本文中的公式均采用爱因斯坦求和约定

设曲面上的曲线$C$: $$ u^i=u^i(s),\ \ i=1,2 $$ 通过$P,Q$两点，$P,Q$对应的参数分别为$s_1,s_2$，过这两点邻近$C$的曲线$\widetilde{C}$为: $$ \widetilde{u}^i=u^i(s)+\varepsilon w^i(s),\ \ i=1,2 $$ 其中，$w^i(s_1)=w^i(s_2)=0(i=1,2)$。

沿曲线$C$在$PQ$间的弧长为 $$ s=\int_{s_1}^{s_2}\mathrm{d}s=\int_{s_1}^{s_2}\sqrt{g_{ij}\frac{\mathrm{d}u^i}{\mathrm{d}s}\frac{\mathrm{d}u^j}{\mathrm{d}s}}\mathrm{d}s=\int_{s_1}^{s_2}\varphi(u^1,u^2,\dot{u}^1,\dot{u}^2)\mathrm{d}s. $$

沿曲线$\widetilde{C}$在$PQ$间的弧长为 $$ \widetilde{s}(\varepsilon)=\int_{s_1}^{s_2}\varphi(u^1+\varepsilon w^1,u^2+\varepsilon w^2,\dot{u}^1+\varepsilon \dot{w}^1,\dot{u}^2+\varepsilon \dot{w}^2)\mathrm{d}s. $$ 要使曲线$C$为曲线簇$\widetilde{C}$中弧长最短的曲线，必须有 $$ \frac{\mathrm{d}\widetilde{s}(\varepsilon)}{\mathrm{d}\varepsilon}|_{\varepsilon=0}=0. $$ 即 $$ \int_{s_1}^{s_2}(\frac{\partial \varphi}{\partial u^1}w^1+\frac{\partial \varphi}{\partial u^2}w^2+\frac{\partial \varphi}{\partial\dot{u}^1}\dot{w}^1+\frac{\partial \varphi}{\partial\dot{u}^2}\dot{w}^2)\mathrm{d}s=0. $$ 即 $$ \int_{s_1}^{s_2}(\frac{\partial \varphi}{\partial u^i}w^i+\frac{\partial \varphi}{\partial\dot{u}^i}\dot{w}^i)\mathrm{d}s=0. $$

由分部积分有 $$ \int_{s_1}^{s_2}\frac{\partial \varphi}{\partial\dot{u}^i}\dot{w}^ids=\int_{s_1}^{s_2}\frac{\partial \varphi}{\partial\dot{u}^i}\mathrm{d}{w^i}=(\frac{\partial \varphi}{\partial\dot{u}^i}w^i)|_{s_1}^{s_2}-\int_{s_1}^{s_2}w^i\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial \varphi}{\partial\dot{u}^i}=-\int_{s_1}^{s_2}w^i\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial \varphi}{\partial\dot{u}^i}. $$ 所以 $$ \int_{s_1}^{s_2}w^i(\frac{\partial \varphi}{\partial\dot{u}^i}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial \varphi}{\partial\dot{u}^i})\mathrm{d}s=0. $$ 由于$w^i$是任意的，得到欧拉方程 $$ \frac{\partial \varphi}{\partial u^i}-\frac{\mathrm{d}}{\mathrm{d}s}\frac{\partial \varphi}{\partial\dot{u}^i}=0,\ \ i=1,2. $$ 由于 $$ \varphi=\sqrt{g_{ij}\frac{\mathrm{d}u^i}{\mathrm{d}s}\frac{\mathrm{d}u^j}{\mathrm{d}s}}, $$

$$ \frac{\partial \varphi}{\partial\dot{u}^i}=\frac{g_{ij}\dot{u}^j}{\sqrt{g_{ij} \dot{u}^i\dot{u}^j}}=g_{ij}\dot{u}^j, $$

$$ \frac{\partial \varphi}{\partial u^i}=\frac{\frac{\partial g_{jk}}{\partial u^i}\dot{u}^j\dot{u}^k}{2\sqrt{g_{ij}\dot{u}^i\dot{u}^j}}=\frac{1}{2}\frac{\partial g_{jk}}{\partial u^i}\dot{u}^j\dot{u}^k. $$

于是欧拉方程化简为 $$ \frac{1}{2}\frac{\partial g_{jk}}{\partial u^i}\dot{u}^j\dot{u}^k-g_{ij}\ddot{u}^j-\frac{\partial g_{ij}}{\partial u^{k}}\dot{u}^k\dot{u}^j=0. $$ 又 $$ [jk,i]=g_{il}\Gamma_{jk}^{l}=\frac{1}{2}(\frac{\partial g_{ij}}{\partial u^k}+\frac{\partial g_{ik}}{\partial u^j}-\frac{\partial g_{jk}}{\partial u^i}), $$ 所以 $$ g_{il}\Gamma_{jk}^{l}\dot{u}^j\dot{u}^k=(\frac{1}{2}\frac{\partial g_{ij}}{\partial u^k}\dot{u}^j\dot{u}^k+\frac{1}{2}\frac{\partial g_{ik}}{\partial u^j}\dot{u}^j\dot{u}^k-\frac{1}{2}\frac{\partial g_{jk}}{\partial u^i}\dot{u}^j\dot{u}^k)=\frac{\partial g_{ij}}{\partial u^{k}}\dot{u}^j\dot{u}^k-\frac{1}{2}\frac{\partial g_{jk}}{\partial u^i}\dot{u}^j\dot{u}^k. $$ 于是欧拉方程可化为 $$ g_{il}(\ddot{u}^l+\Gamma_{jk}^{l}\dot{u}^j\dot{u}^k)=0, $$ 又$g=\det(g_{il})\neq0$，故上式等价于 $$ \ddot{u}^l+\Gamma_{jk}^{l}\dot{u}^j\dot{u}^k=0,\ \ l=1,2. $$ 此即曲面上的测地线方程。