Newsgroups: sci.physics


I wrote:

>The "connection" is an additional restraint that need not
>be imposed to insure the tensor character of the resulting tensor.

John writes:

>Huh?  This makes no sense to me, which is probably related to
>the problem you are having.  The "connection" is not a restraint, or
>constraint.  There are many ways of thinking about connections but the
>most relevant here is that the connection simply IS the entity
>
>   k
> G
>  ij
>
>More precisely, the Christoffel symbol expresses the connection in a
>given coordinate system (or frame).   It's obvious from this point of
>view that one needs a connection to define covariant derivatives.

I don't know but,,,

I found in Spain's "Tensor Calculus" a formulation of a covariant
derivative which starts with the transformation law of the metric
tensor.

(21.1) g'  = (dx /dx')(dx /dx') g  , where the derivatives are partials.
        mn                      ij

Now differentiate wrt x' to get

                o      i   m    j   n    k   o         k
(21.1a) dg'  /dx' = (dx /dx')(dx /dx')(dx /dx')(dg  /dx ) +
          mn                                      ij

                    2 i   m  o    j   n      2 j   n  o    i   m
             g  { (d x /dx'dx')(dx /dx') + (d x /dx'dx')(dx /dx') }
              ij

Subtract this equation from the sum of the two equations obtained by
appropriate (not cyclic) interchange of indices to obtain:

                     i   m    j   n    k   o
(21.2a) [mn,o]' = (dx /dx')(dx /dx')(dx /dx')[ij,k] +

                                  i   o   2 j   m  n
                   .5(g  + g  )(dx /dx')(d x /dx'dx')
                       ij   ji

Appropriate manipulations finally yield the formulae:

          2 r   m  n   ij   r   j            kr   i   m    j   n
(21.5a) (d x /dx'dx')=s' (dx /dx')[mn,i]' - s  (dx /dx')(dx /dx')[ij,k] ,

                           ij      j          j
where s  = g  + g   , and s  s  = d  , where d  = 1 if j=k, else =0.
       ij   ij   ji           ik   k          k

Now differentiate the transformation law of a tensor and replace the 2nd
order partial derivative with the above formulae (21.5a) and you get a
formulation of the covariant derivative.

         i    i   j   ki        n    i   j   i  k
(22.2a) A = dA /dx + s  [nj,k] A = dA /dx + G  A
         ,j                                  kj

But hey! Instead of g, use any non-singular covariant tensor of rank 2 in
(21.1) and you get another G, but the resultant A is still a tensor in that
it obeys the transformation law of tensors, with the new index being a
covariant index. This is what I meant by:

>The "connection" is an additional restraint that need not
>be imposed to insure the tensor character of the resulting tensor.

What I think the "connection" is (correct me if I'm wrong) that which
constrains the choice of G. By imposing the restriction that the inner product
of two tensors has zero variation along an infinitesimal parallel transport.
This leads to the relations:

                    m       m       m   m         m   m
(70b) 2[ij,k] = g  G  + g  G  + g (G - G  ) + g (G - G  )
                 mk ij   km ji   im kj  jk     mj ki  ik

which if g and G are symmetric in their lower indices, reduces G to the
Christoffel symbol of the second kind. The symmetry of g will be given, but
that of G in it's two lower indices is not. In Einstein's "The Meaning of
Relativity", he suggests that G must be symmetric since a parallel
transport along dx1 of dx2 must give the same result as that for dx2
along dx1.

John also writes:

>You misunderstood my point since I wasn't precise enough.  To take the
>covariant derivative of a tensor A in a given direction v at a certain
>point x, all *v* needs to be is a tangent vector at x.  I would write
>this covariant derivative as D_v A(x); in coordinates it's
>
> j
>v A   (x)
>   i;j

I had misunderstood you. Still, why does v need to be a tangent vector?

 j        k  1/2
v A    /(v v )    , here can't v be any non-null vector?
   i;j      k

Getting in way over my head,

--------------------------------------------------------------------
Ric Peregrino   c/o Hewlett Packard Co.    I represent only myself
ric@spd.hp.com  1501 Page Mill Rd Bldg 5M  I may be wrong, maybe so
415-857-7526    Palo Alto, CA 94304        Do DC photons exist?
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Newsgroups: sci.physics