Re: Download a new book on quantum mechanics and relativity.

From: Bilge (dubious_at_radioactivex.lebesque-al.net)
Date: 11/09/04 

Date: 09 Nov 2004 10:24:54 GMT

 Eugene Stefanovich:
>Bilge wrote:
>> Eugene Stefanovich:
>> >Bilge:
>> >
>> >The principle of relativity says that all inertial frames of reference
>> >are physically equivalent. There is not a single word about spacetime,
>> >geometry etc.
>>
>> You must be really dense. By your own definition the transform
>> between two inertial frames can't depend upon any interaction,
>> since an inertial frame is one which is force free. If there is
>> a force acting, the frame isn't inertial.
>
>You are right, inertial frames do not interact. They move uniformly
>along straight lines without rotations. The transformations between
>inertial frames (space and time translations, rotations, boosts) form
>a universal Poincare group, whose properties (the multiplication law,
>or structure of the Lie algebra) do not depend on interactions.
 
  You are still missing the point here. Inertial frames, straight lines,
force, etc., are ill defined concepts in and of themselves. They are
_defined_ by geometry (obviously a straight line has to be a geometric
concept). When you say that the lorentz transforms are ``modified'',
in effect, you are redefining a a non-inertial trajectory to be
an inertial one. That is why your explanation sounds like you're
trying to do general relativity, except in this case, you can't
transform the force away.

>However, when we talk about Lorentz or boost transformations, we
>are talking about how observables of a PHYSICAL SYSTEM look from
>different reference frames. In order to find these transformations,

  ``Appearance'' has no relevance. If I rotate a stick about the y
direction, its projection on the x-axis changes. Rather than pretend the
z-direction doesn't exist and conclude the stick gets shorter, I use the
postulate of relativity, to conclude that the length of a stick does't
depend upon the direction it's oriented, postulate that my geometry
is three dimensional, not two dimensional and construct the coordinate
transforms, such that the length of the stick is invariant, l^2 = x_i x_i,
x -> x_i' = (I_i'i + a_i'i)x_i so that l^2 = x_i' x_i' = x_i x_i.
Relativity is no different apart from the fact that i runs from 0-3
instead of 1-3 so that a_ij has 6 components instead of 3. A rotation
in x-t changes the angle the stick makes on the x and t axes. The slope
\beta = tanh(A) is called the velocity.

>it is not sufficient to know the Poincare group. In order to find
>these transformations we need to build a unitary representation of
>the Poincare group in the Hilbert space of the system.

  Non-relativistic quantum mechanics is poincare invariant, so whatever
it is you think follows from poincare invariance, a finite value for
the constant `c' is not one of those things.

>The Poincare group is unique for all physical systems, however the
>representation
 
  No, it isn't. For example, the ten generators,

    d/dt, \grad, (x_i d/dx_j - x_j d/dx_i) and (t d/dx_i + m x_i)

represent time and space translations, rotations and galilean boosts.
(incorporate the i\hbar's if you want quantum versions). So, poincare
invariance holds, with c->\infinity. If you don't believe the last term
is a boost:

    (-i\hbar t d/dx_i + m x_i)\phi = (tp - mx)\psi = m(t (p/m) - x)\psi

The quantity, p/m is the classical velocity, so I get m(vt - x).
 
>of the Poincare group depends on the physical system and on interactions
>acting in this system. In particular, time translations and boosts
>are represented by interaction-dependent operators which makes time and
>boost evolutions rather non-trivial.
 
  It might be non-trivial, but you're going beyond non-trivial to physically
incorrect. If you define a particular representation, you have to stay
within the context of what you've defined. For example, given the generators
for boosts and translations, [L^ab, p^c], if the commutator is unchanged by
a unitary transform, it's obvious that either or both generators can
be transformed. That doesn't mean any physical results differ. It means
that you have to know what you are measuring.

[...]

>Einstein and all textbooks after him derive Lorentz transformations
>from discussion of propagation of light pulses.

  I referenced bjorken and drell for you. They don't reference light
pulses. Light is an electromagnetic effect. It's speed of propagation
in a theory, depends upon the theory. If you don't assume lorentz
invariance, what is your theoretical rational for assuming the number
`c' is finite? In special relativity, time and space are treated on
equal footing and since `t' is just a coordinate, it's natural to
set c = 1 for the same reason no one defines a 45 degree angle in
the x-y plane as dy = b tan(\theta) dx. Whether or not light propagates
along the 45 degree line in the x-t plane is question that should
be addressed theoretically in a theory of E&M. It has nothing to do
with poincare invariance or lorentz invariance. I can create a poincare
invariant theory in which light does not propagate at `c'.

[...]
>I asked you to give me a proof of Lorentz transformations for
>arbitrary physical system (they are related to boosts, not to
>spacetime displacements, by the way). From your response I gather
>that you cannot do that.
 
  Don't be an imbecile. I demonstrated above that poincare invariance
isn't sufficient to even define your energy-momentum relation, so
if it didn't come from invariance under a spacetime translation,
where _did_ it come from? The energy and momentum form a locally
conserved current if the lagrangian is invariant under spacetime
displacements. I realize you have a personal issue with lagrangians,
but that doesn't alleviate you from having to prove the relationships
you lift from derivations using lagrangians if you are going to
declare the lagrangian formalism invalid.
  
[...]
>> >What about trajectories of two charges (let's make them macroscopic
>> >to avoid your objection that quantum particles do not have
>> >trajectories)?
>>
>> Make it anything you want.
>
>This was an example of dynamical time-dependent system.
>I haven't heard from you how QED is going to approach it.
 
  I just told you below.

>> positron pairs from gamma conversion in lead so that the gamma energy
>> could be determined by the radius of curvature of the electrons and
>> positrons spiralling in the magnetic field. There's nothing magic about
>> it. It doesn't require any real sophisticated physics to do this.
>
>That's my point. Simple classical trajectories are beyond reach of QED.

  Don't be dense. You seem to have a mental block against the rather
general and widely held concept that an approximation to an exact solution
is still an approximate solution and seem to think that the only
approximations allowed are the ones that are the least physically
appropriate to the solution being sought. I can obtain classical E&M as
the lowest order approximation to qed. Therefore, it's a valid
approximation, even if I don't muster all of the machinery of feynman
diagrams to do the calculation. Since a particle trajectory is a classical
concept, the appropriate way to calculate one is via the classical
reduction of qed to classical E&M. Throwing math at a problem without
first figuring out what physics is involved is silly. Do you believe it's
necessary to solve a problem with spherical boundary conditions using
cylindrical coordinates and cartesian coordinates to prove you should
calculate the same numbers for a given coordinate?

[...]
>> >See eq. (11.1) for the current operator
>>
>> Post it or don't discuss it.
>
>
> j_{el}^{\mu}(x) =-e \psi^{\dag}(x) \gamma^0\gamma^{\mu} \psi(x)

   That's somewhat of a dilemna, eugene. First of all, you have stated on
numerous occasions that your theory does not contain the quantity \rho. I
don't know what your view of the (three-) current density J is, but that
expression definitely contians \rho:

            __ __
    \rho = \psi\gamma^0\psi, \psi = \gamma^0\psi^{\dagger}

  See, the basic problem I have, is that you deny these quantities exist
in your theory, and then you not only say the quantities exist in your
theory, but the expressions you claim represent those quantities turn out
to be the same ones used in theories to which you object, based upon the
kinds of quantities you say don't have any physical meaning. If you wonder
why I am rude, this sort of sophistry is precisely why. Every time I say
something about a charge density, your theory doesn't have one. When I say
that's problem, then all of a sudden, your theory has one, and it just
happens to be the same one found in every textbook. When I the one of two
or more mutually contradictory answers I'm given depends upon the question
I ask, I tend to get the impression that the person is trying to bull***
me, hoping I won't notice. I consider that rude - even more so than
outright abuse, since at least outright abuse isn't an attempt at
deception.

  In order to prove charge conservation, you need to show explicitly
that (1) the current is conserved, and the integral over all space
is a constant so that the time derivative of the integral of the
charge density is zero. That's usually easy to do, except that you
obviously cannot assume that j^u is invariant under a lorentz transform,
because you have additional terms in your lorentz transforms that
correspond to the interactions.

[...]

>> >See subsection 9.2.3 (after eq. (9.49)) for the proof that
>> >interaction commutes with the operators of total momentum and total
>> >angular momentum. So, these two quatities are conserved.
>>
>> Post it.
>
>Each interaction term has this general form
>
> V = \int dq'_1 ... dq'_N dq_1 ... dq_M
> D(q'_1, ..., q'_N; q_1, ..., q_M )
> \delta (\sum_{i=1}^{N} q'_i - \sum_{j=1}^{M} q_j)
> a^{\dag}_{q'_1} ... a^{\dag}_{q'_N} a_{q_1} ... a_{q_M}.
>
>where a^{\dag} are creation operators, a are annihilation operators;
>q_i are particle momenta (spin labels are omitted for simplicity).
>The delta function
>\delta (\sum_{i=1}^{N} q'_i - \sum_{j=1}^{M} q_j)
>guarantees that the sums of annihilated momenta and created momenta are
> equal, so the total momentum is conserved. The coefficient function
>D(...) is a rotationally invariant function of momenta. This guarantees
>that V commutes with the angular momentum operator.
>
>Note also that there is a theorem (see section 4.2 in Weinberg's vol. 1)
>that any operator in the Fock space can be represented as a sum of
>operators written above.
 
  That obviously requires you to find and interpret D(...) to not
just satisfy momentum conservation, but to satisfy the expression you
want to use in relating p and E. Also, you still keep forgetting that
those theorems you quote aren't likely to be valid if you insist that
real physics is representation dependent.

[...]
>>
>> It's called gauge invariance. A massless photon can't carry a
>> charge. You claimed you understood all of that. Apparently not.
>> If you don't understand it, I'll explain it (again). The last time
>> I posted something, you decided you already knew all about it.
>> Which is it eugene? Do you or do you not understand what a gauge
>> theory is? Personally, I don't think you do and I don't think
>> you want to know, either.
>
>You are right, I don't understand gauge theory in spite of many years
>of trying. I spent a lot of efforts trying to understand what is the
> physical meaning of fields, canonically conjugate momenta, their
>phases, and gauge invariance.
 
  Then why did you refer to what I wrote a while back in terms of the
unitary transformtion of p_u as if it were obvious and inconsequential
and that I didn't know what physics it contained?

>Now I made up my mind. I concluded that all this has no physical
>meaning at all. It is just a mathematical trick at the end of which
>we somewhat mysteriously obtain "correct" Hamiltonian of QED or
>standard model.

  Well, that's unfortunate, because in my opinion, the problem with your
theory is precisely that your misinterpretation of the ``mathematical
trick'' of dressing your charges. Personally, I think your motivation
has little to do with E&M and everything to do with having an excuse
to say special relativity is wrong. Otherwise, you wouldn't be so adamantly
opposed to suggestions that your idea about boosts is simply a misinter-
pretation of your theory. You seen to prefer having the entire theory
tied to your interpretation.

>I write "correct" in quotes,
>because this Hamiltonian H is just partly correct. The S-matrix computed
>with H is infinite. One needs to make two rounds of corrections in order
>to arrive to a physically sensible Hamiltonian. First round is
>renormalization. The Hamiltonian H' with renormalization counterterms
>yields exact S-matrix, but this Hamiltonian is not good either.
>First, it is infinite (depends on infinite masses and charges).
>Second, it contains unphysical trilinear interactions responsible for
>processes of particle creation and annihilation never observed in
>nature, like
>
>1 -> 1 + 1
>0 -> 1 + 1 + 1,
>
>etc. The second round of corrections is "dressing transformation" which
>produces a finite Hamiltonian H'' without unphysical interaction terms.
>
>I would prefer a derivation which directly produces the final good
>Hamiltonian H'', however, I do not know such derivation. So, in the
>absence of better solution, I am forced to accept this rather ugly
>3-step procedure
 
  Non-observation of something is not a legitimate reason to discard
terms that the theory says exist.

>gauge theory Hamiltonian H -> renormalization theory Hamiltonian H' ->
>"dressed particle" Hamiltonian H''
>
>[...]

>> That's why I said you do not prove charge is conserved. You admit
>> that you assume it.
>
>The fact that charge is conserved in my theory is proven inside my
>theory.
 
  But you don't prove it. In fact, you don't even give a reason for
charge to exist. If I called your charge lepton and baryon number,
it would mean the same thing.
 
>The fact that charge is conserved in nature is an experimental fact.
 
  But theories are supposed to explain experimental facts and if you
say something is the charge, you are supposed to explain why it's the
charge you claim. To put it another way, angular momentum, momentum,
energy, baryon number, lepton number, etc. are all conserved charges.
Each is associated with a symmetry. There is a conserved charge associated
with invariance under a rotation. That charge is called angular momentum.
In a gauge theory, the electrical charge is the charge associated with
invariance under a U(1) phase transformation (or actually, it's the
integral of a conserved current.) The four-momentum is the conserved
current associated with invariance under spacetime displacements.
The conserved charge is the integral of that current, and is called
the mass.
 
>I am not trying to prove this fact. I am not accepting your "proof"
>either. Because your proof involves fields and gauge invariance, which
>do not make physical sense to me.
 
  It really doesn't matter if it makes sense to you. It makes sense to
enough people that the standard model was based upon gauge invariance.
It's really quite straight forward, too. Invariance under U(1), SU(2)
and SU(3) phase transforms gives the electromagnetic, weak and strong
currents.

>> QED is obviously built on well established experimental facts.
>> It has the distinction of being confirmed by experimental facts
>> better than any other scientific theory ever proposed.
>
>I fully agree. With one exception that QED can say nothing about
>time evolution of states and observables.
 
  What you call observables, so far, have been things which quantum
mechanics does not call observables, so I don't see that as being
a problem.

>> You can keep on waiting. I already told you to post whatever you
>> think is relevant.
>
>In order to conform with the charge conservation law the general
>interaction operator V written above must satisfy simple conditions,
>for example, a terms having operator structure
>
>a^{\dag}a^{\dag}a^{\dag}aa
>
>is forbidden. It describes creation of an additional electron in
>collision of two electrons. This process does not conserve charge.
 
  So what allows you to discard those terms on theoretical grounds?
Simply discarding terms you don't like is ok for phenomenological
models which are fit from data (optical models, for example), but
not for a fundanental theory.
 
>However, the operator
>
>a^{\dag}b^{\dag}a^{\dag}a^{\dag}aa
>
>is allowed. It describes creation of an electron-positron pair
>in collision of two electrons; a and b denote operators of electrons
>and positrons, respectively).

>> Post it.
>
>These interactions look rather ugly in ASCII codes. Don't be so lazy.
>Print out the relevant portion of the book, or if you despise my book so
>much, you can read section 83 (Breit's equation) in "Quantum
>electrodynamics" by Berestetskii, Livshits, and Pitaevskii
  
  I have no difficulty whatsoever deriving the spin interaction for
myself. I don't need to look it up. That isn't the issue. Even the
dressing transformation in your theory is not the issue as far as it
goes. Your interpretation of that is the issue. First of all, if you
don't accept the concept of a B field, then it's rather hard to explain
a term like S.B, which is, after all, the term relevant to a stern-gerlach
experiment. Of course, you don't _have_ to use B, since a B-field is
produced by charged particle dynamics, but the fact that you believe
your theory refutes lorentz invariance, doesn't allow you to simply
use results derived from assumptions that include lorentz invariance.
Reading articles in which the assumptions include an assumption you
explicitly reject, doesn't address the question here.

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