Re: The symmetry of the metric tensor

From: Ken S. Tucker (dynamics_at_vianet.on.ca)
Date: 09/20/04 

Date: 20 Sep 2004 13:17:27 -0700

dubious@radioactivex.lebesque-al.net (Bilge) wrote in message news:<slrnckthpl.fep.dubious@radioactivex.lebesque-al.net>...
> Ken S. Tucker:
> >Bilge:
> >> Uh, there is a great deal implied. In particular, the metric is
> >> locally the minkowski metric, which implies something you should
> >> try to figure out regarding the symmetric and anti-symmetric components.
> >
> >You'll need to clarify your view of the problem, we don't
> >see one. That one may transform away all fields at a single
> >point makes no difference, the field requires two points to be
> >defined.
>
> Ok, since what you just said has nothing to do with the question I
> asked you to answer, I'll assume you don't understand the difference
> between real and imaginary numbers or the difference between an orthogonal
> and unitary.

It's Monday...
 
> > Here's an example, you'll like it...
> >
> >Simply set the antisymmetrical part of g^14 denoted a^14 to
> >
> > a^14 == q*F^14 = q*E =~ qQ/r^2
>
> The F^uv is insufficient to specify all electromagnetic phenomena,
> for precisely the same reason that E&M is not a geometric symmetry
> like general relativity.

ah, ok..
 
> >We see that Q = E*r^2, is an invariant.
>
> >Obviously this invariant depends on "r > 0" and not a point wherein r=0.
>
> Why is that ken? Plug r = 0 into that equation: Q = Er^2 = 0.
> So, there's no charge at the origin. You shouldn't use equations
> which make specific assumptions and then violate the assumptions.
>
> div E = 4\pi\rho(r-r')
>
> \integral E.dS = integral \rho(r-r') d3r'
>
> E = (1/r^2) \integral \rho(r-r') d3r'
>
> Obviously, the exact field depends upon the details of the charge
> distribution, \rho.

\rho does not apply, it's an old habit from electrical engineering
when charge quantization wasn't important, like a plumber does need
to know water in molecular.
  In theoretical physics use the quantized charge, forget \rho.
 
> > So what happens to that invariant when one transforms to
> >a Minkowski metric?
> >
> > Well Q becomes undefined since r=0.
>
> Not the way you've written it as Q = Er^2. If Q = Er^2, then it
> looks like you can't say anything about Q without first specifying
> E. Obvioulsly, if E is a constant, then Q = 0.

Agreed.

> >OTOH antisymmetrical tensors import a physical reality into the
> >definition of the invariant,
> >
> > a^14 x_1 x_4 == qQ
> >
> >and the x_1 and x_4 are defined by the locations of q and Q.
 
> I'm not about to look at the rest of this until you define the quirky
> notation you use, especially since you seem to use it for the sake of
> being obtuse.

Me obtusive? For this diatribe, set x_1 =r =ct = x_4.

>As far as I'm concerned, if your indicies run 1-4, x^4 means
> ict, since that is how I've seen it defined. If you mean something other
> than ict or x^0, then define it. If you mean x^0 and you are just using
> x^4 in some misguided quest to be unique, then I'm not going to read it.
> That uniqueness conflicts with my desire to not waste my time figuring out
> notation for no particular reason. If you mean ict, define it.

You know I never use imaginary numbers in relativity, but my
habit of using x_4 for time is hard to break. The new notation
adopted by Conception Dynamics has reserved x_0 for compactified
dimensional notation and has re=adopted x_4 for time, that
renders x_0 = time obsolete, but when in Rome use square wheels,
good for parking..

> I was tempted to stop reading when you wrote down F^14, since you did
> not tell me what F is. If you want to be understood, tell people what
> you mean and define your terms. F is a letter. It's not a tensor until
> you say it's a tensor and what it represents.

Right I shall repost with complete definitions.

Thanks
Ken

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