Re: metric tensor 'operator' does not commute...

From: Igor Khavkine (igor.kh_at_gmail.com)
Date: 01/25/05 

Date: Tue, 25 Jan 2005 16:34:52 +0000 (UTC)

On Sun, 23 Jan 2005 14:56:40 +0000, qmagick wrote:

> I was thinking about the metric and if you could create a operator for it
> ala QM style. I came up with one and want to see what people think. I have
> also commuted it with a couple of other operators such as the position
> operator, momentum operator and itself, the metric operator I created.
>
> Here is the trick. The metric is defined (well, one definition anyways) as
>
> g_mn = (dy_k/dx_m) (dy_k/dx_n) (1)

Or rather

  g_mn = (dy^k/dx^m) (dy^l/dx^n) g_kl,

where g_kl is the metric in the y coordinate system and g_mn is the metric
in the x coordinate system.

[...]

> So I played around with the following 'operators' to see if I could get to
> the metric in some way:
>
> u^n_j = (d/dx_j) y_n (2)

Before you go any further, you should answer this: What is y_n? Is it
arbitrary? If yes, then why is there an arbitrary parameter in your
definition of the metric. If not, then what determines it?

Igor