ポアンカレ計量による距離ではX軸に沿った移動では線分が最短にならない

計量による距離

$$\begin{eqnarray} g_{(x,y)}\left(\boldsymbol{v},\boldsymbol{w}\right)&=& \frac{\left\langle \boldsymbol{v},\boldsymbol{w}\right\rangle}{y^2} \;\cdots\;ポアンカレ計量,\;\left\langle \boldsymbol{x},\boldsymbol{y} \right\rangle=\sum_{i=1}^n{x_i y_i}は標準内積 \\\gamma\left(t\right)&=&\left(x(t),y(t)\right)\;\cdots\;曲線\gamma,\;tによる媒介表示,\;始点\gamma(a),\;終点\gamma(b) \\\gamma^\prime\left(t\right)&=&\left(x^\prime(t),y^\prime(t)\right)\;\cdots\;(x(t),y(t))での接ベクトル \\L{\left(\gamma\right)}&=&\int_a^b \sqrt{g\left(\gamma^\prime\left(t\right),\gamma^\prime\left(t\right)\right)} \;\cdots\;\gammaの計量による距離 \\&=&\int_a^b \sqrt{\frac{\left\langle \gamma^\prime\left(t\right),\gamma^\prime\left(t\right)\right\rangle}{y^2}}\mathrm{d}t \\&=&\int_a^b \frac{\sqrt{\left\langle \gamma^\prime\left(t\right),\gamma^\prime\left(t\right)\right\rangle}}{\sqrt{y^2}}\mathrm{d}t \\&=&\int_a^b \frac{\left||\gamma^\prime\left(t\right)\right||}{y}\mathrm{d}t \;\cdots\;\left||x\right||=\sqrt{\left\langle \boldsymbol{x},\boldsymbol{x} \right\rangle} \end{eqnarray}$$

\((\cos{\left(\theta\right)}, \sin{\left(\theta\right)})\)から \((-\cos{\left(\theta\right)}, \sin{\left(\theta\right)})\)までの線分で移動(ただし\(\theta\in \left(0, \frac{\pi}{2}\right)\))

$$\begin{eqnarray} \gamma_3(t)&=&\left((x_1-x_0)t+x_0,(y_1-y_0)t+y_0 \right) \\&=&\left( (-\cos{\left(\theta\right)}-\cos{\left(\theta\right)})t+\cos{\left(\theta\right)}, (\sin{\left(\theta\right)}-\sin{\left(\theta\right)})t+\sin{\left(\theta\right)} \right) \\&=&\left((1-2t)\cos{\left(\theta\right)},\sin{\left(\theta\right)}\right) \\\gamma_3^\prime(t)&=&\left(-2\cos{\left(\theta\right)}, 0\right) \\L_3=L\left(\gamma_3\right)&=&\int_0^1 \frac{\left||\gamma_3^\prime\left(t\right)\right||}{y(t)}\mathrm{d}t \\&=&\int_0^1 \frac{\sqrt{ \left(-2\cos{\left(\theta\right)}\right)^2 +0^2 }}{\sin{\left(\theta\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{2\cos{\left(\theta\right)}}{\sin{\left(\theta\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{2}{\tan{\left(\theta\right)}}\mathrm{d}t \\&=&\frac{2}{\tan{\left(\theta\right)}}\int_0^1\mathrm{d}t \\&=&\frac{2}{\tan{\left(\theta\right)}}[t]_0^1 \\&=&\frac{2}{\tan{\left(\theta\right)}}[1-0] \\&=&\frac{2}{\tan{\left(\theta\right)}}\cdot 1 \\&=&\frac{2}{\tan{\left(\theta\right)}} \end{eqnarray}$$

\((\cos{\left(\theta\right)}, \sin{\left(\theta\right)})\)から \((-\cos{\left(\theta\right)}, \sin{\left(\theta\right)})\)まで，原点を中心とした半径1の円弧で移動(ただし \( \theta\in \left(0,\frac{\pi}{2}\right) \))

$$\begin{eqnarray} \gamma_4(t)&=&\left(\cos{\left(\theta+(\pi-2\theta)t\right)}, \sin{\left(\theta+(\pi-2\theta)t\right)} \right) \\\gamma_4^\prime(t)&=&\left( -(\pi-2\theta)\sin{\left(\theta+(\pi-2\theta)t\right)}, (\pi-2\theta)\cos{\left(\theta+(\pi-2\theta)t\right)} \right) \\L_4=L\left(\gamma_4\right)&=&\int_0^1 \frac{\left||\gamma_4^\prime\left(t\right)\right||}{y(t)}\mathrm{d}t \\&=&\int_0^1 \frac{\sqrt{ \left( -(\pi-2\theta)\sin{\left(\theta+(\pi-2\theta)t\right)} \right)^2 +\left( (\pi-2\theta)\cos{\left(\theta+(\pi-2\theta)t\right)} \right)^2 }} {\sin{\left(\theta+(\pi-2\theta)t\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{\sqrt{(\pi-2\theta)^2\left( \sin^2{\left(\theta+(\pi-2\theta)t\right)} +\cos^2{\left(\theta+(\pi-2\theta)t\right)} \right) }} {\sin{\left(\theta+(\pi-2\theta)t\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{\sqrt{(\pi-2\theta)^2\cdot 1}} {\sin{\left(\theta+(\pi-2\theta)t\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{\pi-2\theta} {\sin{\left(\theta+(\pi-2\theta)t\right)}}\mathrm{d}t \\&=&\int_0^1 \frac{\alpha} {\sin{\left(\beta+\alpha t\right)}}\mathrm{d}t \\&=&\int_\beta^{\alpha+\beta} \frac{\cancel{\alpha}} {\sin{\left(u\right)}}\frac{1}{\cancel{\alpha}}\mathrm{d}u \\&=&\left[\ln{\left|\tan{\left(\frac{u}{2}\right)}\right|}\right]_\beta^{\alpha+\beta} \;\cdots\;\int \frac{1}{\sin{\left(x\right)}} \mathrm{d}x= \ln{\left|\tan{\left(\frac{x}{2}\right)}\right|}+C \\&=&\left[\ln{\left|\tan{\left(\frac{u}{2}\right)}\right|}\right]_\theta^{\pi-2\theta+\theta} \\&=&\left[\ln{\left|\tan{\left(\frac{u}{2}\right)}\right|}\right]_\theta^{\pi-\theta} \\&=&\ln{\left|\tan{\left(\frac{\pi-\theta}{2}\right)}\right|} -\ln{\left|\tan{\left(\frac{\theta}{2}\right)}\right|} \\&=&\ln{\left| \frac{\tan{\left(\frac{\pi-\theta}{2}\right)}}{\tan{\left( \frac{\theta}{2}\right)}}\right|} \\&=&\ln{\left| \frac{\cot{\left(\frac{\theta}{2}\right)}}{\tan{\left( \frac{\theta}{2}\right)}}\right|} \\&=&\ln{\left| \frac{1}{\tan^2{\left( \frac{\theta}{2}\right)}} \right|} \\&=&\ln{\left| \tan^{-2}{\left( \frac{\theta}{2}\right)} \right|} \\&=&-2\ln{\left| \tan{\left( \frac{\theta}{2}\right)} \right|} \end{eqnarray}$$

\(L_3-L_4\)

$$\begin{eqnarray} L_3-L_4 &=&\left\{\frac{2}{\tan{\left(\theta\right)}}\right\} -\left\{-2\ln{\left(\tan{\left(\frac{\theta}{2}\right)}\right)}\right\} \\&=&2\left\{ \frac{1}{\tan{\left(\theta\right)}}+\ln{\left(\tan{\left(\frac{\theta}{2}\right)}\right)}\right\} \\\frac{\mathrm{d}}{\mathrm{d}\theta}\left(L_3-L_4\right) &=&2\left\{ \frac{-1}{\sin^2{\left(\theta\right)}} +\frac{1}{2\cos{\left(\frac{\theta}{2}\right)}\sin{\left(\frac{\theta}{2}\right)}} \right\} \\&=&2\left\{ \frac{-1}{\sin^2{\left(\theta\right)}} +\frac{1}{\sin{\left(\theta\right)}} \right\} \\&=&2\left\{ \frac{-1}{\sin^2{\left(\theta\right)}} +\frac{1}{\sin{\left(\theta\right)}}\frac{\sin{\left(\theta\right)}}{\sin{\left(\theta\right)}} \right\} \\&=&2\left\{ \frac{-1}{\sin^2{\left(\theta\right)}} +\frac{\sin{\left(\theta\right)}}{\sin^2{\left(\theta\right)}} \right\} \\&=&2\left\{ \frac{\sin{\left(\theta\right)}-1}{\sin^2{\left(\theta\right)}} \right\}\leq0\;\cdots\; 分母は常に正，分子は\theta\in \left(0,\frac{\pi}{2}\right)なので(-1,0)であり，全体では常に負となる． \\\lim_{\theta \rightarrow \frac{\pi}{2}-0}L_3-L_4&=&0 \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2024/11/lim-x2-cotx.html}{ \lim_{\theta \rightarrow \frac{\pi}{2}-0}{\frac{1}{\tan{\left(\theta\right)}}}=0} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2024/12/lim-x2-lntanx2.html}{\lim_{\theta \rightarrow \frac{\pi}{2}-0}{\ln{\left(\tan{\left(\frac{\theta}{2}\right)}\right)}}=0} \\よって \\L_3-L_4&\geq&0\;\cdots\;，\theta\in \left(0,\frac{\pi}{2}\right)では微分が常に負(減少関数)であり，\theta=\frac{\pi}{2}で0なので，範囲内では常に正． \end{eqnarray}$$

まとめ

ポアンカレ計量では水平の線分より，\(x\)軸から離れる側となる円弧に沿った経路の方が短いことになる．
これは境界\(x\)軸\(y=0\)に近いほど計量中にある\(\frac{1}{y^2}\)の効果で値が大きくなり，距離としては長くなることが影響した結果．