Re: h bar and Lorentz transformation

From: Non Ame (noname_at_nospam.net)
Date: 03/22/05 

Date: Tue, 22 Mar 2005 10:00:31 +0000 (UTC)

"Ken S. Tucker" <dynamics@vianet.on.ca> writes:

>Non Ame wrote:
>> "Ken S. Tucker" <dynamics@vianet.on.ca> writes:
>>
>> >Hi non ame,
>> >You should take up your arguments with others
>> >who post to this NG, to acquire a wider perspective.
>>
>> >For instance, post some of your questions to
>> >David McAnally.
>>
>> Translation: My comments were beyond Ken's capability to comprehend.

>And I'm lazy!

>>He
>> really believes that any statement which satisfies General Covariance
>is a
>> law of nature in General Relativity. He also really believes that
>quantum
>> observables must commute.

>A more complete explanation uses "nonsymmetrical
>metrics", and I figure your eyes glaze over when
>you read that.

Not at all. I just want to learn about it from somebody who is actually
competent, which you are not.

Now that you have shot all your credibility on Quantum Mechanics, I would
hope that you won't discuss Quantum Theory, any more.

>Consider the following, (based on the introduced
>metric),

>A_u B_v = - A_v B_u

Do you have examples for A_u, B_u?

>and A_u B_u =/= 0 , summation suppressed.

>So now, take an antisymmetrical tensor F_uv
>and define it,

>a*F_uv = A_u B_v .

Is F_uv supposed to be the EM field?

>Defining s_uv to be symmetrical metrics,
>((I use s_uv in this notation, I'm following
>John Moffat)).

>g_uv = s_uv + a*F_uv .

>That's a hard equation to crack, (for me).

>For fun, let's outer multiply by U^v,
>and obtain by association,

>U_u = U*_u + f_u

>((U*_u = s_uv U^v)).

>Without further ado, I'll sit down on "a"
>and call that the center of the universe,
>so that, (see GR1916, eq.65+),

It is presumably central on the basis that it commutes with everything.

>((I'm invoking the relativity of acceleration
>at this point to crack the problem)),

>f_u=0 and

Again with f_u = 0.

>U_u = U*_u.

>That's a powerful statement where association
>is concerned, because either g_uv or s_uv
>can associate.
>A requirement of association is the vanishing
>of the covariant derivative,

>g_uv;w = 0 = s_uv;w and leads to,

>a*F_uv;w =0.

>Please note Maxwell is alive and well in the
>partial diff,

The above equation is not what Maxwell's Equations say.

>a*F_uv,w =/=0

>= GAMMA^k_uw (a*F_kv) + GAMMA^k_vw (a*F_uk)

>relative to "a".

>Regards
>Ken S. Tucker
>kxsxt

Relevant Pages

  • Re: h bar and Lorentz transformation
    ... "Ken S. Tucker" writes: ... >> really believes that any statement which satisfies General Covariance ... >> observables must commute. ... Now that you have shot all your credibility on Quantum Mechanics, ...
    (sci.physics.relativity)
  • Re: h bar and Lorentz transformation
    ... "Ken S. Tucker" writes: ... Translation: My comments were beyond Ken's capability to comprehend. ... really believes that any statement which satisfies General Covariance is a ... observables must commute. ...
    (sci.physics.particle)
  • Re: h bar and Lorentz transformation
    ... "Ken S. Tucker" writes: ... Translation: My comments were beyond Ken's capability to comprehend. ... really believes that any statement which satisfies General Covariance is a ... observables must commute. ...
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