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

From: Eugene Stefanovich (eugenev_at_synopsys.com)
Date: 10/08/04 

Date: Fri, 08 Oct 2004 10:00:40 -0700

Ken S. Tucker wrote:
> Eugene Stefanovich <eugenev@synopsys.com> wrote in message news:<41645359.5070605@synopsys.com>...
>
>>Ken S. Tucker wrote:
>>
>>>Eugene Stefanovich <eugenev@synopsys.com> wrote in message news:<41631613.4020305@synopsys.com>...
>>>
>>>
>>>>Ken S. Tucker wrote:
>>>
>>>
>>>>>A problem is in (8.2.3) the "position operator".
>>>>>You've talked about the issue, and decided to merge
>>>>>CJS Theorem with Bakamjian-Thomas, using FTL.
>>>>>...
>>>>
>>>>Could you please be more specific. What is the problem in 8.2.3?
>>>>What's wrong with position operator? This subsection is before
>>>>CJS theorem, Bakamjian-Thomas approach and FTL interactions.
>>>>How it is related to them?
>>>
>>>
>>>Well you are having a problem with the vector "r" I'll denote
>>>"r>", that defines the distances between particles.
>>>To solve the problem you choose to place the particles at
>>>the same point in time by instanteous potential change,
>>>so they have no separation in "time", by using FTL.
>>
>>I think I can perform measurements of positions of two
>>particles at the same time instant and form a vector
>>"r>" connecting these two positions. What's wrong with that?
>
>
> Prove it. (you can't that's your error).

The possibility of measurement of observables (including position) is
a postulate in my approach (Postulate I in subsection 3.2.1).
The possibility of simultaneous measurement of observables of two
particles is also a postulate (postulate O(II) in subsection 8.1.1).
I think that you cannot even start to talk about physics without first
assuming that observables of particles can be measured. There is nothing
more fundamental than that.

>
>
>>Such measurements should be possible independent on the
>>speed of propagation of interaction. They are possible
>>even if particles do not interact at all.
>
>
> "should", I'm getting out my *dingbat*.

See above

>
>
>>>That's a BIG MISTAKE!!! because magnetic forces, which are
>>>based on space AND time, need to interact at "c" ie. with
>>>a "retarded potential".
>>>
>>>To prove your theory you will need to demonstrate to us how
>>>magnetic forces arise using FTL potentials.
>>>[...]
>>
>>That's easy, the magnetic potential
>
>
> How do you define Magnetic potential without Maxwell's
> help, go right ahead give me your best A_u and then F_uv
> and then answer is,
>
> f_u = q*F_uv U^v = 0 ?
>
>
>>of interaction between
>>two charged particles is given by eq. (12.28).
>>This is
>>the well-known Darwin's potential. Note that in contrast
>>to (often used) Biot-Savart interaction law, the Darwin's
>>potential does not violate the 3rd Newton's law
>>(action=reaction), i.e., it conserves the total momentum.
>>There is a lot of discussions on this subject in the
>>literature. If you are interested I can give you references.
>
>
> Just answer my direct question.

I directly told you how I defined magnetic potential.
It is in eq. (12.28). It is obtained in several steps:

1. Take usual Hamiltonian of QED in the Coulomb gauge
2. Perform renormalization
3. Perform dressing transformation in the 2nd perturbation order
4. Extract electron-proton part of the resulting potential
    in the c^{-2} approximation.

Your formula
above has no relation to my approach, because there are no A_u and F_uv
in my approach.

>
>
>
>>>No joke, both, the Poincare group is for finite velocity,
>>
>>I disagree. My theory is an example in which Poincare group
>>properties are respected and interactions are instantaneous.
>>I haven't found a hole in my arguments, so I still stick to my
>>position.
>
>
> So your getting sticky are you...well then, your FTL=oo
> translations make you everwhere at once, Right?

I don't understand what you mean by that. Could you elaborate,
please.

>
>
>>>>>PS: Here's something you may find interesting...
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>>If we take an electron with some momentum m0 and then accelerate it in
>>>>>>an electric field to some momentum m1 emitting photons of total momentum
>>>>>>mp can we say that m0 = m1 + mp?
>>>>>>Or is some momentum transferred directly to the charging apparatus?
>>>>>
>>>>>
>>>>>Hi Oz et al...
>>>>>I would venture (cautiously) an explanation this way.
>>>>>Start with a hydrogen atom (proton + electron), and then
>>>>>let the electron and proton move closer and emit a photon.
>>>>> In context of Oz's question we may call the proton, the
>>>>>"charging apparatus".
>>>>> After this has happened, the proton and electron had
>>>>>an equal relative change in potential, electrostatically and
>>>>>magnetically, based on two assumptions,
>>>>>
>>>>>1) The electron moved the same distance to the proton as
>>>>>the proton is to the electron, (electrostatically equal change).
>>>>>
>>>>>2) The rate of rotation of the electron around the proton
>>>>>is the same as the proton around the electron (magnetic =).
>>>>>
>>>>>So an electromagnetic diagram describing the changes of the
>>>>>relatively emitting particles looks like,
>>>>>
>>>>>p+ ===> . <====e-
>>>>>
>>>>>But these are opposite charges in opposite directions,
>>>>>in terms of same charges in same directions becomes,
>>>>>
>>>>>P+===>. ====>e+
>>>>>or
>>>>>P-<====.<====e-
>>>>>
>>>>>and these vectors are the relatively incremented
>>>>>electromagnetic potentials of the proton and electron.
>>>>> A diagram that summarizes this effect should look
>>>>>like this,
>>>>>
>>>>>e-......
>>>>> |......e-
>>>>>
>>>>> ~~~~~~~~> photon
>>>>>
>>>>> |......p+
>>>>>p+.....
>>>>>
>>>>>1st state.......>.2nd state
>>>>>
>>>>>My reason for these diagrams is to show that the emission
>>>>>of radiation is due to the relative motion of the apparatus (p+)
>>>>>and the electron.
>>>>>
>>>>>Referring to Oz's question, "m0 = m1 + mp?"
>>>>> The term, "momentum" alone carries relativistic components.
>>>>>Perhaps the invariant energy would be easier to start with.
>>>>>
>>>>>In 1st state, Invariant Energy = Invariant (p1) + Invariant (e1).
>>>>>
>>>>>In 2nd state, Invariant Energy = Invariant (p2) + Invariant (e2)
>>>>>+ Invariant (photon)
>>>>>
>>>>>The invariant energies of p2 and e2 are less than p1 and e1
>>>>>because each surrenders equal electromagnetic potential to
>>>>>create the photon's invariant energy.
>>>>>
>>>>>And to Oz's question..." can we say that m0 = m1 + mp?"
>>>>>
>>>>>I would say yes (?) if these quantities are invariants.
>>>>>
>>>>>Regards and thanx...
>>>>>Ken S. Tucker
>>>>
>>>>What this has to do with our discussion? I thought that light
>>>>emission by hydrogen atom is better described by quantum mechanics,
>>>
>>>
>>>Quantum mechanics is not a description, it's an accounting tool.
>>
>>I disagree with you on that. Quantum mechanics tells you everything
>>there is to measure. All mechanistic "descriptions", like the one
>>you gave above are just fantasies. They involve things (like the value
>>of potential on the electron in the hydrogen atom) which are
>>impossible to measure and
>>which, therefore do not belong to physics. Electrons are described by
>>wavefunctions, and your example ignores that.
>
>
> Na, wavefunctions ain't physics, it's like being an actuary
> at a death site. i.e. cause of death is attributable to lack
> of life.
>
> Just apply a covariant derivative to the product of potentials
> in a charge couple
>
> (phi_u phi_v);w = 0
> a b
>
> that's a deduction from Einstein's Law.
> Oh spooky, might need to read.
>
> If you knew squat about GR you would also know
>
> 0 = (g^uv phi_u phi_v);w =0 since g^uv;w =0.
> a b
> and
>
> 0 = (phi*phi);w
> a b
>
> Integrate that
>
> $ 0 dw = constant ~~~~~~> photon
>
> Anybody just notice a quantum fly by Eugene's head?
>
> Ken

Your description of hydrogen atom without quantum mechanics
sounds very revolutionary to me, but I don't see the relevance
to our discussion. If you are suggestng that QM should be abandoned
in favor of GR, I disagree with you completely. If you are not
suggesting that, then I prefer to stay within commonly accepted
bounds and discuss hydrogen atom in terms of QM.

Eugene.

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