Roberto, you got the math right, but were slightly off the mark in actually proving the argument. In particular, you can't
use the result of [14.26], because that's the very thing we're trying to demonstrate!
Instead, look at the equation at the top of your page 2 (of the attachment).
As you later point out, if
u is chosen to be an affine parameter,
R is a constant along the curve and thus the RHS of this equation vanishes, leaving you with an equation defining
(=
dt/du) in terms of
t,
g, and derivatives of
g.
Now consider that along the affine geodesic
is zero. Expanding the expression for the covariant derivative (connection derivative) of a vector
t along
t yeilds:
and since that is zero, you can substitute your
g expression to replace the partial derivative term, yielding the
definition of
, just as it is in [14.26], as required!
[Actually, there's a little subtlety here: What you actually get is an expression for
that's true for any vector
t at any point (since you can always construct an affine geodesic through any point in any direction).
It's a simple matter to show, however, that this is sufficient to completely define
when it's symmetric in it's two bottom indices: Just note that due to the symmetry, given
any two arbitrary vectors
s and
t,
st =
uu-
vv where
s = u + v,
t = u - v.]
