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V. Parallel Transport
The geodesic equation (5.13) may be derived by extremizing the
action (5.10). This variational problem is
solved by considering infinitesimal variations of the path:
\[{\mathbf{x}} \to {\mathbf{x}}
+ \delta {\mathbf{x}}\quad {\mathbf{u}} \to {\mathbf{u}} + \delta
{\mathbf{u}}\quad \delta {\mathbf{u}}: = \frac{{d(\delta
{\mathbf{x}})}}{{d\lambda }}\]
Then the first-order variation of the path-length is given
by
\[\delta s[{\mathbf{x}}(\lambda
)] = \int\limits_1^2 {d\lambda } \delta L(\lambda) = \int\limits_1^2
{d\lambda \frac{1}{L}} {\mathbf{u}} \cdot \frac{{d(\delta
{\mathbf{x}})}}{{d\lambda }} \]
with Lagrangian $L(\lambda) := \sqrt {\left| {{\mathbf{u}}(\lambda
) \cdot {\mathbf{u}}(\lambda )} \right|}$. It is convenient to
change the parametrization from $\lambda$ to the proper time $\tau$
by defining $d\tau /d\lambda = L$. Replacing $d\lambda$ by $d\tau$
and then integrating by parts, one obtains the variational equation
for the extremal path:
\[\delta s[{\mathbf{x}}(\tau
)] = \left. {\frac{{d{\mathbf{x}}(\tau )}}{{d\tau }} \cdot \delta
{\mathbf{x}}} \right|_1^2 - \int\limits_1^2 {d\tau } \left[
{\frac{d}{{d\tau }}\frac{{d{\mathbf{x}}(\tau )}}{{d\tau }}} \right]
\cdot \delta {\mathbf{x}} = 0\]
To meet the requirement that the action is stationary for
arbitrary variations that vanish at the two endpoints, the integrand
in the second term must be set to zero:
\[\frac{d}{{d\tau
}}\frac{{d{\mathbf{x}}(\tau )}} {{d\tau }} = {\mathbf{v}} \cdot
D{\mathbf{v}} = 0 \quad {\mathbf{v}}: = \frac{{d{\mathbf{x}}(\tau
)}}{{d\tau }}\]
So by Hamilton's action principle, one finds that the geodesic
equation is the Euler–Lagrange equation of the action
integral (5.10). The compactness of the above derivation may be
compared with the equivalent derivation in the tensor formalism of
GRT; see e.g.
Wikipedia: Geodesics in general relativity.