（九）二阶张量的不变量
本文主要内容如下：
- 1. 张量的标量不变量
- 2. 二阶张量的主不变量
- 3. 二阶张量的矩
1. 张量的标量不变量
张量 T \bold T T自身不随坐标的改变而改变，张量 T \bold T T通过与自身、度量张量 G \bold G G或置换张量 ϵ \epsilon ϵ进行张量的代数运算可以得到一系列标量，这些标量也不随坐标的改变而改变，将其称之为张量的（标量）不变量。如，
G : T = G ⋅ ⋅ T = T i ∙ i = T ∙ i i = t r T T : T = T i ∙ j T ∙ j i T ⋅ ⋅ T = T i ∙ j T j ∙ i = T ∙ j i T ∙ i j = t r ( T ⋅ T ) ϵ i j k ϵ p q l T ∙ p i T ∙ q j T ∙ l k ⋯ \begin{aligned} &\bold{G}:\bold{T}=\bold{G}\cdot\cdot\bold{T}=T_i^{\bullet i}=T^i_{\bullet i}=tr{\bold{T}}\\\ \\ &\bold{T}:\bold{T}=T_i^{\bullet j}T^i_{\bullet j}\\\ \\ &\bold{T}\cdot\cdot\bold{T}=T_i^{\bullet j}T_j^{\bullet i}=T^i_{\bullet j}T^j_{\bullet i}=tr(\bold{T}\cdot\bold{T})\\\ \\ &\epsilon_{ijk}\epsilon^{pql}T^{i}_{\bullet p}T^{j}_{\bullet q}T^{k}_{\bullet l}\\\ \\ &\cdots \end{aligned} G:T=G⋅⋅T=Ti∙i=T∙ii=trTT:T=Ti∙jT∙jiT⋅⋅T=Ti∙jTj∙i=T∙jiT∙ij=tr(T⋅T)ϵijkϵpqlT∙piT∙qjT∙lk⋯
2. 二阶张量的主不变量
在二阶张量 T \bold T T众多的不变量中，将如下三个不变量称作二阶张量 T \bold T T的主不变量：
C 1 T = δ p i T ∙ i p = T ∙ i i = T ∙ 1 1 + T ∙ 2 2 + T ∙ 3 3 = t r ( T ) = G : T = G ⋅ ⋅ T C 2 T = 1 2 δ p q i j T ∙ i p T ∙ j q = 1 2 ( T ∙ i i T ∙ j j − T ∙ i j T ∙ j i ) = ∣ T ∙ 1 1 T ∙ 2 1 T ∙ 1 2 T ∙ 2 2 ∣ + ∣ T ∙ 1 1 T ∙ 3 1 T ∙ 1 3 T ∙ 3 3 ∣ + ∣ T ∙ 2 2 T ∙ 3 2 T ∙ 2 3 T ∙ 3 3 ∣ C 3 T = 1 3 ! δ p q l i j k T ∙ i p T ∙ j q T ∙ k l = 1 6 T ∙ i i T ∙ j j T ∙ k k − 1 2 T ∙ k k T ∙ i j T ∙ j i + 1 3 T ∙ i k T ∙ j i T ∙ k j = 1 3 ! ( T ∙ i p T ∙ j q T ∙ k l e p q l ) e i j k = 1 3 ! ( T ∙ 1 p T ∙ 2 q T ∙ 3 l e p q l e i j k ) e i j k = 1 3 ! ( T ∙ 1 p T ∙ 2 q T ∙ 3 l e p q l ) ( e i j k e i j k ) = d e t ( T ) = ∣ T ∙ 1 1 T ∙ 2 1 T ∙ 3 1 T ∙ 1 2 T ∙ 2 2 T ∙ 3 2 T ∙ 1 3 T ∙ 2 3 T ∙ 3 3 ∣ \begin{aligned} & \mathscr{C}_1^{T}= \delta^i_pT^p_{\bullet i} =T^i_{\bullet i} =T^1_{\bullet 1}+T^2_{\bullet 2}+T^3_{\bullet 3} =tr(\bold T) =\bold{G}:\bold{T} =\bold{G}\cdot\cdot\bold{T} \\\ \\& \mathscr{C}_2^{T} =\frac{1}{2}\delta^{ij}_{pq}T^p_{\bullet i}T^q_{\bullet j} =\frac{1}{2}(T^i_{\bullet i}T^j_{\bullet j}-T^j_{\bullet i}T^i_{\bullet j}) =\begin{vmatrix} T^1_{\bullet 1} & T^1_{\bullet 2} \\\ \\ T^2_{\bullet 1} & T^2_{\bullet 2} \end{vmatrix} +\begin{vmatrix} T^1_{\bullet 1} & T^1_{\bullet 3} \\\ \\ T^3_{\bullet 1} & T^3_{\bullet 3} \end{vmatrix} +\begin{vmatrix} T^2_{\bullet 2} & T^2_{\bullet 3} \\\ \\ T^3_{\bullet 2} & T^3_{\bullet 3} \end{vmatrix} \\\ \\& \mathscr{C}_3^{T} =\frac{1}{3!}\delta^{ijk}_{pql}T^p_{\bullet i}T^q_{\bullet j}T^l_{\bullet k} =\frac{1}{6}T^i_{\bullet i}T^j_{\bullet j}T^k_{\bullet k}-\frac{1}{2}T^k_{\bullet k}T^j_{\bullet i}T^i_{\bullet j}+\frac{1}{3}T^k_{\bullet i}T^i_{\bullet j}T^j_{\bullet k} =\frac{1}{3!}(T^p_{\bullet i}T^q_{\bullet j}T^l_{\bullet k}e_{pql})e^{ijk} =\frac{1}{3!}(T^p_{\bullet 1}T^q_{\bullet 2}T^l_{\bullet 3}e_{pql}e_{ijk})e^{ijk} =\frac{1}{3!}(T^p_{\bullet 1}T^q_{\bullet 2}T^l_{\bullet 3}e_{pql})(e_{ijk}e^{ijk}) =det(\bold T) =\begin{vmatrix} T^1_{\bullet 1} & T^1_{\bullet 2} & T^1_{\bullet 3}\\\ \\ T^2_{\bullet 1} & T^2_{\bullet 2} & T^2_{\bullet 3}\\\ \\ T^3_{\bullet 1} & T^3_{\bullet 2} & T^3_{\bullet 3} \end{vmatrix} \end{aligned} C1T=δpiT∙ip=T∙ii=T∙11+T∙22+T∙33=tr(T)=G:T=G⋅⋅TC2T=21δpqijT∙ipT∙jq=21(T∙iiT∙jj−T∙ijT∙ji)=∣ ∣T∙11 T∙12T∙21T∙22∣ ∣+∣ ∣T∙11 T∙13T∙31T∙33∣ ∣+∣ ∣T∙22 T∙23T∙32T∙33∣ ∣C3T=3!1δpqlijkT∙ipT∙jqT∙kl=61T∙iiT∙jjT∙kk−21T∙kkT∙ijT∙ji+31T∙ikT∙jiT∙kj=3!1(T∙ipT∙jqT∙klepql)eijk=3!1(T∙1pT∙2qT∙3lepqleijk)eijk=3!1(T∙1pT∙2qT∙3lepql)(eijkeijk)=det(T)=∣ ∣T∙11 T∙12 T∙13T∙21T∙22T∙23T∙31T∙32T∙33∣ ∣可知，二阶张量 T \bold T T的主不变量 C 1 、 C 2 、 C 3 \mathscr{C}_1、\mathscr{C}_2、\mathscr{C}_3 C1、C2、C3分别为张量的矩阵 τ 3 \tau_{3} τ3 的一、二、三阶主子式之和。可进一步证明存在如下关系式：
( 1 ) [ T ∙ u ⃗ v ⃗ w ⃗ ] + [ u ⃗ T ∙ v ⃗ w ⃗ ] + [ u ⃗ v ⃗ T ∙ w ⃗ ] = C 1 T [ u ⃗ v ⃗ w ⃗ ] ( 2 ) [ T ∙ u ⃗ T ∙ v ⃗ w ⃗ ] + [ u ⃗ T ∙ v ⃗ T ∙ w ⃗ ] + [ T ∙ u ⃗ v ⃗ T ∙ w ⃗ ] = C 2 T [ u ⃗ v ⃗ w ⃗ ] ( 3 ) [ T ∙ u ⃗ T ∙ v ⃗ T ∙ w ⃗ ] = C 3 T [ u ⃗ v ⃗ w ⃗ ] \begin{aligned} &(1) \begin{bmatrix}\bold{T}\bullet\vec{u} & \vec{v} & \vec{w}\end{bmatrix}+ \begin{bmatrix}\vec{u} & \bold{T}\bullet\vec{v} & \vec{w}\end{bmatrix}+ \begin{bmatrix}\vec{u} & \vec{v} & \bold{T}\bullet\vec{w}\end{bmatrix}= \mathscr{C}_1^T\begin{bmatrix}\vec{u} & \vec{v} & \vec{w}\end{bmatrix}\\\ \\ &(2) \begin{bmatrix}\bold{T}\bullet\vec{u} & \bold{T}\bullet\vec{v} & \vec{w}\end{bmatrix}+ \begin{bmatrix}\vec{u} & \bold{T}\bullet\vec{v} & \bold{T}\bullet\vec{w}\end{bmatrix}+ \begin{bmatrix}\bold{T}\bullet\vec{u} & \vec{v} & \bold{T}\bullet\vec{w}\end{bmatrix}= \mathscr{C}_2^T\begin{bmatrix}\vec{u} & \vec{v} & \vec{w}\end{bmatrix}\\\ \\ &(3) \begin{bmatrix}\bold{T}\bullet\vec{u} & \bold{T}\bullet\vec{v} & \bold{T}\bullet\vec{w}\end{bmatrix}= \mathscr{C}_3^T\begin{bmatrix}\vec{u} & \vec{v} & \vec{w}\end{bmatrix} \end{aligned} (1)[T∙uvw]+[uT∙vw]+[uvT∙w]=C1T[uvw](2)[T∙uT∙vw]+[uT∙vT∙w]+[T∙uvT∙w]=C2T[uvw](3)[T∙uT∙vT∙w]=C3T[uvw]第三式的证明参见张量行列式的几何意义说明，前两式证明如下：
[ T ∙ u ⃗ v ⃗ w ⃗ ] + [ u ⃗ T ∙ v ⃗ w ⃗ ] + [ u ⃗ v ⃗ T ∙ w ⃗ ] = ( T ∙ m i u m ) v j w k ϵ i j k + u i ( T ∙ m j v m ) w k ϵ i j k + u i v j ( T ∙ m k w m ) ϵ i j k = u i v j w k ( T ∙ i m ϵ m j k + T ∙ j m ϵ i m k + T ∙ k m ϵ i j m ) = 1 2 u i v j w k [ ( T ∙ i m ϵ m j k + T ∙ j m ϵ i m k + T ∙ k m ϵ i j m ) + ( T ∙ i m ϵ k m j + T ∙ j m ϵ k i m − T ∙ k m ϵ m j i ) ] = 1 2 u i v j w k ( T ∙ p m ϵ m q l δ i p δ j q δ k l + T ∙ p m ϵ q m l δ j p δ i q δ k l + T ∙ p m ϵ q l m δ k p δ i q δ j l + T ∙ p m ϵ q m l δ i p δ k q δ j l + T ∙ p m ϵ q l m δ j p δ k q δ i l − T ∙ p m ϵ m q l δ k p δ j q δ i l ) = 1 2 u i v j w k ( T ∙ p m ϵ m q l ) ( δ i p δ j q δ k l − δ j p δ i q δ k l + δ k p δ i q δ j l − δ i p δ k q δ j l + δ j p δ k q δ i l − δ k p δ j q δ i l ) = 1 2 u i v j w k T ∙ p m ( ϵ m q l ϵ p q l ) ϵ i j k = u i v j w k T ∙ p m ϵ i j k δ m p = T ∙ m m ( u i v j w k ϵ i j k ) = C 1 T [ u ⃗ v ⃗ w ⃗ ] [ T ∙ u ⃗ T ∙ v ⃗ w ⃗ ] + [ u ⃗ T ∙ v ⃗ T ∙ w ⃗ ] + [ T ∙ u ⃗ v ⃗ T ∙ w ⃗ ] = ( T ∙ m i u m ) ( T ∙ n j v n ) w k ϵ i j k + u i ( T ∙ m j v m ) ( T ∙ n k w n ) ϵ i j k + ( T ∙ m i u m ) v j ( T ∙ n k w n ) ϵ i j k = u i v j w k ( T ∙ i m T ∙ j n ϵ m n k + T ∙ j m T ∙ k n ϵ i m n + T ∙ i m T ∙ k n ϵ m j n ) = 1 2 u i v j w k ( T ∙ i m T ∙ j n ϵ m n k + T ∙ j m T ∙ k n ϵ i m n + T ∙ i m T ∙ k n ϵ m j n ) + ( T ∙ i n T ∙ j m ϵ n m k + T ∙ j n T ∙ k m ϵ i n m + T ∙ i n T ∙ k m ϵ n j m ) = 1 2 u i v j w k ( T ∙ p m T ∙ q n ϵ m n l δ i p δ j q δ k l + T ∙ p m T ∙ q n ϵ l m n δ j p δ k q δ i l + T ∙ p m T ∙ q n ϵ m l n δ i p δ k q δ j l + T ∙ q n T ∙ p m ϵ n m l δ j p δ i q δ k l + T ∙ q n T ∙ p m ϵ l n m δ k p δ j q δ i l + T ∙ q n T ∙ p m ϵ n l m δ k p δ i q δ j l ) = 1 2 u i v j w k T ∙ p m T ∙ q n ϵ m n l ( δ i p δ j q δ k l + δ j p δ k q δ i l − δ i p δ k q δ j l − δ j p δ i q δ k l − δ k p δ j q δ i l + δ k p δ i q δ j l ) = 1 2 u i v j w k T ∙ p m T ∙ q n ( ϵ m n l ϵ p q l ) ϵ i j k = ( u i v j w k ϵ i j k ) ( 1 2 T ∙ p m T ∙ q n δ m n p q ) = C 2 T [ u ⃗ v ⃗ w ⃗ ] \begin{aligned} &\quad \begin{bmatrix}\bold{T}\bullet\vec{u} & \vec{v} & \vec{w}\end{bmatrix}+ \begin{bmatrix}\vec{u} & \bold{T}\bullet\vec{v} & \vec{w}\end{bmatrix}+ \begin{bmatrix}\vec{u} & \vec{v} & \bold{T}\bullet\vec{w}\end{bmatrix} \\\ \\ &=(T^i_{\bullet m}u^m)v^jw^k\epsilon_{ijk}+u^i(T^j_{\bullet m}v^m)w^k\epsilon_{ijk}+u^iv^j(T^k_{\bullet m}w^m)\epsilon_{ijk} \\\ \\ &=u^iv^jw^k(T^m_{\bullet i}\epsilon_{mjk}+T^m_{\bullet j}\epsilon_{imk}+T^m_{\bullet k}\epsilon_{ijm}) \\\ \\ &=\frac{1}{2}u^iv^jw^k[(T^m_{\bullet i}\epsilon_{mjk}+T^m_{\bullet j}\epsilon_{imk}+T^m_{\bullet k}\epsilon_{ijm})+(T^m_{\bullet i}\epsilon_{kmj}+T^m_{\bullet j}\epsilon_{kim}-T^m_{\bullet k}\epsilon_{mji})] \\\ \\ &=\frac{1}{2}u^iv^jw^k (T^m_{\bullet p}\epsilon_{mql}\delta^{p}_{i}\delta^{q}_{j}\delta^{l}_{k}+ T^m_{\bullet p}\epsilon_{qml}\delta^{p}_{j}\delta^{q}_{i}\delta^{l}_{k}+ T^m_{\bullet p}\epsilon_{qlm}\delta^{p}_{k}\delta^{q}_{i}\delta^{l}_{j}+ T^m_{\bullet p}\epsilon_{qml}\delta^{p}_{i}\delta^{q}_{k}\delta^{l}_{j}+ T^m_{\bullet p}\epsilon_{qlm}\delta^{p}_{j}\delta^{q}_{k}\delta^{l}_{i}- T^m_{\bullet p}\epsilon_{mql}\delta^{p}_{k}\delta^{q}_{j}\delta^{l}_{i})\\\ \\ &=\frac{1}{2}u^iv^jw^k(T^m_{\bullet p}\epsilon_{mql}) (\delta^{p}_{i}\delta^{q}_{j}\delta^{l}_{k}- \delta^{p}_{j}\delta^{q}_{i}\delta^{l}_{k}+ \delta^{p}_{k}\delta^{q}_{i}\delta^{l}_{j}- \delta^{p}_{i}\delta^{q}_{k}\delta^{l}_{j}+ \delta^{p}_{j}\delta^{q}_{k}\delta^{l}_{i}- \delta^{p}_{k}\delta^{q}_{j}\delta^{l}_{i})\\\ \\ &=\frac{1}{2}u^iv^jw^kT^m_{\bullet p}(\epsilon_{mql}\epsilon^{pql})\epsilon_{ijk}\\\ \\ &=u^iv^jw^kT^m_{\bullet p}\epsilon_{ijk}\delta^p_m\\\ \\ &=T^m_{\bullet m}(u^iv^jw^k\epsilon_{ijk})\\\ \\ &=\mathscr{C}_1^T\begin{bmatrix}\vec{u} & \vec{v} & \vec{w}\end{bmatrix} \\\ \\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &\quad \begin{bmatrix}\bold{T}\bullet\vec{u} & \bold{T}\bullet\vec{v} & \vec{w}\end{bmatrix}+ \begin{bmatrix}\vec{u} & \bold{T}\bullet\vec{v} & \bold{T}\bullet\vec{w}\end{bmatrix}+ \begin{bmatrix}\bold{T}\bullet\vec{u} & \vec{v} & \bold{T}\bullet\vec{w}\end{bmatrix}\\\ \\ &=(T^i_{\bullet m}u^m)(T^j_{\bullet n}v^n)w^k\epsilon_{ijk}+ u^i(T^j_{\bullet m}v^m)(T^k_{\bullet n}w^n)\epsilon_{ijk}+ (T^i_{\bullet m}u^m)v^j(T^k_{\bullet n}w^n)\epsilon_{ijk}\\\ \\ &=u^iv^jw^k(T^m_{\bullet i}T^n_{\bullet j}\epsilon_{mnk}+ T^m_{\bullet j}T^n_{\bullet k}\epsilon_{imn}+ T^m_{\bullet i}T^n_{\bullet k}\epsilon_{mjn})\\\ \\ &=\frac{1}{2}u^iv^jw^k(T^m_{\bullet i}T^n_{\bullet j}\epsilon_{mnk}+ T^m_{\bullet j}T^n_{\bullet k}\epsilon_{imn}+ T^m_{\bullet i}T^n_{\bullet k}\epsilon_{mjn})+ (T^n_{\bullet i}T^m_{\bullet j}\epsilon_{nmk}+ T^n_{\bullet j}T^m_{\bullet k}\epsilon_{inm}+ T^n_{\bullet i}T^m_{\bullet k}\epsilon_{njm})\\\ \\ &=\frac{1}{2}u^iv^jw^k (T^m_{\bullet p}T^n_{\bullet q}\epsilon_{mnl}\delta^p_i\delta^q_j\delta^l_k+ T^m_{\bullet p}T^n_{\bullet q}\epsilon_{lmn}\delta^p_j\delta^q_k\delta^l_i+ T^m_{\bullet p}T^n_{\bullet q}\epsilon_{mln}\delta^p_i\delta^q_k\delta^l_j+ T^n_{\bullet q}T^m_{\bullet p}\epsilon_{nml}\delta^p_j\delta^q_i\delta^l_k+ % T^n_{\bullet q}T^m_{\bullet p}\epsilon_{lnm}\delta^p_k\delta^q_j\delta^l_i+ T^n_{\bullet q}T^m_{\bullet p}\epsilon_{nlm}\delta^p_k\delta^q_i\delta^l_j)\\\ \\ &=\frac{1}{2}u^iv^jw^kT^m_{\bullet p}T^n_{\bullet q}\epsilon_{mnl}( \delta^p_i\delta^q_j\delta^l_k+ \delta^p_j\delta^q_k\delta^l_i- \delta^p_i\delta^q_k\delta^l_j- \delta^p_j\delta^q_i\delta^l_k- \delta^p_k\delta^q_j\delta^l_i+ \delta^p_k\delta^q_i\delta^l_j)\\\ \\ &=\frac{1}{2}u^iv^jw^kT^m_{\bullet p}T^n_{\bullet q}(\epsilon_{mnl}\epsilon^{pql})\epsilon_{ijk}\\\ \\ &=(u^iv^jw^k\epsilon_{ijk})(\frac{1}{2}T^m_{\bullet p}T^n_{\bullet q}\delta^{pq}_{mn})\\\ \\ &=\mathscr{C}_2^T\begin{bmatrix}\vec{u} & \vec{v} & \vec{w}\end{bmatrix} \end{aligned} [T∙uvw]+[uT∙vw]+[uvT∙w]=(T∙mium)vjwkϵijk+ui(T∙mjvm)wkϵijk+uivj(T∙mkwm)ϵijk=uivjwk(T∙imϵmjk+T∙jmϵimk+T∙kmϵijm)=21uivjwk[(T∙imϵmjk+T∙jmϵimk+T∙kmϵijm)+(T∙imϵkmj+T∙jmϵkim−T∙kmϵmji)]=21uivjwk(T∙pmϵmqlδipδjqδkl+T∙pmϵqmlδjpδiqδkl+T∙pmϵqlmδkpδiqδjl+T∙pmϵqmlδipδkqδjl+T∙pmϵqlmδjpδkqδil−T∙pmϵmqlδkpδjqδil)=21uivjwk(T∙pmϵmql)(δipδjqδkl−δjpδiqδkl+δkpδiqδjl−δipδkqδjl+δjpδkqδil−δkpδjqδil)=21uivjwkT∙pm(ϵmqlϵpql)ϵijk=uivjwkT∙pmϵijkδmp=T∙mm(uivjwkϵijk)=C1T[uvw][T∙uT∙vw]+[uT∙vT∙w]+[T∙uvT∙w]=(T∙mium)(T∙njvn)wkϵijk+ui(T∙mjvm)(T∙nkwn)ϵijk+(T∙mium)vj(T∙nkwn)ϵijk=uivjwk(T∙imT∙jnϵmnk+T∙jmT∙knϵimn+T∙imT∙knϵmjn)=21uivjwk(T∙imT∙jnϵmnk+T∙jmT∙knϵimn+T∙imT∙knϵmjn)+(T∙inT∙jmϵnmk+T∙jnT∙kmϵinm+T∙inT∙kmϵnjm)=21uivjwk(T∙pmT∙qnϵmnlδipδjqδkl+T∙pmT∙qnϵlmnδjpδkqδil+T∙pmT∙qnϵmlnδipδkqδjl+T∙qnT∙pmϵnmlδjpδiqδkl+T∙qnT∙pmϵlnmδkpδjqδil+T∙qnT∙pmϵnlmδkpδiqδjl)=21uivjwkT∙pmT∙qnϵmnl(δipδjqδkl+δjpδkqδil−δipδkqδjl−δjpδiqδkl−δkpδjqδil+δkpδiqδjl)=21uivjwkT∙pmT∙qn(ϵmnlϵpql)ϵijk=(uivjwkϵijk)(21T∙pmT∙qnδmnpq)=C2T[uvw]
根据上述推导过程可知：
( 1 ) T ∙ i m ϵ m j k + T ∙ j m ϵ i m k + T ∙ k m ϵ i j m = C 1 T ϵ i j k ( 2 ) T ∙ i m T ∙ j n ϵ m n k + T ∙ j m T ∙ k n ϵ i m n + T ∙ i m T ∙ k n ϵ m j n = C 2 T ϵ i j k ( 3 ) T ∙ i m T ∙ j n T ∙ k l ϵ m n l = C 3 T ϵ i j k \begin{aligned} &(1) \ T^m_{\bullet i}\epsilon_{mjk}+ T^m_{\bullet j}\epsilon_{imk}+ T^m_{\bullet k}\epsilon_{ijm} =\mathscr{C}_1^T\epsilon_{ijk}\\\ \\ &(2) \ T^m_{\bullet i}T^n_{\bullet j}\epsilon_{mnk}+ T^m_{\bullet j}T^n_{\bullet k}\epsilon_{imn}+ T^m_{\bullet i}T^n_{\bullet k}\epsilon_{mjn} =\mathscr{C}_2^T\epsilon_{ijk}\\\ \\ &(3) \ T^m_{\bullet i}T^n_{\bullet j}T^l_{\bullet k}\epsilon_{mnl} =\mathscr{C}_3^T\epsilon_{ijk} \end{aligned} (1) T∙imϵmjk+T∙jmϵimk+T∙kmϵijm=C1Tϵijk(2) T∙imT∙jnϵmnk+T∙jmT∙knϵimn+T∙imT∙knϵmjn=C2Tϵijk(3) T∙imT∙jnT∙klϵmnl=C3Tϵijk
此外，对于正则张量的线性变换还有如下的Nanson公式 —— 描述了映射前后面元的变化情况
( T ∙ u ⃗ ) × ( T ∙ v ⃗ ) = C 3 T ( T T ) − 1 ( u ⃗ × v ⃗ ) (\bold{T}\bullet\vec{u})\times(\bold{T}\bullet\vec{v})=\mathscr{C}_3^T(\bold{T}^T)^{-1}(\vec{u}\times\vec{v}) (T∙u)×(T∙v)=C3T(TT)−1(u×v)证明如下：
(法一)
T T ∙ [ ( T ∙ u ⃗ ) × ( T ∙ v ⃗ ) ] = [ ( T T ) p ∙ q g ⃗ p g ⃗ q ] ∙ [ ( T ∙ m i u m ) ( T ∙ n j v n ) ϵ i j k g ⃗ k ] = ( T ∙ p q g ⃗ p g ⃗ q ) ∙ [ ( T ∙ m i u m ) ( T ∙ n j v n ) ϵ i j k g ⃗ k ] = ( T ∙ m i T ∙ n j T ∙ p k ϵ i j k ) u m v n g ⃗ p = C 3 T ( u m v n ϵ m n p g ⃗ p ) = C 3 T ( u ⃗ × v ⃗ ) \begin{aligned} & \quad \bold{T}^T\bullet[(\bold{T}\bullet\vec{u})\times(\bold{T}\bullet\vec{v})]\\\ \\ &=[(T^T)_p^{\bullet q}\vec{g}^p\vec{g}_q]\bullet[(T^i_{\bullet m}u^m)(T^j_{\bullet n}v^n)\epsilon_{ijk}\vec{g}^k]\\\ \\ &=(T^q_{\bullet p}\vec{g}^p\vec{g}_q)\bullet[(T^i_{\bullet m}u^m)(T^j_{\bullet n}v^n)\epsilon_{ijk}\vec{g}^k]\\\ \\ &=(T^i_{\bullet m}T^j_{\bullet n}T^k_{\bullet p}\epsilon_{ijk})u^mv^n\vec{g}^p\\\ \\ &=\mathscr{C}_3^T(u^mv^n\epsilon_{mnp}\vec{g}^p)\\\ \\ &=\mathscr{C}_3^T(\vec{u}\times\vec{v}) \end{aligned} TT∙[(T∙u)×(T∙v)]=[(TT)p∙qgpgq]∙[(T∙mium)(T∙njvn)ϵijkgk]=(T∙pqgpgq)∙[(T∙mium)(T∙njvn)ϵijkgk]=(T∙miT∙njT∙pkϵijk)umvngp=C3T(umvnϵmnpgp)=C3T(u×v)则：
( T ∙ u ⃗ ) × ( T ∙ v ⃗ ) = ( T T ) − 1 ∙ [ C 3 T ( u ⃗ × v ⃗ ) ] = C 3 T ( T T ) − 1 ∙ ( u ⃗ × v ⃗ ) (\bold{T}\bullet\vec{u})\times(\bold{T}\bullet\vec{v}) =(\bold{T}^T)^{-1}\bullet[\mathscr{C}_3^T(\vec{u}\times\vec{v})] =\mathscr{C}_3^T(\bold{T}^T)^{-1}\bullet(\vec{u}\times\vec{v}) (T∙u)×(T∙v)=(TT)−1∙[C3T(u×v)]=C3T(TT)−1∙(u×v)(法二)
对任意矢量 w ⃗ \vec{w} w有： ( T ∙ w ⃗ ) ∙ [ ( T ∙ u ⃗ ) × ( T ∙ v ⃗ ) ] = C 3 T w ⃗ ∙ ( u ⃗ × v ⃗ ) = w ⃗ ∙ T T ∙ [ ( T ∙ u ⃗ ) × ( T ∙ v ⃗ ) ] ⟹ w ⃗ ∙ { T T ∙ [ ( T ∙ u ⃗ ) × ( T ∙ v ⃗ ) ] − C 3 T ( u ⃗ × v ⃗ ) } = 0 (\bold{T}\bullet\vec{w})\bullet[(\bold{T}\bullet\vec{u}) \times (\bold{T}\bullet\vec{v} )] = \mathscr{C}_3^T \vec{w}\bullet(\vec{u} \times \vec{v})= \vec{w}\bullet\bold{T}^T\bullet[(\bold{T}\bullet\vec{u}) \times (\bold{T}\bullet\vec{v} )] \\\ \\ \Longrightarrow \vec{w}\bullet\{\bold{T}^T\bullet[(\bold{T}\bullet\vec{u}) \times (\bold{T}\bullet\vec{v} )]-\mathscr{C}_3^T (\vec{u} \times \vec{v})\}=0 (T∙w)∙[(T∙u)×(T∙v)]=C3Tw∙(u×v)=w∙TT∙[(T∙u)×(T∙v)] ⟹w∙{TT∙[(T∙u)×(T∙v)]−C3T(u×v)}=0由 w ⃗ \vec{w} w 任意性知： T T ∙ [ ( T ∙ u ⃗ ) × ( T ∙ v ⃗ ) ] − C 3 T ( u ⃗ × v ⃗ ) = 0 \bold{T}^T\bullet[(\bold{T}\bullet\vec{u}) \times (\bold{T}\bullet\vec{v} )]-\mathscr{C}_3^T (\vec{u} \times \vec{v})=0 TT∙[(T∙u)×(T∙v)]−C3T(u×v)=0由于 T \bold T T正则，故：
( T ∙ u ⃗ ) × ( T ∙ v ⃗ ) = C 3 T ( T T ) − 1 ( u ⃗ × v ⃗ ) (\bold{T}\bullet\vec{u}) \times (\bold{T}\bullet\vec{v} )=\mathscr{C}_3^T(\bold{T}^T)^{-1} (\vec{u} \times \vec{v}) (T∙u)×(T∙v)=C3T(TT)−1(u×v)
3. 二阶张量的矩
n n n个二阶张量 T \bold T T依次点积得到的二阶张量的迹称作二阶张量 T \bold T T的 n n n阶矩 C n ∗ T \mathscr{C}_n^{*T} Cn∗T。如：
C 1 ∗ T = t r ( T ) = T ∙ i i C 2 ∗ T = t r ( T ∙ T ) = T ∙ j i T ∙ i j C 3 ∗ T = t r ( T ∙ T ∙ T ) = T ∙ j i T ∙ k j T ∙ i k \begin{aligned} &\mathscr{C}_1^{*T} =tr(\bold{T})=T^i_{\bullet i} \\\ \\ &\mathscr{C}_2^{*T}=tr(\bold{T \bullet T})=T^i_{\bullet j}T^j_{\bullet i} \\\ \\ &\mathscr{C}_3^{*T}=tr(\bold{T \bullet T \bullet T})=T^i_{\bullet j}T^j_{\bullet k}T^k_{\bullet i} \end{aligned} C1∗T=tr(T)=T∙iiC2∗T=tr(T∙T)=T∙jiT∙ijC3∗T=tr(T∙T∙T)=T∙jiT∙kjT∙ik可以证明，高于三阶的矩可由一、二、三阶矩得到。另外，二阶张量的一、二、三阶矩与其主不变量之间并不独立，可以互相转化:
{ C 1 ∗ T = C 1 T C 2 ∗ T = ( C 1 T ) 2 − 2 C 2 T C 3 ∗ T = ( C 1 T ) 3 − 3 C 1 T C 2 T + 3 C 3 T ⟸ { C 1 T = T ∙ i i = C 1 ∗ T ( C 1 T ) 2 − 2 C 2 T = T ∙ i i T ∙ j j − 2 ⋅ 1 2 ( T ∙ i i T ∙ j j − T ∙ i j T ∙ j i ) = T ∙ i j T ∙ j i = C 2 ∗ T ( C 1 T ) 3 − 3 C 1 T C 2 T + 3 C 3 T = T ∙ i i T ∙ j j T ∙ k k − 3 2 T ∙ k k ( T ∙ i i T ∙ j j − T ∙ i j T ∙ j i ) + 3 ( 1 6 T ∙ i i T ∙ j j T ∙ k k − 1 2 T ∙ k k T ∙ i j T ∙ j i + 1 3 T ∙ i k T ∙ j i T ∙ k j ) = T ∙ i k T ∙ j i T ∙ k j = C 3 ∗ T { C 1 T = C 1 ∗ T C 2 T = 1 2 [ ( C 1 ∗ T ) 2 − C 2 ∗ T ] C 3 T = 1 6 ( C 1 ∗ T ) 3 − 1 2 C 1 ∗ T C 2 ∗ T + 1 3 C 3 ∗ T ⟸ { C 1 ∗ T = T ∙ i i = C 1 T 1 2 [ ( C 1 ∗ T ) 2 − C 2 ∗ T ] = 1 2 ( T ∙ i i T ∙ j j − 2 T ∙ j i T ∙ i j ) = C 2 T 1 6 ( C 1 ∗ T ) 3 − 1 2 C 1 ∗ T C 2 ∗ T + 1 3 C 3 ∗ T = 1 6 T ∙ i i T ∙ j j T ∙ k k − 1 2 T ∙ k k T ∙ i j T ∙ j i + 1 3 T ∙ i k T ∙ j i T ∙ k j = C 3 T \begin{aligned} & \begin{cases} \mathscr{C}_1^{*T}= \mathscr{C}_1^{T}\\\ \\ \mathscr{C}_2^{*T}= (\mathscr{C}_1^{T})^2-2\mathscr{C}_2^{T}\\\ \\ \mathscr{C}_3^{*T}= (\mathscr{C}_1^{T})^3-3\mathscr{C}_1^{T}\mathscr{C}_2^{T}+3\mathscr{C}_3^{T} \end{cases} \Longleftarrow \begin{cases} \mathscr{C}_1^{T}= T^i_{\bullet i}=\mathscr{C}_1^{*T}\\\ \\ (\mathscr{C}_1^{T})^2-2\mathscr{C}_2^{T} =T^i_{\bullet i}T^j_{\bullet j}-2\cdot\frac{1}{2}(T^i_{\bullet i}T^j_{\bullet j}-T^j_{\bullet i}T^i_{\bullet j}) =T^j_{\bullet i}T^i_{\bullet j} =\mathscr{C}_2^{*T}\\\ \\ (\mathscr{C}_1^{T})^3-3\mathscr{C}_1^{T}\mathscr{C}_2^{T}+3\mathscr{C}_3^{T} =T^i_{\bullet i}T^j_{\bullet j}T^k_{\bullet k}-\frac{3}{2}T^k_{\bullet k}(T^i_{\bullet i}T^j_{\bullet j}-T^j_{\bullet i}T^i_{\bullet j})+3(\frac{1}{6}T^i_{\bullet i}T^j_{\bullet j}T^k_{\bullet k}-\frac{1}{2}T^k_{\bullet k}T^j_{\bullet i}T^i_{\bullet j}+\frac{1}{3}T^k_{\bullet i}T^i_{\bullet j}T^j_{\bullet k}) =T^k_{\bullet i}T^i_{\bullet j}T^j_{\bullet k} =\mathscr{C}_3^{*T} \end{cases} \\\ \\ & \begin{cases} \mathscr{C}_1^{T}= \mathscr{C}_1^{*T}\\\ \\ \mathscr{C}_2^{T}= \frac{1}{2}[(\mathscr{C}_1^{*T})^2-\mathscr{C}_2^{*T}]\\\ \\ \mathscr{C}_3^{T}= \frac{1}{6}(\mathscr{C}_1^{*T})^3-\frac{1}{2}\mathscr{C}_1^{*T}\mathscr{C}_2^{*T}+\frac{1}{3}\mathscr{C}_3^{*T} \end{cases} \Longleftarrow \begin{cases} \mathscr{C}_1^{*T}= T^i_{\bullet i}=\mathscr{C}_1^{T}\\\ \\ \frac{1}{2}[(\mathscr{C}_1^{*T})^2-\mathscr{C}_2^{*T}] =\frac{1}{2}(T^i_{\bullet i}T^j_{\bullet j}-2T^i_{\bullet j}T^j_{\bullet i}) =\mathscr{C}_2^{T}\\\ \\ \frac{1}{6}(\mathscr{C}_1^{*T})^3-\frac{1}{2}\mathscr{C}_1^{*T}\mathscr{C}_2^{*T}+\frac{1}{3}\mathscr{C}_3^{*T} =\frac{1}{6}T^i_{\bullet i}T^j_{\bullet j}T^k_{\bullet k}-\frac{1}{2}T^k_{\bullet k}T^j_{\bullet i}T^i_{\bullet j}+\frac{1}{3}T^k_{\bullet i}T^i_{\bullet j}T^j_{\bullet k} =\mathscr{C}_3^{T} \end{cases} \end{aligned} ⎩ ⎨ ⎧C1∗T=C1T C2∗T=(C1T)2−2C2T C3∗T=(C1T)3−3C1TC2T+3C3T⟸⎩ ⎨ ⎧C1T=T∙ii=C1∗T (C1T)2−2C2T=T∙iiT∙jj−2⋅21(T∙iiT∙jj−T∙ijT∙ji)=T∙ijT∙ji=C2∗T (C1T)3−3C1TC2T+3C3T=T∙iiT∙jjT∙kk−23T∙kk(T∙iiT∙jj−T∙ijT∙ji)+3(61T∙iiT∙jjT∙kk−21T∙kkT∙ijT∙ji+31T∙ikT∙jiT∙kj)=T∙ikT∙jiT∙kj=C3∗T⎩ ⎨ ⎧C1T=C1∗T C2T=21[(C1∗T)2−C2∗T] C3T=61(C1∗T)3−21C1∗TC2∗T+31C3∗T⟸⎩ ⎨ ⎧C1∗T=T∙ii=C1T 21[(C1∗T)2−C2∗T]=21(T∙iiT∙jj−2T∙jiT∙ij)=C2T 61(C1∗T)3−21C1∗TC2∗T+31C3∗T=61T∙iiT∙jjT∙kk−21T∙kkT∙ijT∙ji+31T∙ikT∙jiT∙kj=C3T