obviousness of metric tensor 'operator' does not commute...

qmagick_at_yahoo.com
Date: 01/30/05 

Date: Sun, 30 Jan 2005 08:41:51 +0000 (UTC)

Hontas F. Farmer III wrote:
> I have not done the math to confirm what you have written. Takeing you at
> your word....
>
> The exercise you have performed is a confirmation of Heisenberg's
> uncertainty principle. I ask you what is the differece between the metric
> and position. In a sense they are the same thing. So of course they
> commute.

I am not so sure. The metric is supposed to be a function of position
in the classical view, but I do not see why you could not for instance
fourier transform it into a function of the momentum, then would it be
as obvious that it should commute? Don't know.

In any event, I am pretty sure that there exist a number of ways of
looking at this idea. For instance you can define the metric tensor as
the set of functions that solves for the geodesic equation which is
intimately related to the momentum via: dp^n/ds + Gamma^n_ab p^a
(dx^b/ds) = 0. The question is not so simple as one might suppose.
-- NPC

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