The geodesic equation is

$$ {d^2 x^\mu \over {ds}^2}+\Gamma^\mu {}_{\alpha \beta}{d x^\alpha \over ds}{d x^\beta \over ds}=0\ $$

for some scalar parameter of motion s and connection coefficients of the second kind $\Gamma^\mu {}_{\alpha \beta}$. It's straightforward to show this reduces to

$$ \frac{d^2 x^n}{{dt}^2} = -\Gamma^n{}_{00} $$

for n in [1, 2, 3] when rewritten in terms of $t\equiv x^0$ and in the limit of velocities $\ll c$. This negative makes sense when substituting $\Gamma^\mu {}_{00}=-\frac 1 2 g^{\mu\sigma}\partial_\sigma g_{00}$ (static field limit), and then $g_{00}=1+2\phi$ (weak field limit, $\eta_{\mu\nu}=\operatorname{diag}(1, -1, -1, -1)$).

But if we forget that, and instead consider that $\Gamma^n {}_{00}=\frac{\partial \mathbf{e}_0}{\partial x^0}\cdot \mathbf{e}^n$, then intuitively why is this negative? Shouldn't a test particle moving in $\mathbf{e}_0$ accelerate in the direction $\mathbf{e}_0$ changes?

Edit: It was the dual. It makes more sense to consider $\Gamma_{kij} = \frac{\partial \mathbf{e}_{i}}{\partial x^j} \cdot \mathbf{e}^{m} g_{mk} = \frac{\partial \mathbf{e}_{i}}{\partial x^j} \cdot \mathbf{e}_{k}$, whose sign is flipped by $g^{nn}$.


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