Re: Tom Van Flandern and Newtonian Gravity

From: Bilge (dubious_at_radioactivex.lebesque-al.net)
Date: 09/16/04 

Date: Thu, 16 Sep 2004 11:02:46 -0000

Eugene Stefanovich:
>Bilge:
>> 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.
>
>We are walking in circles repeating the same arguments. You keep
>telling me that my approach is wrong because it is different from
>the traditional approach.

  I'm not telling you that your approach is wrong because it's different.
I'm telling you it's wrong because you make two mutually inconsistent
statements about what you've done. The first is that the choice of gauge
has no physical effect. The second is that your results don't depend upon
any choice of gauge. The first is correct. The second is a result of
your misunderstanding the first and denial that working in the coulomb
gauge doesn't constitute choosing a gauge.

  You chose to work in the coulomb gauge. When you perform a boost you
have one of two choices. (1) You perform the gauge transform necessary
due to the fact that the coulomb gauge is not covariant, or (2) you
end up with results that depend upon your choice of the coulomb gauge
in a particular coordinate system which is not consistent with simul-
taneously stating that a choice of gauge can't have physical consequences.

  You can't have it both ways. If there were no physical consequences
associated with choosing the coulomb gauge, you wouldn't need to
use the coulomb gauge to obtain them. You cannot assert that you
``don't use gauges'', as you put it, since you clearly did when
constructing your hamiltonian. What you mean is that you ignore
the fact that the coulomb gauge is not covariant, and pretend that
it is because you don't like the word ``gauge''.

>Yes, it is different, but it does not make my approach wrong.

  The fact that you claim that a choice of gauge has no physical
consequences and derive physical consequences from your choice
of gauge proves you wrong. If you were right, choosing the
coulomb gauge wouldn't be the basis of the physical results
you derive.

>You can prove me wrong in 2 ways: 1)
>show that my predictions disagree with experiment; or 2) show
>that my approach has internal logical inconsistencies.
 
  Well, you might as well have not bothered writing (2), since I've
been doing that using your own claims. However, you refuse to discuss
anything related to your even own statements. I also think (1) is
out, since I've already pointed out an experiment. You don't want to
discuss the experiment either.

>However, before we discuss that, we must agree on some basic stuff,
>like ordinary quantum mechanics. So far we disagree on two things:
>operators of observables not commuting with the Hamiltonian
>and interpretation of probabilities in quantum mechanics (see below).

  Observables are defined as hermitian operators and measurements
are the eigenvalues of those operators. Here's a simple example.
The operator x is an hermitian operator. The operator p is an
hermitian operator. If both x and p were simultaneously observables,
then xp and px would each be hermitian operators with xp = px.

  Another example. S_y is hermitian. S_z is hermitian. If both S_z
and S_y were simultaneously observable, then in some basis, S_yS_z
would be observable which would imply S_y S_z was hermitian.
I can write S_yS_z using the the representation in which S_z is
diagonal to get,

            [0 i] If you diagnolize this operator, the
 S_y S_z = [ ], eigenvalues are pure imaginary, hence
            [i 0] S_yS_z is not hermitian and therefore
                        not simultaneously observable.

[...]

>This is not a random statement. This is exactly the point we are
>discussing. You say that operators of observables must commute with
>the Hamiltonian.

  I wouldn't be so picky if you hadn't insisted upon it. Normally, I hate
being pedantic if I know what someone means and can figure it out.
Previously, I attempted to make a point concrete, using the terminology
rather loosely (and pointed that out, too), but you objected to that and
requested I not use terminology loosely. So, I'm now insisting that you
comply with your own request. That will eliminate any excuses based on
claiming not to know what I mean.

>I say, this is not true. Even in the case of one free
>particle there is operator of position which does not commute with
>the Hamiltonian. This is reflected in an observable fact: position
>changes with time.

  You have a number of misconceptions regarding quantum mechanics.
The wavefunction is a function over configuration space, not spacetime.
Expectation values are not eigenvalues. If you measure a position,
the momentum is indeterminate. If you measure the momentum, the position
is indeterminate. The wavefunction (or an operator) evolves with time.
That means the probability of finding a particle in various locations
changes with time which is not the same thing as the position of a
particle changing with time.

>If in your theory all operators of observables commute with the
>Hamiltonian, then nothing changes with time, there is no dynamics.

  Well, apart from the time evolution of the wavefunction, strictly
speaking, that's correct. However, that presupposes you have solved
the full hamiltonian and found the exact wavefunctions, which is
almost never possible and is why you split H into H_0 + H'. It
would be pointless to do perturbation theory if the interaction was
already diagonal in the basis of the unperturbed wavefunctions.

  The dynamics to which you refer are the transitions between the
unperturbed states, so obviously the dynamics are in the terms
like <f|H'|i> where the initial and final states are different,
i.e., the off-diagonal terms of O.

>I may agree with you that particle interferes with itself.
>But still, one electron will produce just one dot on the photographic
>plate.

   That is completely irrelevant to your claim that an ensemble has
something to do with the observables. The probability of the electron
landing in some particular location is independent of where the previous
electron landed or the next electron will land.

>In order to see the interference picture you need to shoot
>many electrons through the slits.

 So? An interference pattern isn't an eigenstate or even an expectation
value of an electron.
 
>Then you'll see that there are
>regions on the plate with many dots (constructive interference),
>and there are regions with a few dots (destructive interference).
>There is no way you can observe the interference picture with just one
>electron.

 That's all well and good, but isn't relevant to anything.

>Operators of observables (with their eigenvectors and eigenvalues)
>describe the measuring apparatus. The state vector of the system (or wave
>function) describes the state in which the system was prepared. You are
>right to say that probability is not an observable, there is no Hermitian
>operator associated with probability. But probability is routinely
>measured in experiments. It is a MEASURABLE quantity.

  You are mixing several different concepts, which are related but not
equivalent.

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