Re: Tom Van Flandern and Newtonian Gravity
From: Bilge (dubious_at_radioactivex.lebesque-al.net)
Date: 09/08/04
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Date: Wed, 08 Sep 2004 06:22:10 -0000
Eugene:
>Bilge:
>> Sure you can. The propagation speed of the interaction in qed is
>> limited by the derivation of the field itself. The requirement
>> that the _free_particle_ dirac lagrangian be both lorentz invariant
>> and locally gauge invariant is what requires the addition of the
>> interaction, j.A. That only is true if A^u transforms precisely
>> the same way as the lagrangian and if it propagates instantaneously,
>> then that is not true.
>
>What you are saying has no direct relationship to the speed of
>propagation of interactions.
Sure it does.
>I do not accept your arguments about quantum fields and currents,
>because they are not observable.
Quantum fields are as ``observable'' as a wavefunction. What isn't
observable about a current?
>In order to convince me you ought to
>solve the dynamical problem with two interacting particles. You
>should demonstrate that when trajectory of one particle gets disturbed
>then the second particle will feel this disturbance after time R/c.
Why is that even necessary? The derivation is based upon the
explicit requirement that A_u transform properly, so it's impossible
for it to transform otherwise.
>I bet that QED simply cannot do that, because in order to solve
>time-dependent problems the theory must have a well-defined Hamiltonian.
>There is no well-defined finite Hamiltonian in QED.
Since the qed lagrangian (density) is well defined, the hamiltonian
(density) is well defined via, H = p\Psidot - L, where p is the canonical
momentum. The hamiltonian is just the integral of that over d3x.
>> The creation and anihilation operators operate on fields.
>
>This is not correct. Creation and annihilation operators act on
>the state vector of the system, not on fields.
The creation and anihilation operators operate on state vectors in a
first quantized theory, i.e., quantizing the schrodinger equation.
In this case, the photon is quantized.
>Creation operator adds one particle to the state of the system.
>Annihilation operator removes one particle (if such particle existed).
First quantization of the harmonic oscillator gives you quantized
energy levels. If you want to quantize the energy of the oscillator,
then using a state vector is fine. If you want to turn those into
particles, you have to quantize the state vector itself. That is a
field. Choosing a representation does not turn it back into a
state vector, even if it looks like one.
[...]
>> Which only reinforces my belief that what you have done is mistake
>> the gauge freedom in a theory formulated in a particular non-covariant
>> gauge, for instantaneous propagation.
>I don't think so. In my approach I can calculate trajectories of two
>interacting charged particles, and I can see that if trajectory of
>one particle is perturbed, the other particle will adjust its
>trajectory instantaneously (see fig. 2 in my paper "Is Minkowski
>Space-Time...").
Of course. That is precisely my point. To put it in very loose
terminolgy, local gauge invariance is what allows electrons to be
electrons and act independently of what other electrons in the
universe are doing. The photon is the way charges reconcile
phases over (possibly) infinite distances.
>I also demonstrated that this instantaneous signalling does not
>contradict causality, because boost transformations of trajectories
>are interaction-dependent. They ought to be interaction-dependent,
>because otherwise Currie-Jordan-Sudarshan theorem forbids any interaction.
At the risk of repeating myself, again, qed is not a ``stand alone''
theory. It's part of the standard model. The standard model, taken
as a whole, is believed to be asymptotically free. Therefore, the
difficulty lies in your refusal to accept the fact that qed by itself
might be expected to blow up once other interactions need to be included.
Try applying your hamiltonian to an enantiomer pair. Do the left-handed
and right-handed states have the same energy? In real life, they don't.
>These arguments involve observable properties, so they should be
>independent on the choice of gauge.
At the risk of repeating myself, I'll say it again. That is the
entire point.
>If you think I made a mistake, could you please point the exact place.
I think the problem is that you don't understand the point of
gauge invariance.
[...]
>> You can call it a wavefunction or a field, it doesn't matter, since
>> nothing depends upon it, specifically. The S is a completely generic
>> scalar function used as a phase factor, \exp(-iS). For the purposes
>> here S is a function of the spacetime variables, x^u. Beyond that
>> it doesn't matter _what_ S is, because the entire purpose of the
>> exercise is to identify it.
>
>I think, there is a big difference between wave function \psi (or
>state vector in the Fock space) and the field \Psi in quantum
>field theory.
Not really. If you write, \psi\exp(iS) p \psi\exp(-iS), what difference
does it make to the end result, if you redefine the operator p as p ==
\exp(-iS)p\exp(-iS), and treat \psi as whatever suits you? After all,
field theory is just second quantization. In addition, fock space is the
starting point for field theory. When you quantize everything to obtain
creation and anihilation operators, you _are_ doing field theory. The
creation and destruction operators you are using are _not_ the same thing
as those operators for the 1 dim harmonic oscillator.
>\psi is a complex function. Its square gives probabilities
>for various observables: number of particles, their momenta, spins,
>positions, etc. You cannot modify \psi without affecting its physical
>content: the most you can do is to multiply \psi by a constant
>factor.
What ``rule'' reqiures the phase to be a constant? The requirement
that |\Psi|^2 be a probaility density doesn't constrain changing
the phase by a constant. Indeed, the phase is not a constant, since
\psi is usually of the form \psi = A \exp(ik.r-iwt).
>On the other hand, quantum field \Psi is an abstract
>operator function of four variables. It does not have any physical
>content. Its only purpose is to produce interaction Hailtonian via
>formula
> ___
>V = e\Psi \gamma^u \Psi A_u (1)
By that same argument, the same is true of the wavefunction. Have you
ever measured a wavefunction? Strictly speaking, the only observables
are the eigenvalues of operators.
[...]
>to select momentum-dependent factors in front of products of
>creation and annihilation operators in V, so that relativistic
>invariance of the theory is satisfied. Using creation and
>annihilation operators "packaged" in the fields \Psi and
>A_u in (1) ensures that interaction V is relativistically
>invariant (i.e. H = H_0 + V satisfies Poincare commutation
I suggest looking at the canonical momentum, p_u == p_u + eA_u.
It goes all the way back to classical physics. Compare with the
phase invariance in qed.
[...]
>> That isn't true. A gauge transform is just a change of phase and
>> you most certainly have phases, since the number operator and the
>> phase define an uncertainty relation just like p and x.
>
>I don't understand. In my approach I do have the particle number
>operator. It is expressed through creation and annihilation operators
>of particles, e.g.
>
>N = \int dp a^+(p)a(p)
>
>What is phase?
The difference between \exp(i(kx-wt)) and \exp(i(kx-wt)+\phi(x))
Like, for a traveling wave, \psi = b_0 \exp(ik.x-wt). The argument
of the exponential is a phase.
>Why do I need it? How phase defines the uncertainty relation?
x = a\exp(i\phi) + (a'+)\exp(-i\phi)
etc. There is a phase operator there.
>> Your operators operate on something, and that something is not a
>> particle.
>
>All operators act on the state vector of the system |\Psi >.
>This is a vector in the Fock space.
I think you have a misonception about fock space, since fock space is
the basis for the field theory hamiltonian.
[...]
>I agree that one can add a constant to V without affecting physics.
>To correct myself: forces are observable. Force acting on particle 1
>is given by the commutator of the momentum operator p1 with the
>potential energy V
>
>f_1 = [p_1, V]
>
>In classical physics, force is the gradient of the potential energy.
Forces aren't observables either. The gradient is not an hermitian
operator. i\grad is hermitian.
>> It's defined in _a_ frame and only as a relative quantity. Since your
>> working in the coulomb gauge, you have to perform a change of gauge
>> to A^u if you change frames so that j^u A_u isn't frame dependent.
>>
>> I think what you've done is something like change frames, so that
>>
>> j^u A_u -> j'^u A'_u + terms
>>
>> gets cancelled by terms added to the rest of the new hamiltonian,
>> and attributed the additional terms to the lorentz transforms.
>> Is that about right?
>
>I am loosing you here. I have told you that there are no fields in
>my approach: there are particles with their measurable momenta, spins,
>positions, etc,; there is a Fock space with sectors having any number of
>particles from zero to infinity;
You just don't call them fields. You get particle states from second
quantization, ergo, you are employing field theory.
[...]
>> This will be the third time I've said this and you happened to ignore
>> and delete it each time. Since it specifically proves your statement
>> _false_ and you keep repeating that statement, it really pisses me off.
>> LOOK UP AHARANOV-BOHM EFFECT. THERE IS NO ELECTRIC FIELD OR MAGNETIC
>> FIELD ANYWHERE IN THE ELECTRON'S PATH.
>
>I prefer not to discuss Aharonov-Bohm effect here.
Since you can't seem to get the idea that all electromagnetic
effects cannot be explained by E and B, the aharanov-bohm effect
is _the_ _main_ point. It's tied to the potentials, A and V.
>For a couple of
>reasons. First, this is a complex multiparticle effect: there are many
>electrons in the solenoid moving under the influence of external force.
>This is a complicated problem.
All that is relevant is that there is _no_ E or B field anywhere
that the electrons are.
>I haven't though about it yet in
>the framework of my approach. So I do not have any opinion. Can we stick
>to something simple, like two charged particles, or hydrogen atom?
>Second, I do not see what is the relevance of A-B to our discussion
>of quantum fields.
The potential in qed originates from the gauge covariant derivative,
D_u = d_u + ieA_u which enforces the gauge invariance under a transformation
\Psi -> \Psi\exp(-iS). The phase in the aharanov-bohm effect is given
by, \Delta S = e\integral A^u dx_u, identify S in the phase transform-
ation, \exp(-iS), esp. as the first order change to the wavefunction,
\exp(-iS) = (1 - iS).
[...]
>> >2. It does not predict something that is not observable
>> > (QED predicts virtual particles which are not observable)
>>
>> I've already addressed this ad infinitum. That argument borders
>> on stupidity, since you have certainly never ``observed'' a single
>> particle predicted by _any_ theory in any way that isn't inextricably
>> linked to the very processes you say aren't observable.
>
>I don't agree. Observation of particles do not depend on our theories.
What you identify as a ``particle'', does. Not only does it depend
upon a theory, but the theory we have says that the particle you observe
depends on the experiment you perform. For example, are the neutral
kaons the K0 and K0-bar or K_long and K_short? What about photons?
What we call observables are eigenstates. What we measure depends upon
the basis selected by the experiment. If we measure p, x is completely
indeterminate (i.e., unobservable). If we measure x, p is completely
indeterminate. Both p and x are observables, but not in the same
experiment. When me measure photons, we measure particles. When we
perform an interference experiment, we measure relative phases, not
photons.
>There are visible tracks of particles in bubble chambers, in
>photoemulsions, etc. All these tracks were left by real particles
>with positive mass. Nobody have ever seen a track left by virtual
>particle with imaginary mass.
Nobody has ever ever seen a track by a quark or gluon either.
I'm not suggesting that quarks and internal lines have anything in
common in that regard, only that your idea of ``observable'' is
naive. A symmetry means something is unobservable. For a rotation,
the angular position is unobservable, so chaning \Theta to \Theta'
doesn't change any physics, i.e., there's no eigenvalue equation,
\Theta\Psi = \theta\psi, if \psi is rotationally invariant. There
is a conservation law because of it though.
>> If you are
>> using ``observable'' in the quantum mechanical sense, then your
>> argument is pure sophistry, since the virtual particles are the
>> result of applying quantum theory.
>
>In my opinion, virtual particles are the result of wrongly interpreted
>application of quantum theory. It is wrong to interpret internal
>lines in Feynman diagrams as some really existing particles.
Why not? Internal lines can produce real radiation and it matters
that the radiation originated there. I'm suggesting that you are
being very naive regarding what you call observable, since you aren't
talking about eigenstates, exclusively. Have you ever seen an electric
field? You've only seen the effects of an electric field. I claim that
there is no more reality to E and B than the virtual photons you haven't
seen.
>External lines do describe really existing particles which leave tracks
>in bubble chambers and photoemulsions.
Again, I suggest that your concept of ``observing a particle'' is naive.
What you see are the effects of a particle. Have you ever seen an electron
or have you just seen the ionization that something we call an electron
produced? By ``seeing'', do you mean that something produced a chemical
reaction in your retina? What mediated the fields that produced the
reaction?
[...]
>> At waht level do you predict this violation of gauss' law?
>
>Please do not substitute one effect (speed of propagation of force)
>with another (Gauss' law).
I'm not. If your electric field propagates instantaneously, then
obviously you can see a force at infinity from a charge ``now'',
which cannot be reconciled with the receipt of a signal regarding
the existence of that charge for an infinite amount of time.
[...]
>I predict, this will happen immediately. Traditional approach says that
>this willhappen after time R/c. Let's experiment decide.
I think experiment has decided, since maxwell's equations haven't
been disproven.
>
>> Don't you think it's also a little strange to need a different field
>> for light than for static fields?
>
>Yes, I agree, this looks like a step back: Maxwell's unification
>of electric and magnetic forces and light was a nice idea. My approach
>destroys this unification: electric and magnetic forces are described
>by certain terms in the potential energy operator.
[...]
>But Maxwell's unified theory has its problems: if you describe
>light as electromagnetic wave you can never explain the photoelectric
>effect.
Actually, that's only partially correct, as the problem lies in trying
to interpret maxwell's equations as classical waves. Maxwell's equations
do not describe classical waves. In a classical wave, you can observe the
amplitude and phase of the wave directly. Interference is secondary effect
resulting from ripples in a medium. You can't observe those ``ripples'' in
an electromagnetic wave. You observe the mod^2. Absolute phases are
unobservable. All you can observe are phase differences. You cannot
``count oscillations'' between the source and destination point. Maxwell's
equations contain the seeds of quantum theory, but were interpreted
incorrectly.
[...]
>(although one cannot prove that from Einstein's postulates, because
>the second postulate applies only to light pulses). I disagree with
>Einstein when he generalizes Lorentz transformations to systems with
>interactions. This generalization is not based on any evidence.
>Actually, this generalization contradicts relativistic invariance!
I addressed that. Coordinate transformations are related to the
geometric aspects of the manifold. If you want to discuss the geometry
associated with E&M, then you can either use fiber bundles, in which
case, dA is the connection on a principle bundle. You can do the
entire gauge theory thing in terms of differential geometry. The point
being that the electromagnetic field requires ``attaching'' the fibre
bundle to the manifold.
[...]
>>
>> >The interaction is between the real photon and real electrons in
>> >the wire.
>>
>> The photon has no charge. Is this a new force or something?
>
>There is electron-photon interaction in Compton scattering
Sure. But in brueckner scattering, the photons never interact
with any charges. If I wanted to provide an example in which any
observable charge participated directly, I would have. In this
case, there is no real e+/e- pair at any vertex.
>> Well unfortunately, photons have no charge, so I can't write down
>> any expression like F = qQ/r^2.
>
>For electron-electron interaction we can draw a bunch of
>Feynman diagrams with internal lines interpreted as virtual particles
>(your approach) or we can define a potential V = q1q2/r + relativistic
>corrections + radiative corrections (my approach) which will yield
>the same results for scattering.
>The same for electron-photon interaction (i.e., Compton scattering):
>we can draw a bunch of diagrams (your approach) or we can find
>an electron-photon potential (my approach). I haven't derived
>this potential yet, but I am sure it exist, and I have a clear plan
>how to derive it.
You must not know what brueckner scattering is. I can't see how
you would even do it your way. You would need to create an interaction
between the static B or E fields and the photon.
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