, I can only draw dressed lines.
In this formulation scattering amplitudes can still be expanded in
terms of Feynman diagrams, but only dressed propagators will be used.
Wherever Feynman diagrams will have internal lines and loops,
integration over the momenta of these internal lines is still required.
These internal lines whose momenta must be integrated over are commonly
referred to as "virtual particles". In this formulation, there are no
bare particles, but still there are virtual particles. Conclusion: bare
particles have nothing to do with virtual particles and vice versa.
And, I'm afraid I'm repeating myself, "virtual particles" are not
particles at all!

Quote:
According to classical electromagnetic theory, electromagnetic waves
(= real photons in my language) can be created only by charges in
accelerated motion. Now, let us take a van-der-Graaf generator with
two
highly charged metallic balls. There is no movement of charges
accelerated or otherwise. So, there are no electromagnetic waves in
this
case. If you put the generator in a dark room, place photographic
plates everywhere around it, and wait for 100 years, you'll not find
a
single blackened spot on any of the plates. So, there was no single
real photon in the room. However, there was a huge electrostatic
attraction between the two
balls. How are you going to apply your statement "Electromagnetic
waves
== electromagnetic interaction" in this case?

Yes, I will apply that statement. And if you will indeed perform the
experminet with a van der Graaf generator, you will see spots on the
plates. Well, perhaps the photographic paper is not sensitive for low
enough fequencies, in that case I suggest a simple radio reciever. You
say that electromagnetic radiation is associated with accelerating
charges. But tell me, how do you charge up a van der Graaf generator
without accelerating any charges? Even if you assume that the electric
fields have settled into a static configuration, the statement that
this interaction can still be explained with electromagnetic waves is
nothing but the statement that the electric field can be written in a
Fourier decomposition in terms of different frequency components.

If you've ever witnessed interference in radio or TV signals, have you
ever wondered why it is called "static"?

And if you think of the thought experiment that I propose to you (take
to relatively stationary particles, perturb one of them at event (x,t)
and measure the event (x',t') where the second particle will start
feeling the effects) you'll see that particle one cannot be perturbed
without suffering acceleration and thus emitting radiation. There is
nothing mysterious that the space-time separation between events (x,t)
and (x',t') can only be light-like.

Quote:
As I understand, you think that the Coulomb interaction was
transmitted by some material carrier (field) which cannot be detected
by itself (it doesn't blacken photographic plates, it doesn't leave
traces in bubble chambers, etc). In other words, the field is a form
of matter quite different from real particles (photons, electrons,
neutrinos, etc). I am not talking about the transverse
electromagnetic
field which freely propagates in space and, in my language, is
equivalent to the flow of real particles photons. I am talking about
the static Coulomb field. The only way to detect such a field is by
its
action on charged particles.

Blackening photographic plates is not the only way to detect a field.
Your eyes, your radio, and charged particles in the ionosphere do a
good job as well. The electromagnetic field is detectable and changes
in the electromagnetic field and the field itself *are* electromagnetic
waves, that is can they can be decomposed in oscillating Fourier modes.

Quote:
My interpretation is that there is no such field. The interactions
between charged particles can be described more economically by
using direct instantaneous interparticle potentials
(Coulomb and magnetic). I understand that currently there is no
convincing experiment that allows us to distinguish between our two
interpretations. So, I admit that your point of view has a right to
exist. All I am asking from you is to admit that my point of view
does not
violate any sacred physical principle, does not contradict
experiment,
and also has a right for existence. Let future experiments decide who
was right.

Your interpretation is at odds with not only with the conventional one,
but from the above I can also see that you do not recognize well
established experimental evidence. So far I have not seen you propose
an experiment that would be able to falsify your hypothesis. Care to
speculate?

Igor
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Arnold Neumaier
science forum Guru


Joined: 24 Mar 2005
Posts: 379

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

Eugene Stefanovich wrote:
Quote:
Igor Khavkine wrote:

There is absolutely no relation between bare particles and what are
called "virtual particles". The latter are simply not particles, they
are squiggles on paper.

There is a relation. When you draw a real electron in QED you first draw
a line corresponding to a "bare" electron. Then you add lines of virtual
photons that begin and end on the bare line. Then you add small
virtual electron-positron loops to the virtual photon lines, etc.
So, loosely speeking, in QED real electron = bare electron + coat of
virtual particles.

Not 'in QED' but only 'in standard perturbative QED'.
One can do QED in many ways, and depending on how it is done,
what is virtual is very different. In NRQED, which is the version
of QED used for high agguracy calculations of the Lamb shift
(and hence responsible for the supreme trust in QED),
there are no virtual particles at all.


Arnold Neumaier
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Eugene Stefanovich
science forum Guru


Joined: 24 Mar 2005
Posts: 519

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

Arnold Neumaier wrote:
Quote:
Eugene Stefanovich wrote:

Arnold Neumaier wrote:


Eugene Stefanovich wrote:


Arnold Neumaier wrote:


I wonder how _you_ predict a relativistic quantum Boltzmann equation
for an plasma in an electromagnetic field, which is surely part of QED.

No problem, RQD approach has a well-defined finite Hamiltonian in which
particles interact via direct instantaneous potentials, just as in
ordinary non-relativistic quantum mechanics. This Hamiltonian can be
directly plugged into the time evolution operator, or diagonalized to
find the stationary states. With this Hamiltonian, you can define the
partition function exp(-\beta H) and all kinds of thermodynamical and
statistical quantities just as in ordinary quantum statistics.

The Boltzmann equation is not the Boltzmann distribution but a kinetic
equation for the 1-particle phase space density. If you would be able
to reproduce the qantum Boltzmann equation of Klimontovich with your
formalism, I'd be very surprised. He is doing real dynamical work with
QED, and gets manifestly covariant equations that are useful in practice.

Repeat this feat before making again your exaggerated claims!

Are you saying then that in order to get non-zero-temperature dynamics
one needs to have something else beside the Hamiltonian?


Instead of the Hamiltonian one can also have the action. This are
two equivalent starting point and the latter is much superior in the
relativistic case.

Try to compute the hydrodynamic limit of your theory and see
whether you can get it into the form of the relativistic perfect
fluid in an electromagnetic field. (There is no freedom here -
you must arrive exactly at these equations!) You'll see that you have
_much_ more difficulty in your formulation than if you stay always
on the action level model.

You say that your theory is equivalent. But it is _not_ equivalent in
terms of complexity of computations, where yours is much worse.



OK, there is no point for me to argue. I have a very vague idea about
what Schwinger, Keldysh, and Klimontovich did in this area, so I better
roll up my sleeves and go study before answering you.

Quote:


So, in principle, the two theories could be distinguished
by their predictions of observable effects. However, I doubt that
experimental observations of plasma's dynamics are sensitive to
radiative corrections.


If there were a different prediction presented with some authority
(which in particular involves removing your superluminal assertions)
some experimentalists would probably be motivated to push for a
discrimination.

Could you please give me references to works where "closed path"
dynamical predictions were compared with experiment. It would be
especially useful if some quantum field effects (radiative corrections)
were involved and if the studied system was simple (not plasma
or solid state, etc.) and at zero temperature.

Eugene Stefanovich.
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Arnold Neumaier
science forum Guru


Joined: 24 Mar 2005
Posts: 379

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

Eugene Stefanovich wrote:

Quote:
Arnold Neumaier wrote:

All renormalization until today follows the same pattern,
including yours. One does certain formal computations at
finite cutoff and at some point where it no longer harms
moves the cutoff to infinity, being left with approximate
formulas at some (fixed or variable) loop order which no
longer contain a cutoff and have finite values.

In Tomonaga-Schwinger-Feynman theory the cutoff-dependent
counterterms are
added to the Hamiltonian. When the cutoff momentum is finite, everything
is finite: the Hamiltonian, the counterterms, the S-matrix. However,
when the cutoff goes to infinity
the conterterms do not have a finite limit. That's what I call
"the Hamiltonian is infinite". The S-matrix does have a definite limit,
but the Hamiltonian does not.

This is only because you ignore wave function renormalization,
which is of course necessary to get finite matrix elements
for the Hamiltonian.

The limiting Hamiltonian is perturbatively well-defined in the
physical Hilbert space obtained as limit of renormalized Hilbert
spaces at finite cutoff, as the cutoff goes to infinity.
For this, it is best to work in the functional Schroedinger picture.

Maybe I should be even more precise: The Hamiltonian in T-S-F theory
is a polynomial function of creation and annihilation operators of
(bare) particles. In the limit of infinite cutoff the coefficients
of the terms in this polynom do not tend to finite values.

But this is irrelevant. Nobody uses these coefficients, and they have
no physical meaning. What _is_ relevant is that the renormalized
matrix elements
<phi|H|psi> = lim_{Lambda to inf} <phi(Lambda)|H(Lambda)|psi(Lambda)>
exist, since these contain the physics. This can be done in any of
a number of different representations. People currently prefer the
front point form for doing actual calculations, though work by Klink
and his collaborators are exploring the covariant point form.

You are simply exploring instead the instant form.
Which may be useful if you get far enough.
But it does not justify your claim that this is the only
correct way of doing finite dynamical calculations.


Quote:
(I simply say that they are infinite). RQD approach suggests to
apply a unitary transformation to this Hamiltonian (you can also
consider this as a unitary redefinition of particles, i.e., the
transition from bare to dressed particles) which makes the coefficients
in the Hamiltonian finite (in the limit of infinite cutoff).

This is discussed at length in my recent article hep-th/0503076
where I use a simple model of QFT as an example. I would appreciate
any comments/opinions/ideas about this paper.

I'd have appreciated if you had acknowledged the discussions with me
that made you aware of the work by Glazek and Wilson.


Arnold Neumaier
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Joseph Lazio
science forum beginner


Joined: 16 May 2005
Posts: 18

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: non-GR theories of gravity Reply with quote

Quote:
"m" == melroysoares <melroysoares@hotmail.com> writes:

m> Joseph Lazio wrote:
Quote:
"PH" == Phillip Helbig---remove CLOTHES to reply
helbig@astro.multiCLOTHESvax.de> writes:
PH> Personally, I think there is nothing mysterious about dark

PH> matter. After all, why should everything be so convenient so
PH> as to glow so that an astronomer can see it?

Quote:
Particularly given that we know of one kind of dark matter already,
neutrinos. They appear to have some mass and they do not interact
via the electromagnetic force (i.e., they are dark). If there's
one kind of dark matter, why not another?

m> except that neutrino is a hot-dark matter candidate

People seem to be misinterpreting my comment. With respect to the
current state of cosmology, there are at least three options:

a) Our census of the Universe is incomplete;
b) There's new physics; or
c) All of the above.

I think that the most simple explanation is that GR is the correct
form of gravity on macroscopic scales and that we have not detected
all of the particles in the Universe.

Even for luminous matter, depending upon how you split things, there
are at least two different ways to describe it: baryonic (i.e.,
protons, neutrons, etc.) and leptons (electrons, muons, etc.). Of
course, some leptons are also dark matter, notably neutrinos.

m> and we knew in the 80's itself that hot dark matter dominated
m> universe is more or less ruled out.

Yes, but that doesn't rule out the possibility that there are other
forms of dark matter, e.g., cold dark matter.

m> note that besides dark matter (which implied that on galactic
m> scales gravity is more attractive as compared to Newtonian gravity)
m> you also have to account for dark energy (which implies that on
m> cosmologica scales gravity is repulsive) IMO in my opinion a theory
m> which accounts for both ought to be taken seriously.

Of course, GR does both quite perfectly. There's no requirement
within GR that mass be luminous, and the addition of a cosmological
constant is one way of explaining dark energy.

--
Lt. Lazio, HTML police | e-mail: jlazio@patriot.net
No means no, stop rape. | http://patriot.net/%7Ejlazio/
sci.astro FAQ at http://sciastro.astronomy.net/sci.astro.html
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PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: non-GR theories of gravity Reply with quote

Joseph Lazio wrote: > >>>>> "PH" == Phillip Helbig---remove CLOTHES to
reply <helbig@astro.multiCLOTHESvax.de> writes: > > PH> Personally, I
think there is nothing mysterious about dark matter. > PH> After all,
why should everything be so convenient so as to glow so > PH> that an
astronomer can see it? > > Particularly given that we know of one kind
of dark matter already, > neutrinos. They appear to have some mass and
they do not interact via > the electromagnetic force (i.e., they are
dark). If there's one kind > of dark matter, why not another? > except
that neutrino is a hot-dark matter candidate and we knew in the 80's
itself that hot dark matter dominated universe is more or less ruled
out. note that besides dark matter (which implied that on galactic
scales gravity is more attractive as compared to Newtonian gravity)
you also have to account for dark energy (which implies that on
cosmologica scales gravity is repulsive) IMO in my opinion a theory
which accounts for both ought to be taken seriously. I don't know if
Bekenstein's TeVS theory can account for the acceleration of the
universe.

However Mannheim's conformal gravity theory does account for both.
see
http://www.mit.edu/people/cabi/blog/2005/02/mannheims-conformal-gravity.html
for some discussion on this.

Melroy
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Eugene Stefanovich
science forum Guru


Joined: 24 Mar 2005
Posts: 519

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

Igor Khavkine wrote:

Quote:

Guessing the form of the quantum Hamiltonian is called quantization.
Afterward, you keep the Hamiltonian *operator* the same while its
*matrix elements* may change because the basis in which they are written
is may change. This is true regardless of how many times you claim that
your unitary transformation changes the Hamiltonian.

There are two equivalent approaches to dressing. One approach promoted
by Shirokov and Shebeko is, as you said, to keep the Hamilton operator
of QED the same while change the definition of particles (from bare
particles
to dressed particles) and rewrite the Hamiltonian in terms of dressed
particle operators. Another approach is to keep the original particle
operators (just call them dressed) and to apply the unitary dressing
transformation directly to the Hamiltonian. Both approaches are
completely
equivalent (just as the Schroedinger and Heisenberg pictures of quantum
mechanics are equivalent). This is discussed in subsection 12.1.10 of
my book. If you prefer to use the language of the first approach, this
is OK with me. The confusion arose because I used the language of the
2nd approach in my book. I say that the unitary dressing transformation
changes the Hamiltonian because the functional dependence of the
Hamiltonian on the dressed particle operators is different from
the functional dependence of the original QED Hamiltonian on the bare
particle operators.

Hopefully, these remarks clear up the confusion.

Quote:

You are right, I do need bare particles BEFORE the correct Hamiltonian is
written. I need them to go through the 3 steps mentioned above. I hate to
do that, but I do not have other choice at the moment. My point was that I
do not need bare or virtual particles AFTER the RQD Hamiltonian is
written.


There is absolutely no relation between bare particles and what are
called "virtual particles". The latter are simply not particles, they
are squiggles on paper.

There is a relation. When you draw a real electron in QED you first draw
a line corresponding to a "bare" electron. Then you add lines of virtual
photons that begin and end on the bare line. Then you add small
virtual electron-positron loops to the virtual photon lines, etc.
So, loosely speeking, in QED real electron = bare electron + coat of
virtual particles.

Quote:


If you say that interaction is retarded, then, in order to comply with
conservation laws (for momentum and angular momentum, in particular) you
should admit that there is some time interval during which the momentum
transferring between two interacting particles is en route between the
particles. There should be some material carrier which takes this momentum
from one particles, keeps it for some time, and then gives it to the other
particle. There should be some "degrees of freedom" associated with this
carrier. You can call this carrier "virtual particles" or "fields", I do
not care.


I call this carrier the electromagnetic field. And virtual particles have
nothing to do with it.


The important thing is that there should be some material stuff
which can possess momentum, angular momentum, energy, etc, but which so
far has escaped direct observations. The whole point of the RQD approach
is to reformulate the theory in such a way that this mysterious stuff is
not present in the theory just as it is not seen in experiments.


You continually make claims against virtual particles, believing that
these become claims against QED. I can only conclude that you are
confused about what they are. Do you understand that claims against
virtual particles have no impact on QED? Do you understand that virtual
particles are figments of an interpretation of perturbative QFT
calculations? Do you understand that bare particles have nothing to do
with virtual particles, even though they both have "particles" in their
name? Do you understand that virtual particles have nothing to do the
electromagnetic field?


To my knowledge, the retarded character of electromagnetic interactions
has not been confirmed by experiment either. (I am not talking here about
the propagation of EM waves, of course)


You seek experimental evidence, yet you disregard the very evidence that
is in front of your eyes (literally). Electromagnetic waves ==
electromagnetic interaction. This evidence is insufficient for you, only
because you have convinced yourself so.

According to classical electromagnetic theory, electromagnetic waves
(= real photons in my language) can be created only by charges in
accelerated motion. Now, let us take a van-der-Graaf generator with two
highly charged metallic balls. There is no movement of charges
accelerated or otherwise. So, there are no electromagnetic waves in this
case. If you put the generator in a dark room, place photographic
plates everywhere around it, and wait for 100 years, you'll not find a
single blackened spot on any of the plates. So, there was no single
real photon in the room. However, there was a huge electrostatic
attraction between the two
balls. How are you going to apply your statement "Electromagnetic waves
== electromagnetic interaction" in this case?

As I understand, you think that the Coulomb interaction was
transmitted by some material carrier (field) which cannot be detected
by itself (it doesn't blacken photographic plates, it doesn't leave
traces in bubble chambers, etc). In other words, the field is a form
of matter quite different from real particles (photons, electrons,
neutrinos, etc). I am not talking about the transverse electromagnetic
field which freely propagates in space and, in my language, is
equivalent to the flow of real particles photons. I am talking about
the static Coulomb field. The only way to detect such a field is by its
action on charged particles.

My interpretation is that there is no such field. The interactions
between charged particles can be described more economically by
using direct instantaneous interparticle potentials
(Coulomb and magnetic). I understand that currently there is no
convincing experiment that allows us to distinguish between our two
interpretations. So, I admit that your point of view has a right to
exist. All I am asking from you is to admit that my point of view
does not
violate any sacred physical principle, does not contradict experiment,
and also has a right for existence. Let future experiments decide who
was right.

Eugene Stefanovich.
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Igor Khavkine
science forum Guru


Joined: 01 May 2005
Posts: 607

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

On Fri, 11 Mar 2005 08:31:26 +0000, Eugene Stefanovich wrote:

Quote:
I agree, I do not have any idea how to write the RQD Hamiltonians in terms
of dressed particles from first principles. In order to get to this
Hamiltonian one should go through a painful set of steps: canonical gauge
quantization -> renormalization -> dressing. On this way you'll meet a
bunch of objects which, in my opinion, have no any relevance to physics.
These are bare and virtual particles, quantum fields, gauges, etc. They
play useful mathematical roles at some stages of the above derivation, but
they are just parts of a 3-step mathematical trick. This mathematical
trick is designed to guess the correct form of the Hamiltonian, and it
reaches this goal, however the physical significance of this procedure
remains obscure to me. IMHO physics starts AFTER we wrote the correct
(dressed particle) form of the Hamiltonian. Everything we did BEFORE that
(Lagrangians, action, quantum fields, bare particles, gauges, etc.) is no
longer relevant after the Hamiltonian is written. With this Hamiltonian we
can answer any question having relevance to observations.

It is nice that you agree with my corrections to your language. I only
hope that this time you actually stop saying things that require
correction.

You are throwing a large number of objects into the same waste basket.
Just because some of them, say quantum fields, seem obscure to you, they
need not be to other people.

Guessing the form of the quantum Hamiltonian is called quantization.
Afterward, you keep the Hamiltonian *operator* the same while its
*matrix elements* may change because the basis in which they are written
is may change. This is true regardless of how many times you claim that
your unitary transformation changes the Hamiltonian.

The path that has lead you the a renormalized Hamiltonian allows you to
trace back your steps and recover the correct classical limit.


Quote:
You are right, I do need bare particles BEFORE the correct Hamiltonian is
written. I need them to go through the 3 steps mentioned above. I hate to
do that, but I do not have other choice at the moment. My point was that I
do not need bare or virtual particles AFTER the RQD Hamiltonian is
written.

There is absolutely no relation between bare particles and what are
called "virtual particles". The latter are simply not particles, they
are squiggles on paper.

Quote:
If you say that interaction is retarded, then, in order to comply with
conservation laws (for momentum and angular momentum, in particular) you
should admit that there is some time interval during which the momentum
transferring between two interacting particles is en route between the
particles. There should be some material carrier which takes this momentum
from one particles, keeps it for some time, and then gives it to the other
particle. There should be some "degrees of freedom" associated with this
carrier. You can call this carrier "virtual particles" or "fields", I do
not care.

I call this carrier the electromagnetic field. And virtual particles have
nothing to do with it.

Quote:
The important thing is that there should be some material stuff
which can possess momentum, angular momentum, energy, etc, but which so
far has escaped direct observations. The whole point of the RQD approach
is to reformulate the theory in such a way that this mysterious stuff is
not present in the theory just as it is not seen in experiments.

You continually make claims against virtual particles, believing that
these become claims against QED. I can only conclude that you are
confused about what they are. Do you understand that claims against
virtual particles have no impact on QED? Do you understand that virtual
particles are figments of an interpretation of perturbative QFT
calculations? Do you understand that bare particles have nothing to do
with virtual particles, even though they both have "particles" in their
name? Do you understand that virtual particles have nothing to do the
electromagnetic field?

Quote:
To my knowledge, the retarded character of electromagnetic interactions
has not been confirmed by experiment either. (I am not talking here about
the propagation of EM waves, of course)

You seek experimental evidence, yet you disregard the very evidence that
is in front of your eyes (literally). Electromagnetic waves ==
electromagnetic interaction. This evidence is insufficient for you, only
because you have convinced yourself so.

Igor
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Eugene Stefanovich
science forum Guru


Joined: 24 Mar 2005
Posts: 519

PostPosted: Sun May 08, 2005 6:46 pm    Post subject: Re: How real are the "Virtual" partticles? Reply with quote

Arnold Neumaier wrote:
Quote:
Eugene Stefanovich wrote:

Arnold Neumaier wrote:


Eugene Stefanovich wrote:


I am reading some papers trying to grasp the idea of the "closed path
formalism". I cannot understand the following thing: In quantum
theory the time evolution is described by the standard time
evolution operator, i.e., the exponent of the Hamiltonian.
Why someone needs to invent the new "closed path formalism"?

Because relativistic field theory is defined in terms of
actions which are covariant, and picking a Hamiltonian destroys
the manifest symmetry and makes everything look messy.


This is not the case if you
are interested in time evolution. Then you need an explicit expression
for the Hamiltonian (and for 9 other generators of the Poincare group,


There is such an expression for all generators of the Poincare group
in every P-invariant field theory, no matter which form of dynamics
is used.

Agreed. But these expressions destroy manifest covariance.
In the instant form dynamics, boost generators contain interaction
terms, therefore particle observables do not transform via linear
Lorentz formulas wrt to boosts. In the point form dynamics,
generators of space translations contain interaction terms, so particle
observables do not obey simple familiar transformation rules wrt
translations. In the front form dynamics, translations, rotations, and
boosts are interaction-dependent. Whichever form of dynamics you choose,
you'll find deviations from manifest covariance in some of inertial
transformations.

Eugene Stefanovich
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