測地線
となる。上式に g k a {\displaystyle g^{ka}} を掛け合わせ a に関して足しあわせを行うと、
δ i k ( x ̈ i − x ̇ i s ̈ s ̇ ) + 1 2 g k a ( ∂ g a i ∂ x j + ∂ g a j ∂ x i − ∂ g i j ∂ x a ) x ̇ i x ̇ j = x ̈ k − x ̇ k s ̈ s ̇ + 1 2 g k a ( ∂ g a i ∂ x j + ∂ g a j ∂ x i − ∂ g i j ∂ x a ) x ̇ i x ̇ j = 0 {\displaystyle \delta _{i}^{k}\left({\ddot {x}}^{i}-{\dot {x}}^{i}{\frac {\ddot {s}}{\dot {s}}}\right)+{\frac {1}{2}}g^{ka}\left({\frac {\partial g_{ai}}{\partial x^{j}}}+{\frac {\partial g_{aj}}{\partial x^{i}}}-{\frac {\partial g_{ij}}{\partial x^{a}}}\right){\dot {x}}^{i}{\dot {x}}^{j}={\ddot {x}}^{k}-{\dot {x}}^{k}{\frac {\ddot {s}}{\dot {s}}}+{\frac {1}{2}}g^{ka}\left({\frac {\partial g_{ai}}{\partial x^{j}}}+{\frac {\partial g_{aj}}{\partial x^{i}}}-{\frac {\partial g_{ij}}{\partial x^{a}}}\right){\dot {x}}^{i}{\dot {x}}^{j}=0}