測地線方程組推導

令 $\gamma(t)$ 為一 $C^\infty$ -affine manifold $(M^n, \nabla)$ 上一 $C^\infty$ -曲線，我們定義當

\nabla_{\frac{d\gamma}{dt}} \frac{d\gamma}{dt}=0

對所有 $t$ 成立時， $\gamma$ 為一測地線。只要參考切向量場沿著某一曲線如何被視為平行定義的推廣就可以直觀的看出測地線的幾何意義。

藉由座標 $\gamma(t) = (x^i(t))$ ，可將 $\gamma$ 表達為

\frac{d\gamma}{dt} = \frac{d x^i}{dt} \frac{\partial}{\partial x^i}

故推導當 $\gamma$ 為一測地線時，有方程式

\begin{aligned} 0 &= \nabla_{\frac{d x^i}{dt} \partial_i}{\frac{d x^j}{dt} \partial x_j} \\ &= \frac{d x^i}{dt} \textcolor{red}{\nabla_{\partial_i}{\frac{d x^j}{dt} \partial_j}} \quad\quad\quad \text{by linear} \\ &= \frac{d x^i}{dt} (\textcolor{red}{(\partial_i \frac{d x^j}{dt}) \partial_j + \frac{d x^j}{dt} \nabla_{\partial_i}{\partial_j}}) \quad \text{by Leibniz} \\ &= \frac{d x^i}{dt}(\partial_i \frac{d x^j}{dt}) \partial_j + \textcolor{blue}{\frac{d x^i}{dt}\frac{d x^j}{dt} \nabla_{\partial_i}{\partial_j}} \\ &= \textcolor{blue}{\frac{d x^i}{dt}\frac{d x^j}{dt} \nabla_{\partial_i}{\partial_j}} + \textcolor{green}{\frac{d x^i}{dt}}(\textcolor{green}{\partial_i} \frac{d x^j}{dt}) \partial_j \\ &= \frac{d x^i}{dt}\frac{d x^j}{dt} \nabla_{\partial_i}{\partial_j} + \textcolor{green}{\frac{d}{dt}}(\frac{d x^j}{dt}) \partial_j \quad \text{by chain rule} \\ &= \frac{d x^i}{dt}\frac{d x^j}{dt} \nabla_{\partial_i}{\partial_j} + \frac{d^2 x^{\textcolor{red}{j}}}{dt^2} \partial_{\textcolor{red}{j}} \\ &= \frac{d x^i}{dt}\frac{d x^j}{dt} \nabla_{\partial_i}{\partial_j} + \frac{d^2 x^k}{dt^2} \partial_k \\ &= \frac{d x^i}{dt}\frac{d x^j}{dt} \Gamma^k_{ij}\textcolor{red}{\partial_k} + \frac{d^2 x^k}{dt^2} \textcolor{red}{\partial_k} \quad \text{by } \nabla_{\partial_i}\partial_j = \Gamma^k_{ij}\partial_k \\ &= (\frac{d x^i}{dt}\frac{d x^j}{dt} \Gamma^k_{ij} + \frac{d^2 x^k}{dt^2}) \textcolor{red}{\partial_k} \end{aligned}

因此測地線方程組就是指

\frac{d^2 x^k}{dt^2} + \frac{d x^i}{dt}\frac{d x^j}{dt} \Gamma^k_{ij} = 0, \quad \forall k=1,\dots,n