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?