📂幾何学

測地線座標パッチマッピングとクリストッフェル記号

定理¹

$$ \left[ g_{ij} \right] = \begin{bmatrix} 1 & 0 \\ 0 & h^{2} \end{bmatrix} \quad (h \gt 0) $$

つまり、$\mathbf{x}$のクリストッフェル記号は以下の通りで、下記のもの以外はすべて$0$だ。

$$ \Gamma_{22}^{1} = -hh_{1},\quad \Gamma_{12}^{2} = \Gamma_{21}^{2} = \dfrac{h_{1}}{h},\quad \Gamma_{22}^{2} = \dfrac{h_{2}}{h} $$

この時、$(u^{1}, u^{2})$は$U$の座標であり、$h_{i} = \dfrac{\partial h}{\partial u^{i}}$である。

証明

証明に先立って、必要な計算を行おう。$g_{11} = \left\langle\mathbf{x}_{1}, \mathbf{x}_{1}\right\rangle = 1$なので、

$$ \dfrac{\partial g_{11}}{\partial u_{i}} = \dfrac{\partial \left\langle\mathbf{x}_{1}, \mathbf{x}_{1}\right\rangle}{\partial u_{i}} = 2\left\langle\mathbf{x}_{1i}, \mathbf{x}_{1}\right\rangle = 0 $$

$$ \implies \left\langle\mathbf{x}_{11}, \mathbf{x}_{1}\right\rangle = 0 \quad \text{ and } \quad \left\langle\mathbf{x}_{12}, \mathbf{x}_{1}\right\rangle = 0 \tag{1} $$

同様に$g_{12} = \left\langle\mathbf{x}_{1}, \mathbf{x}_{2}\right\rangle = 0$なので、

$$ \dfrac{\partial g_{12}}{\partial u_{i}} = \dfrac{\partial \left\langle\mathbf{x}_{1}, \mathbf{x}_{2}\right\rangle}{\partial u_{i}} = \left\langle\mathbf{x}_{1i}, \mathbf{x}_{2}\right\rangle + \left\langle\mathbf{x}_{1}, \mathbf{x}_{2i}\right\rangle = 0 $$

ここで$i=1$の場合を見ると、$\left\langle \mathbf{x}_{11}, \mathbf{x}_{2}\right\rangle + \left\langle\mathbf{x}_{1}, \mathbf{x}_{21}\right\rangle = 0$で、$(1)$によって$\left\langle \mathbf{x}_{11}, \mathbf{x}_{2} \right\rangle = 0$になる。$i=2$の場合は、

$$ \left\langle\mathbf{x}_{12}, \mathbf{x}_{2}\right\rangle + \left\langle\mathbf{x}_{1}, \mathbf{x}_{22}\right\rangle = 0 \tag{2} $$

また、$h = \sqrt{g_{22}} = \sqrt{\left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle}$なので、

$$ h_{i} = \dfrac{\partial h}{\partial u^{i}} = \dfrac{\partial \sqrt{\left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle}}{\partial u^{i}} = \dfrac{1}{2\sqrt{\left\langle \mathbf{x}_{2}, \mathbf{x}_{2} \right\rangle}} 2\left\langle \mathbf{x}_{2i}, \mathbf{x}_{2} \right\rangle = \dfrac{1}{h}\left\langle \mathbf{x}_{2i}, \mathbf{x}_{2} \right\rangle $$

$$ \implies \left\langle \mathbf{x}_{21}, \mathbf{x}_{2} \right\rangle = hh_{1} \quad \text{ and } \quad \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle = hh_{2}\tag{3} $$

クリストッフェル記号

$$ \Gamma_{ij}^{k} := \sum \limits_{l=1}^{2} \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle g^{lk} $$

今、上で計算したものをクリストッフェル記号の定義に代入するだけで結果が得られる。

$$ \begin{align*} \Gamma_{11}^{1} &= \left\langle \mathbf{x}_{11}, \mathbf{x}_{1} \right\rangle g^{11} + \left\langle \mathbf{x}_{11}, \mathbf{x}_{2} \right\rangle g^{21} \\ &= 0 \cdot 1 + 0 \cdot 0 = 0 \end{align*} $$

$$ \begin{align*} \Gamma_{12}^{1} = \Gamma_{21}^{1} &= \left\langle \mathbf{x}_{12}, \mathbf{x}_{1} \right\rangle g^{11} + \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle g^{21} \\ &= 0 \cdot 1 + \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle \cdot 0 = 0 \end{align*} $$

$$ \begin{align*} \Gamma_{22}^{1} &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle g^{11} + \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle g^{21} \\ &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle \cdot 1 + \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle \cdot 0 \\ &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle = - \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle & \text{ by } (2)\\ &= - hh_{1} & \text{ by } (3)\\ \end{align*} $$

$$ \begin{align*} \Gamma_{11}^{2} &= \left\langle \mathbf{x}_{11}, \mathbf{x}_{1} \right\rangle g^{12} + \left\langle \mathbf{x}_{11}, \mathbf{x}_{2} \right\rangle g^{22} \\ &= 0 \cdot 0 + 0 \cdot \dfrac{1}{h^{2}} = 0 \\ \end{align*} $$

$$ \begin{align*} \Gamma_{12}^{2} = \Gamma_{21}^{2} &= \left\langle \mathbf{x}_{12}, \mathbf{x}_{1} \right\rangle g^{12} + \left\langle \mathbf{x}_{12}, \mathbf{x}_{2} \right\rangle g^{22} \\ &= 0 \cdot 0 + hh_{1} \cdot \dfrac{1}{h^{2}} \\ &= \dfrac{h_{1}}{h} \\ \end{align*} $$

$$ \begin{align*} \Gamma_{22}^{2} &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle g^{12} + \left\langle \mathbf{x}_{22}, \mathbf{x}_{2} \right\rangle g^{22} \\ &= \left\langle \mathbf{x}_{22}, \mathbf{x}_{1} \right\rangle \cdot 0 + hh_{2} \cdot \dfrac{1}{h^{2}} \\ &= \dfrac{h_{2}}{h} \\ \end{align*} $$

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