Re: Tom Van Flandern and Newtonian Gravity

From: Bilge (dubious_at_radioactivex.lebesque-al.net)
Date: 09/15/04 

Date: Wed, 15 Sep 2004 06:20:40 -0000

 Eugene Stefanovich:
>
>
>Bilge wrote:
>> Eugene Stefanovich:
>> >Bilge wrote:
>> >>
>> >> (2) Forces are out, since forces are not quantum observables,
>> >
>> >Forces are not out. Forces are quantum observables. There is Hermitian
>> >operator corresponding to force.
>>
>> Oh? And would that operator be diagnolizable in the force representation
>> of some hilbert space? Certainly it can't be simultaneously diagnolized
>> with H, otherwise the operator wouldn't be a force, or else your
>> hamiltonian wouldn't be the energy. Last time I checked, forces change the
>> energy of a particle, so whatever you think it is, it doesn't commute
>> with H.

>This operator is Hermitian and diagonalizable. However, it does not
>commute with H. So what?

  So, it isn't diagonolizable in the same basis as H.

>I presume you are not trying to say that only observables commuting
>with H can be observed.

   I'm saying precisely that. ``Observable'' has a specific meaning.
The operators which are simultaneously diagnolizable in a given basis
are the observables are called ``observables'' for that basis. Two
operators can only be diagnolized in the same basis if they commute,
so if they don't commute, for one of the operators, the basis vectors
aren't eigenvectors.

>This will leave out positions, and even momenta of interacting particles.

  Well, then that is rather unfortunate if you were counting on that.

[...]
>> That's irrelevant. Maxwell's theory is well tested and the only
>> way to compare your theory to standard E&M is express one in terms
>> of the other in order to pin down the differences. Since you didn't
>> like the way I did it, you do it.
>
>I don't want to compare my theory with E&M (QED) because the latter
>has a heavy load of garbage: fields, gauges, etc. My theory has no
>counterparts for those things.

  My claim is that your prediction of instantaneous signal propagation
is precisely because you don't have those things.

>I want to compare experimentally measurable predictions of both
>theories: scattering cross-sections, Lamb's shifts, etc. My claim
>is that both theories predict exactly the same numbers for these
>effects. My other claim is that my theory can predict time evolution
>of observables, and QED can't.

  Obviously that isn't true. Your theory predicts instantaneous signal
propagation. (Call it anything you like if you don't like the word
signal. I'm not going to deal with semantics bull*** any longer).

>> That happens to an experiment with a well-known result which addresses
>> the precisely the point here. It is essential, since it's a direct
>> measurement of something which is affected _only_ by the potentials.
>> If gauge invariant potentials are mathematical artifacts, the effect can't
>> exist.
>
>The fact that Aharonov-Bohm effect was explained from the point of view
>of potentials does not immediately mean that this point of view is
>entirely correct. I may suspect that there are other explanations.
>I am just not ready to discuss them now.
 
  Let me know when you do.

>> I explicitly said the opposite. It's you who discard gauge invariance
>> and think the lorentz group and the homogeneous poincare group are
>> different groups.
>
>I am talking about the same Lorentz group and Poincare group as
>everybody else.

  No, you aren't. Everybody else doesn't believe the lorentz group is
non-linear and different from the homogeneous poincare group. Everybody
else doesn't think the lorent transforms are something other than
coordinate transforms.

>The same multiplication law, the same commutation
>relations in the Lie algebra, etc. I have different approach to
>the _representations_ of these group, i.e., how they act on state
>vectors and observables.
 
  A different representation can't change a linear transform into
a non-linear one.

>The traditional approach claims (without proof) that certain groups of
>observables called 4-vectors (position-time, momentum-energy,
>etc.) must transform in a universal linear fashion.
 
   The proof is simple and I've told you how to do it at least twice.
Use noether's theorem.

>I disagree and say
>that transformations are linear and universal only for total observables
>(e.g., total momentum and total energy of any system do transform as
>4-vector). The transformations for observables of subsystems are not
>linear and universal, they depend on interactions.Gauge invariance has
>nothing to do with it.
 
  Look, I'm not going to keep repeating the same thing. The reason you
think some non-linearity is involved is precisely because you claim
``gauge invariance has nothing to do with it''. Gauge invariance has
everything to do with the fact that you get instantaneous propagation
of the interaction.

  Prove gauge invariance is not your problem by using the lorentz gauge
and see if you get the same result you get by using the coulomb gauge
and ignoring the fact that the coulomb is not covariant.

>> Don't you think you're going to have to redefine the spin, too? The spin
>> tensor is derived as part of the proof of covariance, defining the lorentz
>> transforms the way most people define them. The metric and spin tensor are
>> the anti-commutator and commutator of the dirac matrices, which are
>> obtained from proving the manifest covariance of the dirac equation. [cf
>> ``Relativistic Quantum Mechanics'', Bjorken, J, and Drell, S.]
>
>Sure, I have a different definition of spin. The derivation usually goes
>together with the derivation of the Newton-Wigner position operator.
>Just a sample of relevant papers:
 
  Which isn't invariant under a boost.

[...]
>> I wouldn't care, except for the fact that you've also ruled out
>> discussing a hamiltonian that contains anything normally associated
>> with E&M.
>
>My Hamiltonian contains as leading terms 1/r Coulomb interaction between
>charged particles and Darwin's potential responsible for the magnetic
>interaction between charged particles. So, it has some association
>with "normal" E&M.
 
  I didn't say you had a problem with your use of potentials. What I
said, is that I can't use the same concept or else you reject the
argument as irrelevant.

>> Since I can't talk about any physical quantities, I'm can't see how
>> ``Real physical quantities'', means any of the quantities most people
>> consider real and physical.
>
>Again, real physical quantities are positions, momenta, energies,
>and spin projections of particles (or bound states, like deuteron).
>These physical quantities are measured in experiments (photoemulsions,
>Stern-Gerlach apparatus, mass-spectrometers, etc.). I prefer to discuss
>physics in terms of these quantities. Not in terms of scalar and vectors
>potentials or even electric and magnetic fields which no one ever
>observed.
 
  (1) Am I then supposed to equate ``observable'' with only that which
      human eyes can detect?

  (2) Your list of things which can be observed is inconsistent.
      No one has ever seen ``spin'' or a ``spin projection.''
      Never mind that what is observed in a stern-gerlach experiment
      are the signals from detectors located at some point which is
      off the symmetry axis of the apparatus. The direction of the
      asymmetry is _defined_ by the direction the magnetic field
      points. But you don't believe magnetic fields exist.

  (3) Defining an observable as something you can see with your
      eyes is so naive that it's unacceptable as anything remotely
      related to science. I assume you expect your theory to
      describe 10 MeV gamma rays as well as green light, so you're
      being hypocritical about that, too.

>> Stop telling me that, since you object to anything derived from
>> quantum field theory and can't seem to relate your result to anything
>> derived from quantum field theory.
>
>I do have some contact with QFT: I use the same Fock space, I use
>the same annihilation and creation operators, I use non-interacting
>fields to construct the same Hamiltonian as in QED in the Coulomb gauge.
>After this step I drop fields and proceed with calculations in terms
>of creation and annihilation operators.
 
  It really doesn't matter, since you are the only one allowed to use
the `F' word. I have no way to connect anything, observable or other-
wise, to your theory, since the problems to which I refer are due
precisely to what you did and what I can't discuss.

[...]
>without them. Up to this point my approach and standard QED are the
>same. There is a difference in rhetoric, but the physical results are
>all the same.
 
  Obviously the physical results are not the same, since qed doesn't
predict the instantaneous propagation of signals and your theory
does.

>
>The significant difference appear when I do unitary "dressing"
>transformation of the QED Hamiltonian to obtain finite "dressed
>particle" Hamiltonian which can be used for studying dynamics.

  That isn't the part I have been referring to.

>> >I use normal non-interacting electron-positron and photon fields to
>> >construct the Hamiltonian of QED in the Coulomb gauge.
>>
>> I'm not sure why it's surprising that your lorentz transforms need
>> correction terms, since the coulomb gauge isn't covariant.
>
>The difference between our approaches is in understanding of the meaning
>of the word "covariant". I understand it as follows. From the relativity

   yada yada yada....

[...]
>
>You undestand "covariant" as universal linear interaction-independent
>transformations via Lorentz formulas.

  No, the difference is that I consider the lorentz group to be
a transformation on the spacetime variables. You consider it to
be a replacement for a gauge transformation to fix the coulomb
gauge so you can ignore the fact that the coulomb gauge is not
covariant. You predict instantaneous propagation of an interaction.
I don't. Guess why?
 
>I bet you cannot prove this statement. From my perspective, this
>statement is an approximation.

  Your perspective is wrong. It's trivial to prove the lorentz
transforms are linear and prove that the conserved quantity
which correspond to an infinitesimal spacetime displacement
is the four-momentum. Since all of that is determined by a
transformation in the neighborhood of the identity, it's
all linear.

[...]
>> That is how you define a density operator, not the probability density.
>> Ensembles are irrelevant. Observables are eigenstates of an hermitian
>> operator and in your case in particular, those operators all have to
>> commute the hamiltonian, since you've chosen to work with a hamiltonian.
>
>1) Operators of observables don't have to commute with Hamiltonian.
 
  That's only true if your hamiltonian isn't an observable. Look it
up in _any_ quantum mechanics book. If you have a set of basis
vectors, the observables for that basis all commute.
 
>There are observables which depend on time, e.g., position of a free
>particle, or momentum of a particle interacting with other particles.
>These observables do not commute with Hamiltonian.
 
  Stop throwing out random statements. Everytime I try to narrow the
thread down to a few, focussed concepts, you expand it back with
a lot of additional non-sense, which so far is incorrect besides having
nothing to do with the thread.

>> >In an experiment two things are measured: 1) the value of
>> >observable (eigenvalue of an hermitian operator) and 2) how often
>> >the measurement of this value is repeated in the ensemble.
>>
>> Are you claiming I can't measure a single particle?
>
>Yes, you can measure a single particle, but such a measurement
>does not allow you to fully characterize the state of the particle.
 
  You are very confused regarding quantum mechanics.
 
>You can shoot an electron through two holes and obtain one blackened
>spot on the photographic plate on the other side of the screen. This
>one measurement will not show you the interference picture.
>Only after you shot many electrons you can notice that there
>is a interference picture on the photographic plate. Quantum mechanics
>requires repetition of identical experiments and accumulation of
>statistics.
 
  You are very confused regarding even the double slit. The particle
passing through the slits, interferes with itself. This is easy to
illustrate, too. If the particle did not interfere with itself,
by passing particles one at a time, you would end up with the
sum of two single slit diffraction patterns, not a double slit
diffraction pattern.

>> >This is probability described by the wave function. This is foundation
>> >of quantum theory.
>>
>> No, it isn't. Quantum mechanics is probabilistic, not statistical.
>> Density operators are statistical and deal with ensembles, (hence
>> the name ``statistical operator''). There is a difference and few
>> if any, statistical interpretations are viable interpretations.
>> If quantum theory were statistical, the epr experiment would get
>> a different result.
>
>OK, you said "Quantum mechanics is probabilistic". How do you suppose
>this probability is measured?

  So? What's your point? We're talking about eigenvalues and observables.
The probability is not an observable. Seriously, if you really believe
all of this stuff you've said about quantum mechanics, there's not much
point in discussing anything. I took it for granted that you were
familiar with basic quantum theory, but it seems like you've gotten
lost in the formalism without any physical connection to what it
means in terms of experiments.

Quantcast