任意の面に働く表面力vectorの成分を求める

$ \pmb{t}:=\pmb{\sigma}\cdot\pmb{n}

単位vectorだとする

面に垂直な成分

$ t_\top:=\pmb{t}\cdot\pmb{n}=\pmb{\sigma}:\pmb{n}\pmb{n}=\sum_i\sigma_i(\pmb{n}\cdot\pmb{p}_i)^2

面に平行な成分

$ t_{//}:=|\pmb{t}-\pmb{t}\cdot\pmb{n}\pmb{n}|=|(\pmb{\sigma}-(\pmb{t}\cdot\pmb{n})\pmb{I})\cdot\pmb{n}|

$ =\left|\sum_i\left(\sigma_i-\sum_j\sigma_j(\pmb{n}\cdot\pmb{p}_j)^2\right)\pmb{p}_i\pmb{p}_i\cdot\pmb{n}\right|

$ =\left|\sum_{i,j}\left(\sigma_i-\sigma_j(\pmb{n}\cdot\pmb{p}_j)^2\right)(\pmb{n}\cdot\pmb{p}_i)\pmb{p}_i\right|

$ =\sqrt{\sum_{i,j}\left(\sigma_i-\sigma_j(\pmb{n}\cdot\pmb{p}_j)^2\right)^2(\pmb{n}\cdot\pmb{p}_i)^2}

$ =\sqrt{\sum_{i,j}\left(\sigma_i^2-2\sigma_i\sigma_j(\pmb{n}\cdot\pmb{p}_j)^2+\sigma_j^2(\pmb{n}\cdot\pmb{p}_j)^4\right)(\pmb{n}\cdot\pmb{p}_i)^2}

どうすりゃええねん

少し前の式に戻ってから展開し直す

$ =\left|\sum_{i,j}\left(\sigma_i-\sigma_j(\pmb{n}\cdot\pmb{p}_j)^2\right)(\pmb{n}\cdot\pmb{p}_i)\pmb{p}_i\right|

$ =\left|\sum_{i,j}\left(\sigma_i(\pmb{n}\cdot\pmb{p}_j)^2-\sigma_j(\pmb{n}\cdot\pmb{p}_j)^2\right)(\pmb{n}\cdot\pmb{p}_i)\pmb{p}_i\right|

$ =\left|\sum_{i,j}(\sigma_i-\sigma_j)(\pmb{n}\cdot\pmb{p}_j)^2(\pmb{n}\cdot\pmb{p}_i)\pmb{p}_i\right|

$ =\sqrt{\sum_{i,j}(\sigma_i-\sigma_j)^2(\pmb{n}\cdot\pmb{p}_j)^4(\pmb{n}\cdot\pmb{p}_i)^2}

3次元の場合

$ =\sqrt{(\sigma_0-\sigma_1)^2(\pmb{n}\cdot\pmb{p}_1)^4(\pmb{n}\cdot\pmb{p}_0)^2+(\sigma_1-\sigma_0)^2(\pmb{n}\cdot\pmb{p}_0)^4(\pmb{n}\cdot\pmb{p}_1)^2+(\sigma_1-\sigma_2)^2(\pmb{n}\cdot\pmb{p}_1)^4(\pmb{n}\cdot\pmb{p}_2)^2+(\sigma_1-\sigma_2)^2(\pmb{n}\cdot\pmb{p}_2)^4(\pmb{n}\cdot\pmb{p}_1)^2+(\sigma_2-\sigma_0)^2(\pmb{n}\cdot\pmb{p}_0)^4(\pmb{n}\cdot\pmb{p}_2)^2+(\sigma_0-\sigma_2)^2(\pmb{n}\cdot\pmb{p}_2)^4(\pmb{n}\cdot\pmb{p}_0)^2}

$ =\sqrt{(\sigma_0-\sigma_1)^2(\pmb{n}\cdot\pmb{p}_1)^2(\pmb{n}\cdot\pmb{p}_0)^2((\pmb{n}\cdot\pmb{p}_0)^2+(\pmb{n}\cdot\pmb{p}_1)^2)+(\sigma_1-\sigma_2)^2(\pmb{n}\cdot\pmb{p}_1)^2(\pmb{n}\cdot\pmb{p}_2)^2((\pmb{n}\cdot\pmb{p}_2)^2+(\pmb{n}\cdot\pmb{p}_1)^2)+(\sigma_2-\sigma_0)^2(\pmb{n}\cdot\pmb{p}_0)^2(\pmb{n}\cdot\pmb{p}_2)^2((\pmb{n}\cdot\pmb{p}_2)^2+(\pmb{n}\cdot\pmb{p}_0)^2)}

$ =\sqrt{(\sigma_0-\sigma_1)^2(\pmb{n}\cdot\pmb{p}_0)^2(\pmb{n}\cdot\pmb{p}_1)^2(|\pmb{n}|^2-(\pmb{n}\cdot\pmb{p}_2)^2)+(\sigma_1-\sigma_2)^2(\pmb{n}\cdot\pmb{p}_1)^2(\pmb{n}\cdot\pmb{p}_2)^2(|\pmb{n}|^2-(\pmb{n}\cdot\pmb{p}_0)^2)+(\sigma_2-\sigma_0)^2(\pmb{n}\cdot\pmb{p}_2)^2(\pmb{n}\cdot\pmb{p}_0)^2(|\pmb{n}|^2-(\pmb{n}\cdot\pmb{p}_1)^2)}

これ以上展開できそうにないな……

$ =\sqrt{(\sigma_0-\sigma_1)^2(\pmb{n}\cdot\pmb{p}_0)^2(1-(\pmb{n}\cdot\pmb{p}_0)^2-(\pmb{n}\cdot\pmb{p}_2)^2)(1-(\pmb{n}\cdot\pmb{p}_2)^2)}\\\overline{+((\sigma_0-\sigma_1)+(\sigma_2-\sigma_0))^2(1-(\pmb{n}\cdot\pmb{p}_0)^2-(\pmb{n}\cdot\pmb{p}_2)^2)(\pmb{n}\cdot\pmb{p}_2)^2(1-(\pmb{n}\cdot\pmb{p}_0)^2)}\\\overline{+(\sigma_2-\sigma_0)^2(\pmb{n}\cdot\pmb{p}_2)^2(\pmb{n}\cdot\pmb{p}_0)^2((\pmb{n}\cdot\pmb{p}_0)^2+(\pmb{n}\cdot\pmb{p}_2)^2)}

$ =\sqrt{\sigma_+^2n_0^2(1-n_0^2-n_2^2)(1-n_2^2)+(\sigma_++\sigma_-)^2n_2^2(1-n_0^2-n_2^2)(1-n_0^2)+\sigma_-^2n_2^2n_0^2(n_2^2+n_0^2)}

$ =\sqrt{\sigma_+^2n_0^2(1-2n_2^2+n_2^4-n_0^2+n_0^2n_2^2)+(\sigma_++\sigma_-)^2n_2^2(1-2n_0^2+n_0^4-n_2^2+n_0^2n_2^2)+\sigma_-^2n_2^4n_0^2+\sigma_-^2n_2^2n_0^4}

$ =\sqrt{\sigma_+^2(n_0^2-2n_2^2n_0^2+n_2^4n_0^2-n_0^4+n_0^4n_2^2+n_2^2-2n_0^2n_2^2+n_2^2n_0^4-n_2^4+n_0^2n_2^4)+(\sigma_++\sigma_-)^2n_2^2(1-2n_0^2+n_0^4-n_2^2+n_0^2n_2^2)+\sigma_-^2n_2^4n_0^2+\sigma_-^2n_2^2n_0^4}

2次元の場合

$ =\sqrt{(\sigma_0-\sigma_1)^2(\pmb{n}\cdot\pmb{p}_0)^2(\pmb{n}\cdot\pmb{p}_1)^2(|\pmb{n}|^2-0)}

$ =|(\sigma_0-\sigma_1)(\pmb{n}\cdot\pmb{p}_0)(\pmb{n}\cdot\pmb{p}_1)|

2024-01-07 16:48:11 非ッッッッ常に複雑な式になっているが、本来ならもっと簡単になるはずなんだけどな……

$ t_{//}=\bm\sigma:\bm n\bm n

$ = n_0^2\sigma_{00}+2n_0n_1\sigma_{01}+n_1^2\sigma_{11}

$ \bm t_\top=\bm\sigma\cdot\bm n-\bm\sigma:\bm n\bm n\bm n

$ = (\bm\sigma\cdot\bm n)\cdot(\bm I-\bm n\bm n)

$ = n_i\sigma_{ij}(\delta_{jk}-n_jn_k)

$ = (n_0\sigma_{00}+n_1\sigma_{10})(\delta_{0k}-n_0n_k)+(n_0\sigma_{01}+n_1\sigma_{11})(\delta_{1k}-n_1n_k)

$ = -n_0^2\sigma_{00}n_k+n_0\delta_{0k}\sigma_{00}+\sigma_{01}(n_1\delta_{0k}+n_0\delta_{1k}-2n_0n_1n_k)+n_1\delta_{1k}\sigma_{11}-n_1^2\sigma_{11}n_k

$ = n_0\delta_{0k}\sigma_{00}+\sigma_{01}(n_1\delta_{0k}+n_0\delta_{1k})+n_1\delta_{1k}\sigma_{11}-n_kt_\top

$ =(n_0\sigma_{00}+n_1\sigma_{10})\delta_{0k}+(n_0\sigma_{01}+n_1\sigma_{11})\delta_{1k}-n_kt_\top

$ = \bm n\cdot\bm\sigma-t_\top\bm n

$ = \bm n\cdot(\bm\sigma-t_\top\bm I)

いやこれは当然の結果だ

$ |\bm t_\top|^2=\bm n\cdot(\bm\sigma-t_\top\bm I)\cdot(\bm\sigma-t_\top\bm I)\cdot\bm n

$ = \bm n\cdot(\bm\sigma^2-2t_\top\bm\sigma+{t_\top}^2)\cdot\bm n

$ =\bm\sigma^2:\bm n\bm n-2{t_\top}^2+{t_\top}^2

$ = \bm\sigma^2:\bm n\bm n-{t_\top}^2

$ = n_0^2(\sigma_{00}^2+\sigma_{01}^2)+2n_0n_1\sigma_{01}(\sigma_{00}+\sigma_{11})+n_1^2(\sigma_{11}^2+\sigma_{01}^2)-{t_\top}^2

$ n_0=\cos\theta,n_1=\sin\theta

$ p:=\frac12(\sigma_{00}+\sigma_{11}),q:=\frac12(\sigma_{00}-\sigma_{11})

$ t_{//}=p+q\cos2\theta

$ |\bm t_\top|^2=(\cos\theta)^2(p+q)^2+(\sin\theta)^2(p-q)^2-{t_\top}^2

$ =p^2+q^2+2pq\cos2\theta-p^2-2pq\cos2\theta-q^2(\cos2\theta)^2

$ = q^2-q^2(\cos2\theta)^2

$ =(q\sin2\theta)^2

同様に3次元対称tensorで調べる

$ t_{//}=\bm\sigma:\bm n\bm n

$ = n_0^2\sigma_{00}+2n_0n_1\sigma_{01}+n_1^2\sigma_{11}+2n_1n_2\sigma_{12}+n_2^2\sigma_{22}+2n_2n_0\sigma_{20}

$ \bm t_\top=\bm\sigma\cdot\bm n-\bm\sigma:\bm n\bm n\bm n

$ = (\bm\sigma\cdot\bm n)\cdot(\bm I-\bm n\bm n)

$ = \bm n\cdot(\bm\sigma-t_\top\bm I)

$ |\bm t_\top|^2=\bm\sigma^2:\bm n\bm n-{t_\top}^2

$ =\sigma_{ik}\sigma_{kj}n_in_j-{t_\top}^2

$ =n_0^2(\sigma_{00}^2+\sigma_{01}^2+\sigma_{02}^2)+2n_0n_1(\sigma_{00}\sigma_{01}+\sigma_{01}\sigma_{11}+\sigma_{02}\sigma_{21})+n_1^2(\sigma_{11}^2+\sigma_{12}^2+\sigma_{10}^2)+2n_1n_2(\sigma_{10}\sigma_{02}+\sigma_{11}\sigma_{12}+\sigma_{12}\sigma_{22})+n_2^2(\sigma_{22}^2+\sigma_{20}^2+\sigma_{21}^2)+2n_2n_0(\sigma_{20}\sigma_{00}+\sigma_{21}\sigma_{10}+\sigma_{22}\sigma_{20})-{t_\top}^2

$ t_{//}=n_0^2\sigma_{00}+n_1^2\sigma_{11}+n_2^2\sigma_{22}

$ |\bm t_\top|^2=n_0^2\sigma_{00}^2+n_1^2\sigma_{11}^2+n_2^2\sigma_{22}^2-{t_{//}}^2

$ n_0=\sin\theta\cos\phi,n_1=\sin\theta\sin\phi,n_2=\cos\theta

$ p_{ij}:=\frac12(\sigma_{ii}+\sigma_{jj}),q_{ij}:=\frac12(\sigma_{ii}-\sigma_{jj})

$ t_{//}=n_0^2\sigma_{00}+n_1^2\sigma_{11}+n_2^2\sigma_{22}

$ =n_0^2\left(\sigma_{00}-\frac13(\sigma_{00}+\sigma_{11}+\sigma_{22})\right)+n_1^2\left(\sigma_{11}-\frac13(\sigma_{00}+\sigma_{11}+\sigma_{22})\right)+n_2^2\left(\sigma_{22}-\frac13(\sigma_{00}+\sigma_{11}+\sigma_{22})\right)+3\cdot\frac13(\sigma_{00}+\sigma_{11}+\sigma_{22})

$ =\frac23n_0^2(q_{01}-q_{20})+\frac23n_1^2(q_{12}-q_{01})+\frac23n_2^2(q_{20}-q_{12})+p_{01}+p_{12}+p_{20}

$ \because\sigma_{00}-\frac13(\sigma_{00}+\sigma_{11}+\sigma_{22})=\frac13(\sigma_{00}-\sigma_{11})+\frac13(\sigma_{00}-\sigma_{22})=\frac23q_{01}-\frac23q_{20}

$ |\bm t_\top|^2=n_0^2\sigma_{00}^2+n_1^2\sigma_{11}^2+n_2^2\sigma_{22}^2-{t_\top}^2

$ = \left(\frac23n_0(q_{01}-q_{20})\right)^2+\left(\frac23n_1(q_{12}-q_{01})\right)^2+\left(\frac23n_2(q_{20}-q_{12})\right)^2+\frac23(n_0^2\sigma_{00}+n_1^2\sigma_{11}+n_2^2\sigma_{22})(p_{01}+p_{12}+p_{20})-\frac19(p_{01}+p_{12}+p_{20})^2-{t_\top}^2

$ \because\frac49(q_{01}-q_{20})^2=\left(\sigma_{00}-\frac13(\sigma_{00}+\sigma_{11}+\sigma_{22})\right)^2=\sigma_{00}^2-\frac23\sigma_{00}(p_{01}+p_{12}+p_{20})+\frac19(p_{01}+p_{12}+p_{20})^2

$ = \left(\frac23n_0(q_{01}-q_{20})\right)^2+\left(\frac23n_1(q_{12}-q_{01})\right)^2+\left(\frac23n_2(q_{20}-q_{12})\right)^2+\frac23t_{//}(p_{01}+p_{12}+p_{20})-\frac19(p_{01}+p_{12}+p_{20})^2-{t_\top}^2

$ = \left(\frac23n_0(q_{01}-q_{20})\right)^2+\left(\frac23n_1(q_{12}-q_{01})\right)^2+\left(\frac23n_2(q_{20}-q_{12})\right)^2-\left({t_{//}}-\frac13(p_{01}+p_{12}+p_{20})\right)^2

$ t_{//}=\frac23n_0^2(q_{01}-q_{20})+\frac23n_1^2(q_{12}-q_{01})+\frac23n_2^2(q_{20}-q_{12})+p_{01}+p_{12}+p_{20}

$ |\bm t_\top|^2= \left(\frac23n_0(q_{01}-q_{20})\right)^2+\left(\frac23n_1(q_{12}-q_{01})\right)^2+\left(\frac23n_2(q_{20}-q_{12})\right)^2-\left({t_{//}}-\frac13(p_{01}+p_{12}+p_{20})\right)^2

$ 1=n_0^2+n_1^2+n_2^2

$ \iff\begin{pmatrix}\frac23(q_{01}-q_{20})&\frac23(q_{12}-q_{01})&\frac23(q_{20}-q_{12})\\\left(\frac23(q_{01}-q_{20})\right)^2&\left(\frac23(q_{12}-q_{01})\right)^2&\left(\frac23(q_{20}-q_{12})\right)^2\\1&1&1\end{pmatrix}\begin{pmatrix}n_0^2\\n_1^2\\n_2^2\end{pmatrix}=\begin{pmatrix}-3\sigma_m\\(t_{//}-\sigma_m)^2\\1\end{pmatrix}

$ \iff\begin{pmatrix}\frac23(q_{01}-q_{20})&\frac23(q_{12}-q_{01})&\frac23(q_{20}-q_{12})\\0&\frac43q_{10}(q_{12}-q_{01})&\frac43q_{20}(q_{20}-q_{12})\\1&1&1\end{pmatrix}\begin{pmatrix}n_0^2\\n_1^2\\n_2^2\end{pmatrix}=\begin{pmatrix}-3\sigma_m\\(t_{//}-\sigma_m)^2+2\sigma_m(q_{01}-q_{20})\\1\end{pmatrix}

$ \because\frac23(q_{12}-q_{01})-\frac23(q_{01}-q_{20})=\frac23(q_{10}-2q_{01})=2q_{10}

$ \because\frac23(q_{20}-q_{12})-\frac23(q_{01}-q_{20})=\frac23(2q_{20}-q_{02})=2q_{20}

$ \iff\begin{pmatrix}0&2q_{10}&2q_{20}\\0&\frac43q_{10}(q_{12}-q_{01})&\frac43q_{20}(q_{20}-q_{12})\\1&1&1\end{pmatrix}\begin{pmatrix}n_0^2\\n_1^2\\n_2^2\end{pmatrix}=\begin{pmatrix}-3\sigma_m-\frac23(q_{01}-q_{20})\\(t_{//}-\sigma_m)^2+2\sigma_m(q_{01}-q_{20})\\1\end{pmatrix}

$ \iff\begin{pmatrix}0&2q_{10}&2q_{20}\\0&0&4q_{20}q_{21}\\1&1&1\end{pmatrix}\begin{pmatrix}n_0^2\\n_1^2\\n_2^2\end{pmatrix}=\begin{pmatrix}-3\sigma_m-\frac23(q_{01}-q_{20})+\frac23(q_{12}-q_{01})\left(3\sigma_m+\frac23(q_{01}-q_{20})\right)\\(t_{//}-\sigma_m)^2+2\sigma_m(q_{01}-q_{20})\\1\end{pmatrix}

$ =\begin{pmatrix}\left(\frac23(q_{12}-q_{01})-1\right)\left(3\sigma_m+\frac23(q_{01}-q_{20})\right)\\(t_{//}-\sigma_m)^2+2\sigma_m(q_{01}-q_{20})\\1\end{pmatrix}

$ \because\frac23(q_{20}-q_{12})-\frac23(q_{12}-q_{01})=\frac23(q_{21}-2q_{12})=2q_{21}

$ \iff\begin{pmatrix}0&1&\frac{q_{20}}{q_{10}}\\0&0&4q_{20}q_{21}\\1&1&1\end{pmatrix}\begin{pmatrix}n_0^2\\n_1^2\\n_2^2\end{pmatrix}=\begin{pmatrix}\frac1{q_{10}}\left(\frac23(q_{12}-q_{01})-1\right)\left(3\sigma_m+\frac23(q_{01}-q_{20})\right)\\(t_{//}-\sigma_m)^2+2\sigma_m(q_{01}-q_{20})\\1\end{pmatrix}

$ \begin{pmatrix}a^2&b^2&c^2\\a&b&c\\1&1&1\end{pmatrix}[\bm A]=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}

$ \iff \begin{pmatrix}0&b^2-a^2&c^2-a^2\\0&b-a&c-a\\1&1&1\end{pmatrix}[\bm A]=\begin{pmatrix}1&0&-a^2\\0&1&-a\\0&0&1\end{pmatrix}

$ \iff \begin{pmatrix}0&0&(c+a-(b+a))(c-a)\\0&b-a&c-a\\1&1&1\end{pmatrix}[\bm A]=\begin{pmatrix}1&-(b+a)&-a^2+(b+a)a\\0&1&-a\\0&0&1\end{pmatrix}

$ \iff \begin{pmatrix}0&0&(c-b)(c-a)\\0&1&\frac{c-a}{b-a}\\1&1&1\end{pmatrix}[\bm A]=\begin{pmatrix}1&-(b+a)&ab\\0&\frac{1}{b-a}&\frac{-a}{b-a}\\0&0&1\end{pmatrix}

$ \iff \begin{pmatrix}0&0&(c-b)(c-a)\\0&1&\frac{c-a}{b-a}\\1&0&\frac{b-c}{b-a}\end{pmatrix}[\bm A]=\begin{pmatrix}1&-(b+a)&ab\\0&\frac{1}{b-a}&\frac{-a}{b-a}\\0&-\frac{1}{b-a}&\frac{b}{b-a}\end{pmatrix}

$ \iff \begin{pmatrix}0&0&1\\0&1&\frac{c-a}{b-a}\\1&0&\frac{b-c}{b-a}\end{pmatrix}[\bm A]=\begin{pmatrix}\frac1{c-b}\frac1{c-a}&-\frac1{c-b}\frac{b+a}{c-a}&\frac b{c-b}\frac a{c-a}\\0&\frac{1}{b-a}&\frac{-a}{b-a}\\0&-\frac{1}{b-a}&\frac{b}{b-a}\end{pmatrix}

$ \iff \begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}[\bm A]=\begin{pmatrix}\frac1{c-b}\frac1{c-a}&-\frac1{c-b}\frac{b+a}{c-a}&\frac b{c-b}\frac a{c-a}\\-\frac1{c-b}\frac1{b-a}&\frac{1}{b-a}+\frac1{c-b}\frac{b+a}{b-a}&\frac{-a}{b-a}-\frac b{c-b}\frac a{b-a}\\-\frac1{b-a}\frac1{a-c}&-\frac{1}{b-a}-\frac1{b-a}\frac{b+a}{c-a}&\frac{b}{b-a}+\frac b{b-a}\frac a{c-a}\end{pmatrix}

$ =\begin{pmatrix}\frac1{c-b}\frac1{c-a}&-\frac1{c-b}\frac{b+a}{c-a}&\frac b{c-b}\frac a{c-a}\\-\frac1{c-b}\frac1{b-a}&\frac1{c-b}\frac{c+a}{b-a}&-\frac c{c-b}\frac a{b-a}\\-\frac1{b-a}\frac1{a-c}&-\frac1{b-a}\frac{b+c}{c-a}&\frac b{b-a}\frac c{c-a}\end{pmatrix}

$ =\begin{pmatrix}-\frac1{b-c}\frac1{c-a}&\frac1{b-c}\frac{b+a}{c-a}&-\frac b{b-c}\frac a{c-a}\\-\frac1{a-b}\frac1{b-c}&\frac1{a-b}\frac{a+c}{b-c}&-\frac a{a-b}\frac c{b-c}\\-\frac1{c-a}\frac1{a-b}&\frac1{c-a}\frac{c+b}{a-b}&-\frac c{c-a}\frac b{a-b}\end{pmatrix}

$ =\frac{-1}{(a-b)(b-c)(c-a)}\begin{pmatrix}a-b&-(a-b)(a+b)&ab(a-b)\\c-a&-(c-a)(c+a)&ca(c-a)\\b-c&-(b-c)(c+b)&bc(b-c)\end{pmatrix}

$ \iff [\bm A]=\frac{-1}{(a-b)(b-c)(c-a)}\begin{pmatrix}b-c&-(b-c)(c+b)&bc(b-c)\\c-a&-(c-a)(c+a)&ca(c-a)\\a-b&-(a-b)(a+b)&ab(a-b)\end{pmatrix}

$ \begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}^{-1}=\left(\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}\begin{pmatrix}a^2&b^2&c^2\\a&b&c\\1&1&1\end{pmatrix}\right)^{-1}

$ = \begin{pmatrix}a^2&b^2&c^2\\a&b&c\\1&1&1\end{pmatrix}^{-1}\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}

$ =\frac{-1}{(a-b)(b-c)(c-a)}\begin{pmatrix}bc(b-c)&-(b-c)(c+b)&b-c\\ca(c-a)&-(c-a)(c+a)&c-a\\ab(a-b)&-(a-b)(a+b)&a-b\end{pmatrix}

$ a-b=2q_{01},b-c=2q_{12},c-a=2q_{20}

$ a+b=-c,b+c=-a,c+a=-b

$ \begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}^{-1}=\frac{-1}{8q_{01}q_{12}q_{20}}\begin{pmatrix}2q_{12}bc&2q_{12}a&2q_{12}\\2q_{20}ca&2q_{20}b&2q_{20}\\2q_{01}ab&2q_{01}c&2q_{01}\end{pmatrix}

$ =\frac{-1}{4q_{01}q_{12}q_{20}}\begin{pmatrix}q_{12}bc&q_{12}a&q_{12}\\q_{20}ca&q_{20}b&q_{20}\\q_{01}ab&q_{01}c&q_{01}\end{pmatrix}

$ \therefore \begin{pmatrix}n_0^2\\n_1^2\\n_2^2\end{pmatrix}=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}^{-1}\begin{pmatrix}1\\t_{//}-3\sigma_m\\|\bm t_\top|^2+(t_{//}-\sigma_m)^2\end{pmatrix}

$ = -\frac1{4q_{01}q_{12}q_{20}}\begin{pmatrix}q_{12}bc+q_{12}a(t_{//}-3\sigma_m)+q_{12}(|\bm t_\top|^2+(t_{//}-\sigma_m)^2)\\q_{20}ca+q_{20}b(t_{//}-3\sigma_m)+q_{20}(|\bm t_\top|^2+(t_{//}-\sigma_m)^2)\\q_{01}ab+q_{01}c(t_{//}-3\sigma_m)+q_{01}(|\bm t_\top|^2+(t_{//}-\sigma_m)^2)\end{pmatrix}

何かがおかしい気もするが……

$ \therefore \begin{pmatrix}n_0^2\\n_1^2\\n_2^2\end{pmatrix}=\begin{pmatrix}1&1&1\\\sigma_{00}&\sigma_{11}&\sigma_{22}\\\sigma_{00}^2&\sigma_{11}^2&\sigma_{22}^2\end{pmatrix}^{-1}\begin{pmatrix}1\\t_{//}\\|\bm t_\top|^2+{t_{//}}^2\end{pmatrix}

$ = -\frac1{4q_{01}q_{12}q_{20}}\begin{pmatrix}q_{12}\sigma_{11}\sigma_{22}-2q_{12}p_{12}t_{//}+q_{12}(|\bm t_\top|^2+{t_{//}}^2)\\q_{20}\sigma_{22}\sigma_{00}-2q_{20}p_{20}t_{//}+q_{20}(|\bm t_\top|^2+{t_{//}}^2)\\q_{01}\sigma_{00}\sigma_{11}-2q_{01}p_{01}t_{//}+q_{01}(|\bm t_\top|^2+{t_{//}}^2)\end{pmatrix}

$ = -\frac1{4q_{01}q_{12}q_{20}}\begin{pmatrix}q_{12}|\bm t_\top|^2+q_{12}(t_{//}-p_{12})^2-q_{12}{p_{12}}^2+q_{12}\sigma_{11}\sigma_{22}\\q_{20}|\bm t_\top|^2+q_{20}(t_{//}-p_{20})^2-q_{20}{p_{20}}^2+q_{20}\sigma_{22}\sigma_{00}\\q_{01}|\bm t_\top|^2+q_{01}(t_{//}-p_{01})^2-q_{01}{p_{01}}^2+q_{01}\sigma_{00}\sigma_{11}\end{pmatrix}

$ = -\frac1{4q_{01}q_{12}q_{20}}\begin{pmatrix}q_{12}(t_{//}-p_{12})^2+q_{12}|\bm t_\top|^2-{q_{12}}^3\\q_{20}(t_{//}-p_{20})^2+q_{20}|\bm t_\top|^2-{q_{20}}^3\\q_{01}(t_{//}-p_{01})^2+q_{01}|\bm t_\top|^2-{q_{01}}^3\end{pmatrix}

やっぱりこっちのほうがすっきりする