\( \newcommand{\be}{\begin{equation}}
\newcommand{\ee}{\end{equation}} \newcommand{\ba}{\begin{array}}
\newcommand{\ea}{\end{array}} \newcommand{\bea}{\begin{eqnarray}}
\newcommand{\eea}{\end{eqnarray}} \newcommand{\bean}{\begin{eqnarray*}}
\newcommand{\eean}{\end{eqnarray*}} \newcommand{\la}{\label}
\newcommand{\nn}{\nonumber} \newcommand{\half}{{\scriptstyle \frac{1}{2}}}
\newcommand{\third}{{\scriptstyle \frac{1}{3}}}
\newcommand{\bli}[2]{\begin{list}{#1}{\itemsep=0.0cm \topsep=0.0cm \partopsep=0.0cm
#2}} \newcommand{\eli}{\end{list}} \newtheorem{problem}{Problem}[chapter]
\newcommand{\bprob}{\begin{problem}} \newcommand{\eprob}{\end{problem}}\)
V. Parallel Transport
Consider
a smooth spacetime curve ${\mathbf{x}}(\lambda )$ parametrized
by a parameter $\lambda $. The tangent vectors to this curve are given
by
\[ {\mathbf{u}}(\lambda ) : =
\frac{{d{\mathbf{x}}}}{{d\lambda }} = \frac{{\partial {\mathbf{x}}}}
{{\partial {x^\mu }}}\frac{{d{x^\mu }}}{{d\lambda }} = {u^\mu }{\partial _\mu
}{\mathbf{x}} = {u^\mu }(\lambda ){{\mathbf{g}}_\mu }({\mathbf{x}}(\lambda
)) \]
with respect to the coordinate basis along the curve.
For a
vector ${\mathbf{Y}}(\lambda ): = {\mathbf{Y}}({\mathbf{x}}(\lambda ))$, the directional
coderivative along the curve is defined as
\[ \frac{{D{\mathbf{Y}}}}{{d\lambda
}}: = {u^\mu }{D_\mu }{\mathbf{Y}} = {\mathbf{u}} \cdot D{\mathbf{Y}} \]
This is a natural generalization of ${D_\mu }: = {{\mathbf{g}}_\mu } \cdot
D$, the coderivative along the basis vector ${{\mathbf{g}}_\mu }$. In the
literature the directional derivative ${\mathbf{u}} \cdot D$ is often denoted
as (bold) ${\nabla _{\mathbf{u}}}$.
Any
vector ${\mathbf{Y}}$ can be expanded in terms of the basis vectors of the coordinate
frame as ${\mathbf{Y}} = {Y^\nu }{{\mathbf{g}}_\nu }$. In view of (1.17)
the coordinate form of (5.2) is then
\[ {u^\mu }{D_\mu }{\mathbf{Y}} =
{u^\mu }({\partial _\mu }{Y^\nu } + \Gamma _{\mu \kappa }^\nu {Y^\kappa })
{{\mathbf{g}}_\nu } \]
If the same vector is expanded with respect to the local tetrad frame
$\{ {{\mathbf{e}}_a}\} $, the connection is determined by the bivector (4.4):
\[ {u^\mu }{D_\mu }{\mathbf{Y}} =
({u^\mu }{\partial _\mu }{Y^a}) {{\mathbf{e}}_a} + {u^\mu }{{\mathbf{\omega
}}_\mu } \cdot {\mathbf{Y}} \]