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（スミルノフ訳書４、369頁）\\[0.2cm]
§130. ~~ Gauss~の第一基本形式\\
\\
問~~~~~曲面S~:~(x(u,v), y(u,v),z(u,v))\\[0.2cm]
~~~~~~~S~上の曲線(~x(u(t),v(t)), ~y(u(t),v(t)),~z(u(t),v(t))~)\\[0.2cm]
が与えられているとき、\\[0.2cm]
~~~~~この曲線の線素~ds~の２乗ds^2を求めよ。\\[0.2cm]
解~~~~ds^2=dx^2+dy^2+dz^2=
\left( \displaystyle\genfrac{}{}{0.5pt}{}{\partial x}{\partial u}du
+ \genfrac{}{}{0.5pt}{}{\partial x}{\partial v}dv \right)^2+
\left(\displaystyle \genfrac{}{}{0.5pt}{}{\partial y}{\partial u}du
+ \genfrac{}{}{0.5pt}{}{\partial y}{\partial v}dv \right)^2+
\left(\displaystyle \genfrac{}{}{0.5pt}{}{\partial z}{\partial u}du
+ \genfrac{}{}{0.5pt}{}{\partial z}{\partial v}dv \right)^2\\
=\left(\left(\displaystyle\genfrac{}{}{0.5pt}{}{\partial x}{\partial u}\right)^2 +
\left(\displaystyle\genfrac{}{}{0.5pt}{}{\partial y}{\partial u}\right)^2+
\left(\displaystyle\genfrac{}{}{0.5pt}{}{\partial z}{\partial u}\right)^2\right)du^2\\
+2\left(\displaystyle\genfrac{}{}{0.5pt}{}{\partial x}{\partial u}
\displaystyle\genfrac{}{}{0.5pt}{}{\partial x}{\partial v}+
\displaystyle\genfrac{}{}{0.5pt}{}{\partial y}{\partial u}
\displaystyle\genfrac{}{}{0.5pt}{}{\partial y}{\partial v}+
\displaystyle\genfrac{}{}{0.5pt}{}{\partial z}{\partial u}
\displaystyle\genfrac{}{}{0.5pt}{}{\partial z}{\partial v}\right)dudv+\\
\left(\left(\displaystyle\genfrac{}{}{0.5pt}{}{\partial x}{\partial v}\right)^2 +
\left(\displaystyle\genfrac{}{}{0.5pt}{}{\partial y}{\partial v}\right)^2+
\left(\displaystyle\genfrac{}{}{0.5pt}{}{\partial z}{\partial v}\right)^2\right)dv^2\\[0.2cm]
（解終り）\\[0.5cm]
問~~~~~\mathbf{r}_u=(x_u,y_u,z_u),~\mathbf{r}_v=(x_v,y_v,z_v),~とおき、
前問のds^2~のdu^2~の係数をＥ、dudv~の係数をＦ、dv^2~の係数をＧ、としたとき、
Ｅ，Ｆ，Ｇ~を\mathbf{r}_u、\mathbf{r}_v~で表せ。\\[0.2cm]
解~~~~Ｅ=x_u^2+y_u^2+z_u^2=\mathbf{r}_u^2\\[0.2cm]
~~~~~~~Ｆ=2(x_ux_v+y_uy_v+z_uz_v)=2\mathbf{r}_u\mathbf{r}_v\\[0.2cm]
~~~~~~~Ｇ=x_v^2+y_v^2+z_v^2=\mathbf{r}_v^2\\[0.2cm]
（解終り）\\[0.5cm]
問~~~~~上のS~の面分dS~を求めよ。\\[0.2cm]
解~~~~dS=\sqrt{\mathbf{r}_u \times \mathbf{r}_v}dudv
=\sqrt{\begin{vmatrix}
y_u & z_u\\
y_v &z_v\\
\end{vmatrix}\\^2+
\begin{vmatrix}
x_u & z_u\\
x_v &z_v\\
\end{vmatrix}\\^2+
\begin{vmatrix}
x_u & y_u\\
x_v &y_v\\
\end{vmatrix}\\^2
}\\[0.2cm]
=\sqrt{\left |\begin{pmatrix}
x_u & y_u&z_u\\
x_v &y_v&z_v\\
\end{pmatrix}\\
\begin{pmatrix}
x_u & x_v\\
y_u &y_v\\
z_u & z_v\\
\end{pmatrix} \right |\\
}dudv
=\sqrt{\left |
\begin{pmatrix}
\mathbf{r}_u^2 & \mathbf{r}_u\mathbf{r}_v\\
\mathbf{r}_v\mathbf{r}_u&\mathbf{r}_v^2\\
\end{pmatrix} \right |\\
}dudv\\[0.2cm]
=\sqrt{EG-F^2}dudv\\[0.2cm]
（解終り）\\[0.5cm]
問~~~~~上のS~の単位法線ベクトル\mathbf{m}~を求めよ。\\[0.2cm]
（解）~~~~\mathbf{m}=\displaystyle\genfrac{}{}{0.5pt}{}
{\mathbf{r}_u \times \mathbf{r}_v}
{\begin{vmatrix}\mathbf{r}_u \times \mathbf{r}_v\end{vmatrix}}\\[0.5cm]
131~~Gauss~の第2~基本形式\\[0.2cm]
Recall~~~~~~~C:~~\mathbf{r}=(x(t),~y(t),~z(t)~) ~~~曲線\\[0.2cm]
~~~~~~接線ベクトル~~\displaystyle\genfrac{}{}{0.5pt}{}{d\mathbf{r}}{ds}
=\mathbf{t}\\[0.2cm]
~~~~ここで\begin{vmatrix}\mathbf{t}\end{vmatrix}=1~~~(\because d\mathbf{r}d\mathbf{r}=
ds^2)\\[0.2cm]
~~~\displaystyle\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds}=\kappa \mathbf{n} \\[0.2cm]
~~~ここで\begin{vmatrix}\mathbf{n}\end{vmatrix}=1~~~(\because となるように~\kappa~ を定義
したのであった）\\[0.2cm]
~~~\mathbf{b}=\mathbf{t} \times \mathbf{n}\\[0.2cm]
~~~\mathbf{t}~と\mathbf{n}~ではる平面をC~の、点P~における接触平面という。\\[0.2cm]
~~~\mathbf{b}~を陪法線、又は従法線という。\\[0.2cm]
~~~また、\displaystyle\genfrac{}{}{0.5pt}{}{1}{\kappa}=\rho~~を曲率半径、
~~\kappa~を曲率という。\\[0.5cm]
Recall~~~\displaystyle\genfrac{}{}{0.5pt}{}{d\mathbf{b}}{ds}~は、
\mathbf{t},~\mathbf{b}~の両方に垂直。即ち\mathbf{n}~に平行。\\[0.2cm]
証明~~~\mathbf{b}\mathbf{b}=1\\[0.2cm]
\therefore ~~~~\displaystyle\genfrac{}{}{0.5pt}{}{d\mathbf{b}}{ds}\mathbf{b}=0\\[0.2cm]
\therefore~~~~ \genfrac{}{}{0.5pt}{}{d\mathbf{b}}{ds} \perp \mathbf{b}\\[0.2cm]
~~~また\\[0.2cm]
~~~~\mathbf{b}=\mathbf{t} \times \mathbf{n}\\[0.2cm]
\therefore~~~~\genfrac{}{}{0.5pt}{}{d\mathbf{b}}{ds}=\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds}
\times \mathbf{n}+\mathbf{t} \times \genfrac{}{}{0.5pt}{}{d\mathbf{n}}{ds}
=\kappa \mathbf{n} \times \mathbf{n}
+\mathbf{t} \times \genfrac{}{}{0.5pt}{}{d\mathbf{n}}{ds}=
\mathbf{t} \times \genfrac{}{}{0.5pt}{}{d\mathbf{n}}{ds}\\[0.2cm]
\therefore~~~~ \genfrac{}{}{0.5pt}{}{d\mathbf{b}}{ds} \perp \mathbf{t}
~~~~~~~~~~~~QED\\[0.5cm]
定義~~~~~上の~Recall~から\\[0.2cm]
~~~~~~~~~~\genfrac{}{}{0.5pt}{}{d\mathbf{b}}{ds}=\tau \mathbf{n}\\[0.2cm]
となるが、この~\tau ~を曲線C~の点P~における捩率という。\\[0.5cm]
命題~~~~~Sが曲面で~~~~~~S=(x(u,v), y(u,v), z(u,v)~)~と表されている。\\[0.2cm]
~~~~~~~~~\mathbf{r}はS~上の曲線で~~~\mathbf{r}=(x(u(t),v(t)), y(u(t),v(t)), z(u(t),v(t))~)~
と表されている。\\[0.2cm]
~~~~~~~~P~は\mathbf{r}~上の一点。\\[0.2cm]
~~~~~~~~\mathbf{m}~はS~上のP~における単位法線ベクトル。\\[0.2cm]
~~~~~~~~\mathbf{n}~は\mathbf{r}~上のP~における単位法線ベクトル。\\[0.2cm]
~~~~~~~~\phi~ は~\mathbf{n}~と\mathbf{m}~とのなす角\\[0.2cm]
のとき、次の式が成立する。\\[0.2cm]
~~~~~~~~\genfrac{}{}{0.5pt}{} {cos \phi}{\rho}
=-\genfrac{}{}{0.5pt}{}{d\mathbf{r}d\mathbf{m}}{ds^2}~~~~~...(46_1)\\[0.2cm]
証明~~~~~~\mathbf{t}\mathbf{m}=0\\[0.2cm]
~~~~~~~~~\therefore ~~\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds}\mathbf{m}
+\mathbf{t}\genfrac{}{}{0.5pt}{}{d\mathbf{m}}{ds}=0\\[0.2cm]
~~~~~~~~~~~~~~\genfrac{}{}{0.5pt}{}{1}{\rho}\mathbf{n}\mathbf{m}
+\genfrac{}{}{0.5pt}{}{d\mathbf{r}}{ds}\genfrac{}{}{0.5pt}{}{d\mathbf{m}}{ds}=0\\[0.2cm]
~~~~~~~~~~~~~~\genfrac{}{}{0.5pt}{} {cos \phi}{\rho}
=-\genfrac{}{}{0.5pt}{}{d\mathbf{r}d\mathbf{m}}{ds^2}~~~~~~~~~~~QED\\[0.5cm]
問（上の命題の意味を知るための練習）\\[0.2cm]
~~~中心が(0,0,0)~半径が~a~の球の表面を、次のパラメータ平面~K~で表す。\\[0.2cm]
~~~~~K=\left \{~(v,u)：0<v<2 \pi,~-\genfrac{}{}{0.5pt}{}{\pi}{2}<u<\genfrac{}{}{0.5pt}{}{\pi}{2} \right\}
\\[0.2cm]
~~~~~~~x=a \cos{u} \cos{v}\\[0.2cm]
~~~~~~~y=a \cos{u} \sin{v}\\[0.2cm]
~~~~~~~z=a \sin{v}\\[0.2cm]
このとき、次の問に答えよ。\\[0.2cm]
~~~~~~1)~~~~点~P~\left (x \left (\genfrac{}{}{0.5pt}{}{\pi}{4},
\genfrac{}{}{0.5pt}{}{\pi}{4}\right),
y \left (\genfrac{}{}{0.5pt}{}{\pi}{4},\genfrac{}{}{0.5pt}{}{\pi}{4}\right),
z\left(\genfrac{}{}{0.5pt}{}{\pi}{4},\genfrac{}{}{0.5pt}{}{\pi}{4}\right)
\right)~の法線方向を求めよ。\\[0.2cm]
~~~~~~2）~~~~点~P~を通る曲線~\mathbf{r}~
\left (x \left (v,\genfrac{}{}{0.5pt}{}{\pi}{4}\right),
y \left (v,\genfrac{}{}{0.5pt}{}{\pi}{4}\right),
z\left(v,\genfrac{}{}{0.5pt}{}{\pi}{4}\right) \right)~(0<v<2 \pi)の点~P~における
法線方向を求めよ。\\[0.2cm]
解~~~1)~~を解く。\\[0.2cm]
~~~\mathbf{r}_u=(-a\sin u \cos v, -a \sin u \sin v, a \cos u)\\[0.2cm]
~~~~~~~~~~~~~~\mathbf{r}_v=(-a\cos u \sin v, a \cos u \cos v, 0)\\[0.2cm]
\mathbf{r}_u \times \mathbf{r}_v=
\left(
\begin{vmatrix}
-a \sin u \sin v&a \cos u\\
a \cos u \cos v& 0\end{vmatrix} ,~
-\begin{vmatrix}
-a \sin u \cos v&a \cos u\\
-a \cos u \sin v& 0\end{vmatrix} ,~
\begin{vmatrix}
-a \sin u \cos v&-a \sin u \sin v\\
-a \cos u \sin v&a \cos u \cos v \end{vmatrix}
\right) \\[0.2cm]
=(-a^2 \cos {u}^2 \cos v,~ -a^2 \cos {u}^2 \sin v, ~-a^2 \sin {u} \cos u)\\[0.2cm]
これに~v=\genfrac{}{}{0.5pt}{}{\pi}{4},~u=\genfrac{}{}{0.5pt}{}{\pi}{4}~を
代入して、\\[0.2cm]
\mathbf{r}_u \times \mathbf{r}_v=
\left(-\genfrac{}{}{0.5pt}{}{a^2}{2\sqrt{2}},~-\genfrac{}{}{0.5pt}{}{a^2}{2\sqrt{2}},
~-\genfrac{}{}{0.5pt}{}{a^2}{2} \right)\\[0.2cm]
\therefore~~ \begin{vmatrix}\mathbf{r}_u \times \mathbf{r}_v\end{vmatrix}
=\genfrac{}{}{0.5pt}{}{a^2}{\sqrt{2}}\\[0.2cm]
\therefore ~~\mathbf{m}=\genfrac{}{}{0.5pt}{}
{\mathbf{r}_u \times \mathbf{r}_v}
{\begin{vmatrix}\mathbf{r}_u \times \mathbf{r}_v\end{vmatrix}}
=\left(-\genfrac{}{}{0.5pt}{}{1}{2},~-\genfrac{}{}{0.5pt}{}{1}{2},~
-\genfrac{}{}{0.5pt}{}{1}{\sqrt{2}}\right)~~~~~~1)~の答。\\[0.2cm]
~~~~~2)~~を解く。そのためにまづ、弧長~s~を求める。\\[0.2cm]
~~~~~~\mathbf{r}=\left(\genfrac{}{}{0.5pt}{}{a}{\sqrt{2}}\cos v,~
\genfrac{}{}{0.5pt}{}{a}{\sqrt{2}}\sin v,
~\genfrac{}{}{0.5pt}{}{a}{\sqrt{2}} \right)\\[0.2cm]
~\therefore \genfrac{}{}{0.5pt}{}{d\mathbf{r}}{dv}
=\left(-\genfrac{}{}{0.5pt}{}{a}{\sqrt{2}}\sin v,~
\genfrac{}{}{0.5pt}{}{a}{\sqrt{2}}\cos v,~0 \right)\\[0.2cm]
\therefore~~ds^2=\genfrac{}{}{0.5pt}{}{d\mathbf{r}}{dv}
\genfrac{}{}{0.5pt}{}{d\mathbf{r}}{dv}=\genfrac{}{}{0.5pt}{}{a^2}{2}\\[0.2cm]
\therefore~~s=\int_0^v\sqrt{\genfrac{}{}{0.5pt}{}{a^2}{2}}\,dv=
\genfrac{}{}{0.5pt}{}{a}{\sqrt2}v\\[0.2cm]
\therefore~~v=\genfrac{}{}{0.5pt}{}{\sqrt2}{a}s\\[0.2cm]
もとの曲線の方程式を弧長~s~で表せば、\\[0.2cm]
~~~~\mathbf{r}=\left(\genfrac{}{}{0.5pt}{}{a}{2}
\cos{\genfrac{}{}{0.5pt}{}{\sqrt2}{a}s},~
\genfrac{}{}{0.5pt}{}{a}{2}
\sin{\genfrac{}{}{0.5pt}{}{\sqrt2}{a}s},~\genfrac{}{}{0.5pt}{}{a}{\sqrt2} \right)\\[0.2cm]
~~~これを~s~で微分したものが接線の方向~\mathbf{t}。即ち、\\[0.2cm]
~~~\mathbf{t}=\genfrac{}{}{0.5pt}{}{d\mathbf{r}}{ds}
=\left(-\sin{\genfrac{}{}{0.5pt}{}{\sqrt2}{a}s},
~\cos{\genfrac{}{}{0.5pt}{}{\sqrt2}{a}s},~0 \right)\\[0.2cm]
~~~これを~s~で微分したものが~ \kappa \mathbf{n}。即ち、\\[0.2cm]
~~~\kappa \mathbf{n}=\left(-\genfrac{}{}{0.5pt}{}{\sqrt2}{a}
\cos{\genfrac{}{}{0.5pt}{}{\sqrt2}{a}s},~
-\genfrac{}{}{0.5pt}{}{\sqrt2}{a}
\sin{\genfrac{}{}{0.5pt}{}{\sqrt2}{a}s},~0 \right) \\[0.2cm]
v=\genfrac{}{}{0.5pt}{}{\pi}{4}~のときs=\genfrac{}{}{0.5pt}{}{a}{\sqrt2}
\genfrac{}{}{0.5pt}{}{\pi}{4}~。これを代入して、\\[0.2cm]
~~~~~~\mathbf{t} =\left(-\genfrac{}{}{0.5pt}{}{1}{\sqrt2},~
\genfrac{}{}{0.5pt}{}{1}{\sqrt2},~0 \right)\\[0.2cm]
~~~~~~\kappa \mathbf{n}=\left(-\genfrac{}{}{0.5pt}{}{1}{a},~
-\genfrac{}{}{0.5pt}{}{1}{a},~0 \right)\\[0.2cm]
~~~~\therefore~~\kappa^2=\genfrac{}{}{0.5pt}{}{2}{a^2}\\[0.2cm]
~~~~\therefore~~\kappa=\genfrac{}{}{0.5pt}{}{\sqrt2}{a}\\[0.2cm]
~~~~\therefore~~\mathbf{n}=\left(-\genfrac{}{}{0.5pt}{}{1}{\sqrt2},~
-\genfrac{}{}{0.5pt}{}{1}{\sqrt2},~0 \right)~~~~~（解おわり）\\[0.2cm]
以上、地道に計算して答を出したのだが、点~P~は球面上の点であり、
与えられた曲線は緯度線であることを考慮すれば、地道な計算なしで
答は出る。曲面に対する法線~\mathbf{m}~と、曲線に対する法線~\mathbf{n}~
の違いはこの問によってはっきりする筈である。\\[0.5cm]
問~~~~前命題の~d\mathbf{r}の~\mathbf{r}~を曲面~S~上を表す\mathbf{r}~と考えて、
-d\mathbf{r}d\mathbf{m}~を~u, v~で表せ。\\[0.2cm]
解~~~~~-d\mathbf{r}d\mathbf{m}=-(\mathbf{r}_udu+\mathbf{r}_vdv)
(\mathbf{m}_udu+\mathbf{m}_vdv)\\[0.2cm]
=(-\mathbf{r}_u\mathbf{m}_u)du^2+(-\mathbf{r}_u\mathbf{m}_v
-\mathbf{r}_v\mathbf{m}_u)dudv+(-\mathbf{r}_v\mathbf{r}_v)dv^2
~~~~~~~~~~（解終り）\\[0.2cm]
定義~~~~~上のdu^2、dudv、dv^2~の係数をそれぞれL、2M、N~とおき、即ち\\[0.2cm]
~~~~~~~~L=-\mathbf{r}_u\mathbf{m}_u\\[0.2cm]
~~~~~~~~2M=-\mathbf{r}_u\mathbf{m}_v-\mathbf{r}_v\mathbf{m}_u\\[0.2cm]
~~~~~~~~N=-\mathbf{r}_v\mathbf{r}_v\\[0.2cm]
とおき、\\[0.2cm]
~~~~~~~Ldu^2+2Mdudv+Ndv^2\\[0.2cm]
を、「Gauss~の第二基本形式」という。これを用いれば、(46_1)~は、\\[0.2cm]
~~~~~~\genfrac{}{}{0.5pt}{} {cos \phi}{\rho}
=\genfrac{}{}{0.5pt}{}{Ldu^2+2Mdudv+Ndv^2}{Edu^2+2Fdudv+Gdv^2}~~~~~...~~(48)\\[0.5cm]
命題~~~~L=\genfrac{}{}{0.5pt}{}{\mathbf{r}_{uu}(\mathbf{r}_u\times \mathbf{r}_v)}
{\sqrt{EG-F^2}}~~~~~~~~~~....~~(50_1)\\[0.2cm]
~~~~~~~~M=\genfrac{}{}{0.5pt}{}{\mathbf{r}_{uv}(\mathbf{r}_u\times \mathbf{r}_v)}
{\sqrt{EG-F^2}}~~~~~~~~~~....~~(50_2)\\[0.2cm]
~~~~~~~~N=\genfrac{}{}{0.5pt}{}{\mathbf{r}_{vv}(\mathbf{r}_u\times \mathbf{r}_v)}
{\sqrt{EG-F^2}}~~~~~~~~~~....~~(50_3)\\[0.2cm]
証明~~~~~~(50_1)~を示す。\\[0.2cm]
~~~~~~~~~~\mathbf{r}_u\mathbf{m}=0\\[0.2cm]
~~~~~~~~\therefore~~~\mathbf{r}_{uu}\mathbf{m}+\mathbf{r}_u\mathbf{m}_u=0\\[0.2cm]
~~~~~~~~\therefore~~~-\mathbf{r}_u\mathbf{m}_u=\mathbf{r}_{uu}\mathbf{m}
=\genfrac{}{}{0.5pt}{}{\mathbf{r}_{uu}(\mathbf{r}_u\times \mathbf{r}_v)}
{\sqrt{EG-F^2}}\\[0.2cm]
~~~~~~~~この左辺は~L~であるから、これで示された。\\[0.2cm]
~~~~~~次に、(50_2)~を示す。\\[0.2cm]
~~~~~~~~\mathbf{r}_{uv}\mathbf{m}+\mathbf{r}_u\mathbf{m}_v=0\\[0.2cm]
~~~~~~~~\therefore~~~\mathbf{r}_u\mathbf{m}_v=-\mathbf{r}_{uv}\mathbf{m}\\[0.2cm]
~~~また、~~~~~~~~\mathbf{r}_v\mathbf{m}=0\\[0.2cm]
~~~~~~~~\therefore~~~\mathbf{r}_{uv}\mathbf{m}+\mathbf{r}_v\mathbf{m}_u=0\\[0.2cm]
~~~~~~~~\therefore~~~\mathbf{r}_v\mathbf{m}_u=-\mathbf{r}_{uv}\mathbf{m}\\[0.2cm]
~~~~~~~~\therefore~~~M=-\genfrac{}{}{0.5pt}{}{1}{2}
(\mathbf{r}_v\mathbf{m}+\mathbf{r}_v\mathbf{m}_u)=\mathbf{r}_{uv}\mathbf{m}
=\mathbf{r}_{uv}\genfrac{}{}{0.5pt}{}{(\mathbf{r}_u\times \mathbf{r}_v)}{\sqrt{EG-F^2}}
\\[0.2cm]
~~~~~~次に、(50_3)~を示す。\\[0.2cm]
~~~~~~~~\mathbf{r}_{vv}\mathbf{m}+\mathbf{r}_v\mathbf{m}_v=0\\[0.2cm]
~~~~~~~~\therefore~~~-\mathbf{r}_v\mathbf{m}_v=\mathbf{r}_{vv}\mathbf{m}
=\genfrac{}{}{0.5pt}{}{\mathbf{r}_{vv}(\mathbf{r}_u\times \mathbf{r}_v)}
{\sqrt{EG-F^2}}\\[0.2cm]
~~~~~~~~この左辺は~N~であるから、これで示された。~~~~~~~~QED\\[0.2cm]
問~~~~~~~特に~S:~曲面が~z=f(x,y)~で表されているとき、E, F, G, L, M, N~を、\\[0.2cm]
~~~~~~~p=\genfrac{}{}{0.5pt}{}{\partial f}{\partial x},~
~~~~~~~q=\genfrac{}{}{0.5pt}{}{\partial f}{\partial y},~
~~~~~~~r=\genfrac{}{}{0.5pt}{}{\partial^2 f}{\partial x \partial x},~
~~~~~~~s=\genfrac{}{}{0.5pt}{}{\partial^2 f}{\partial x \partial y},~
~~~~~~~t=\genfrac{}{}{0.5pt}{}{\partial^2 f}{\partial y \partial y}\\[0.2cm]
~で表せ。\\[0.2cm]
解~~~~~S~は、\mathbf{r}=(x,~y,~f(x,y))~~~~~u=x, ~v=y~~~と表せる。\\[0.2cm]
~~~~\therefore~~~\genfrac{}{}{0.5pt}{}{\partial \mathbf{r} }{\partial u}=(1,~0,~p)\\[0.2cm]
~~~~~~~~~~~\genfrac{}{}{0.5pt}{}{\partial \mathbf{r} }{\partial v}=(0,~1,~q)\\[0.2cm]
~~~~~~~~~~~\genfrac{}{}{0.5pt}{}{\partial \mathbf{r} }{\partial u \partial u}=(0,~0,~r)\\[0.2cm]
~~~~~~~~~~~\genfrac{}{}{0.5pt}{}{\partial \mathbf{r} }{\partial u \partial v}=(0,~0,~s)\\[0.2cm]
~~~~~~~~~~~\genfrac{}{}{0.5pt}{}{\partial \mathbf{r} }{\partial v \partial v}=(0,~0,~t)\\[0.2cm]
~~~~\therefore~~~E=\mathbf{r}_u^{2}=1+p^2\\[0.2cm]
~~~~~~~~~~~F=\mathbf{r}_u\mathbf{r}_v=pq\\[0.2cm]
~~~~~~~~~~~G=\mathbf{r}_v^2=1+q^2\\[0.2cm]
~~~~~~~~~~~EG-F^2=(1+p^2)(1+q^2)-p^2q^2=1+p^2+q^2\\[0.2cm]
~~~~~~~~~~~\mathbf{r}_u \times \mathbf{r}_v=
\left(\begin{vmatrix}0&p\\1&q\end{vmatrix} ,~
-\begin{vmatrix}1&p\\0&q\end{vmatrix},~
\begin{vmatrix}1&0\\0&1\end{vmatrix}
\right)=(-p,~-q,~1)\\[0.2cm]
~~~~\therefore~~~L=\genfrac{}{}{0.5pt}{}{\mathbf{r}_{uu}(\mathbf{r}_u\times \mathbf{r}_v)}
{\sqrt{EG-F^2}}=\genfrac{}{}{0.5pt}{}{(0,0,r)(-p,-q,1)}{\sqrt{EG-F^2}}
=\genfrac{}{}{0.5pt}{}{r}{\sqrt{EG-F^2}}\\[0.2cm]
同様に、~~~~~M=\genfrac{}{}{0.5pt}{}{\mathbf{r}_{uv}(\mathbf{r}_u\times \mathbf{r}_v)}
{\sqrt{EG-F^2}}=\genfrac{}{}{0.5pt}{}{s}{\sqrt{EG-F^2}}\\[0.2cm]
~~~~~~~~~~~~~N=\genfrac{}{}{0.5pt}{}{\mathbf{r}_{vv}(\mathbf{r}_u\times \mathbf{r}_v)}
{\sqrt{EG-F^2}}=\genfrac{}{}{0.5pt}{}{t}{\sqrt{EG-F^2}}~~~~~~~~\\[0.2cm]
（解終り）\\[0.2cm]
練習問題~~~~1)~双曲放物面~\mathbf{r}=(x, y, xy)~の点(1,1,1)~におけるE,F,G,L,M,N~を求めよ。\\[0.2cm]
~~~~~~~~~~2)~楕円放物面~\mathbf{r}=(x, y, x^2+y^2)~の点(1,1,2)~におけるE,F,G,L,M,N~を求めよ。\\[0.2cm]
答~~~~~1)~E=2,~ F=1, ~G=2, ~L=0, ~M=\genfrac{}{}{0.5pt}{}{1}{\sqrt{3}}, ~N=0~~~~~~~\\[0.2cm]
~~~~~~~2)~E=5,~ F=4, ~G=5, ~L=\genfrac{}{}{0.5pt}{}{2}{3}, ~M=0,
~N=\genfrac{}{}{0.5pt}{}{2}{3}~~~~~~~~答終り\\[0.5cm]
定義~~~~S：曲面\\[0.2cm]
~~~~~~~~P：~S~上の一点\\[0.2cm]
~~~~~~~~\mathbf{m}：~P~における単位法線ベクトル\\[0.2cm]
が与えられているとき\\[0.2cm]
~~~~~~~\mathbf{m}を含む平面で~S~を切った時に出来る曲線を「曲面~S~の点~P~
における直截口（ちょくせつこう）」という。\\[0.5cm]
命題~~~~直截口の点~P~における曲率~\genfrac{}{}{0.5pt}{}{1}{\rho}(=\kappa)~は\\[0.2cm]
\genfrac{}{}{0.5pt}{}{1}{\rho}=±\genfrac{}{}{0.5pt}{}{Ldu^2+2Mdudv+Ndv^2}{Edu^2+2Fdudv+Gdv^2}
\\[0.2cm]
証明~~~\mathbf{m}~と~\mathbf{n}~とのなす角は、0°か、180°である。\\[0.2cm]
従って、(48)~式の ~\cos \phi~は1~或は~-1。即ち、上の式が示された。\\[0.2cm]
~~~~（証明おわり）\\[0.5cm]
定義~~~直截口の曲率半径を~R~と書く。（そして符号も含める。）\\[0.2cm]
~~~即ち、~~~\genfrac{}{}{0.5pt}{}{1}{R}=\genfrac{}{}{0.5pt}{}{Ldu^2+2Mdudv+Ndv^2}
{Edu^2+2Fdudv+Gdv^2}~~~~~~...~(56)\\[0.5cm]
~~~(56)~式の右辺の分子、分母を~du^2~で割って、\genfrac{}{}{0.5pt}{}{dv}{du}
を~t~とおくと、\\[0.2cm]
\genfrac{}{}{0.5pt}{}{1}{R}=\genfrac{}{}{0.5pt}{}{L+2Mt+Nt^2}
{E+2Ft+Gt^2}~~~~~~\\[0.2cm]
~~~この右辺の分母はいかなる~t~の値に対しても~0~にならない。（何故ならば~
Edu^2+2Fdudv+Gdv^2~は弧長だから。）\\[0.2cm]
~~~分子は、t=\genfrac{}{}{0.5pt}{}{dv}{du}~の値によって符号が変化する可能性がある。\\[0.2cm]
~~~符号の変化の可能性を次の三つに分類する。\\[0.2cm]
~~~(1)~プラスならプラスだけ、マイナスならマイナスだけしか取らない場合。
（即ち、分子を~t~に関する二次式と考えた時、判別式がマイナスのとき。）
この時は、この曲面の近傍で曲率の符号が変らないから、
そこで曲面は膨らんでいることが分る。\\[0.2cm]
~~~(2)~プラスにもマイナスにも成り得る場合。（即ち、分子を~t~に関する二次式
と考えた時、判別式がプラスのとき。）
この時は、この曲面の近傍で、曲率がプラス、マイナスの値を
とるので、そこで曲面は鞍点になっていることが分る。\\[0.2cm]
~~~(3)~ある一つの方向で０になり、それ以外は符号を変えない場合。（即ち、分子を~t~に関する二次式
と考えた時、判別式が０のとき。）
この時は、この曲面の近傍のある一方向で曲率が０になり、それ以外は同じ
符号なので、そこで曲面は、円柱のようになっていることが分る。\\[0.2cm]
~~~~~上記のことを考慮に入れ、次のように定義する。\\[0.2cm]
定義~~~1)~M^2-LN<0~のとき、点~P~は楕円的であるという。\\[0.2cm]
~~~~~~~~2)~M^2-LN>0~のとき、点~P~は双曲的であるという。\\[0.2cm]
~~~~~~~~3)~M^2-LN=0~のとき、点~P~は放物的であるという。~~~（定義おわり）\\[0.2cm]
予備工作~~~x,~y~が陰函数表示\\[0.2cm]
~~~~~~~~~f(x,y)=0\\[0.2cm]
されているときの、y~のx~に関する極値を求めるための必要条件は何か。\\[0.2cm]
解~~~~y~が極値（x~に対して）をとるときは、\\[0.2cm]
~~~~~~~~~~~\genfrac{}{}{0.5pt}{}{dy}{dx}=0~~~~が必要。\\[0.2cm]
即ち、\genfrac{}{}{0.5pt}{}{dy}{dx}=\genfrac{}{}{0.5pt}{}
{\genfrac{}{}{0.5pt}{}{\partial f}{\partial x}}
{\genfrac{}{}{0.5pt}{}{\partial f}{\partial y}}~~~~~だから、\\[0.2cm]
\genfrac{}{}{0.5pt}{}{\partial f}{\partial x}=0~~~~が必要。~~~~~（解おわり）\\[0.5cm]
問~~~~直截口の曲率の式~(56)~からR^{-1}の最大値、最小値を求めよ。\\[0.2cm]
解~~~ER^{-1}du^2+2FR^{-1}dudv+GR^{-1}dv^2=Ldu^2+2Mdudv+Ndv^2\\[0.2cm]
~~~~~(L-ER^{-1})du^2+2(M-FR^{-1})dudv+(N-GR^{-1})dv^2=0\\[0.2cm]
~~~dv^2~で割って、\genfrac{}{}{0.5pt}{}{du}{dv}=t~とおいて、\\[0.2cm]
~~~\phi (R^{-1},t)=(L-ER^{-1})t^2+2(M-FR^{-1})t+(N-GR^{-1})\\[0.2cm]
~~~\genfrac{}{}{0.5pt}{}{\partial \phi}{\partial t}=2(L-ER^{-1})t
+2(M-FR^{-1})=0\\[0.2cm]
予備工作を用いて、~~~\genfrac{}{}{0.5pt}{}{\partial \phi}{\partial t}=
2(L-ER^{-1})t + 2(M-FR^{-1})=0~~~とおく。\\[0.2cm]
\therefore~~~~~(L-ER^{-1})\genfrac{}{}{0.5pt}{}{du}{dv}+(M-FR^{-1})=0\\[0.2cm]
また、du^2~で割って、\genfrac{}{}{0.5pt}{}{dv}{du}=t_1~とおいて、\\[0.2cm]
~~~\phi_1 (R^{-1},t_1)=(L-ER^{-1})+2(M-FR^{-1})t_1+(N-GR^{-1})t_1^2\\[0.2cm]
~~~\genfrac{}{}{0.5pt}{}{\partial \phi}{\partial t_1}=2(N-GR^{-1})t_1
+2(M-FR^{-1})=0\\[0.2cm]
\therefore~~~(N-GR^{-1})\genfrac{}{}{0.5pt}{}{dv}{du}+(M-FR^{-1})=0\\[0.2cm]
\genfrac{}{}{0.5pt}{}{du}{dv}~を消去して\\[0.2cm]
~~~~\genfrac{}{}{0.5pt}{}{M-FR^{-1}}{L-ER^{-1}}=
\genfrac{}{}{0.5pt}{}{N-GR^{-1}}{M-FR^{-1}}\\[0.2cm]
\therefore~~~M^2-2FMR^{-1}+F^2R^{-2}=LN-LGR^{-1}-ENR^{-1}+EGR^{-2}=0\\[0.2cm]
\therefore~~~(EG-F^2)\genfrac{}{}{0.5pt}{}{1}{R^2}-(EN-2FM+LG)\genfrac{}{}{0.5pt}{}{1}{R}
+(LN-M^2)=0\\[0.2cm]
~~~この２次方程式の２根~\genfrac{}{}{0.5pt}{}{1}{R_1},~\genfrac{}{}{0.5pt}{}{1}{R_2}
~が答。~~~~~~~~~~~~（解おわり）\\[0.5cm]
定義~~~上の~\genfrac{}{}{0.5pt}{}{1}{R_1},~\genfrac{}{}{0.5pt}{}{1}{R_2}~の積~K~を
「Gauss~の曲率」といい、その平均値~H~を「平均曲率」という。\\[0.2cm]
~~~即ち、~~~K=\genfrac{}{}{0.5pt}{}{1}{R_1R_2}=
\genfrac{}{}{0.5pt}{}{LN-M^2}{EG-F^2}\\[0.2cm]
~~~~~~H=\genfrac{}{}{0.5pt}{}{1}{2} \left( \genfrac{}{}{0.5pt}{}{1}{R_1}
+\genfrac{}{}{0.5pt}{}{1}{R_2} \right)=
\genfrac{}{}{0.5pt}{}{EN-2FM+LG}{2(EG-F^2)}\\[0.2cm]
問~~~(S):~~曲面~~~~~\mathbf{r}=\mathbf{r}(u,v)\\[0.2cm]
~~~~~~(S_1):~~(S)~をもとに、法線方向に距離~n(u,v)~だけ離れたところ\\[0.2cm]
~~~~~~~~~~~~~~~~~~にある曲面\\[0.2cm]
このとき、~(S)~の面積~S~と~(S_1)~の面積~S_1~の差を求めよ。\\[0.2cm]
（但し、~n,~ n_u,~n_v~の2次の項は無視せよ。）\\[0.2cm]
解~~~~(S_1)~を~\mathbf{r}^{(1)}(u,v)~とすれば、\\[0.2cm]
~~~~~~~~~~~~\mathbf{r}^{(1)}=\mathbf{r}+n~\mathbf{m}\\[0.2cm]
~~~~~~~~~~~~\mathbf{r}^{(1)}_u=\mathbf{r}_u+n_u~\mathbf{m}
+n~\mathbf{m}_u\\[0.2cm]
~~~~~~~~~~~~\mathbf{r}^{(1)}_v=\mathbf{r}_v+n_v~\mathbf{m}
+n~\mathbf{m}_v\\[0.2cm]
曲面の面積を求めるために、曲面分を求める。\\[0.2cm]
E_1=\mathbf{r}^{(1)}_u~\mathbf{r}^{(1)}_u=(\mathbf{r}_u+n_u~\mathbf{m}
+n~\mathbf{m}_u)(\mathbf{r}_u+n_u~\mathbf{m}
+n~\mathbf{m}_u)\\[0.2cm]
＝\mathbf{r}_u^{2}+2n_u~\mathbf{r}_u~\mathbf{m}+2n~\mathbf{r}_u~\mathbf{m}_u\\[0.2cm]
=E-2nL\\[0.2cm]
F_1=\mathbf{r}^{(1)}_u~\mathbf{r}^{(1)}_v=\mathbf{r}_u~\mathbf{r}_v+
n_u~\mathbf{m}\mathbf{r}_u+n_v~\mathbf{m}\mathbf{r}_v
+n~(\mathbf{r}_v\mathbf{m}_u+\mathbf{r}_u\mathbf{m}_v)\\[0.2cm]
=F-2nM\\[0.2cm]
G_1=\mathbf{r}^{(1)}_v~\mathbf{r}^{(1)}_v=\mathbf{r}_v\mathbf{r}_v
+2n_v\mathbf{m}\mathbf{r}_v+2n\mathbf{r}_v\mathbf{m}_v\\[0.2cm]
=G-2nN\\[0.2cm]
\therefore~~~E_1G_1-F^2=(E-2nL)(G-2nN)-(F-2nM)^2\\[0.2cm]
~~~~~=EG-2n(EN+GL)-F^2+4nFM\\[0.2cm]
~~~~~=(EG-F^2)-2n(EN-2Fm+GL)\\[0.2cm]
~~~~~=(EG-F^2)(1-4nH)~~~~~~（\because~~平均曲率~H~の定義参照）\\[0.2cm]
~~~\therefore~~~\sqrt{E_1G_1-F_1^2}=\sqrt{EG-F^2}(1-2nH)\\[0.2cm]
~~~~~~~~~~(\because~~(1-4nH)^{-\genfrac{}{}{0.5pt}{}{1}{2}}
=1-\genfrac{}{}{0.5pt}{}{1}{2}(4nH)+ \cdots)\\[0.2cm]
\therefore~~~\iint_{S_1}\sqrt{E_1G_1-{F_1}^2}\,du\,dv
-\iint_{S}\sqrt{EG-F^2}\,du\,dv=-\iint_{S}2nH\sqrt{EG-F^2}\,du\,dv\\[0.2cm]
（解終り）\\[0.5cm]
問~~~上の結果から、面積が極値をとるときの、曲面の条件を述べよ。\\[0.2cm]
解~~~H=0~~~即ち、\\[0.2cm]
~~~~~~EN-2Fm+GL=0~~~~~~（解終り）\\[0.5cm]
問~~~曲面が~z=f(x,y)~と表されているとき、上の条件はどう書けるか。\\[0.2cm]
解~~~E=1+p^2,~F=pq,~G=1+q^2\\[0.2cm]
~~~~~~~L=\genfrac{}{}{0.5pt}{}{r}{\sqrt{EG-F^2}},~
M=\genfrac{}{}{0.5pt}{}{s}{\sqrt{EG-F^2}},~
N=\genfrac{}{}{0.5pt}{}{t}{\sqrt{EG-F^2}}\\[0.2cm]
~~~(1+p^2)t-2pqs+(1+q^2)r=0~~~~~~（解終り）\\[0.5cm]
~~~ここで、Gauss~の公式、~Weingarten~の公式、測地線の方程式を
導く。\\
\\
定義~(Christoffel~の記号)\\[0.2cm]
~~~~~~~\left\{{}^{\,\gamma}_{\alpha\beta}\right\}~~を~Christoffel~の
第二記号といい、次のように定義する。\\[0.2cm]
~~~\begin{pmatrix}g^{11}&g^{12}\\g^{21}&g^{22}\end{pmatrix}=
\genfrac{}{}{0.5pt}{}{1}{g}
\begin{pmatrix}g_{22}&-g_{12}\\-g_{21}&g_{11}\end{pmatrix}\\[0.2cm]
~~~として、\\[0.2cm]
~~~\left\{{}^{\,\gamma}_{\alpha\beta}\right\}=\genfrac{}{}{0.5pt}{}{1}{2}
g^{\gamma1}\left( \genfrac{}{}{0.5pt}{}{\partial g_{1\alpha}}{\partial u^{\beta}}
+\genfrac{}{}{0.5pt}{}{\partial g_{1\beta}}{\partial u^{\alpha}}-
\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^{1}} \right)
+\genfrac{}{}{0.5pt}{}{1}{2}
g^{\gamma2}\left( \genfrac{}{}{0.5pt}{}{\partial g_{2\alpha}}{\partial u^{\beta}}
+\genfrac{}{}{0.5pt}{}{\partial g_{2\beta}}{\partial u^{\alpha}}-
\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^{2}} \right)\\[0.2cm]
と定義する。\\[0.5cm]
命題~~~（Gauss~の公式）\\[0.2cm]
~~~~~~~~~~\genfrac{}{}{0.5pt}{}{\partial\mathbf{x}_1 }
{\partial u}=\mathbf{x}_{11},~
\genfrac{}{}{0.5pt}{}{\partial\mathbf{x}_1 }
{\partial v}=\mathbf{x}_{12}~~~~~~~~~etc.~~~と書く。\\[0.2cm]
すると、次の式が成立する。これらを~Gauss~の公式~という。\\[0.2cm]
\mathbf{x}_{11}=\left\{{}^{\,1}_{11}\right\}\mathbf{x}_{1}
+\left\{{}^{\,2}_{11}\right\}\mathbf{x}_{2}+h_{11}\mathbf{N}\\[0.2cm]
\mathbf{x}_{12}=\left\{{}^{\,1}_{12}\right\}\mathbf{x}_{1}
+\left\{{}^{\,2}_{12}\right\}\mathbf{x}_{2}+h_{12}\mathbf{N}\\[0.2cm]
\mathbf{x}_{22}=\left\{{}^{\,1}_{22}\right\}\mathbf{x}_{1}
+\left\{{}^{\,2}_{22}\right\}\mathbf{x}_{2}+h_{22}\mathbf{N}\\[0.2cm]
証明~~~\mathbf{x}_{11}=\Gamma^{1}_{11}\mathbf{x}_{1}
+\Gamma^{2}_{11}\mathbf{x}_{1}+\Gamma _{11}\mathbf{N}\\[0.2cm]
とおき、\Gamma^{1}_{11}、~\Gamma^{2}_{11},~\Gamma_{11}~を求める。\\[0.2cm]
（\mathbf{x}_{1},~\mathbf{x}_{2},~\mathbf{N}~は一次独立だから、
\mathbf{x}_{11}~はこれらの線形結合で表せる。）\\[0.2cm]
~~~\mathbf{x}_{1}~をかけて、\\[0.2cm]
~~~~~\mathbf{x}_{11}\mathbf{x}_{1}=\Gamma^{1}_{11}g_{11}+
\Gamma^{2}_{11}g_{12}\\[0.2cm]
~~~\mathbf{x}_{2}~をかけて、\\[0.2cm]
~~~~~\mathbf{x}_{11}\mathbf{x}_{2}=\Gamma^{1}_{11}g_{21}+
\Gamma^{2}_{11}g_{22}\\[0.2cm]
\therefore~~~\Gamma^{1}_{11}=\genfrac{}{}{0.5pt}{}
{\begin{vmatrix} \mathbf{x}_{11}\mathbf{x}_{1} & g_{12} \\
\mathbf{x}_{11}\mathbf{x}_{2}&g_{22}
\end{vmatrix}}
{\begin{vmatrix} g_{11}&g_{12}\\g_{21}&g_{22}\end{vmatrix}}
=\genfrac{}{}{0.5pt}{}{g_{22}}{g}\mathbf{x}_{11}\mathbf{x}_{1}
-\genfrac{}{}{0.5pt}{}{g_{12}}{g}\mathbf{x}_{11}\mathbf{x}_{2}
\\[0.2cm]
=g^{11}\mathbf{x}_{11}\mathbf{x}_{1}+g^{12}\mathbf{x}_{11}\mathbf{x}_{2}\\[0.2cm]
~~\mathbf{x}_{11}\mathbf{x}_{1}~を~g~で表すために、~
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}~を計算する。\\[0.2cm]
~~~~~g_{11}=\mathbf{x}_{1}\mathbf{x}_{1}\\[0.2cm]
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}=\mathbf{x}_{11}\mathbf{x}_{1}
+\mathbf{x}_{1}\mathbf{x}_{11}=2\mathbf{x}_{1}\mathbf{x}_{11}\\[0.2cm]
~~~~~\therefore~~~\mathbf{x}_{1}\mathbf{x}_{11}=
\genfrac{}{}{0.5pt}{}{1}{2}\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}\\[0.2cm]
次に、\mathbf{x}_{11}\mathbf{x}_{2}~を~g~で表すために、\\[0.2cm]
\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}+
\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2}~
を計算する。\\[0.2cm]
~~~~~~g_{21}=\mathbf{x}_{2}\mathbf{x}_{1}\\[0.2cm]
\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}=\mathbf{x}_{21}\mathbf{x}_{1}
+\mathbf{x}_{2}\mathbf{x}_{11}\\[0.2cm]
\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}=\mathbf{x}_{21}\mathbf{x}_{1}
+\mathbf{x}_{2}\mathbf{x}_{11}\\[0.2cm]
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2}=\mathbf{x}_{12}\mathbf{x}_{1}
+\mathbf{x}_{1}\mathbf{x}_{21}\\[0.2cm]
~~~~\therefore~~~\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}+
\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2}=
2\mathbf{x}_{2}\mathbf{x}_{11}\\[0.2cm]
~~~~\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}~は、
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}+
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}~
とも書けるから、\\[0.2cm]
求める~\Gamma^{1}_{11}~は、次のように書ける。\\[0.2cm]
~~~\Gamma^{1}_{11}=\genfrac{}{}{0.5pt}{}{1}{2}g^{11}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}+
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1} \right)
+\genfrac{}{}{0.5pt}{}{1}{2}g^{12}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}+
\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2} \right)\\[0.2cm]
~~~~~~~~~=\left\{{}^{\,1}_{11}\right\}\\[0.2cm]
また、~~~\Gamma^{2}_{11}=
\genfrac{}{}{0.5pt}{}{\begin{vmatrix} g_{11}&\mathbf{x}_{11}\mathbf{x}_{1}
\\g_{21}&\mathbf{x}_{11}\mathbf{x}_{2}\end{vmatrix}}{g}=
\genfrac{}{}{0.5pt}{}{g_{11}}{g}\mathbf{x}_{11}\mathbf{x}_{2}
-\genfrac{}{}{0.5pt}{}{g_{21}}{g}\mathbf{x}_{11}\mathbf{x}_{1}\\[0.2cm]
=g^{22}\mathbf{x}_{11}\mathbf{x}_{2}
+g^{21}\mathbf{x}_{11}\mathbf{x}_{1}\\[0.2cm]
~~~次に~\Gamma _{11}~を求める。\\[0.2cm]
~~N~をかけて、\\[0.2cm]
~~~~~~~~~~\mathbf{x}_{11}N=\Gamma _{11}\\[0.2cm]
~~~~~~\therefore~~~\Gamma _{11}=L=h_{11}\\[0.2cm]
~~~~~これで１番目の式が出た。\\[0.2cm]
~~~ここで一般的に~\mathbf{x}_{\gamma}\mathbf{x}_{\alpha \beta}~を
~g~で表すために、
\genfrac{}{}{0.5pt}{}{\partial g_{\gamma \alpha}}{\partial u^{\beta}}+
\genfrac{}{}{0.5pt}{}{\partial g_{\gamma \beta}}{\partial u^{\alpha}}-
\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^{\gamma}}
を計算する。\\[0.2cm]
\genfrac{}{}{0.5pt}{}{\partial g_{\gamma \alpha}}{\partial u^{\beta}}
=\mathbf{x}_{\gamma \beta}\mathbf{x}_{\alpha}+
\mathbf{x}_{\gamma}\mathbf{x}_{\alpha \beta}\\[0.2cm]
\genfrac{}{}{0.5pt}{}{\partial g_{\gamma \beta}}{\partial u^{\alpha}}
=\mathbf{x}_{\gamma \alpha}\mathbf{x}_{\beta}+
\mathbf{x}_{\gamma}\mathbf{x}_{\alpha \beta}\\[0.2cm]
\genfrac{}{}{0.5pt}{}{\partial g_{\gamma \alpha}}{\partial u^{\beta}}
=\mathbf{x}_{\alpha \gamma}\mathbf{x}_{\beta}+
\mathbf{x}_{\alpha}\mathbf{x}_{\beta \gamma}\\[0.2cm]
~~~\therefore ~~~\genfrac{}{}{0.5pt}{}{\partial
g_{\gamma \alpha}}{\partial u^{\beta}}+
\genfrac{}{}{0.5pt}{}{\partial g_{\gamma \beta}}{\partial u^{\alpha}}-
\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^{\gamma}}
=2\mathbf{x}_{\gamma}\mathbf{x}_{\alpha \beta}\\[0.2cm]
以下、上の式の左辺の~1/2~を~\Gamma _{\alpha \beta, \gamma}~とおく。
即ち、\\[0.2cm]
\genfrac{}{}{0.5pt}{}{1}{2}\left( \genfrac{}{}{0.5pt}{}{\partial
g_{\gamma \alpha}}{\partial u^{\beta}}+
\genfrac{}{}{0.5pt}{}{\partial g_{\gamma \beta}}{\partial u^{\alpha}}-
\genfrac{}{}{0.5pt}{}{\partial g_{\alpha \beta}}{\partial u^{\gamma}}
\right)=\Gamma _{\alpha \beta, \gamma}\\[0.2cm]
とおく。（これを~「Christoffel~の第１記号」という。）\\[0.2cm]
~~~この定義を設けると、クリストッフェルの第２記号~
\left\{{}^{\,\gamma}_{\alpha\beta}\right\}~
の定義から、\\[0.2cm]
~~~~~\left\{{}^{\,\gamma}_{\alpha \beta}\right\}=
g^{\gamma 1}\Gamma _{\alpha \beta, 1} +g^{\gamma 2}\Gamma _{\alpha \beta, 2}\\[0.2cm]
となる。\\[0.2cm]
次に、２番目の式を示す。\\[0.2cm]
~~~\mathbf{x}_{12}=\Gamma ^{1}_{12}\mathbf{x}_{1}+
\Gamma ^{2}_{12}\mathbf{x}_{2}+\Gamma _{12}\mathbf{N}\\[0.2cm]
~~\mathbf{x}_{1}~をかけて、\\[0.2cm]
~~~~~~\mathbf{x}_{1}\mathbf{x}_{12}=
\Gamma ^{1}_{12}g_{11}+\Gamma ^{2}_{12}g_{12}\\[0.2cm]
~~\mathbf{x}_{2}~をかけて、\\[0.2cm]
~~~~~~\mathbf{x}_{2}\mathbf{x}_{12}=
\Gamma ^{1}_{12}g_{21}+\Gamma ^{2}_{12}g_{22}\\[0.2cm]
\therefore~~~~\Gamma ^{1}_{12}=\genfrac{}{}{0.5pt}{}{1}{g}~
{\begin{vmatrix} \mathbf{x}_{1}\mathbf{x}_{12} & g_{12} \\
\mathbf{x}_{2}\mathbf{x}_{12}&g_{22}
\end{vmatrix}}
=\genfrac{}{}{0.5pt}{}{g_{22}}{g}\Gamma _{12,1}-
\genfrac{}{}{0.5pt}{}{g_{12}}{g}\Gamma _{12,2}\\[0.2cm]
=g^{11}\Gamma _{12,1}+g^{12}\Gamma _{12,2}
=\left\{{}^{\,1}_{12}\right\}\\[0.2cm]
~~~\Gamma _{12}~を求めるには、両辺に~\mathbf{N}~をかけて、\\[0.2cm]
~~~~~~~~\Gamma _{12}=\mathbf{x} _{12}\mathbf{N}=M=h_{12}\\[0.2cm]
~~~最後に３番目の式を示す。\\[0.2cm]
~~~\mathbf{x}_{22}=\Gamma ^{1}_{22}\mathbf{x}_{1}+
\Gamma ^{2}_{22}\mathbf{x}_{2}+\Gamma _{22}\mathbf{N}\\[0.2cm]
~~\mathbf{x}_{1}~をかけて、\\[0.2cm]
~~~~~~\mathbf{x}_{1}\mathbf{x}_{22}=
\Gamma ^{1}_{22}g_{11}+\Gamma ^{2}_{22}g_{12}\\[0.2cm]
~~\mathbf{x}_{2}~をかけて、\\[0.2cm]
~~~~~~\mathbf{x}_{2}\mathbf{x}_{22}=
\Gamma ^{1}_{22}g_{21}+\Gamma ^{2}_{22}g_{22}\\[0.2cm]
\therefore~~~~\Gamma ^{1}_{22}=\genfrac{}{}{0.5pt}{}{1}{g}~
{\begin{vmatrix} \mathbf{x}_{1}\mathbf{x}_{22} & g_{12} \\
\mathbf{x}_{2}\mathbf{x}_{22}&g_{22}
\end{vmatrix}}
=\genfrac{}{}{0.5pt}{}{g_{22}}{g}\Gamma _{22,1}-
\genfrac{}{}{0.5pt}{}{g_{12}}{g}\Gamma _{22,2}\\[0.2cm]
=g^{11}\Gamma _{22,1}+g^{12}\Gamma _{22,2}
=\left\{{}^{\,1}_{22}\right\}\\[0.2cm]
\Gamma ^{2}_{22}=\genfrac{}{}{0.5pt}{}{1}{g}~
{\begin{vmatrix} g_{11}&\mathbf{x}_{1}\mathbf{x}_{22} \\
g_{21}&\mathbf{x}_{2}\mathbf{x}_{22}
\end{vmatrix}}
=\genfrac{}{}{0.5pt}{}{g_{11}}{g}\Gamma _{22,2}-
\genfrac{}{}{0.5pt}{}{g_{21}}{g}\Gamma _{22,1}\\[0.2cm]
=g^{22}\Gamma _{22,2}+g^{21}\Gamma _{22,1}
=\left\{{}^{\,2}_{22}\right\}\\[0.2cm]
~~~\Gamma _{22}~を求めるには、両辺に~\mathbf{N}~をかけて、\\[0.2cm]
~~~~~~~~\Gamma _{22}=\mathbf{x} _{22}\mathbf{N}=N=h_{22}
~~~~~~~~~~~~~~（証明終）\\[0.5cm]
命題~~~（Weingarten~の公式）\\[0.2cm]
~~~~\mathbf{N}_1=\genfrac{}{}{0.5pt}{}{\partial \mathbf{N}}{\partial u^{1}}
=-(h_{11}g^{11}+h_{12}g^{21})\mathbf{x_1}
-(h_{11}g^{12}+h_{12}g^{22})\mathbf{x_2}\\[0.2cm]
~~~~\mathbf{N}_2=\genfrac{}{}{0.5pt}{}{\partial \mathbf{N}}{\partial u^{2}}
=-(h_{21}g^{11}+h_{22}g^{21})\mathbf{x_1}
-(h_{21}g^{12}+h_{22}g^{22})\mathbf{x_2}\\[0.2cm]
Einstein~の記号を使えば、次のように書ける\\[0.2cm]
~~~~\mathbf{N}_1=
-(h_{1\alpha} g^{\alpha 1})\mathbf{x_1}-(h_{1\alpha} g^{\alpha 2})\mathbf{x_2}
=-(h_{1\alpha} g^{\alpha \beta})\mathbf{x_\beta}\\[0.2cm]
~~~~\mathbf{N}_2=
-(h_{2\alpha} g^{\alpha 1})\mathbf{x_1}-(h_{2\alpha} g^{\alpha 2})\mathbf{x_2}
=-(h_{2\alpha} g^{\alpha \beta})\mathbf{x_\beta}\\[0.2cm]
もっと簡単に、\delta=1,~2~として、\\[0.2cm]
~~~~\mathbf{N}_\delta=-(h_{\delta \alpha}
g^{\alpha \beta})\mathbf{x_\beta}\\[0.2cm]
証明~~~\mathbf{N}\mathbf{N}=1\\[0.2cm]
~~~~~\therefore ~~~\mathbf{N}_{\alpha}\mathbf{N}
+\mathbf{N}\mathbf{N}_{\alpha}=0~~~(\alpha=1,2)\\[0.2cm]
~~~~\therefore~~\mathbf{N}\mathbf{N}_{\alpha}=0~~~(\alpha=1,2)\\[0.2cm]
即ち、\mathbf{N}_{\alpha}~は、\mathbf{x}_{1},~\mathbf{x}_{2}~のはる
平面内にある。故に\mathbf{N}_{1}は、\\[0.2cm]
~~~~~~~~\mathbf{N}_{1}=\Lambda ^{1}_{1}\mathbf{x}_{1}+
\Lambda ^{2}_{1}\mathbf{x}_{2}\\[0.2cm]
~~とおける。\\[0.2cm]
\mathbf{x}_{1}~をかける。\mathbf{x}_{1}\mathbf{N}_{1}=-h_{11}~であるから、\\[0.2cm]
~~~~~-h_{11}=\Lambda ^{1}_{1}g_{11}+\Lambda ^{2}_{1}g_{12}\\[0.2cm]
\mathbf{x}_{2}~をかける。\mathbf{x}_{2}\mathbf{N}_{1}=-h_{21}~であるから、\\[0.2cm]
~~~~~-h_{21}=\Lambda ^{1}_{1}g_{21}+\Lambda ^{2}_{1}g_{22}\\[0.2cm]
これを解いて、
~~~~~\Lambda ^{1}_{1}=\genfrac{}{}{0.5pt}{}{1}{g}
{\begin{vmatrix} -h_{11}&g_{12}\\-h_{12}&g_{22}\end{vmatrix}}
=-\genfrac{}{}{0.5pt}{}{g_{22}}{g}h_{11}+\genfrac{}{}{0.5pt}{}{g_{12}}{g}h_{12}\\[0.2cm]
=-h_{11}g^{11}-h_{12}g^{21}=-h_{1\alpha}g^{\alpha 1}\\[0.2cm]
~~~~~\Lambda ^{1}_{2}=\genfrac{}{}{0.5pt}{}{1}{g}
{\begin{vmatrix} g_{11}&-h_{11}\\g_{21}&-h_{12}\end{vmatrix}}
=-\genfrac{}{}{0.5pt}{}{g_{11}}{g}h_{12}+\genfrac{}{}{0.5pt}{}{g_{21}}{g}h_{11}\\[0.2cm]
=-h_{12}g^{22}-h_{11}g^{12}=-h_{1\alpha}g^{\alpha 2}\\[0.2cm]
次に~\mathbf{N}_{2}~を求める。\mathbf{N}_{2}~も\mathbf{x}_{1}~と\mathbf{x}_{2}~
で張られているから、\\[0.2cm]
~~~~~~~~\mathbf{N}_{2}=\Lambda ^{1}_{2}\mathbf{x}_{1}+
\Lambda ^{2}_{2}\mathbf{x}_{2}\\[0.2cm]
~~とおける。\\[0.2cm]
\mathbf{x}_{1}~をかける。\mathbf{x}_{1}\mathbf{N}_{2}=-h_{21}~であるから、\\[0.2cm]
~~~~~-h_{21}=\Lambda ^{1}_{2}g_{11}+\Lambda ^{2}_{2}g_{12}\\[0.2cm]
\mathbf{x}_{2}~をかける。\mathbf{x}_{2}\mathbf{N}_{2}=-h_{22}~であるから、\\[0.2cm]
~~~~~-h_{22}=\Lambda ^{1}_{2}g_{12}+\Lambda ^{2}_{2}g_{22}\\[0.2cm]
~~~~~\Lambda ^{1}_{2}=\genfrac{}{}{0.5pt}{}{1}{g}
{\begin{vmatrix} -h_{21}&g_{12}\\-h_{22}&g_{22}\end{vmatrix}}
=-\genfrac{}{}{0.5pt}{}{g_{22}}{g}h_{21}+\genfrac{}{}{0.5pt}{}{g_{12}}{g}h_{22}\\[0.2cm]
=-h_{21}g^{11}-h_{22}g^{21}=-h_{2\alpha}g^{\alpha 1}\\[0.2cm]
~~~~~\Lambda ^{2}_{2}=\genfrac{}{}{0.5pt}{}{1}{g}
{\begin{vmatrix} g_{11}&-h_{21}\\g_{21}&-h_{22}\end{vmatrix}}
=-\genfrac{}{}{0.5pt}{}{g_{11}}{g}h_{22}+\genfrac{}{}{0.5pt}{}{g_{21}}{g}h_{21}\\[0.2cm]
=-h_{22}g^{22}-h_{1}g^{12}=-h_{2\alpha}g^{\alpha 2}~~~~~~~~（証明終）\\[0.2cm]
定義~~~S:~\mathbf{X}=\mathbf{X}(u,v)~~~~曲面\\[0.2cm]
~~~~~~~~D:~\mathbf{x}=\mathbf{x}(u(s),v(s))~~~~~~S~上の曲線\\[0.2cm]
~~~~~~~~~\mathbf{t}=\genfrac{}{}{0.5pt}{}{d\mathbf{x}}{ds}=
\genfrac{}{}{0.5pt}{}{\partial \mathbf{x}}{\partial u}
\genfrac{}{}{0.5pt}{}{du}{ds}+
\genfrac{}{}{0.5pt}{}{\partial \mathbf{x}}{\partial v}
\genfrac{}{}{0.5pt}{}{dv}{ds}=\mathbf{x}_1u'+\mathbf{x}_2v'\\[0.2cm]
~~~~~~~\mathbf{u}~を次のように定義する。\\[0.2cm]
~~~~~~~~~\mathbf{u}=\mathbf{N} \times \mathbf{t}\\[0.2cm]
~~~~~（\mathbf{u}~は、曲面~S~の接平面上にある。）\\[0.2cm]
さて、\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds}=\kappa \mathbf{n}~である
ことはフルネー・セレーの定理でやったが、（この~\mathbf{n}~は曲線の接触平面
内にある。つまり、もとの曲面~S~とは無関係。）この~\kappa \mathbf{n}~の、
\mathbf{u}~への正射影の長さを、「測地的曲率」といい、\kappa _g~と書く。\\[0.2cm]
即ち、\\[0.2cm]
~~~~~~~~\kappa _g={\begin{vmatrix} \kappa \mathbf{n}\end{vmatrix}}
\cos {(\mathbf{u},\mathbf{n})}=\kappa ~(\mathbf{u},\mathbf{n})
=(\mathbf{u},\kappa \mathbf{n})=
\left( \mathbf{u},\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds} \right)\\[0.2cm]
\\
問~~~まづ~\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds}~を計算し、
次に~\mathbf{u}=\mathbf{N} \times \mathbf{t}~との内積を作り、
\kappa _g~を~g_{\alpha \beta}~で表せ。\\[0.2cm]
解~~~\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds}=\genfrac{}{}{0.5pt}{}
{d\mathbf{x}_1}{ds}u'+\mathbf{x}_1\genfrac{}{}{0.5pt}{}{du'}{ds}+
\genfrac{}{}{0.5pt}{}
{d\mathbf{x}_2}{ds}v'+\mathbf{x}_2\genfrac{}{}{0.5pt}{}{dv'}{ds}
\\[0.2cm]
=\genfrac{}{}{0.5pt}{}{\partial\mathbf{x}_1}{\partial u}\genfrac{}{}{0.5pt}{}{du}{ds}u'+
\genfrac{}{}{0.5pt}{}{\partial\mathbf{x}_1}{\partial v}\genfrac{}{}{0.5pt}{}{dv}{ds}u'+
\mathbf{x}_1u"+
\genfrac{}{}{0.5pt}{}{\partial\mathbf{x}_2}{\partial u}\genfrac{}{}{0.5pt}{}{du}{ds}v'+
\genfrac{}{}{0.5pt}{}{\partial\mathbf{x}_2}{\partial v}\genfrac{}{}{0.5pt}{}{dv}{ds}v'+
\mathbf{x}_2v"\\[0.2cm]
=\mathbf{x}_{11}(u')^2+\mathbf{x}_{12}u'v'+\mathbf{x}_1u"+
\mathbf{x}_{21}u'v'+\mathbf{x}_{22}(v')^2+\mathbf{x}_2v"\\[0.2cm]
=\mathbf{x}_{11}(u')^2+2\mathbf{x}_{12}u'v'+\mathbf{x}_{22}(v')^2
+\mathbf{x}_1u"+\mathbf{x}_2v"\\[0.2cm]
=\left(\left\{{}^{\,1}_{11}\right\}\mathbf{x}_1
+\left\{{}^{\,2}_{11}\right\}\mathbf{x}_2
+h_{11}\mathbf{N} \right)(u')^2+2
\left(\left\{{}^{\,1}_{12}\right\}\mathbf{x}_1
+\left\{{}^{\,2}_{12}\right\}\mathbf{x}_2
+h_{12}\mathbf{N} \right)u'v'\\[0.2cm]
+\left(\left\{{}^{\,1}_{22}\right\}\mathbf{x}_1
+\left\{{}^{\,2}_{22}\right\}\mathbf{x}_2
+h_{22}\mathbf{N} \right)(v')^2+\mathbf{x}_1u"+\mathbf{x}_2v"\\[0.2cm]
=\left(\left\{{}^{\,1}_{11}\right\}(u')^2+2\left\{{}^{\,1}_{12}\right\}u'v'
+\left\{{}^{\,1}_{22}\right\}(v')^2 +u"\right)\mathbf{x}_1\\[0.2cm]
+\left(\left\{{}^{\,2}_{11}\right\}(u')^2+2\left\{{}^{\,2}_{12}\right\}u'v'
+\left\{{}^{\,2}_{22}\right\}(v')^2 +v"\right)\mathbf{x}_2\\[0.2cm]
+\left(h_{11}(u')^2+2h_{12}u'v'+h_{22}(v')^2 \right)\mathbf{N}\\[0.2cm]
ここで~\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds}~と~(\mathbf{N} \times \mathbf{t})
~との内積を計算する。\\[0.2cm]
(\mathbf{N} \times \mathbf{x}_1)\mathbf{x}_1=0,~
(\mathbf{N} \times \mathbf{x}_2)\mathbf{x}_2=0,~
(\mathbf{N} \times \mathbf{x}_1)\mathbf{N}=0,~
(\mathbf{N} \times \mathbf{x}_2)\mathbf{N}=0\\[0.2cm]
に注意して、\\[0.2cm]
\kappa _g=(\mathbf{N} \times \mathbf{t})\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds}
=[\mathbf{N} \times (\mathbf{x}_1u'+\mathbf{x}_2v')]
\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds}\\[0.2cm]
=(\mathbf{N} \times \mathbf{x}_1)u'\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds}
+(\mathbf{N} \times \mathbf{x}_2)v'\genfrac{}{}{0.5pt}{}{d\mathbf{t}}{ds}\\[0.2cm]
=(\mathbf{N} \times \mathbf{x}_1)u'
\left(\left\{{}^{\,2}_{11}\right\}(u')^2+2\left\{{}^{\,2}_{12}\right\}u'v'
+\left\{{}^{\,2}_{22}\right\}(v')^2 +v"\right)\mathbf{x}_2\\[0.2cm]
+(\mathbf{N} \times \mathbf{x}_2)v'
\left(\left\{{}^{\,1}_{11}\right\}(u')^2+2\left\{{}^{\,1}_{12}\right\}u'v'
+\left\{{}^{\,1}_{22}\right\}(v')^2 +u"\right)\mathbf{x}_1\\[0.2cm]
~~~~(\mathbf{a} \times \mathbf{b})\mathbf{c}
=\mathbf{a}(\mathbf{b} \times \mathbf{c}),~~
(\mathbf{a} \times \mathbf{c})\mathbf{b}
=-\mathbf{a}(\mathbf{b} \times \mathbf{c})~に注意して、\\[0.2cm]
=\mathbf{N}(\mathbf{x}_1 \times \mathbf{x}_2)u'
\left(\left\{{}^{\,2}_{11}\right\}(u')^2+2\left\{{}^{\,2}_{12}\right\}u'v'
+\left\{{}^{\,2}_{22}\right\}(v')^2 +v"\right)
\\[0.2cm]
-\mathbf{N}(\mathbf{x}_1 \times \mathbf{x}_2)v'
\left(\left\{{}^{\,1}_{11}\right\}(u')^2+2\left\{{}^{\,1}_{12}\right\}u'v'
+\left\{{}^{\,1}_{22}\right\}(v')^2 +u"\right)\\[0.2cm]
~~~~~\mathbf{N}=\genfrac{}{}{0.5pt}{}{\mathbf{x}_1 \times \mathbf{x}_2}{\sqrt g},~
~~(\mathbf{x}_1 \times \mathbf{x}_2)(\mathbf{x}_1 \times \mathbf{x}_2)=g~
に注意して、\\[0.2cm]
=\sqrt g\begin{vmatrix} u',&\left(\left\{{}^{\,1}_{11}\right\}(u')^2+
2\left\{{}^{\,1}_{12}\right\}u'v'
+\left\{{}^{\,1}_{22}\right\}(v')^2 +u"\right)
\\ v', &
\left(\left\{{}^{\,2}_{11}\right\}(u')^2+2\left\{{}^{\,2}_{12}\right\}u'v'
+\left\{{}^{\,2}_{22}\right\}(v')^2 +v"\right)
\end{vmatrix}\\[0.2cm]
~~~（解終り）\\[0.2cm]
注意~~~u=u^1,~v=u^2~と書き、アインシュタインの約束を用いれば、綺麗に
次のように書ける。\\[0.2cm]
~~~\kappa _g=
\sqrt g\begin{vmatrix} (u^1)',& \left\{{}^{\,1}_{\alpha \beta }\right\}(u^{\alpha})'
(u^{\beta})'+(u^1)"
\\ (u^2)', &
\left\{{}^{\,2}_{\alpha \beta}\right\}(u^{\alpha})'(u^{\beta})'+(u^2)"
\end{vmatrix}\\[0.2cm]
定義~~\\[0.2cm]
~~~~~~~ \left\{{}^{\,\gamma}_{\alpha \beta }\right\}(u^{\alpha})'
(u^{\beta})'+(u^\gamma)"=0~~(\gamma=1,~2)\\[0.2cm]
を測地線の方程式と言い、曲面上でこの方程式を満たす曲線を測地線という。\\[0.2cm]
問~~~W= \left \{(u,v); 0 \leqq v \leqq 2\pi,~-\genfrac{}{}{0.5pt}{}{\pi}{2}
\leqq \genfrac{}{}{0.5pt}{}{\pi}{2} \right \} \\[0.2cm]
と半径~a~の球とは、\\[0.2cm]
~~~~x_1=a\cos u \cos v,~x_2=a\cos u \sin v,~x_3=a\sin u \\[0.2cm]
で対応している。\\[0.2cm]
~~~W~における測地線を求めよ。\\[0.2cm]
解~~~\mathbf{x}=(a\cos u \cos v,~x_2=a\cos u \sin v,~x_3=a\sin u)\\[0.2cm]
~~~~~~~u=u^1,~v=u^2~とおいて\\[0.2cm]
~~~~~\mathbf{x}_1=(-a\sin u \cos v, -a\sin u \sin v, a\cos u)\\[0.2cm]
~~~~~\mathbf{x}_2=(-a\cos u \sin v, -a\cos u \cos v, 0)\\[0.2cm]
~~~~~g_{11}=a^2\sin^2 u \cos^2 v+a^2 \sin^2 u \sin^2 v+a^2 \cos^2u=a^2\\[0.2cm]
~~~~~g_{12}=a^2\sin u \cos u\sin v \cos v-a^2 \sin u \cos u\sin v \cos v=0\\[0.2cm]
~~~~~g_{22}=a^2\cos^2 u \sin^2 v+a^2\cos^2 u\cos^2 v=a^2\cos^2 u\\[0.2cm]
~~~g=g_{11}g_{22}-g_{12}^2=a^4 \cos^2 u\\[0.2cm]
~~~g^{11}=\genfrac{}{}{0.5pt}{}{g_{22}}{g}=\genfrac{}{}{0.5pt}{}{1}{a^2},~~
g^{12}=0,~~g^{22}=\genfrac{}{}{0.5pt}{}{1}{a^2\cos^2 u}\\[0.2cm]
\left\{{}^{\,1}_{11}\right\}=g^{11}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}
+\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}
-\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1} \right)
+
g^{12}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}
+\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}
-\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1} \right)\\[0.2cm]
=\genfrac{}{}{0.5pt}{}{1}{a^2}\genfrac{}{}{0.5pt}{}{1}{2} \times
0+0\times \genfrac{}{}{0.5pt}{}{1}{2}\times (...)=0\\[0.2cm]
\left\{{}^{\,1}_{12}\right\}=g^{11}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2}
+\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}
-\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1} \right)
+
g^{12}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^2}
+\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^1}
-\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^2} \right)\\[0.2cm]
=\genfrac{}{}{0.5pt}{}{1}{a^2}\genfrac{}{}{0.5pt}{}{1}{2} \times
0+0=0\\[0.2cm]
\left\{{}^{\,1}_{21}\right\}=0\\[0.2cm]
\left\{{}^{\,1}_{22}\right\}=g^{11}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^2}
+\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^2}
-\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^1} \right)
+
g^{12}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2}
+\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2}
-\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2} \right)\\[0.2cm]
=\genfrac{}{}{0.5pt}{}{1}{a^2}\genfrac{}{}{0.5pt}{}{1}{2} \times
\left(-\genfrac{}{}{0.5pt}{}{\partial(a^2\cos^2u)}{\partial u} \right)+0=
\sin u \cos u\\[0.2cm]
\left\{{}^{\,2}_{11}\right\}=g^{21}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}
+\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}
-\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1} \right)
+
g^{22}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}
+\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}
-\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2} \right)\\[0.2cm]
=0\\[0.2cm]
\left\{{}^{\,2}_{12}\right\}=g^{21}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2}
+\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}
-\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1} \right)
+
g^{22}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^2}
+\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^1}
-\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^2} \right)\\[0.2cm]
=\genfrac{}{}{0.5pt}{}{1}{a^2\cos^2 u} \times \genfrac{}{}{0.5pt}{}
{1}{2}\genfrac{}{}{0.5pt}{}{\partial(a^2 \cos^2 u)}{\partial u}
=-\genfrac{}{}{0.5pt}{}{\sin u}{\cos u}\\[0.2cm]
\left\{{}^{\,2}_{21}\right\}=\left\{{}^{\,2}_{12}\right\}
=-\genfrac{}{}{0.5pt}{}{\sin u}{\cos u}\\[0.2cm]
\left\{{}^{\,2}_{22}\right\}=g^{21}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^2}
+\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^2}
-\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^1} \right)
+
g^{22}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2}
+\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2}
-\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2} \right)\\[0.2cm]
=g^{22}\genfrac{}{}{0.5pt}{}{1}{2} \genfrac{}{}{0.5pt}{}
{\partial(a^2 \cos^2 u)}{\partial v}=0\\[0.2cm]
これらを測地線の方程式\\[0.2cm]
~~~~~~~ (u^\gamma)"+\left\{{}^{\,\gamma}_{\alpha \beta }\right\}(u^{\alpha})'
(u^{\beta})'=0~~(\gamma=1,~2)\\[0.2cm]
に代入。\\[0.2cm]
\gamma=1~のとき、\\[0.2cm]
\genfrac{}{}{0.5pt}{}{d^2u}{ds^2}
+\left\{{}^{\,1}_{11}\right\}\genfrac{}{}{0.5pt}{}{du}{ds}\genfrac{}{}{0.5pt}{}{du}{ds}
+2\left\{{}^{\,1}_{12}\right\}\genfrac{}{}{0.5pt}{}{du}{ds}\genfrac{}{}{0.5pt}{}{dv}{ds}
+\left\{{}^{\,1}_{22}\right\}\genfrac{}{}{0.5pt}{}{dv}{ds}\genfrac{}{}{0.5pt}{}{dv}{ds}
=0\\[0.2cm]
\therefore~~~\genfrac{}{}{0.5pt}{}{d^2u}{ds^2}+\sin u \cos u
\left(\genfrac{}{}{0.5pt}{}{dv}{ds}\right)^2=0~~~ ... ~~~ (1)\\[0.2cm]
\gamma=2~のとき、\\[0.2cm]
\genfrac{}{}{0.5pt}{}{d^2v}{ds^2}
+\left\{{}^{\,2}_{11}\right\}\genfrac{}{}{0.5pt}{}{du}{ds}\genfrac{}{}{0.5pt}{}{du}{ds}
+2\left\{{}^{\,2}_{12}\right\}\genfrac{}{}{0.5pt}{}{du}{ds}\genfrac{}{}{0.5pt}{}{dv}{ds}
+\left\{{}^{\,2}_{22}\right\}\genfrac{}{}{0.5pt}{}{dv}{ds}\genfrac{}{}{0.5pt}{}{dv}{ds}
=0\\[0.2cm]
\therefore~~~\genfrac{}{}{0.5pt}{}{d^2v}{ds^2}
-2\genfrac{}{}{0.5pt}{}{\sin u}{\cos u}\genfrac{}{}{0.5pt}{}{du}{ds}
\genfrac{}{}{0.5pt}{}{dv}{ds}=0 ~~~ ...~~~(2) \\[0.2cm]
(2)~から\\[0.2cm]
\genfrac{}{}{0.5pt}{}{\genfrac{}{}{0.5pt}{}{d^2v}{ds^2}}{\genfrac{}{}{0.5pt}{}{dv}{ds}}
=2\genfrac{}{}{0.5pt}{}{\sin u}{\cos u}\genfrac{}{}{0.5pt}{}{du}{ds}\\[0.2cm]
\therefore~~~\log \genfrac{}{}{0.5pt}{}{dv}{ds}=-2\log \cos u + \log c \\[0.2cm]
\therefore~~~\genfrac{}{}{0.5pt}{}{dv}{ds}=\genfrac{}{}{0.5pt}{}{c}{\cos^2 u}\\[0.2cm]
２乗して、また~~ds^2=g_{11}du^2+2g_{12}dudv+g_{22}=a^2du^2+a^2\cos^2 u dv^2~
に気をつけて、\\[0.2cm]
~~~dv^2=\genfrac{}{}{0.5pt}{}{c^2}{\cos^4 u}(a^2du^2+a^2\cos^2 u dv^2)\\[0.2cm]
\therefore~~~\cos^4 u dv^2=a^2c^2du^2+a^2c^2\cos^2 u dv^2\\[0.2cm]
~~~~~~~~~~~dv^2(\cos^4 u -a^2c^2\cos^2 u)=a^2c^2du^2\\[0.2cm]
~~~~~~~~~~~dv=\genfrac{}{}{0.5pt}{}{ac}{\cos u \sqrt{\cos^2 u-a^2c^2}}du
=\genfrac{}{}{0.5pt}{}{1}{\cos u\sqrt{\genfrac{}{}{0.5pt}{}{1}{a^2c^2}
\cos^2 -1}}du\\[0.2cm]
この不定積分を求めるために、次の計算をする。\\[0.2cm]
\genfrac{}{}{0.5pt}{}{d}{dx}[\sin^{-1}(k\tan x)]=\genfrac{}{}{0.5pt}{}
{1}{\sqrt{1-k^2\tan^2 x}}\genfrac{}{}{0.5pt}{}{k}{\cos^2 x}
=\genfrac{}{}{0.5pt}{}{k}{\cos x \sqrt{\cos^2 x-k^2\sin^2 x}}\\[0.2cm]
=\genfrac{}{}{0.5pt}{}{1}{\cos x \sqrt{(\genfrac{}{}{0.5pt}{}{1+k^2}{k^2})
\cos^2 x -1}}\\[0.2cm]
故に、\genfrac{}{}{0.5pt}{}{1+k^2}{k^2}=\genfrac{}{}{0.5pt}{}{1}{a^2c^2}~
とおいて、k~を求めれば、k=\genfrac{}{}{0.5pt}{}{ac}{\sqrt{1-a^2c^2}}\\[0.2cm]
となるから、上の不定積分は、\\[0.2cm]
~~~~v=\sin^{-1} \left({\genfrac{}{}{0.5pt}{}{ac}{\sqrt{1-a^2c^2}}\tan u}
\right)+c\\[0.2cm]
となる。\\[0.2cm]
\therefore~~~ \sin (v-D)=\genfrac{}{}{0.5pt}{}{ac}{\sqrt{1-a^2c^2}}\tan u\\[0.2cm]
~~~~ここで、~~~~\genfrac{}{}{0.5pt}{}{ac}{\sqrt{1-a^2c^2}}=l~~とおけば、\\[0.2cm]
~~~~~u=\tan^{-1}(l \sin (v-D))~~~.....(3)\\[0.2cm]
~~~~~~~~~~~~~~~~~~（解おわり）\\[0.2cm]
問~~~上の測地線は、大円になることを示せ。\\[0.2cm]
解~~~大円の方程式を~u,v~で表してみる。大円は、円の中心を通る平面と円\\[0.2cm]
との交線だから、連立方程式\\[0.2cm]
~~~a_1x_1+a_2x_2+a_3x_3=0\\[0.2cm]
~~~x_1=a\cos u \cos v\\[0.2cm]
~~~x_2=a\cos u \sin v\\[0.2cm]
~~~x_3=a\sin u\\[0.2cm]
を解けばよい。\\[0.2cm]
~~~~~a_1\cos u \cos v+a_2 \cos u \sin v+a_3\sin u=0\\[0.2cm]
~~~~~\cos u(a_1\cos v+a_2\sin v)=-a_3\sin u\\[0.2cm]
-a_3\tan u=\sqrt{a_1^2+a_2^2}
\left(\sin v \genfrac{}{}{0.5pt}{}{a_2}{\sqrt{a_1^2+a_2^2}}
+\cos v \genfrac{}{}{0.5pt}{}{a_1}{\sqrt{a_1^2+a_2^2}}
\right)=k\sin(v-D)\\[0.2cm]
~~~~\therefore~~~\tan u=l \sin (v-D)~~~.....(4)\\[0.2cm]
~~~これにより、円上の測地線は大円に対応していることが分る。\\[0.2cm]
問~~~前問の結果を用いて、\\[0.2cm]
~~~~~~~二点~P_1=(0,\genfrac{}{}{0.5pt}{}{\pi}{4}),
~P_2=(\genfrac{}{}{0.5pt}{}{\pi}{4},\genfrac{}{}{0.5pt}{}{\pi}{4})\\[0.2cm]
の最短距離を求めよ。\\[0.2cm]
解~~~P_1,~P_2~を前問で求めた大円の式~\tan u=l \sin (v-D)~に代入。\\[0.2cm]
~~~~~1=l\sin(-D)~~~.....(5)\\[0.2cm]
~~~~~1=l\sin(\genfrac{}{}{0.5pt}{}{\pi}{4}-D)~~~.....(6)\\[0.2cm]
~~~~\sin (\genfrac{}{}{0.5pt}{}{\pi}{4}-D)+\sin D=0\\[0.2cm]
~~~~2\sin \genfrac{}{}{0.5pt}{}{1}{2}(\genfrac{}{}{0.5pt}{}{\pi}{4}-D+D)
\cos \genfrac{}{}{0.5pt}{}{1}{2}(\genfrac{}{}{0.5pt}{}{\pi}{4}-2D)=0\\[0.2cm]
\therefore~~~ \cos (\genfrac{}{}{0.5pt}{}{\pi}{8}-D)=0\\[0.2cm]
\therefore~~~\genfrac{}{}{0.5pt}{}{\pi}{8}-D=±\genfrac{}{}{0.5pt}{}{\pi}{2}
+2n\pi\\[0.2cm]
~~~~~D=\genfrac{}{}{0.5pt}{}{5\pi}{8}+2n\pi~~or
~~-\genfrac{}{}{0.5pt}{}{3\pi}{8}+2n\pi\\[0.2cm]
\therefore~~~D=-\genfrac{}{}{0.5pt}{}{3\pi}{8}~~~.....(7)\\[0.2cm]
~~~最初の式に代入して~l~を求める。\\[0.2cm]
~~~~~~1=l\sin \genfrac{}{}{0.5pt}{}{3\pi}{8}\\[0.2cm]
~~~~~~~~l=\genfrac{}{}{0.5pt}{}{1}{\sin \genfrac{}{}{0.5pt}{}{3\pi}{8}}
~~~.....(8)\\[0.2cm]
(7)~(8)~を(3)~に代入。\\[0.2cm]
~~~u=\tan^{-1}\left\{\genfrac{}{}{0.5pt}{}{1}{\sin \genfrac{}{}{0.5pt}{}{3\pi}{8}}
\sin (v+\genfrac{}{}{0.5pt}{}{3\pi}{8})\right \}~~~~~（解終り）\\[0.2cm]
\\
問~~~上の問のP_1~とP_2~間の「直線距離」と「大円距離」の比較をせよ。\\[0.2cm]
解~~~直線距離~P_1P_2=a\int_0^{\pi/4}\sqrt{du^2+\cos^2 udv^2}\,dv\\[0.2cm]
~~~~~~=a\int_0^{\pi/4}\sqrt{\genfrac{}{}{0.5pt}{}{1}{2}}\,dv
=a\genfrac{}{}{0.5pt}{}{\sqrt{2}\pi}{8}=a\times 0.55536...\\[0.2cm]
~~~大円距離は、実際に球の大円の方で求める。\\[0.2cm]
~~~~P_1\left(0,\genfrac{}{}{0.5pt}{}{\pi}{4}\right)~の、球における位置は
P_1'\left(\genfrac{}{}{0.5pt}{}{a}{\sqrt2},0,\genfrac{}{}{0.5pt}{}{a}{\sqrt2}\right)\\[0.2cm]
~~~~P_2\left(\genfrac{}{}{0.5pt}{}{\pi}{4},\genfrac{}{}{0.5pt}{}
{\pi}{4}\right)~の、球における位置は~P_2'\left(\genfrac{}{}{0.5pt}{}{a}{2},
\genfrac{}{}{0.5pt}{}{a}{2},\genfrac{}{}{0.5pt}{}{a}{\sqrt2}\right)\\[0.2cm]
~~~~~~大円距離は、a \times <P'_1OP'_2~(radian)~だから、
\cos (<P'_1OP'_2)を求める。\\[0.2cm]
~~~~~\cos (<P'_1OP'_2)=(OP_1'~OP_2')/a^2=\genfrac{}{}{0.5pt}{}{1}{2\sqrt2}
+\genfrac{}{}{0.5pt}{}{1}{2}=\genfrac{}{}{0.5pt}{}{\sqrt 2+2}{4}\\[0.2cm]
~~~~~~即ち、大円距離=a \cos^{-1} \left(\genfrac{}{}{0.5pt}{}{\sqrt 2+2}{4} \right)
=a \times 0.5480 ...\\[0.2cm]
~~~~確かに、直線距離より短い。~~~~~~~（解終り）\\[0.2cm]
\\
問~~~上半平面に、計量\\[0.2cm]
~~~~ds^2=\genfrac{}{}{0.5pt}{}{1}{y^2}(dx^2+dy^2)\\[0.2cm]
がはいっているとき、測地線を求めよ。\\[0.2cm]
解~~~g_{11}=\genfrac{}{}{0.5pt}{}{1}{y^2},~~~g_{12}=g_{21}=0,~~~
g_{22}=\genfrac{}{}{0.5pt}{}{1}{y^2}\\[0.2cm]
~~~g=g_{11}g_{22}-g_{12}^2=\genfrac{}{}{0.5pt}{}{1}{y^4},~~~
~~~g^{11}=\genfrac{}{}{0.5pt}{}{g_{22}}{g}=y^2~~~g^{12}=g^{21}=0\\[0.2cm]
\left\{{}^{\,1}_{11}\right\} =g^{11}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}+
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^1}\right)+
g^{12}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}+
\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2}\right)\\[0.2cm]
=y^2 \genfrac{}{}{0.5pt}{}{1}{2}\left(\genfrac{}{}{0.5pt}{}
{\partial y^2}{\partial x} \right)=0\\[0.2cm]
\left\{{}^{\,1}_{12}\right\} =g^{11}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2}+
\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}\right)+
g^{12}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^2}+
\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^2}\right)\\[0.2cm]
=y^2 \genfrac{}{}{0.5pt}{}{1}{2}\left(\genfrac{}{}{0.5pt}{}
{\partial }{\partial y}\left (\genfrac{}{}{0.5pt}{}{1}{y^2} \right)\right)=
- \genfrac{}{}{0.5pt}{}{1}{y}\\[0.2cm]
\left\{{}^{\,1}_{21}\right\} =\left\{{}^{\,1}_{12}\right\} =
- \genfrac{}{}{0.5pt}{}{1}{y}\\[0.2cm]
\left\{{}^{\,1}_{22}\right\} =g^{11}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^2}+
\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^2}-
\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^1}\right)+
g^{12}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2}+
\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2}-
\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2}\right)\\[0.2cm]
=y^2 \genfrac{}{}{0.5pt}{}{1}{2}\left(\genfrac{}{}{0.5pt}{}{\partial}{\partial x}
\left(\genfrac{}{}{0.5pt}{}{1}{y^2}\right) \right)=0\\[0.2cm]
\left\{{}^{\,2}_{11}\right\} =g^{21}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}+
\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2}\right)+
g^{22}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}+
\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2}\right)\\[0.2cm]
=y^2 \genfrac{}{}{0.5pt}{}{1}{2}\left(-\genfrac{}{}{0.5pt}{}{\partial}{\partial y}
\left(\genfrac{}{}{0.5pt}{}{1}{y^2}\right) \right)=
\genfrac{}{}{0.5pt}{}{1}{y}\\[0.2cm]
\left\{{}^{\,2}_{12}\right\} =g^{21}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{11}}{\partial u^2}+
\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^1}\right)+
g^{22}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^2}+
\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^1}-
\genfrac{}{}{0.5pt}{}{\partial g_{12}}{\partial u^2}\right)\\[0.2cm]
=y^2 \genfrac{}{}{0.5pt}{}{1}{2}\left(\genfrac{}{}{0.5pt}{}{\partial}{\partial x}
\left(\genfrac{}{}{0.5pt}{}{1}{y^2}\right) \right)=0\\[0.2cm]
\left\{{}^{\,2}_{22}\right\} =g^{21}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^2}+
\genfrac{}{}{0.5pt}{}{\partial g_{21}}{\partial u^2}-
\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^1}\right)+
g^{22}\genfrac{}{}{0.5pt}{}{1}{2}
\left(\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2}+
\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2}-
\genfrac{}{}{0.5pt}{}{\partial g_{22}}{\partial u^2}\right)\\[0.2cm]
=y^2 \genfrac{}{}{0.5pt}{}{1}{2}\left(\genfrac{}{}{0.5pt}{}{\partial}{\partial y}
\left(\genfrac{}{}{0.5pt}{}{1}{y^2}\right) \right)=-\genfrac{}{}{0.5pt}{}
{1}{y}\\[0.2cm]
~~~これらを測地線の方程式\\[0.2cm]
~~~~~~~ (u^\gamma)"+\left\{{}^{\,\gamma}_{\alpha \beta }\right\}(u^{\alpha})'
(u^{\beta})'=0~~(\gamma=1,~2)\\[0.2cm]
に代入。\\[0.2cm]
\gamma=1~のとき、\\[0.2cm]
\genfrac{}{}{0.5pt}{}{d^2u}{ds^2}
+\left\{{}^{\,1}_{11}\right\}\genfrac{}{}{0.5pt}{}{du}{ds}\genfrac{}{}{0.5pt}{}{du}{ds}
+2\left\{{}^{\,1}_{12}\right\}\genfrac{}{}{0.5pt}{}{du}{ds}\genfrac{}{}{0.5pt}{}{dv}{ds}
+\left\{{}^{\,1}_{22}\right\}\genfrac{}{}{0.5pt}{}{dv}{ds}\genfrac{}{}{0.5pt}{}{dv}{ds}
=0\\[0.2cm]
\therefore~~~\genfrac{}{}{0.5pt}{}{d^2x}{ds^2}-\genfrac{}{}{0.5pt}{}{2}{y}
\genfrac{}{}{0.5pt}{}{dx}{ds}\genfrac{}{}{0.5pt}{}{dy}{ds}
=0~~~~~~....~~(9)\\[0.2cm]
\gamma=2~のとき、\\[0.2cm]
\genfrac{}{}{0.5pt}{}{d^2v}{ds^2}
+\left\{{}^{\,2}_{11}\right\}\genfrac{}{}{0.5pt}{}{du}{ds}\genfrac{}{}{0.5pt}{}{du}{ds}
+2\left\{{}^{\,2}_{12}\right\}\genfrac{}{}{0.5pt}{}{du}{ds}\genfrac{}{}{0.5pt}{}{dv}{ds}
+\left\{{}^{\,2}_{22}\right\}\genfrac{}{}{0.5pt}{}{dv}{ds}\genfrac{}{}{0.5pt}{}{dv}{ds}
=0~~~~~\\[0.2cm]
\therefore~~~\genfrac{}{}{0.5pt}{}{d^2y}{ds^2}+\genfrac{}{}{0.5pt}{}{1}{y}
\left(\genfrac{}{}{0.5pt}{}{dx}{ds} \right)^2
-\genfrac{}{}{0.5pt}{}{1}{y}\left(\genfrac{}{}{0.5pt}{}{dy}{ds} \right)^2
=0~~~~~~....~~(10)\\[0.2cm]
(9)~から、\\[0.2cm]
~~~~~~~~~~\genfrac{}{}{0.5pt}{}{d^2 x/ds^2}
{dx/ds}-\genfrac{}{}{0.5pt}{}{2}{y}
\genfrac{}{}{0.5pt}{}{dy}{ds}=0\\[0.2cm]
~~~~~\log \genfrac{}{}{0.5pt}{}{dx}{ds}-2\log y=\log c\\[0.2cm]
~~~~~\genfrac{}{}{0.5pt}{}{dx}{ds}=cy^2\\[0.2cm]
~~~~~dx=cy^2 ds\\[0.2cm]
~~~~~dx^2=c^2y^4 ds^2=c^2y^4\genfrac{}{}{0.5pt}{}{dx^2+dy^2}{y^2}\\[0.2cm]
~~~~~dx^2=c^2y^2(dx^2+dy^2)\\[0.2cm]
~~~~~(1-c^2y^2)dx^2=c^2y^2dy^2\\[0.2cm]
~~~~~dx=±\genfrac{}{}{0.5pt}{}{cy}{\sqrt{1-c^2y^2}}dy\\[0.2cm]
~~~~~x=±\genfrac{}{}{0.5pt}{}{1}{c}\sqrt{1-c^2y^2}+D\\[0.2cm]
~~~~~c(x-D)=±\sqrt{1-c^2y^2}\\[0.2cm]
~~~~~c^2(x-D)^2=1-c^2y^2\\[0.2cm]
~~~~~(x-D)^2+y^2=\genfrac{}{}{0.5pt}{}{1}{c^2}\\[0.2cm]
~~~~~これは、x~軸上に中心のある円。即ちこの場合は、その上半分。（解終り）\\[0.2cm]
問~~上の解がもう一方の測地線の方程式~（10）~もみたすことを示せ。\\[0.2cm]
解~~~x-D=\genfrac{}{}{0.5pt}{}{1}{c}\cos \theta,
~~y=\genfrac{}{}{0.5pt}{}{1}{c}\sin \theta\\[0.2cm]
~~dx=-\genfrac{}{}{0.5pt}{}{1}{c}\sin \theta d\theta,~~
dy=\genfrac{}{}{0.5pt}{}{1}{c}\cos \theta d\theta\\[0.2cm]
~~dx^2+dy^2=\genfrac{}{}{0.5pt}{}{1}{c^2}d\theta^2\\[0.2cm]
~~ds^2=\genfrac{}{}{0.5pt}{}{dx^2+dy^2}{y^2}
=\genfrac{}{}{0.5pt}{}{d\theta^2}{sin^2 \theta}\\[0.2cm]
~~\therefore ~~~\genfrac{}{}{0.5pt}{}{d\theta}{ds}=\sin \theta\\[0.2cm]
~~\genfrac{}{}{0.5pt}{}{dx}{ds}=\genfrac{}{}{0.5pt}{}{dx}{d\theta}
\genfrac{}{}{0.5pt}{}{d\theta}{ds}=-\genfrac{}{}{0.5pt}{}{1}{c}
\sin^2 \theta\\[0.2cm]
~~\genfrac{}{}{0.5pt}{}{dy}{ds}=\genfrac{}{}{0.5pt}{}{1}{c}\sin \theta \cos \theta
=\genfrac{}{}{0.5pt}{}{1}{2c}\sin 2\theta\\[0.2cm]
\genfrac{}{}{0.5pt}{}{d^2y}{ds^2}=\genfrac{}{}{0.5pt}{}{d}{ds}
\left( \genfrac{}{}{0.5pt}{}{dy}{ds} \right)=\genfrac{}{}{0.5pt}{}{d}{d\theta}
\left( \genfrac{}{}{0.5pt}{}{dy}{ds} \right) \genfrac{}{}{0.5pt}{}{d\theta}{ds}
=\genfrac{}{}{0.5pt}{}{1}{c}\cos 2\theta \sin\theta\\[0.2cm]
さて、これらを~(10)~の左辺に代入して~0~になるかどうかを調べる。\\[0.2cm]
(10)~の左辺=\genfrac{}{}{0.5pt}{}{1}{c}\cos 2\theta \sin \theta
+\genfrac{}{}{0.5pt}{}{c}{\sin \theta}\genfrac{}{}{0.5pt}{}{1}{c^2}
\sin^4 \theta-\genfrac{}{}{0.5pt}{}{c}{\sin \theta}
\genfrac{}{}{0.5pt}{}{1}{c^2}\cos^2 \theta \sin^2 \theta=0\\[0.2cm]
~~~~~~~~~~~~~~~~~~~~~~~（解終り）\\[0.5cm]
%\left\{{}^{\,\lambda}_{\mu\nu}\right\}
\end{math}
\end{document}