How do I get the Christoffel symbols in the Newtonian limit? I can't seem to understand why $\Gamma_{00}^{\mu}$ is the only term that survives in the Newtonian limit.
For example with $\Gamma_{0i}^{\mu}$ we get 
$$
g_{\nu \mu}\Gamma_{0i}^{\mu}= \frac{1}{2}\left(g_{0\nu,i}+g_{i\nu,0}-g_{0i,\nu}\right)
$$
The middle term vanishes since we have assumed a static field ($g_{\mu \nu,0}=0$), but what about the rest? 
 A: Given the geodesic equation
$\frac{\partial^2 x^\mu}{d\lambda^2}\,+\,\Gamma_{\rho\sigma}^{\ \mu} 
\frac{dx^\rho}{d\lambda} \frac{x^\sigma}{d\lambda}$
It's known to be THREE approximations to do here.


*

*$\frac{\partial x^i}{\partial\tau}\ll c$  (slow velocities)

*Weak gravitational field

*Stationary field (approx).


If you include these, you have all temporal derivatives almost 0. Thus you should only care about the 00 component $(\rho=\sigma=0)$, so you want to check only this symbol.
$\Gamma_{00}^{\ \ \mu}=\frac{1}{2} g^{\mu\lambda} \left( \partial_0g_{0\lambda}+\partial_0g_{\lambda 0}-\partial_\lambda g_{00} \right)$
Condition (3) gives you that the first two terms are void because time derivatives are 0 (stationary). Thereferoe there's only onee term left:
$\Gamma_{00}^{\ \mu}=-\frac{1}{2}\,g^{\mu\lambda}\,\partial_\lambda\,g_{00}$
And finally you have to use (2) (weak field). It says that the metric tensor is almost the one of the flat space (Minkowski). You can write $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ with $|h_{\mu\nu}|\ll1$.
Doing this you get
$\Gamma_{00}^{\ \mu}=-\frac{1}{2}\eta^{\mu\nu}\partial_\lambda\, h_{00}+O(h^2)$
And you neglect the last term for being second order small.
The geodesic equation is now $\frac{d^2 x^\mu}{d\tau^2}\,=\frac{1}{2}\,\eta^{\mu\lambda}\,\partial_\lambda\;\mathcal{h}_{00}\,\left( \frac{dt}{d\tau}\right)^2   $.
The derivative of $t$ respect to $\tau$ is constant. You can see it making $\mu=0$. Only diagonal terms survive (00 component), 
$\dfrac{d^2 t}{d\tau^2}=\frac{1}{2}\ \eta^{00}\;\partial_0\,{h_{00}} \;\left( \dfrac{dt}{d\tau} \right)^2$
but we said that it is stationary, so this derivative is 0. 2nd derivative 0, it means that first derivative is constant, thus $t=a\tau + {const_{void}}$ and $\frac{dt}{d\tau}=constant=1$
Extra:
The other components are
${\dfrac{d^2 x^i}{dt^2}=\frac{1}{2}\,\partial_i \mathcal{h}_{00}}$. 
Comapring with Newton:  $\frac{dx^i}{d\tau}=-\vec\nabla\phi_g$... we just got that
$h_{00}=2\phi_g$ (add units)
and so
$g_{00}=1+2\phi_g)$. There's the link.
A: The third term vanishes because the metric is approximately diagonal in the limit and any $g_{0i}$ terms are zero.
The first term vanishes because spacetime is approximately flat, which means that there is basically no spatial dependence in the metric and any $g_{\mu\nu,i}$ terms (spatial derivatives) vanish.
