= Solution
Vary the <geodesic Lagrangian> before imposing the normalization of <proper time>. With $u^\mu=dx^\mu/d\tau$,
$$
\frac{\partial L}{\partial u^\alpha}=-2g_{\alpha\nu}u^\nu,\qquad
\frac{\partial L}{\partial x^\alpha}=-\partial_\alpha g_{\mu\nu}u^\mu u^\nu.
$$
The <Euler-Lagrange equation> is
$$
g_{\alpha\nu}\frac{du^\nu}{d\tau}
+\partial_\rho g_{\alpha\nu}u^\rho u^\nu
-\frac12\partial_\alpha g_{\mu\nu}u^\mu u^\nu=0,
$$
equivalently the affinely parametrized <geodesic equation>. For the static weak metric, its leading spatial connection coefficient is
$$
\Gamma^i{}_{00}=-\frac12g^{ij}\partial_jg_{00}
=\frac{\partial_i\Phi}{1-2\Phi}\simeq\partial_i\Phi.
$$
All terms with two spatial velocities are suppressed by $v^2$. Therefore $d^2x^i/d\tau^2\simeq-\partial_i\Phi\,(dt/d\tau)^2$.
Because $t$ is cyclic, the same <Euler-Lagrange equation> gives $(1+2\Phi)\,dt/d\tau=\text{constant}$. On converting to coordinate time, the additional term $(d^2t/d\tau^2)(dx^i/dt)$ is of order $v^2\nabla\Phi$ and can be dropped. Hence
$$
\boxed{\frac{d^2\mathbf x}{dt^2}=-\nabla\Phi}
$$
to leading order, the <Newtonian limit of general relativity>. Restoring SI units gives $\Phi=\Phi_{\rm SI}/c^2$, so the acceleration is $-\nabla(c^2\Phi)$. \b[Substituting $L=1$ from proper-time normalization before varying would incorrectly erase the dynamics].
Back to article page
