## 「測地線方程式」の数式の検討

$\frac{d}{d\alpha }\left ( \sqrt{A(\lambda ,\alpha )} \right )= \frac{1}{2 \sqrt{A(\lambda ,\alpha )} }\frac{dA(\lambda ,\alpha ) }{d\alpha } = \frac{1}{2 \sqrt{A(\lambda ,\alpha )} }\frac{d }{d\alpha } \left \{g_{ij}\left ( \tilde{x}^{k} (\lambda )\right )\frac{d\tilde{x}^{i}}{d\lambda }\frac{d\tilde{x}^{j}}{d\lambda } \right \}$
$= \frac{1}{2 \sqrt{A(\lambda ,\alpha )} } \left \{ \frac{d\: g_{ij}\left ( \tilde{x}^{k} (\lambda )\right )}{d\alpha } \frac{d\tilde{x}^{i}}{d\lambda }\frac{d\tilde{x}^{j}}{d\lambda }+g_{ij}\left ( \tilde{x}^{k} (\lambda )\right )\frac{d}{d\alpha }\left ( \frac{d\tilde{x}^{i}}{d\lambda } \right )\frac{d\tilde{x}^{j}}{d\lambda }+g_{ij}\left ( \tilde{x}^{k} (\lambda )\right )\frac{d\tilde{x}^{i}}{d\lambda }\frac{d}{d\alpha }\left ( \frac{d\tilde{x}^{j}}{d\lambda } \right )\right \}$
$= \frac{1}{2 \sqrt{A(\lambda ,\alpha )} } \left \{ \frac{d\: g_{ij}\left ( \tilde{x}^{k} (\lambda )\right )}{d\alpha } \frac{d\tilde{x}^{i}}{d\lambda }\frac{d\tilde{x}^{j}}{d\lambda }+2g_{ij}\left ( \tilde{x}^{k} (\lambda )\right )\frac{d}{d\alpha }\left ( \frac{d\tilde{x}^{i}}{d\lambda } \right )\frac{d\tilde{x}^{j}}{d\lambda }\right \}$

ですが、
$\frac{d\: g_{ij}\left ( \tilde{x}^{k} (\lambda )\right )}{d\alpha } = \frac{\partial \: g_{ij}\left ( \tilde{x}^{k} \right )}{\partial \tilde{x}^{k} }\frac{d\tilde{x}^{k}}{d\alpha }= \frac{\partial \: g_{ij}\left ( \tilde{x}^{k} \right )}{\partial \tilde{x}^{k} }w^{i}$
$\Rightarrow \; \; \frac{d\: g_{ij}\left ( \tilde{x}^{k} (\lambda )\right )}{d\alpha }\bigg|_{\alpha =0}= \frac{\partial \: g_{ij}}{\partial x^{k} }w^{i}$
$\frac{d}{d\alpha }\left ( \frac{d\: \tilde{x}^{i} }{ d\lambda } \right )=\frac{d}{d \lambda}\left ( \frac{d\: \tilde{x}^{i} }{ d\alpha } \right )= \frac{dw^{i}}{d\lambda }$
から、

$\frac{dS}{d\alpha }\bigg |_{\alpha = 0}= \frac{1}{2}\int_{0}^{\lambda _{f}}\frac{1}{\sqrt{ A(\lambda ,0 )}}\left \{ \frac{\partial g_{ij}}{\partial x^{k}}w^{k} \frac{dx^{i}}{d\lambda }\frac{dx^{j}}{d\lambda }+ 2g_{ij}\frac{dw^{i}}{d\lambda }\frac{dx^{j}}{d\lambda } \right \}d\lambda$
$= \int_{0}^{\lambda _{f}}\frac{1}{2\sqrt{ A}} \frac{\partial g_{ij}}{\partial x^{k}}w^{k} \frac{dx^{i}}{d\lambda }\frac{dx^{j}}{d\lambda } d\lambda +\int_{0}^{\lambda _{f}} \frac{1}{\sqrt{ A}} g_{ij}\frac{dw^{i}}{d\lambda }\frac{dx^{j}}{d\lambda } d\lambda$

$\int_{0}^{\lambda _{f}} \frac{1}{\sqrt{ A}} g_{ij}\frac{dw^{i}}{d\lambda }\frac{dx^{j}}{d\lambda } d\lambda= \int_{0}^{\lambda _{f}}\frac{dw^{i}}{d\lambda }\left ( \frac{1}{\sqrt{ A}} g_{ij}\frac{dx^{j}}{d\lambda } \right ) d\lambda$
$= \int_{0}^{\lambda _{f}}\frac{dw^{i}}{d\lambda }\left ( \frac{1}{\sqrt{ A}} g_{ij}\frac{dx^{j}}{d\lambda } \right ) d\lambda = \left [ w^{i} \left ( \frac{1}{\sqrt{ A}} g_{ij}\frac{dx^{j}}{d\lambda } \right ) \right ]_{0}^{\lambda _{f}} - \int_{0}^{\lambda _{f}}w^{i} \frac{d}{d\lambda }\left ( \frac{1}{\sqrt{ A}} g_{ij}\frac{dx^{j}}{d\lambda } \right )d\lambda$
$= - \int_{0}^{\lambda _{f}}w^{i} \frac{d}{d\lambda }\left ( \frac{1}{\sqrt{ A}} g_{ij}\frac{dx^{j}}{d\lambda } \right )d\lambda$

から、

$\frac{dS}{d\alpha }\bigg |_{\alpha = 0}=\int_{0}^{\lambda _{f}} \left \{\frac{1}{2\sqrt{A}}\frac{\partial g_{jk}}{\partial x^{i}} \frac{dx^{j}}{d\lambda }\frac{dx^{k}}{d\lambda }-\frac{d}{d\lambda }\left [\frac{1}{\sqrt{A}}g_{ij}\frac{dx^{j}}{d\lambda } \right ] \right \}w^{i}(\lambda )d\lambda$
$= 0$

となり、被積分関数をゼロとおくと、最後の式が導出できます・

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