Re: Tom Van Flandern and Newtonian Gravity

From: Eugene (eugenev_at_synopsys.com)
Date: 09/05/04 

Date: Sat, 04 Sep 2004 18:43:12 -0700

Bilge wrote:
> Eugene:
> >Bilge wrote:
>
> >> Your statement that qed must be modified to figure out the speed at
> >> which the interaction propagates.
> >
> >I would appreciate if you give me a reference in which QED is used
> >to calculate time dynamics (not just S-matrix elements).
>
> First, how does that relate to your previous comment regarding the
> speed at which the interaction propagates? Second, I'll see what I can
> dig up.

It has very direct relation. Without time dynamics you cannot say
whether your interactions are retarded or instantaneous.

>
> >I have a couple
> >of references where author speaks about calculating the speed of
> >propagation of signals in QED. (e.g., M.I. Shirokov,
> >Uspekhi Fiz. Nauk 124 (1978), p. 697. There should be English
> >translation in Sov. Phys. Uspekhi) I maintain that his approach is
> >wrong. In its current form, QED cannot do that. If you like, we
> >can discuss this paper.
>
> >> You've simply ignored the infinite part and the fact that the
> >> qed hamiltonian may be used to do the same thing.
> >
> >No, I didn't
>
> How do ``clothed operaters'' differ from ``dressed fields'' apart from
> semantics?

In my approach there are no fields. I use free electron and photon
fields to construct interaction Hamiltonian as j.A (in the Coulomb
  gauge). After that I discard fields and operate entirely with
particle creation and annihilation operators. This is sufficient to
predict any desired property: scattering, bound states, time
evolution, etc. "Dressing" transformation a la Greenberg and Schweber
is used to make a unitary transformation of the usual QED Hamiltonian
(with infinite counterterms) to a more physical form in which
1) there are no infinities, 2) the interaction terms are operators like
V(r1, p1, r2, p2) which depend on positions and momenta of particles
taken at the same time instant. In this "dressed" form the theory
describes real (not virtual) particles interacting via instantaneous
forces. All predictions of renormalized QED (S-matrix, Lamb's shifts,
etc.) are preserved

>
> >> >I do not care about gauge invariance, because it is not observable thing.
> >>
> >> I think you're confused regarding what is an ``observable thing''
> >> and what ``observable thing[s]' are due to something which is
> >> unobservable. If a system is rotationally invariant, it's orientation
> >> is an ``unobservable thing'', but the consequence is that a perfectly
> >> ``observable thing'' called angular momentum, is a conserved thing.
> >>
> >> Are you claiming that the substitution,
> >>
> >> \Psi -> \Psi' = \Psi\exp(-iS)
> >>
> >> alters physical observables? Assume S is a function of the spacetime
> >> coordinates. If the phase (i.e., gauge) is to be an ``unobservable
> >> thing'', it had better not appear in a momentum measurement, for
> >> example.
> >
> >What is \Psi here? If \Psi is wave function, then unitary transformation
> >\exp(-iS) changes physical properties of the state described by this
> >wave function.
>
> The whole point is that the field itself originates so that the
> lagrangian (or state or whatever you prefer) does _not_ change
> under that transformation.
>

I still don't understand your point here until you tell me what is
\Psi and what is S

> [...]
> >> Perhaps you should care a little more about gauge invariance, since
> >> what you are describing is precisely what one would expect by ignoring
> >> it. The electric field is _not_ the gradient of a potential. The
> >> electric field includes the time derivative of the vector potential,
> >> too. The scalar potential is meaningless by itself.
> >
> >Pardon my inaccurate expression. I was talking about Coulomb and
> >magnetic potential ENERGY terms in the Hamiltonian. Their gradient is
> >measurable force.
>
> OK, so what? If that is the case, I don't see why you think the
> interaction propagates instantaneously. You are starting with a
> non-covariant gauge (i.e., the coulomb gauge) and if you change
> frames, you _must_ change the gauge at the same time. That change
> of gauge seems to be what you are leaving out.
>

In my approach,there are no fields, therefore there are no gauges.
It operates entirely with directly observable things like particle
positions and momenta and the potential energy of interaction or forces.

> [...]
> >> I'm talking about A^u. You can call it whatever you want. The
> >> propagation velocity is tied to the [some name you choose for
> >> A^u].
> >
> >You are trying to tell me how QED is formulated in textbooks: with
> >potentials A^u, gauges, etc. I always had a problem with these
> >textbook approaches. For example, what is the meaning of A^u?
>
> It's the photon. The j.A term in the lagrangian is the interaction of
> the electromagnetic current, j, with the field, A^u. The term itself is
> derived from the _free_ lagrangian, and the j.A is necessary, whether
> there is an external field or not. The orthogonal linear combination
> of the same two fields which define A^u is called the Z, which forms
> a weak isotriplet with the W+/-.

I agree that fields A^u and currents j are useful mathematical
constructs for building interaction which complies with relativistic
invariance and cluster separability (see Weinberg's book).
They are not needed for anything else. They do not have any physical
meaning. The quantities having more or less direct physical meaning
are particles positions, momenta, spins, Hamiltonian
(or, in general, generators of the representation of the Poincare group
in the Hilbert space of the system). That's all we need to do physics.

>
> >The only useful purpose of A^u is for constructing the interaction
> >in the Hamiltonian
> >
> >V = j_u A^u
>
> I think you've missed the entire point of QED. The point of QED is
> that the term you just wrote down is derived from the free-particle
> dirac lagrangian by insisting it be locally gauge invariant under a
> local U(1) transformation. The fact that the electromagentic field
> is a consequence of local gauge invariance is a big plus, since the
> phase of the electron can't have any physical effects in quantum
> theory. If the phase depends upon the spacetime variables, which it
> must if electrons are electrons as we observe them, then the electro-
> magnetic field (or some other field) is a rather natural result.

I agree that derivation of interaction V = j_u A^u based on
"local gauge invariance" is a nice trick. I don't have this trick
in my theory (you may say this is a deficiency of my approach): I just
plug in this interaction in the Hamiltonian,
express it through particle creation and annihilation operators,
and use it for calculation of observable effects.
I think that "local gauge invariance" is just a mathematical trick.
It leads to correct result, but it does not have any physical basis.
What is "the phase of the field"? What is field itself?
Is it measurable? It seems that we introduced some dummy unobservable
variables in the theory, and then we are trying to get physical
consequences by demanding that no observable effect depends on these
dummy variables. That looks very suspicious to me. Anyway, all
this business with "local gauge invariance" does not seem important
to me. What is important is that we somehow obtained correct
interaction Hamiltonian which can be used to do physics.

>
> >After the Hamiltonian was constructed as a function of particle
> >creation and annihilation operators, we can discard A^u and obtain
> >any information we need (dynamics, scattering, bound states, etc.)
> >just as in usual quantum mechanics, i.e., from the Hamiltonian.
>
> My point has nothing to do with what you can or can't do in that
> regaard. My point is that (1) your rant against qed is unfounded,
> but primarily, (2) your idea regarding invariance. In nuclear physics,
> we use lots of models that are useful for getting numbers even though
> everyone believes that nuclear forces are mediated by meson exchange,
> and ultimately are a manifestation of qcd. There is no experimental
> evidence for meson exchange currents and no one has ever come up with
> a model of even the deuteron using qcd, yet in spite of the fact that
> the nuclear shell model is adequate to explain nuclei for which a
> calculation is tractable, no one considers the shell model hamiltonian
> to be a fundamental model of nuclear processes. So, I'm not really
> sure what your point is with respect to qed. If you've found an
> effective hamiltonian which is simpler, fine. It's the rest of what
> you say that goes with it that I think is incorrect and I think the
> reason you have drawn the conclusions you have, is that you have
> mistaken a physical effect for what is gauge freedom.
>

My model has a few nice properties which distinguishes it from
the nuclear shell model and allows me to call it "theory"
1. It is exact
2. It does not predict something that is not observable
   (QED predicts virtual particles which are not observable)
3. It is mathematically consistent (QED is not consistent because
   of ultraviolet divergences)

The ultimate answer about which model/theory is right should
be provided by experiment. I predict that Coulomb force
between two charged particles propagates instantly. In principle, this
can be measured. If somebody performs a convincing experiment
showing that this force propagates with the speed of light, then
I'll believe in fields, virtual particles, physical vacuum, etc.

>
> >> [...]
> >> >into account this fact, the you'll see that instantaneous interaction
> >> >between two charged particles does not contradict causality.
> >>
> >> Sure it does. If a force is exerted faster than light, I can use that
> >> force to transmit a message.
> >
> >Yes, we can transmit message faster than light. So what?
>
> Let me restate that, since I used the speed of light to mean the
> null ray in minkowski space and the photon mass is an experimental
> issue, not a relativistic one. You can't send messages between
> points with spacelike intervals.

The taboo on superluminal messages is based on application
of universal linear Lorentz transformation. If you accept that
boost transformations may depend on interaction, then this
argument vanishes and instantaneous Coulomb and magnetic interactions
no longer contradict causality

>
> >How this contradicts causality? My approach predicts that
> >(if dynamical character of boosts
> >is taken into account) the effect does not precedes the cause in
> >all reference frames. So, faster-than-light propagation of interaction
> >does not contradict causality.
>
> Then light doesn't propagate at `c' in your model. Two points which have
> a spacelike separation, can't be time ordered - by definition. If you can
> propagate a signal on a spacelike interval, not only is that acausal by
> definition, but you could use that to violate the uncertainty relations
> since position and momentum commutes on a spacelike interval and you
> would be in conflict with the epr experiments.
>
> I think you would encounter less opposition if simply presented
> what you had without trying to say that it's relativistically correct.
> It's not relativistically correct. In special relativity, spacetime
> is a flat manifold and the lorentz transforms are linear transforms
> and interactions do not affect the manifold/change the transformations.
> E&M can't do that anyway. The interaction is based upon U(1), not
> an orthogonal transform. Personally, I think you've mistaken gauge
> freedom for a physical effect.

My approach does not agree with special relativity in which boost
transformations are postulated to be universal and exactly equal to
linear Lorentz formulas for all systems independent on their
composition and interactions. Nevertheless, my approach is
relativistically invariant: transformations between the points of
view of different observers are described by a unitary representation
of the Poincare group in the Hilbert space. This is what I call
relativistic invariance: the laws of physics are the same in
all inertial frames of reference.

Einstein's assumption about universality of boost transformations and
underlying 4D space-time is just an approximation in my opinion. When
special relativity
promotes this approximation to the rank of law it meets a paradox:
"no-interaction" theorem by Currie-Jordan-Sudarshan. World lines
of interacting paerticles cannot transform according to Lorentz
formulas.

>
> >> [...]
> >
> >Are you saying that virtual particles can be observed?
>
> I'm saying that to the extent that you can consider an effect as
> being due to any term or group of terms in any equation, virtual
> particles are no less ``real'' than anything else.
>
> >Take two charged spheres and place them next to each other.
> >According to the current wisdom, there should be a plenty of
> >virtual particles flying around, but no measuring device can
> >register them. Why? Because virtual particles exist only on paper.
>
> Look up brueckner scattering. One of the six diagrams is,
>
> ~ ~
> ~ ~
> ~------~
> | |
> | |
> ~------~
> ~ ~
> ~ ~
>
> The ``box'' represents creation of two virtual e+/e- pairs. The four
> external lines are photons, two propagating and two virtual. The
> virtual photons are supplied by a static magnetic field. The other
> two lines are the ingoing and outgoing photons scattered from the
> static field.
>
> Nowhere does the scattered photon interact directly with a charge
> and since photons are not charged, photons cannot interact directly
> with each other.

You forgot to include in your diagram real electrons which are moving
in the wire creating this "static magnetic field". The interaction is
between the real photon and real electrons in the wire. All the internal
lines in Feynman diagrams do not correspond to anything observable.
The theory can be reformulated, so that these internal lines disappear.

>
> In addition, what I am saying regarding virtual particles, is that
> relegating them to ``mere'' mathematical terms is inconsistent with
> the fact that one treats lots of mathematical terms as being real
> based upon essentially the same criteria: calculating explicit effects
> which depend upon those terms. Finally, let me sum all of that up
> with a comment made by by t'hooft and veltmann. I'll paraphrase
> as best I can recall, which was, ``we subscribe to the growing
> belief that there is more truth in the diagrams than the underlying
> formalism.''
>
  I think such belief persisted because so far there was no alternative
theory which would explain all experiments as well as QED. Now there is
such alternative theory. It has certain advantages (see above).
It predicts some new effects, which are not so difficult to measure.
I am going to use all my power (very limited, indeed) to convince
experimentalists to perform such experiments. Then we'll see.

Eugene.

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