# Why is the Christoffel symbol in the geodesic equation for a test particle negative?

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 $$g^{nn}$$.