巽流体力学の行間埋め　第２章

２．流体の運動と力

1. 式(2･5)の導出
▼ 詳細を開く...2025/ 8/16作成
\begin{align*} \frac{D}{Dt}A(\boldsymbol{x}_0,t) &= \frac{\partial}{\partial t}A(\boldsymbol{x},t)+\frac{Dx_1(\boldsymbol{x},t)}{Dt}\frac{\partial}{\partial x_1}A(\boldsymbol{x},t)+\frac{Dx_2(\boldsymbol{x},t)}{Dt}\frac{\partial}{\partial x_2}A(\boldsymbol{x},t)+\frac{Dx_3(\boldsymbol{x},t)}{Dt}\frac{\partial}{\partial x_3}A(\boldsymbol{x},t) \\ \\ &= \frac{\partial}{\partial t}A(\boldsymbol{x},t)+u_1\frac{\partial}{\partial x_1}A(\boldsymbol{x},t)+u_2\frac{\partial}{\partial x_2}A(\boldsymbol{x},t)+u_3\frac{\partial}{\partial x_3}A(\boldsymbol{x},t)&&...\text{p.15よりLagrange式記述では}\frac{D\boldsymbol{x}}{Dt}=\boldsymbol{u}\text{になる} \\ \\ &= \frac{\partial}{\partial t}A(\boldsymbol{x},t)+u_i\frac{\partial}{\partial x_i}A(\boldsymbol{x},t)&&...\text{p.15下の縮約記法より} \\ \\ &= \frac{\partial}{\partial t}A(\boldsymbol{x},t)+ \left( \begin{array}{cccc} u_1 \\ u_2 \\ u_3 \end{array} \right) \cdot \left( \begin{array}{cccc} \frac{\partial}{\partial x_1}A(\boldsymbol{x},t) \\ \frac{\partial}{\partial x_2}A(\boldsymbol{x},t) \\ \frac{\partial}{\partial x_3}A(\boldsymbol{x},t) \end{array} \right) \\ \\ &= \frac{\partial}{\partial t}A(\boldsymbol{x},t)+ \left( \begin{array}{cccc} u_1 \\ u_2 \\ u_3 \end{array} \right)\cdot \text{grad} A(\boldsymbol{x},t) \\ \\ &= \left\{\frac{\partial}{\partial t}+\boldsymbol{u}\cdot \text{grad}\right\} A(\boldsymbol{x},t) \\ \\ \end{align*} と導出できる。


2. 式(2･9)の導出
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p.17下部より \begin{align*} \text{grad}(\boldsymbol{A}\cdot\boldsymbol{B})-\text{rot}(\boldsymbol{A}\times\boldsymbol{B}) &= (\boldsymbol{A}\cdot\text{grad})\boldsymbol{B}+(\boldsymbol{B}\cdot\text{grad})\boldsymbol{A}+\boldsymbol{A}\times\text{rot}\boldsymbol{B}+\boldsymbol{B}\times\text{rot}\boldsymbol{A} +(\boldsymbol{A}\cdot\text{grad})\boldsymbol{B}-(\boldsymbol{B}\cdot\text{grad})\boldsymbol{A}-\boldsymbol{A}\text{div}\boldsymbol{B}+\boldsymbol{B}\text{div}\boldsymbol{A} \\ \\ &= 2(\boldsymbol{A}\cdot\text{grad})\boldsymbol{B}+\boldsymbol{A}\times\text{rot}\boldsymbol{B}+\boldsymbol{B}\times\text{rot}\boldsymbol{A} -\boldsymbol{A}\text{div}\boldsymbol{B}+\boldsymbol{B}\text{div}\boldsymbol{A} \\ \\ \Leftrightarrow (\boldsymbol{A}\cdot\text{grad})\boldsymbol{B}&=\frac{1}{2}\left\{ \text{grad}(\boldsymbol{A}\cdot\boldsymbol{B})-\text{rot}(\boldsymbol{A}\times\boldsymbol{B})-\boldsymbol{A}\times\text{rot}\boldsymbol{B}-\boldsymbol{B}\times\text{rot}\boldsymbol{A}+\boldsymbol{A}\text{div}\boldsymbol{B}-\boldsymbol{B}\text{div}\boldsymbol{A} \right\} \end{align*} と導出できる。これに対して\(\boldsymbol{A}=\boldsymbol{u},\boldsymbol{B}=\boldsymbol{A}\)と置換して式(2･7)に代入することで得られる。


3. 式(2･11)の導出
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式(2･9)に\(\boldsymbol{A}=\boldsymbol{u}\)を代入すると \begin{align*} \frac{D\boldsymbol{u}}{Dt} &= \frac{\partial \boldsymbol{u}}{\partial t}+\frac{1}{2}\left\{\text{grad}(\boldsymbol{u}\cdot\boldsymbol{u})+\text{rot}\boldsymbol{u}\times\boldsymbol{u}+\text{rot}\boldsymbol{u}\times\boldsymbol{u}-\text{rot}(\boldsymbol{u}\times\boldsymbol{u})+\boldsymbol{u}\text{div}\boldsymbol{u}-\boldsymbol{u}\text{div}\boldsymbol{u}\right\} \\ \\ &= \frac{\partial \boldsymbol{u}}{\partial t}+\frac{1}{2}\left\{\text{grad}(\boldsymbol{u}\cdot\boldsymbol{u})-\text{rot}(\boldsymbol{u}\times\boldsymbol{u})+2\text{rot}\boldsymbol{u}\times\boldsymbol{u}\right\} \\ \\ &= \frac{\partial \boldsymbol{u}}{\partial t}+\frac{1}{2}\left\{\text{grad}(\boldsymbol{u}\cdot\boldsymbol{u})+2\text{rot}\boldsymbol{u}\times\boldsymbol{u}\right\}&&...\boldsymbol{A}\times\boldsymbol{A}=0\text{となるから} \\ \\ &= \frac{\partial \boldsymbol{u}}{\partial t}+\frac{1}{2}\left\{\text{grad}(u_1^2+u_2^2+u_3^2)+2\text{rot}\boldsymbol{u}\times\boldsymbol{u}\right\}& \\ \\ &= \frac{\partial \boldsymbol{u}}{\partial t}+\frac{1}{2}\left\{\text{grad}|\boldsymbol{u}|^2+2\text{rot}\boldsymbol{u}\times\boldsymbol{u}\right\}& \\ \\ &= \frac{\partial \boldsymbol{u}}{\partial t}+\text{grad}\left(\frac{1}{2}|\boldsymbol{u}|\right)^2+\text{rot}\boldsymbol{u}\times\boldsymbol{u}& \\ \\ &= \frac{\partial \boldsymbol{u}}{\partial t}+\text{grad}\left(\frac{1}{2}|\boldsymbol{u}|\right)^2-\boldsymbol{u}\times\text{rot}\boldsymbol{u}&&...\boldsymbol{A}\times\boldsymbol{B}=-\boldsymbol{B}\times\boldsymbol{A}\text{だから} \\ \\ \end{align*} と導出できる。


4. 式(2･13)の導出
▼ 詳細を開く...2025/ 8/16作成
平行であることから定数\(k\)を用いて \begin{align*} d\boldsymbol{x}=k \boldsymbol{u}(\boldsymbol{x},t) \end{align*} と書くことができる。このとき各成分を比較すると \begin{align*} dx_i=k u_i(\boldsymbol{x},t) \\ \\ \Rightarrow k=\frac{dx_i}{u_i(\boldsymbol{x},t)} \end{align*} となり、これはどの\(i\)に対しても成立するため式(2･13)が得られる。


5. 式(2･23)の導出
▼ 詳細を開く...2025/ 8/16作成
\begin{align*} \delta u_i &= u_i^{\prime}-u_i \\ \\ &= u_i(\boldsymbol{x}+\delta\boldsymbol{x})-u_i(\boldsymbol{x})&&...\text{p.22中部より} \\ \\ &= u_i(x_1+\delta x_1,x_2+\delta x_2,x_3+\delta x_3)-u_i(x_1,x_2,x_3)& \\ \\ &= u_i(x_1,x_2,x_3)+\frac{\partial u_i}{\partial x_1}\delta x_1+\frac{\partial u_i}{\partial x_2}\delta x_2+\frac{\partial u_i}{\partial x_3}\delta x_3+\ldots-u_i(x_1,x_2,x_3)&&...\text{多変数関数のテイラー展開より} \\ \\ &= \frac{\partial u_i}{\partial x_1}\delta x_1+\frac{\partial u_i}{\partial x_2}\delta x_2+\frac{\partial u_i}{\partial x_3}\delta x_3+\ldots&\\ \\ &\simeq \frac{\partial u_i}{\partial x_j}\delta x_j&&...\text{p.15下の縮約の規約より}\\ \\ \end{align*}


6. 式(2･28)の導出
▼ 詳細を開く...2025/ 8/16作成
式(2･24)の中で対称の部分を抜き出して計算する。 \begin{align*} \delta u_1 &= \frac{1}{2}\left(\frac{\partial u_1}{\partial x_j}+\frac{\partial u_j}{\partial x_1}\right)\delta x_j \\ \\ &= \frac{1}{2}\left(\frac{\partial u_1}{\partial x_1}+\frac{\partial u_1}{\partial x_1}\right)\delta x_1+\frac{1}{2}\left(\frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}\right)\delta x_2+\frac{1}{2}\left(\frac{\partial u_1}{\partial x_3}+\frac{\partial u_3}{\partial x_1}\right)\delta x_3&&...\text{p.15下の縮約の規約より} \\ \\ &= \frac{1}{2}\left(e_{11}+e_{11}\right)\delta x_1+\frac{1}{2}\left(e_{12}+e_{21}\right)\delta x_2+\frac{1}{2}\left(e_{13}+e_{31}\right)\delta x_3&\\ \\ &= \frac{1}{2}\left(e_{12}+e_{21}\right)\delta x_2&&...e_{12}=e_{21}\text{以外で値を持たないため}\\ \\ &= e_{12}\delta x_2&&...e_{21}=e_{12}\\ \\ \end{align*} と導出できる。また、 \begin{align*} \delta u_2 &= \frac{1}{2}\left(\frac{\partial u_2}{\partial x_j}+\frac{\partial u_j}{\partial x_2}\right)\delta x_j \\ \\ &= \frac{1}{2}\left(\frac{\partial u_2}{\partial x_1}+\frac{\partial u_1}{\partial x_2}\right)\delta x_1+\frac{1}{2}\left(\frac{\partial u_2}{\partial x_2}+\frac{\partial u_2}{\partial x_2}\right)\delta x_2+\frac{1}{2}\left(\frac{\partial u_2}{\partial x_3}+\frac{\partial u_3}{\partial x_2}\right)\delta x_3&&...\text{p.15下の縮約の規約より} \\ \\ &= \frac{1}{2}\left(e_{21}+e_{12}\right)\delta x_1+\frac{1}{2}\left(e_{22}+e_{22}\right)\delta x_2+\frac{1}{2}\left(e_{23}+e_{32}\right)\delta x_3&\\ \\ &= \frac{1}{2}\left(e_{12}+e_{21}\right)\delta x_1&&...e_{12}=e_{21}\text{以外で値を持たないため}\\ \\ &= e_{12}\delta x_1&&...e_{21}=e_{12}\\ \\ \end{align*} と \begin{align*} \delta u_3 &= \frac{1}{2}\left(\frac{\partial u_3}{\partial x_j}+\frac{\partial u_j}{\partial x_3}\right)\delta x_j \\ \\ &= \frac{1}{2}\left(\frac{\partial u_3}{\partial x_1}+\frac{\partial u_1}{\partial x_3}\right)\delta x_1+\frac{1}{2}\left(\frac{\partial u_3}{\partial x_2}+\frac{\partial u_2}{\partial x_3}\right)\delta x_2+\frac{1}{2}\left(\frac{\partial u_3}{\partial x_3}+\frac{\partial u_3}{\partial x_3}\right)\delta x_3&&...\text{p.15下の縮約の規約より} \\ \\ &= \frac{1}{2}\left(e_{31}+e_{13}\right)\delta x_1+\frac{1}{2}\left(e_{32}+e_{23}\right)\delta x_2+\frac{1}{2}\left(e_{33}+e_{33}\right)\delta x_3&\\ \\ &= 0&&...e_{12}=e_{21}\text{以外で値を持たないため}\\ \\ \end{align*} も導出できる。


7. p.25中段の\(\omega_l\)の導出
▼ 詳細を開く...2025/ 8/16作成
式(2･32)の右辺に\(\epsilon_{jil}\)をかけると \begin{align*} \epsilon_{jik}\epsilon_{jil}w_k &= (\delta_{ii}\delta_{kl}-\delta_{il}\delta_{ki})w_k&&...\text{p.25中段より} \\ \\ &= \left((\delta_{11}+\delta_{22}+\delta_{33})\delta_{kl}-(\delta_{1l}\delta_{k1}+\delta_{2l}\delta_{k2}+\delta_{3l}\delta_{k3})\right)w_k&&...\text{p.15下の縮約のルールより} \\ \\ &= \left(3\delta_{kl}-(\delta_{1l}\delta_{k1}+\delta_{2l}\delta_{k2}+\delta_{3l}\delta_{k3})\right)w_k&&...\text{クロネッカーのデルタより} \\ \\ &= \left(3\delta_{1l}-(\delta_{1l}\delta_{11}+\delta_{2l}\delta_{12}+\delta_{3l}\delta_{13})\right)w_1+\left(3\delta_{2l}-(\delta_{1l}\delta_{21}+\delta_{2l}\delta_{22}+\delta_{3l}\delta_{23})\right)w_2+\left(3\delta_{3l}-(\delta_{1l}\delta_{31}+\delta_{2l}\delta_{32}+\delta_{3l}\delta_{33})\right)w_3&&...\text{p.15下の縮約のルールより} \\ \\ &= \left(3\delta_{1l}-\delta_{1l}\right)w_1+\left(3\delta_{2l}-\delta_{2l}\right)w_2+\left(3\delta_{3l}-\delta_{3l}\right)w_3&&...\text{クロネッカーのデルタより} \\ \\ &= 2\delta_{1l}w_1+2\delta_{2l}w_2+2\delta_{3l}w_3 \\ \\ \end{align*} が得られる。左辺も同様にすると \begin{align*} \left(\frac{\partial u_i}{\partial x_j}-\frac{\partial u_j}{\partial x_i}\right)\epsilon_{jil} &= \left(\frac{\partial u_1}{\partial x_j}-\frac{\partial u_j}{\partial x_1}\right)\epsilon_{j1l}+\left(\frac{\partial u_2}{\partial x_j}-\frac{\partial u_j}{\partial x_2}\right)\epsilon_{j2l}+\left(\frac{\partial u_3}{\partial x_j}-\frac{\partial u_j}{\partial x_3}\right)\epsilon_{j3l} \\ \\ &= \left(\frac{\partial u_1}{\partial x_1}-\frac{\partial u_1}{\partial x_1}\right)\epsilon_{11l}+\left(\frac{\partial u_2}{\partial x_1}-\frac{\partial u_1}{\partial x_2}\right)\epsilon_{12l}+\left(\frac{\partial u_3}{\partial x_1}-\frac{\partial u_1}{\partial x_3}\right)\epsilon_{13l} \\ &+ \left(\frac{\partial u_1}{\partial x_2}-\frac{\partial u_2}{\partial x_1}\right)\epsilon_{21l}+\left(\frac{\partial u_2}{\partial x_2}-\frac{\partial u_2}{\partial x_2}\right)\epsilon_{22l}+\left(\frac{\partial u_3}{\partial x_2}-\frac{\partial u_2}{\partial x_3}\right)\epsilon_{23l} \\ &+ \left(\frac{\partial u_1}{\partial x_3}-\frac{\partial u_3}{\partial x_1}\right)\epsilon_{31l}+\left(\frac{\partial u_2}{\partial x_3}-\frac{\partial u_3}{\partial x_2}\right)\epsilon_{32l}+\left(\frac{\partial u_3}{\partial x_3}-\frac{\partial u_3}{\partial x_3}\right)\epsilon_{33l} \\ \\ &= 0+\left(\frac{\partial u_2}{\partial x_1}-\frac{\partial u_1}{\partial x_2}\right)\epsilon_{12l}+\left(\frac{\partial u_3}{\partial x_1}-\frac{\partial u_1}{\partial x_3}\right)\epsilon_{13l} \\ &+ \left(\frac{\partial u_1}{\partial x_2}-\frac{\partial u_2}{\partial x_1}\right)\epsilon_{21l}+0+\left(\frac{\partial u_3}{\partial x_2}-\frac{\partial u_2}{\partial x_3}\right)\epsilon_{23l} \\ &+ \left(\frac{\partial u_1}{\partial x_3}-\frac{\partial u_3}{\partial x_1}\right)\epsilon_{31l}+\left(\frac{\partial u_2}{\partial x_3}-\frac{\partial u_3}{\partial x_2}\right)\epsilon_{32l}+0 \\ \\ &= \left(\frac{\partial u_3}{\partial x_2}-\frac{\partial u_2}{\partial x_3}\right)\epsilon_{23l}+ \left(\frac{\partial u_2}{\partial x_3}-\frac{\partial u_3}{\partial x_2}\right)\epsilon_{32l}\\ &+ \left(\frac{\partial u_1}{\partial x_3}-\frac{\partial u_3}{\partial x_1}\right)\epsilon_{31l}+\left(\frac{\partial u_3}{\partial x_1}-\frac{\partial u_1}{\partial x_3}\right)\epsilon_{13l} \\ &+ \left(\frac{\partial u_2}{\partial x_1}-\frac{\partial u_1}{\partial x_2}\right)\epsilon_{12l}+\left(\frac{\partial u_1}{\partial x_2}-\frac{\partial u_2}{\partial x_1}\right)\epsilon_{21l} \\ \\ \end{align*} が得られる。Eddingtonの記号の性質を用いて、\(l=1,2,3\)のそれぞれの時に\(0\)にならずに残る項を比較すると \begin{align*} &\left\{ \begin{array}{l} 2\omega_1\delta_{11}&=\left(\frac{\partial u_3}{\partial x_2}-\frac{\partial u_2}{\partial x_3}\right)\epsilon_{231}+ \left(\frac{\partial u_2}{\partial x_3}-\frac{\partial u_3}{\partial x_2}\right)\epsilon_{321} \\ \\ 2\omega_2\delta_{22}&=\left(\frac{\partial u_1}{\partial x_3}-\frac{\partial u_3}{\partial x_1}\right)\epsilon_{312}+\left(\frac{\partial u_3}{\partial x_1}-\frac{\partial u_1}{\partial x_3}\right)\epsilon_{132} \\ \\ 2\omega_3\delta_{33}&=\left(\frac{\partial u_2}{\partial x_1}-\frac{\partial u_1}{\partial x_2}\right)\epsilon_{123}+\left(\frac{\partial u_1}{\partial x_2}-\frac{\partial u_2}{\partial x_1}\right)\epsilon_{213} \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow \left\{ \begin{array}{l} 2\omega_1&=\left(\frac{\partial u_3}{\partial x_2}-\frac{\partial u_2}{\partial x_3}\right)-\left(\frac{\partial u_2}{\partial x_3}-\frac{\partial u_3}{\partial x_2}\right) \\ \\ 2\omega_2&=\left(\frac{\partial u_1}{\partial x_3}-\frac{\partial u_3}{\partial x_1}\right)-\left(\frac{\partial u_3}{\partial x_1}-\frac{\partial u_1}{\partial x_3}\right) \\ \\ 2\omega_3&=\left(\frac{\partial u_2}{\partial x_1}-\frac{\partial u_1}{\partial x_2}\right)-\left(\frac{\partial u_1}{\partial x_2}-\frac{\partial u_2}{\partial x_1}\right) \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow \left\{ \begin{array}{l} \omega_1&=\left(\frac{\partial u_3}{\partial x_2}-\frac{\partial u_2}{\partial x_3}\right)&=\epsilon_{ji1}\frac{\partial u_i}{\partial x_j} \\ \\ \omega_2&=\left(\frac{\partial u_1}{\partial x_3}-\frac{\partial u_3}{\partial x_1}\right)&=\epsilon_{ji2}\frac{\partial u_i}{\partial x_j} \\ \\ \omega_3&=\left(\frac{\partial u_2}{\partial x_1}-\frac{\partial u_1}{\partial x_2}\right)&=\epsilon_{ji3}\frac{\partial u_i}{\partial x_j} \\ \\ \end{array} \right. \\ \\ \end{align*} が得られる。


8. 式(2･35)の導出
▼ 詳細を開く...2025/ 8/16作成
式(2･24)の中で反対称の部分を抜き出して計算する。 \begin{align*} \delta u_1 &= \frac{1}{2}\left(\frac{\partial u_1}{\partial x_j}-\frac{\partial u_j}{\partial x_1}\right)\delta x_j \\ \\ &= \frac{1}{2}\left(\frac{\partial u_1}{\partial x_1}-\frac{\partial u_1}{\partial x_1}\right)\delta x_1+\frac{1}{2}\left(\frac{\partial u_1}{\partial x_2}-\frac{\partial u_2}{\partial x_1}\right)\delta x_2+\frac{1}{2}\left(\frac{\partial u_1}{\partial x_3}-\frac{\partial u_3}{\partial x_1}\right)\delta x_3&&...\text{p.15下の縮約の規約より} \\ \\ &= 0-\frac{1}{2}\omega_3\delta x_2+\frac{1}{2}\omega_2\delta x_3&\\ \\ &= -\frac{1}{2}\omega_3\delta x_2&&...\omega_3\text{以外で値を持たないため}\\ \\ \end{align*} と導出できる。また、 \begin{align*} \delta u_2 &= \frac{1}{2}\left(\frac{\partial u_2}{\partial x_j}-\frac{\partial u_j}{\partial x_2}\right)\delta x_j \\ \\ &= \frac{1}{2}\left(\frac{\partial u_2}{\partial x_1}-\frac{\partial u_1}{\partial x_2}\right)\delta x_1+\frac{1}{2}\left(\frac{\partial u_2}{\partial x_2}-\frac{\partial u_2}{\partial x_2}\right)\delta x_2+\frac{1}{2}\left(\frac{\partial u_2}{\partial x_3}-\frac{\partial u_3}{\partial x_2}\right)\delta x_3&&...\text{p.15下の縮約の規約より} \\ \\ &= \frac{1}{2}\omega_3\delta x_1+0-\frac{1}{2}\omega_1\delta x_3&\\ \\ &= \frac{1}{2}\omega_3\delta x_1&&...\omega_3\text{以外で値を持たないため}\\ \\ \end{align*} と \begin{align*} \delta u_3 &= \frac{1}{2}\left(\frac{\partial u_3}{\partial x_j}-\frac{\partial u_j}{\partial x_3}\right)\delta x_j \\ \\ &= \frac{1}{2}\left(\frac{\partial u_3}{\partial x_1}-\frac{\partial u_1}{\partial x_3}\right)\delta x_1+\frac{1}{2}\left(\frac{\partial u_3}{\partial x_2}-\frac{\partial u_2}{\partial x_3}\right)\delta x_2+\frac{1}{2}\left(\frac{\partial u_3}{\partial x_3}-\frac{\partial u_3}{\partial x_3}\right)\delta x_3&&...\text{p.15下の縮約の規約より} \\ \\ &= \frac{1}{2}\omega_2\delta x_1+\frac{1}{2}\omega_1\delta x_2+0&\\ \\ &= 0&&...\omega_3\text{以外で値を持たないため}\\ \\ \end{align*} も導出できる。


9. 式(2･35)の導出
▼ 詳細を開く...2025/ 8/16作成
式(2･13)の導出と同様。


10. 式(2･50)の式変形の導出
▼ 詳細を開く...2025/ 8/16作成
\begin{align*} \boldsymbol{p}(\boldsymbol{n})\delta S+\boldsymbol{p}(-\boldsymbol{e}_j)\delta S_j &= \boldsymbol{p}(\boldsymbol{n})\delta S+\boldsymbol{p}(-\boldsymbol{e}_1)\delta S_1+\boldsymbol{p}(-\boldsymbol{e}_2)\delta S_2+\boldsymbol{p}(-\boldsymbol{e}_3)\delta S_3&&...\text{p.15下の縮約の記法より} \\ \\ &= \boldsymbol{p}(\boldsymbol{n})\delta S+\boldsymbol{p}(-\boldsymbol{e}_1)\{(\boldsymbol{n}\cdot\boldsymbol{e}_1)\delta S\}+\boldsymbol{p}(-\boldsymbol{e}_2)\{(\boldsymbol{n}\cdot\boldsymbol{e}_2)\delta S\}+\boldsymbol{p}(-\boldsymbol{e}_3)\{(\boldsymbol{n}\cdot\boldsymbol{e}_3)\delta S\}&&...(1) \\ \\ &= \{\boldsymbol{p}(\boldsymbol{n})\delta S+\boldsymbol{p}(-\boldsymbol{e}_1)(\boldsymbol{n}\cdot\boldsymbol{e}_1)+\boldsymbol{p}(-\boldsymbol{e}_2)(\boldsymbol{n}\cdot\boldsymbol{e}_2)+\boldsymbol{p}(-\boldsymbol{e}_3)(\boldsymbol{n}\cdot\boldsymbol{e}_3)\}\delta S&\\ \\ &= \{\boldsymbol{p}(\boldsymbol{n})\delta S+\boldsymbol{p}(-\boldsymbol{e}_j)(\boldsymbol{n}\cdot\boldsymbol{e}_j)\}\delta S&&...\text{p.15下の縮約の記法より} \\ \\\\ \\ \end{align*} と導出できる。

(1)では、\(\boldsymbol{e}_1,\boldsymbol{e}_2,\boldsymbol{e}_3\)がそれぞれ垂直であることから、\(\boldsymbol{n}\)から\(\boldsymbol{e}_j\)への射影が\(\delta S_j\)と\(\delta S\)の比になること\((\boldsymbol{e}_j\cdot \boldsymbol{n}=\delta S_i/\delta S)\)を用いた。