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問~~~計量~~~ds^2=\displaystyle\genfrac{}{}{0.5pt}{}{1}{y^2}(dx^2+dy^2)\\[0.2cm]
が入っているとこ、次のそれぞれの二点の最短距離を求めよ。\\[0.2cm]
~~~(1)~~C(0,1)~とA\left(0,~\genfrac{}{}{0.5pt}{}{1}{3}\right)\\[0.2cm]
~~~(2)~A\left(0,~\genfrac{}{}{0.5pt}{}{1}{3}\right)~と~
B\left(\genfrac{}{}{0.5pt}{}{4}{5},~\genfrac{}{}{0.5pt}{}{3}{5}\right)\\[0.2cm]
~~~(3)~C(0,1)~と~B\left(\genfrac{}{}{0.5pt}{}{4}{5},~\genfrac{}{}{0.5pt}{}{3}{5}\right)
\\[0.2cm]
解~~~(1)~この場合は、最短距離は直線であるから、\\[0.2cm]
~~~~d(A,C)=\int _{1/3}^{1} \,ds=\int _{1/3}^{1} \genfrac{}{}{0.5pt}{}{1}{y}
\sqrt{ds^2+dy^2}=\int _{1/3}^{1} \genfrac{}{}{0.5pt}{}{dy}{y}=
\left[ \log y \right] _{1/3}^{1}=\log 3= 1.0986...\\[0.2cm]
(2)~この場合は、与えられた二点を通り、中心が~x~軸上にある円が最短\\[0.2cm]
距離であるから、与えられた二点の垂直二等分線と~x~軸の交点を求める。\\[0.2cm]
~~~x^2+\left(y-\genfrac{}{}{0.5pt}{}{1}{3}\right)^2
=\left(x-\genfrac{}{}{0.5pt}{}{4}{5}\right)^2
+\left(y-\genfrac{}{}{0.5pt}{}{3}{5}\right)^2\\[0.2cm]
~~~~~~~9x+3y=5\\[0.2cm]
~~~y＝0~を代入。\\[0.2cm]
~~~円の中心は\left(\genfrac{}{}{0.5pt}{}{5}{9},~0\right)\\[0.2cm]
~~~半径を求めるために、点~(0,1/3)~と中心の距離の二乗を求める。\\[0.2cm]
~~~\left(\genfrac{}{}{0.5pt}{}{5}{9} \right)^2
+\left(\genfrac{}{}{0.5pt}{}{1}{3} \right)^2
=\genfrac{}{}{0.5pt}{}{34}{81}\\[0.2cm]
即ち、中心が\left(\genfrac{}{}{0.5pt}{}{5}{9},~0\right)~~半径~
\genfrac{}{}{0.5pt}{}{\sqrt{34}}{9}~の円。\\[0.2cm]
円の中心から
\left(\genfrac{}{}{0.5pt}{}{4}{5},~\genfrac{}{}{0.5pt}{}{3}{5}\right)
へ向かうベクトルと~x~軸の正の方向とのなす角を~\alpha\\[0.2cm]
~とし、同じく円の中心から
\left(0,~\genfrac{}{}{0.5pt}{}{1}{3}\right)
へ向かうベクトルと~x~軸の正の方向とのなす角を~\beta~とすると、\\[0.2cm]
~~~\tan \alpha=\genfrac{}{}{0.5pt}{}{27}{11},~~~
\tan \beta=-\genfrac{}{}{0.5pt}{}{3}{5}\\[0.2cm]
あとで必要になるので、\tan \genfrac{}{}{0.5pt}{}{\alpha}{2},~~
\tan \genfrac{}{}{0.5pt}{}{\beta}{2}~~も求めておく。\\[0.2cm]
~~~\tan \alpha=\genfrac{}{}{0.5pt}{}{2\tan {\alpha/2}}{1-\tan^2 {\alpha/2}}\\[0.2cm]
であるから、\tan {\alpha/2}=x~とおいて、\\[0.2cm]
~~~\genfrac{}{}{0.5pt}{}{27}{11}=\genfrac{}{}{0.5pt}{}{2x}{1-x^2}\\[0.2cm]
~~~これを解いて、\\[0.2cm]
~~~~~x=\genfrac{}{}{0.5pt}{}{-11±\sqrt{850}}{27}\\[0.2cm]
今の場合~x~は正だから、\\[0.2cm]
~~~~~x=\tan \genfrac{}{}{0.5pt}{}{\alpha}{2}＝
\genfrac{}{}{0.5pt}{}{-11+\sqrt{850}}{27}\\[0.2cm]
同様にして、\\[0.2cm]
~~~\tan \genfrac{}{}{0.5pt}{}{\beta}{2}
=\genfrac{}{}{0.5pt}{}{5+\sqrt{34}}{3}\\[0.2cm]
~~~d(A,B)=\int \genfrac{}{}{0.5pt}{}{1}{y}\sqrt{dx^2+dy^2}\\[0.2cm]
~~~=\int _\alpha^\beta \genfrac{}{}{0.5pt}{}{9}{\sqrt{34}\sin \theta}
\sqrt{\left( \genfrac{}{}{0.5pt}{}{\sqrt{34}}{9} \right)^2
(\sin^2 \theta+ \cos^2 \theta)}d\theta\\[0.2cm]
=\int _\alpha^\beta \genfrac{}{}{0.5pt}{}{d\theta}{\sin \theta}
=\left[ \log \tan \genfrac{}{}{0.5pt}{}{\theta}{2}\right] _\alpha^\beta
=\log \tan \genfrac{}{}{0.5pt}{}{\beta}{2}
-\log \tan \genfrac{}{}{0.5pt}{}{\alpha}{2}\\[0.2cm]
=\log \left(\genfrac{}{}{0.5pt}{}{5+\sqrt{34}}{3} \right)
-\log \left( \genfrac{}{}{0.5pt}{}{-11+\sqrt{850}}{27}\right)
=1.6806997...\\[0.2cm]
(3)~CB~の垂直二等分線を求める。\\[0.2cm]
~~~~~~~x^2+(y-1)^2=\left( x-\genfrac{}{}{0.5pt}{}{4}{5}\right)^2
+\left( y-\genfrac{}{}{0.5pt}{}{3}{5}\right)^2\\[0.2cm]
~~~~~2x-y=0\\[0.2cm]
~~~~~~y=0~のとき~x=0~~~~~即ち、円の中心は原点。\\[0.2cm]
~~~(2)~と同様に角度~\alpha,~\beta~をとる。\\[0.2cm]
~~~\tan {\alpha/2}=1/3,~~~\tan {\beta/2}=1\\[0.2cm]
~~~d(CB)=\int \genfrac{}{}{0.5pt}{}{1}{y} \sqrt{dx^2+dy^2}
=\int _{\alpha}^{\beta}\genfrac{}{}{0.5pt}{}{d\theta}{\sin \theta}
=\left[\log \tan \theta/2 \right] _\alpha^\beta\\[0.2cm]
=\log \tan \beta/2-\log \tan \alpha/2=\log 1- \log{1/3}
=\log 3=1.0986...\\[0.2cm]
~~~~~~~（解終り）\\[0.5cm]
定理~~~曲面上の二点を結ぶ最短距離の曲線弧が存在するならば、\\[0.2cm]
~~~~それは測地線の方程式を満たす。\\[0.2cm]
解~~~~証明の中で使うので予め次の計算をしておく。\\[0.2cm]
~~~F(\epsilon)=g_{\alpha \beta}(u^{(1)}(t)+
\epsilon \lambda^{(1)}(t),~u^{(2)}(t)+\epsilon \lambda^{(2)}(t))
\left(\dot{u}^{\alpha}(t)+\epsilon \dot{\lambda}^{\alpha}(t) \right)\\[0.2cm]
\left(\dot{u}^{\beta}(t)+\epsilon \dot{\lambda}^{\beta}(t) \right)
\\[0.2cm]
のとき、\\[0.2cm]
\left.\genfrac{}{}{0.5pt}{}{dF(\epsilon)}{d\epsilon}\right|_{\epsilon=0}
~~~~~（を求めておく。）\\[0.2cm]
=\left.\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^1}
\genfrac{}{}{0.5pt}{}{d(u^1(t)+\epsilon\lambda^1(t))}{d\epsilon}
\left(\dot{u}^{\alpha}(t)+\epsilon \dot{\lambda}^{\alpha}(t) \right)
\left(\dot{u}^{\beta}(t)+\epsilon \dot{\lambda}^{\beta}(t) \right)
\right|_{\epsilon=0}
\\[0.2cm]
+\left.\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^2}
\genfrac{}{}{0.5pt}{}{d(u^2(t)+\epsilon\lambda^2(t))}{d\epsilon}
\left(\dot{u}^{\alpha}(t)+\epsilon \dot{\lambda}^{\alpha}(t) \right)
\left(\dot{u}^{\beta}(t)+\epsilon \dot{\lambda}^{\beta}(t) \right)
\right|_{\epsilon=0}
\\[0.2cm]
+\left.g_{\alpha \beta}(u^{(1)}(t)+
\epsilon \lambda^{(1)}(t),~u^{(2)}(t)+\epsilon \lambda^{(2)}(t))
\genfrac{}{}{0.5pt}{}{d\left(\dot{u}^{\alpha}(t)+
\epsilon \dot{\lambda}^{\alpha}(t) \right)}{d\epsilon}
\left(\dot{u}^{\beta}(t)+\epsilon \dot{\lambda}^{\beta}(t) \right)
\right|_{\epsilon=0}\\[0.2cm]
+\left.g_{\alpha \beta}(u^{(1)}(t)+
\epsilon \lambda^{(1)}(t),~u^{(2)}(t)+\epsilon \lambda^{(2)}(t))
\left(\dot{u}^{\alpha}(t)+\epsilon \dot{\lambda}^{\alpha}(t) \right)
\genfrac{}{}{0.5pt}{}{d\left(\dot{u}^{\beta}(t)+
\epsilon \dot{\lambda}^{\beta}(t) \right)}{d\epsilon}
\right|_{\epsilon=0}\\[0.2cm]
=\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^1}
\lambda^1 \dot{u}^{\alpha} \dot{u}^{\beta}
+\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^2}
\lambda^2 \dot{u}^{\alpha} \dot{u}^{\beta}
+g_{\alpha \beta}\dot{\lambda}^{\alpha}\dot{u}^{\beta}
+g_{\alpha \beta}\dot{u}^{\alpha}\dot{\lambda}^{\beta}\\[0.2cm]
=\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^{\gamma}}
\lambda^{\gamma} \dot{u}^{\alpha} \dot{u}^{\beta}
+2g_{\gamma \beta}\dot{\lambda}^{\gamma}\dot{u}^{\beta}\\[0.2cm]
この結果を後で使う。\\[0.2cm]
また、F(0)=g_{\alpha \beta}\dot{u}^{\alpha}\dot{u}^{\beta}\\[0.2cm]
は、自明。\\[0.2cm]
~~~さて証明。\\[0.2cm]
曲面上の定点~P~から~Q~までの曲線を~C~とし、\\[0.2cm]
C: \mathbf{X}(t)=\mathbf{X}(u^1(t), u^2(t))~~~\\[0.2cm]
とする。~~~t ~\epsilon ~[0,1] ~としてよい。\\[0.2cm]
曲面の第一基本量を~g_{\alpha \beta}~とすれば、C~の長さs(C)~は、\\[0.2cm]
~~~s(C)=\int ^1_0 \sqrt{g_{\alpha \beta}
(u^1(t),u^2(t))\dot{u}^{\alpha}\dot{u}^{\beta}}dt\\[0.2cm]
両端~P,Q~を固定して、他の部分を少し移動させた曲線~C'~は、\\[0.2cm]
~~~C':~\mathbf{X}(t)=\mathbf{X}(u^1(t)+\epsilon \lambda^1(t),
u^2(t)+\epsilon \lambda^2(t))~~~\\[0.2cm]
~~~但し、~~~\lambda^{\alpha}(0)=\lambda^{\alpha}(1)=0~~~
(\alpha=1,2)\\[0.2cm]
s(C')=\int_0^1 \sqrt{g_{\alpha \beta}(u^{(1)}(t)+
\epsilon \lambda^{(1)}(t),~u^{(2)}(t)+\epsilon \lambda^{(2)}(t))
\left(\dot{u}^{\alpha}(t)+\epsilon \dot{\lambda}^{\alpha}(t) \right)\\[0.2cm]
\left(\dot{u}^{\beta}(t)+\epsilon \dot{\lambda}^{\beta}(t) \right)}dt
~~~この積分の被積分函数の根号の中味は先に計算した~F(\epsilon)~である
から、\\[0.2cm]
=\int_0^1\sqrt{F(\epsilon)}dt\\[0.2cm]
~~s(C')~を~\epsilon~の函数と考えたとき、\epsilon~がゼロのとき極小値をとる
ことが必要。\\[0.2cm]
\left.\genfrac{}{}{0.5pt}{}{dS(C')}{d\epsilon}\right|_{\epsilon=0}
=\int_0^1\genfrac{}{}{0.5pt}{}{1}{2\sqrt{F(0)}}
\left(\left.\genfrac{}{}{0.5pt}{}{dF(\epsilon)}{d\epsilon}\right|_
{\epsilon=0}\right)dt\\[0.2cm]
=\int_0^1 \genfrac{}{}{0.5pt}{}{1}{2\sqrt{g_{\alpha \beta}\dot{u}^{\alpha}
\dot{u}^{\beta}}}\left(\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}
{\partial \dot{u}^{\gamma}} \lambda^{\gamma}
\dot{u}^{\alpha}\dot{u}^{\beta}+2g_{\gamma \beta}
\dot{\lambda}^{\gamma}\dot{u}^{\beta}\right)dt\\[0.2cm]
ここでパラメター~t~をはじめの曲線弧~C~の弧長にとれば、\\[0.2cm]
~~~g_{\alpha \beta}\dot{u}^{\alpha}\dot{u}^{\beta}=1\\[0.2cm]
であるから、\\[0.2cm]
~~~\left. \genfrac{}{}{0.5pt}{}{dS(C')}{d\epsilon}\right|_{\epsilon=0}
=\int_0^1\left(\genfrac{}{}{0.5pt}{}{1}{2}
\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^{\gamma}}
\dot{u}^{\alpha}\dot{u}^{\beta}\lambda^{\gamma}
+g_{\gamma \beta}\dot{\lambda}^{\gamma}\dot{u}^{\beta}\right)
ds\\[0.2cm]
~~~第二項に部分積分を施して、\\[0.2cm]
=\int_0^1\left(\genfrac{}{}{0.5pt}{}{1}{2}
\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^{\gamma}}
\dot{u}^{\alpha}\dot{u}^{\beta}\lambda^{\gamma}\right)ds
+\left[\lambda^{\gamma}g_{\gamma \beta}\dot{u}^{\beta}\right]_0^1
-\int_0^1\gamma^{\lambda}\genfrac{}{}{0.5pt}{}{d}{ds}
(g_{\gamma \beta}\dot{u}^{\beta})ds\\[0.2cm]
~~~この第２項は函数~\lambda~の性質から~0~。\\
=\int_0^1 \left[
\genfrac{}{}{0.5pt}{}{1}{2}
\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^{\gamma}}
\dot{u}^{\alpha}\dot{u}^{\beta}-
\genfrac{}{}{0.5pt}{}{d}{ds}
(g_{\gamma \beta}\dot{u}^{\beta})\right]\lambda^{\gamma}ds
S(C)~が最短距離であるから、
\left.\genfrac{}{}{0.5pt}{}{dS(C')}{d\epsilon}\right|_{\epsilon=0}\\
\therefore \genfrac{}{}{0.5pt}{}{d}{ds}(g_{\gamma \beta}
\dot{u}^{\beta})-\genfrac{}{}{0.5pt}{}{1}{2}
\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^{\gamma}}
\dot{u}^{\alpha}\dot{u}^{\beta}=0\\[0.2cm]
\therefore g_{\gamma \beta} \ddot{u}^{\beta}
+\left( \genfrac{}{}{0.5pt}{}{\partial g_{\gamma \beta}}
{\partial u^{\alpha}}\dot{u}^{\alpha}\right)\dot{u}^{\beta}
-\genfrac{}{}{0.5pt}{}{1}{2}\genfrac{}{}{0.5pt}{}
{\partial g_{\alpha \beta}}{\partial u^{\gamma}}
\dot{u}^{\alpha}\dot{u}^{\beta}=0\\[0.2cm]
\therefore g_{\gamma \beta} \ddot{u}^{\beta}
+\left( \genfrac{}{}{0.5pt}{}{\partial g_{\gamma \beta}}
{\partial u^{\alpha}}
-\genfrac{}{}{0.5pt}{}{1}{2}\genfrac{}{}{0.5pt}{}
{\partial g_{\alpha \beta}}{\partial u^{\gamma}}
\right)\dot{u}^{\alpha}\dot{u}^{\beta}=0\\[0.2cm]
ここで、\\[0.2cm]
\left( \genfrac{}{}{0.5pt}{}{\partial g_{\gamma \beta}}
{\partial u^{\alpha}}
-\genfrac{}{}{0.5pt}{}{1}{2}\genfrac{}{}{0.5pt}{}
{\partial g_{\alpha \beta}}{\partial u^{\gamma}}
\right)\dot{u}^{\alpha}\dot{u}^{\beta}=
\left( \genfrac{}{}{0.5pt}{}{\partial g_{\gamma 1}}
{\partial u^{1}}
-\genfrac{}{}{0.5pt}{}{1}{2}\genfrac{}{}{0.5pt}{}
{\partial g_{11}}{\partial u^{\gamma}}
\right)\dot{u}^{1}\dot{u}^{1}
+\left( \genfrac{}{}{0.5pt}{}{\partial g_{\gamma 2}}
{\partial u^{1}}
-\genfrac{}{}{0.5pt}{}{1}{2}\genfrac{}{}{0.5pt}{}
{\partial g_{12}}{\partial u^{\gamma}}
\right)\dot{u}^{1}\dot{u}^{2}\\[0.2cm]
+\left( \genfrac{}{}{0.5pt}{}{\partial g_{\gamma 1}}
{\partial u^{2}}
-\genfrac{}{}{0.5pt}{}{1}{2}\genfrac{}{}{0.5pt}{}
{\partial g_{21}}{\partial u^{\gamma}}
\right)\dot{u}^{2}\dot{u}^{1}
+\left( \genfrac{}{}{0.5pt}{}{\partial g_{\gamma 2}}
{\partial u^{2}}
-\genfrac{}{}{0.5pt}{}{1}{2}\genfrac{}{}{0.5pt}{}
{\partial g_{22}}{\partial u^{\gamma}}
\right)\dot{u}^{2}\dot{u}^{2}\\[0.2cm]
=[11,\gamma]\dot{u}^{1}\dot{u}^{1}
+2[12,\gamma]\dot{u}^{1}\dot{u}^{2}
[22,\gamma]\dot{u}^{2}\dot{u}^{2}
=[\alpha \beta,\gamma]\dot{u}^{\alpha}\dot{u}^{\beta}\\[0.2cm]
であるから、\\[0.2cm]
\therefore~~g_{\gamma \beta} \ddot{u}^{\beta}
+[\alpha \beta, \gamma]\dot{u}^{\alpha}\dot{u}^{\beta}=0
~~~~~(\gamma=1,2)\\[0.2cm]
\therefore~~g_{11}\ddot{u}^1+g_{12}\ddot{u}^2
=-[\alpha \beta,1]\dot{u}^{\alpha}\dot{u}^{\beta}\\[0.2cm]
~~~~~~g_{21}\ddot{u}^1+g_{22}\ddot{u}^2
=-[\alpha \beta,2]\dot{u}^{\alpha}\dot{u}^{\beta}\\[0.2cm]
\therefore\\[0.2cm]
\begin{pmatrix}
g_{11} & g_{12}\\
g_{21} & g_{22}\\
\end{pmatrix}
\begin{pmatrix}
\ddot{u}^1\\
\ddot{u}^1\\
\end{pmatrix}
=
\begin{pmatrix}
-[\alpha \beta, 1]\\
-[\alpha \beta, 2]\\
\end{pmatrix}
\dot{u}^{\alpha}\dot{u}^{\beta}\\[0.2cm]
逆行列を左からかけて、\\[0.2cm]
\begin{pmatrix}
\ddot{u}^1\\
\ddot{u}^1\\
\end{pmatrix}
=-
\begin{pmatrix}
g^{11} & g^{12}\\
g^{21} & g^{22}\\
\end{pmatrix}
\begin{pmatrix}
-[\alpha \beta, 1]\\
-[\alpha \beta, 2]\\
\end{pmatrix}
\dot{u}^{\alpha}\dot{u}^{\beta}\\[0.2cm]
即ち、\\[0.2cm]
\ddot{u}^1=-\left\{{}^{\,1}_{\alpha\beta}\right\}
\dot{u}^{\alpha}\dot{u}^{\beta}\\[0.2cm]
\ddot{u}^2=-\left\{{}^{\,2}_{\alpha\beta}\right\}
\dot{u}^{\alpha}\dot{u}^{\beta}\\[0.2cm]
これは測地線の方程式。~~~~~~~~~（証明終り）\\[0.2cm]
%\left\{{}^{\,\lambda}_{\mu\nu}\right\}
\end{math}
\end{document}