🚀 go-pugleaf

Regarding tensor calculus, GR and symmetry. A covariant derivative exists
which depends only on the metric and it's first derivatives. Einstein
approaches the problem with physical arguments and the covariant derivative
so defined has more degrees of freedom. The connection between the
displacement field and the metric depends on the zero variation of an
inner product along a parallel displacement. But, mathematically, this 
is not necessary if one follows the formulation outlined by Spain in
"Tensor Calculus". Also, the implications of symmetry will differ for
these two approaches. What follows is a long mathematical description of
the two approaches to a covariant derivative, as well as some modifications
of the equations used in the references to allow for asymmetry.

Consider pg. 71 of Einstein's "The Meaning of Relativity".
Eq. (69) shows a definition of the Christoffel symbol of the 1st kind.

                        j         i         k
(69) [ij,k] = .5(dg  /dx + dg  /dx - dg  /dx ) , where the d/dx are partials.
                   ik        jk        ij

This definition is fine, unless g has an asymmetric part. If instead
one defines this as

                         j         i         k
(69a) [ij,k] = .5(dg  /dx + dg  /dx - dg  /dx ) ,
                    ik        kj        ij

then the symmetry of g is apparent in the 1st two indices of [ij,k].
The 2nd term on the RHS has indices j & k swapped compared to (69). This
in no way affects the results for the symmetric metric, however, it is
advantageous for the anti-symmetric part of the metric.

Eq. (67) leads to an unnumbered eq. on pg. 71. Call it (67a).

              k      m       m    i j  k
(67a) (dg  /dx - g  G  - g  G  ) A A dx = 0
         ij       im jk   jm ik

If instead of the square of the magnitude of the vector A, consider the
inner product of the contravariant vectors A & B. This leads to:

              k      m       m    i j  k       i j  k
(67b) (dg  /dx - g  G  - g  G  ) A B dx = W   A B dx = 0
         ij       im jk   mj ik            ijk

The arbitrary vectors A, B & dx make W a tensor. If this inner product's
variation is assumed to be zero, as it is here, then this leads to:

                    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

However, if you proceed with the formation of the covariant derivative,
not as Einstein does on pg. 72, but as Spain does in "Tensor Calculus"
on pp. 28-31, keeping in mind the possible asymmetry in g, one gets
a covariant derivative that depends only on g and it's 1st derivatives.
I'll show some of the intermediate steps, since this reference may not
be readily available. Start with the transformation law for g.

                i   m    j   n
(21.1) g'  = (dx /dx')(dx /dx') g  , where the derivatives are partials.
        mn                      ij
                       o
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 the
formulation of the covariant derivative for a general metric. Nowhere is
the displacement field used by Einstein required. In other words, the
formulation of the covariant derivative does not depend on the zero
variation of an inner product along a parallel displacement, but depends
only on the metric (s depends on g) and it's 1st derivatives ([ij,k]).

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

Compared to Einstein's formulation which gives:

          i    i   j   i  k
(71a)    A = dA /dx + G  A  , noticed the swapped j & k of G.
          ,j           kj

But what about the tensor W, defined in (67b) above? Even if it is set
to zero and g is symmetric, there can still exist an anti-symmetric part
of G in (70b). Einstein's formulation of covariant differentiation is
different than that of Spain's in the term multiplying A in the 2nd term
on the RHS of (71a) and (22.2a); namely G is not identical to S. Further, 
if W is not zero, then these two formulations differ even more.

As an example, in Spain's formulation, this term will by symmetric if g
is symmetric, but in Einstein's formulation, this is not necessarily
the case. Consider a two dimensional metric with g = identity. You'll
find that the solutions to (70b) are:

 1    2    1    2        1     2     2     1
G  = G  = G  = G  = 0 , G  = -G   , G  = -G   , whereas all S = 0.
 11   22   12   21       21    11    12    22

So there appears to be some arbitrariness to the covariant derivative.
Of course Einstein also makes the assumption for the symmetrical field,
that G must be symmetric in it's two lower indices, which ends this 
problem. But to study an asymmetric field, this assumption must be
dropped, so this simple 2D example illustrates a major difference
between the two approaches.

Now I ask the net, assuming it's read this far, for it's infinite wisdom.
What gives? How can a 2D flat space have a covariant derivative that
differs from the normal derivative, and in an arbitrary fashion with two
degrees of freedom? Why not just apply the laws of physics with Spain's
covariant derivative at all times? In this case, the only way to get
an asymmetric S is to have an asymmetric g. 

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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 not
415-857-7526    Palo Alto, CA 94304        Do DC photons exist?
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