Re: The symmetry of the metric tensor

From: Bilge (dubious_at_radioactivex.lebesque-al.net)
Date: 09/20/04 

Date: Mon, 20 Sep 2004 11:29:41 -0000

 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.

> 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.

>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.

> 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.

>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. 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.

  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.

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