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Topic: Misner, Thorne and Wheeler: 11.4 Parallel Transport Around a
Closed Curve

Replies: 18   Last Post: Oct 21, 2013 12:09 AM

 Messages: [ Previous | Next ]
 Hetware Posts: 148 Registered: 4/13/13
Re: Misner, Thorne and Wheeler: 11.4 Parallel Transport Around a
Closed Curve

Posted: Apr 17, 2013 7:38 PM

On 4/16/2013 10:19 PM, Hetware wrote:
> The whole section is just one long example, so it's difficult to explain
> it all. If you have the book, it will be easier to just read the
> section there.
>
> I posted images of the 5 pages here, or educational purposes only:
>
> http://www.speakeasy.org/~hattons/mtw/
>
> They construct a small deviant quadrilateral from elements of two
> arbitrary vector fields u and v by first by displacing by u da and
> evaluating v db at the tip of u da. Then they displace by v db and
> evaluate u da at the tip of v db. That produces a quadrilateral that
> isn't quite closed. They close it with [v,u]da db.
>
> All good and well. Then they transport a vector A around the curve.
>
> "With the route now specified, the vector A is to be transported around
> it. One way to do this, "geometrical construction" by the method of
> Schild's ladder applied over and over, is the foundation for planning a
> possible experiment. For planning an abstract and coordinate-free
> calculation (the present line of action), introduce a "fiducial field,"
> only to take it away at the end of the calculation. Plan: conceive of
> A, not as a localized vector, defined solely at the start of the trip,
> but as a vector field (defined throughout the trip). Purpose: To provide
> a standard of referenc (comparison of A transported from the origin with
> A at the place in question.)..."
>
> They then give a longish description of the steps which might be
> summarized as:
>
> delta A =
> A(u da)_;v db
> - A_;v db
> - A(v db)_;u da
> + A_;u da
> + A_;[v,u] da db
>
> where A on the rhs of this expression is the A field.
>
> It seems clear that A^(field) evaluated at the beginning and end will
> have the same value. That's follows from the smoothness requirement. I
> believe the operation on the rhs applied to the parallel transported A
> will be zero.
>
> The end result is supposedly the change in A parallel transported around
> the closed circuit. To me the end result is confusion.
>
> Why does this work?

Perhaps someone in sci.math can help.