To see how the latter reduces to the former, we assume that the test particle's velocity is approximately zero d x β β d τ τ ≈ ≈ ( d t d τ τ , 0 , 0 , 0 ) and thus d d t ( d t d τ τ ) ≈ ≈ 0 and that the metric and its derivatives are approximately static and that the squares of deviations from the Minkowski metric are negligible. Applying these simplifying assumptions to the spatial components of the geodesic equation gives d 2 x i d t 2 ≈ ≈ − − Γ Γ 00 i where two factors of ⁠ dt / dτ ⁠ have been divided out. This will reduce to its Newtonian counterpart, provided Φ Φ , i ≈ ≈ Γ Γ 00 i = 1 2 g i α α ( g α α 0 , 0 + g 0 α α , 0 − − g 00 , α α ) .