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  • 二阶张量矩阵表示
    2022-08-16 17:24:20

    1. 二阶张量的代数运算与矩阵的代数运算

    4.1. 二阶张量的相等、加(减)、数乘

    二阶张量的相等、加(减)、数乘运算与矩阵相等、加(减)、数乘运算一 一对应;

    1.2. 二阶张量的缩并

    与二阶张量的缩并相关的运算为求二阶张量的迹 t r ( T ) tr(\bold{T}) tr(T)
    t r ( T ) = T i j g i j = T i ∙ i = T 1 ∙ 1 + T 2 ∙ 2 + T 3 ∙ 3 = T ∙ i i = T ∙ 1 1 + T ∙ 2 2 + T ∙ 3 3 = T i j g i j tr(\bold{T}) =T^{ij}g_{ij} =T_i^{\bullet i} =T_1^{\bullet 1}+T_2^{\bullet 2}+T_3^{\bullet 3} =T^i_{\bullet i} =T^1_{\bullet 1}+T^2_{\bullet 2}+T^3_{\bullet 3} =T_{ij}g^{ij} tr(T)=Tijgij=Tii=T11+T22+T33=Tii=T11+T22+T33=Tijgij显然,与之相关的矩阵运算为求方阵 τ 2 、 τ 3 \tau_2、\tau_3 τ2τ3 的迹

    1.3. 二阶张量与矢量的点积 —— 线性变换

    w ⃗ = T ∙ u ⃗ ⟺ w i = T ∙ j i u j ⟺ [ w 1 w 2 w 3 ] = [ 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 ] [ u 1 u 2 u 3 ] ⟺ w ⃗ = τ 3 u ⃗   t ⃗ = u ⃗ ∙ T ⟺ t i = u j T j ∙ i ⟺ [ t 1 t 2 t 3 ] = [ 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 ] [ u 1 u 2 u 3 ] ⟺ t ⃗ = τ 2 u ⃗ \vec{w}=\bold{T}\bullet\vec{u} \Longleftrightarrow w^i=T^{i}_{\bullet j}u^j \Longleftrightarrow \begin{bmatrix}w^1\\\\w^2\\\\w^3\end{bmatrix} =\begin{bmatrix} 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{bmatrix} \begin{bmatrix}u^1\\\\u^2\\\\u^3\end{bmatrix} \Longleftrightarrow \vec{w}=\tau_{3}\vec{u} \\\ \\ \vec{t}=\vec{u}\bullet\bold{T} \Longleftrightarrow t^i=u^jT^{\bullet i}_{j} \Longleftrightarrow \begin{bmatrix}t^1\\\\t^2\\\\t^3\end{bmatrix} =\begin{bmatrix} T_{1}^{\bullet 1} & T_{2}^{\bullet 1} & T_{3}^{\bullet 1} \\ \\ T_{1}^{\bullet 2} & T_{2}^{\bullet 2} & T_{3}^{\bullet 2} \\ \\ T_{1}^{\bullet 3} & T_{2}^{\bullet 3} & T_{3}^{\bullet 3} \end{bmatrix} \begin{bmatrix}u^1\\\\u^2\\\\u^3\end{bmatrix} \Longleftrightarrow \vec{t}=\tau_{2}\vec{u} w =Tu wi=Tjiuj w1w2w3 = T11T12T13T21T22T23T31T32T33 u1u2u3 w =τ3u  t =u Tti=ujTji t1t2t3 = T11T12T13T21T22T23T31T32T33 u1u2u3 t =τ2u 显然,二阶张量与矢量的左、右点积一般不等: T ∙ u ⃗ ≠ u ⃗ ∙ T \bold{T}\bullet\vec{u}\ne\vec{u}\bullet\bold{T} Tu =u T且有:
    ( τ T ) 3 u ⃗ = τ 2 u ⃗ ⟺ T T ∙ u ⃗ = u ⃗ ∙ T (\tau^T)_3\vec{u}=\tau_{2}\vec{u} \Longleftrightarrow \bold{T}^T\bullet\vec{u}=\vec{u}\bullet\bold{T} (τT)3u =τ2u TTu =u T那么有:
    N ∙ u ⃗ = u ⃗ ∙ N ( N 为对称二阶张量 )   Ω ∙ u ⃗ = − u ⃗ ∙ Ω ( Ω 为反对称二阶张量 ) \bold{N}\bullet\vec{u}=\vec{u}\bullet\bold{N} \qquad(\bold{N}为对称二阶张量)\\\ \\ \bold{\Omega}\bullet\vec{u}=-\vec{u}\bullet\bold{\Omega} \qquad(\bold{\Omega}为反对称二阶张量) Nu =u N(N为对称二阶张量) Ωu =u Ω(Ω为反对称二阶张量)与矩阵与列向量的乘法相同,二阶张量可将任意向量映射为其它的向量,故也将二阶张量与矢量的点积称作线性变换。另外,任意对称二阶张量也对应着一个二次型,即:
    x ⃗ ∙ N ∙ x ⃗ = N : x ⃗ x ⃗ = N i j x i x j = [ x 1 x 2 x 3 ] [ N 11 N 12 N 13 N 21 N 22 N 23 N 31 N 32 N 33 ] [ x 1 x 2 x 3 ] = x ⃗ T N 1 x ⃗ \vec{x}\bullet\bold{N}\bullet\vec{x} =\bold{N}:\vec{x}\vec{x} =N_{ij}x^ix^j =\begin{bmatrix}x^1 & x^2 & x^3\end{bmatrix} \begin{bmatrix} N_{11} & N_{12} & N_{13} \\ \\ N_{21} & N_{22} & N_{23} \\ \\ N_{31} & N_{32} & N_{33} \end{bmatrix} \begin{bmatrix}x^1 \\\\ x^2 \\\\ x^3\end{bmatrix} =\vec{x}^TN_{1}\vec{x} x Nx =N:x x =Nijxixj=[x1x2x3] N11N21N31N12N22N32N13N23N33 x1x2x3 =x TN1x

    1.4. 二阶张量与二阶张量的点积

    二阶张量的点积采用分量形式有:
    C i j = A i ∙ k B k j = A i k B ∙ j k ⟺ C 1 = [ C i j ] = [ A 1 ∙ 1 A 2 ∙ 1 A 3 ∙ 1 A 1 ∙ 2 A 2 ∙ 2 A 3 ∙ 2 A 1 ∙ 3 A 2 ∙ 3 A 3 ∙ 3 ] T [ B 11 B 12 B 13 B 21 B 22 B 23 B 31 B 32 B 33 ] = A 2 T B 1 = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] [ B ∙ 1 1 B ∙ 2 1 B ∙ 3 1 B ∙ 1 2 B ∙ 2 2 B ∙ 3 2 B ∙ 1 3 B ∙ 2 3 B ∙ 3 3 ] = A 1 B 3   C i ∙ j = A i ∙ k B k ∙ j = A i k B k j ⟺ C 2 = [ C i ∙ j ] = [ A 1 ∙ 1 A 2 ∙ 1 A 3 ∙ 1 A 1 ∙ 2 A 2 ∙ 2 A 3 ∙ 2 A 1 ∙ 3 A 2 ∙ 3 A 3 ∙ 3 ] [ B 1 ∙ 1 B 2 ∙ 1 B 3 ∙ 1 B 1 ∙ 2 B 2 ∙ 2 B 3 ∙ 2 B 1 ∙ 3 B 2 ∙ 3 B 3 ∙ 3 ] = A 2 B 2 = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] [ B 11 B 12 B 13 B 21 B 22 B 23 B 31 B 32 B 33 ] = A 1 B 4   C ∙ j i = A ∙ k i B ∙ j k = A i k B k j ⟺ C 3 = [ C ∙ j i ] = [ A ∙ 1 1 A ∙ 2 1 A ∙ 3 1 A ∙ 1 2 A ∙ 2 2 A ∙ 3 2 A ∙ 1 3 A ∙ 2 3 A ∙ 3 3 ] [ B ∙ 1 1 B ∙ 2 1 B ∙ 3 1 B ∙ 1 2 B ∙ 2 2 B ∙ 3 2 B ∙ 1 3 B ∙ 2 3 B ∙ 3 3 ] = A 3 B 3 = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] [ B 11 B 12 B 13 B 21 B 22 B 23 B 31 B 32 B 33 ] = A 4 B 1   C i j = A ∙ k i B k j = A i k B k ∙ j ⟺ C 2 = [ C i j ] = [ A ∙ 1 1 A ∙ 2 1 A ∙ 3 1 A ∙ 1 2 A ∙ 2 2 A ∙ 3 2 A ∙ 1 3 A ∙ 2 3 A ∙ 3 3 ] [ B 11 B 12 B 13 B 21 B 22 B 23 B 31 B 32 B 33 ] = A 3 B 4 = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] [ B 1 ∙ 1 B 2 ∙ 1 B 3 ∙ 1 B 1 ∙ 2 B 2 ∙ 2 B 3 ∙ 2 B 1 ∙ 3 B 2 ∙ 3 B 3 ∙ 3 ] T = A 4 B 2 T C_{ij}=A_{i}^{\bullet k}B_{k j}=A_{ik}B^k_{\bullet j} \Longleftrightarrow C_{1}=[C_{ij}] =\begin{bmatrix} A_{1}^{\bullet 1} & A_{2}^{\bullet 1} & A_{3}^{\bullet 1} \\\\ A_{1}^{\bullet 2} & A_{2}^{\bullet 2} & A_{3}^{\bullet 2} \\\\ A_{1}^{\bullet 3} & A_{2}^{\bullet 3} & A_{3}^{\bullet 3} \end{bmatrix}^T \begin{bmatrix} B_{11} & B_{12} & B_{13} \\\\ B_{21} & B_{22} & B_{23} \\\\ B_{31} & B_{32} & B_{33} \end{bmatrix} =A_{2}^TB_{1} =\begin{bmatrix} A_{11} & A_{12} & A_{13} \\\\ A_{21} & A_{22} & A_{23} \\\\ A_{31} & A_{32} & A_{33} \end{bmatrix} \begin{bmatrix} B^1_{\bullet 1} & B^1_{\bullet 2} & B^1_{\bullet 3} \\\\ B^2_{\bullet 1} & B^2_{\bullet 2} & B^2_{\bullet 3} \\\\ B^3_{\bullet 1} & B^3_{\bullet 2} & B^3_{\bullet 3} \end{bmatrix} =A_{1}B_{3}\\\ \\ %%%%%%%%%%%%%%%% C_{i}^{\bullet j}=A_{i}^{\bullet k}B_k^{\bullet j}=A_{ik}B^{kj} \Longleftrightarrow C_{2}=[C_{i}^{\bullet j}] =\begin{bmatrix} A_{1}^{\bullet 1} & A_{2}^{\bullet 1} & A_{3}^{\bullet 1} \\\\ A_{1}^{\bullet 2} & A_{2}^{\bullet 2} & A_{3}^{\bullet 2} \\\\ A_{1}^{\bullet 3} & A_{2}^{\bullet 3} & A_{3}^{\bullet 3} \end{bmatrix} \begin{bmatrix} B_{1}^{\bullet 1} & B_{2}^{\bullet 1} & B_{3}^{\bullet 1} \\\\ B_{1}^{\bullet 2} & B_{2}^{\bullet 2} & B_{3}^{\bullet 2} \\\\ B_{1}^{\bullet 3} & B_{2}^{\bullet 3} & B_{3}^{\bullet 3} \end{bmatrix} =A_{2}B_{2} =\begin{bmatrix} A_{11} & A_{12} & A_{13} \\\\ A_{21} & A_{22} & A_{23} \\\\ A_{31} & A_{32} & A_{33} \end{bmatrix} \begin{bmatrix} B^{11} & B^{12} & B^{13} \\\\ B^{21} & B^{22} & B^{23} \\\\ B^{31} & B^{32} & B^{33} \end{bmatrix} =A_{1}B_{4}\\\ \\ %%%%%%%%%%%%%%%% C_{\bullet j}^{i}=A^{i}_{\bullet k}B^k_{\bullet j}=A^{ik}B_{kj} \Longleftrightarrow C_{3}=[C_{\bullet j}^{i}] =\begin{bmatrix} A^{1}_{\bullet 1} & A^{1}_{\bullet 2}& A^{1}_{\bullet 3} \\\\ A^{2}_{\bullet 1} & A^{2}_{\bullet 2}& A^{2}_{\bullet 3} \\\\ A^{3}_{\bullet 1} & A^{3}_{\bullet 2}& A^{3}_{\bullet 3} \end{bmatrix} \begin{bmatrix} B^{1}_{\bullet 1} & B^{1}_{\bullet 2}& B^{1}_{\bullet 3} \\\\ B^{2}_{\bullet 1} & B^{2}_{\bullet 2}& B^{2}_{\bullet 3} \\\\ B^{3}_{\bullet 1} & B^{3}_{\bullet 2}& B^{3}_{\bullet 3} \end{bmatrix} =A_{3}B_{3} =\begin{bmatrix} A^{11} & A^{12} & A^{13} \\\\ A^{21} & A^{22} & A^{23} \\\\ A^{31} & A^{32} & A^{33} \end{bmatrix} \begin{bmatrix} B_{11} & B_{12} & B_{13} \\\\ B_{21} & B_{22} & B_{23} \\\\ B_{31} & B_{32} & B_{33} \end{bmatrix} =A_{4}B_{1}\\\ \\ %%%%%%%%%%%%%%%% C^{ij}=A^{i}_{\bullet k}B^{kj}=A^{ik}B_k^{\bullet j} \Longleftrightarrow C_{2}=[C^{ij}] =\begin{bmatrix} A^{1}_{\bullet 1} & A^{1}_{\bullet 2}& A^{1}_{\bullet 3} \\\\ A^{2}_{\bullet 1} & A^{2}_{\bullet 2}& A^{2}_{\bullet 3} \\\\ A^{3}_{\bullet 1} & A^{3}_{\bullet 2}& A^{3}_{\bullet 3} \end{bmatrix} \begin{bmatrix} B^{11} & B^{12} & B^{13} \\\\ B^{21} & B^{22} & B^{23} \\\\ B^{31} & B^{32} & B^{33} \end{bmatrix} =A_{3}B_{4} =\begin{bmatrix} A^{11} & A^{12} & A^{13} \\\\ A^{21} & A^{22} & A^{23} \\\\ A^{31} & A^{32} & A^{33} \end{bmatrix} \begin{bmatrix} B_{1}^{\bullet 1} & B_{2}^{\bullet 1} & B_{3}^{\bullet 1} \\\\ B_{1}^{\bullet 2} & B_{2}^{\bullet 2} & B_{3}^{\bullet 2} \\\\ B_{1}^{\bullet 3} & B_{2}^{\bullet 3} & B_{3}^{\bullet 3} \end{bmatrix}^T =A_{4}B_{2}^T %%%%%%%%%%%%%%%% Cij=AikBkj=AikBjkC1=[Cij]= A11A12A13A21A22A23A31A32A33 T B11B21B31B12B22B32B13B23B33 =A2TB1= A11A21A31A12A22A32A13A23A33 B11B12B13B21B22B23B31B32B33 =A1B3 Cij=AikBkj=AikBkjC2=[Cij]= A11A12A13A21A22A23A31A32A33 B11B12B13B21B22B23B31B32B33 =A2B2= A11A21A31A12A22A32A13A23A33 B11B21B31B12B22B32B13B23B33 =A1B4 Cji=AkiBjk=AikBkjC3=[Cji]= A11A12A13A21A22A23A31A32A33 B11B12B13B21B22B23B31B32B33 =A3B3= A11A21A31A12A22A32A13A23A33 B11B21B31B12B22B32B13B23B33 =A4B1 Cij=AkiBkj=AikBkjC2=[Cij]= A11A12A13A21A22A23A31A32A33 B11B21B31B12B22B32B13B23B33 =A3B4= A11A21A31A12A22A32A13A23A33 B11B12B13B21B22B23B31B32B33 T=A4B2T
    由两个二阶张量点乘与矩阵乘法的对应关系也可得到:张量点积不可随意更换次序,即
    A ∙ B ≠ B ∙ A \bold{A}\bullet\bold{B}\ne\bold{B}\bullet\bold{A} AB=BA另外,可借助张量点积与矩阵乘法的对应关系推导出如下类似于矩阵乘法与矩阵转置关系的张量点积与张量转置的关系式
    ( A 3 B 3 ) T = ( B 3 ) T ( A 3 ) T = ( B T ) 2 ( A T ) 2 ⟺ ( A ∙ B ) T = B T ∙ A T (A_3B_3)^T=(B_3)^T(A_3)^T=(B^T)_2(A^T)_2 \Longleftrightarrow (\bold{A}\bullet\bold{B})^T=\bold{B}^T\bullet\bold{A}^T (A3B3)T=(B3)T(A3)T=(BT)2(AT)2(AB)T=BTAT张量点积的行列式与张量行列式的关系式 d e t ( A 3 B 3 ) = d e t ( A 3 ) d e t ( B 3 ) ⟺ d e t ( A ∙ B ) = d e t ( A ) d e t ( B ) det(A_3B_3)=det(A_3)det(B_3) \Longleftrightarrow det(\bold{A}\bullet\bold{B})=det(\bold{A})det(\bold{B}) det(A3B3)=det(A3)det(B3)det(AB)=det(A)det(B)

    2. 正则二阶张量与可逆矩阵

    若二阶张量 T \bold T T 的矩阵可逆,则称二阶张量正则,反之称作退化的二阶张量
    显然,二阶张量正则的充要条件为二阶张量的行列式不为零
    则有: T \bold T T如果正则,那么 T T \bold T^T TT也正则
    另外,根据正则张量、张量点积与可逆矩阵、矩阵乘法的对应关系可知:对于正则的二阶张量 T \bold{T} T,必唯一存在正则的二阶张量 T − 1 \bold{T}^{-1} T1,使得:
    T ∙ T − 1 = T − 1 ∙ T = G ( 1 ) \bold{T}\bullet\bold{T}^{-1}=\bold{T}^{-1}\bullet\bold{T}=\bold{G}\qquad(1) TT1=T1T=G(1) T − 1 \bold{T}^{-1} T1称作正则张量 T \bold{T} T的逆张量,且逆张量的矩阵等于原张量矩阵的逆,即
    ( τ − 1 ) 3 = ( τ 3 ) − 1 (\tau^{-1})_3=(\tau_3)^{-1} (τ1)3=(τ3)1则,
    d e t [ ( τ − 1 ) 3 ] = d e t [ ( τ 3 ) − 1 ] = 1 d e t ( τ 3 ) ⟺ d e t ( T − 1 ) = 1 d e t ( T ) det[(\tau^{-1})_3]=det[(\tau_3)^{-1}]=\frac{1}{det(\tau_3)} \Longleftrightarrow det(\bold{T}^{-1})=\frac{1}{det(\bold{T})} det[(τ1)3]=det[(τ3)1]=det(τ3)1det(T1)=det(T)1此外,进一步根据(1)式可得:
    ( T − 1 ) − 1 = T   G T = ( T ∙ T − 1 ) T = ( T − 1 ) T ∙ T T = G = ( T T ) − 1 ∙ T T (\bold{T}^{-1})^{-1}=\bold{T}\\\ \\ \bold{G}^T =(\bold{T}\bullet\bold{T}^{-1})^T =(\bold{T}^{-1})^T\bullet\bold{T}^T =\bold{G} =(\bold{T}^T)^{-1}\bullet\bold{T}^T (T1)1=T GT=(TT1)T=(T1)TTT=G=(TT)1TT由逆张量的唯一性可知:
    ( T T ) − 1 = ( T − 1 ) T (\bold{T}^T)^{-1}=(\bold{T}^{-1})^T (TT)1=(T1)T若二阶张量线性变换可逆,则:
    w ⃗ = τ 3 u ⃗ ⟺ w ⃗ = T ∙ u ⃗   u ⃗ = ( τ 3 ) − 1 w ⃗ = ( τ − 1 ) 3 w ⃗ ⟺ u ⃗ = T − 1 ∙ w ⃗ \vec{w}=\tau_3\vec{u} \Longleftrightarrow \vec{w}=T\bullet\vec{u}\\\ \\ \vec{u}=(\tau_3)^{-1}\vec{w}=(\tau^{-1})_3\vec{w} \Longleftrightarrow \vec{u}=T^{-1}\bullet\vec{w} w =τ3u w =Tu  u =(τ3)1w =(τ1)3w u =T1w

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  • 本文主要内容如下: 1. 二阶张量矩阵表示 2. 二阶张量转置与矩阵转置 3. 二阶张量的行列式与矩阵的行列式

    1. 二阶张量的矩阵表示

    任意二阶张量含有九个分量,正好可以通过三阶方阵进行表示,但二阶张量具有四种不同的分量形式,不同的分量对应于不同的方阵:

    τ 1 = [ T i j ] = [ T 11 T 12 T 13 T 21 T 22 T 23 T 31 T 32 T 33 ]   τ 2 = [ T i ∙ j ] = [ 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 ]   τ 3 = [ T ∙ j i ] = [ 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 ]   τ 4 = [ T i j ] = [ T 11 T 12 T 13 T 21 T 22 T 23 T 31 T 32 T 33 ] %%%%%%%%%%% \tau_1=[T_{ij}]= \begin{bmatrix} T_{11} & T_{12} & T_{13} \\ \\ T_{21} & T_{22} & T_{23} \\ \\ T_{31} & T_{32} & T_{33} \end{bmatrix} \\\ \\ %%%%%%%%%%%% \tau_2=[T_{i}^{\bullet j}]= \begin{bmatrix} T_{1}^{\bullet 1} & T_{2}^{\bullet 1} & T_{3}^{\bullet 1} \\ \\ T_{1}^{\bullet 2} & T_{2}^{\bullet 2} & T_{3}^{\bullet 2} \\ \\ T_{1}^{\bullet 3} & T_{2}^{\bullet 3} & T_{3}^{\bullet 3} \end{bmatrix} \\\ \\ %%%%%%%%%%%% \tau_3=[T^{i}_{\bullet j}]= \begin{bmatrix} 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{bmatrix} \\\ \\ %%%%%%%%%%%% \tau_4=[T^{ij}]= \begin{bmatrix} T^{11} & T^{12} & T^{13} \\ \\ T^{21} & T^{22} & T^{23} \\ \\ T^{31} & T^{32} & T^{33} \end{bmatrix} τ1=[Tij]= T11T21T31T12T22T32T13T23T33  τ2=[Tij]= T11T12T13T21T22T23T31T32T33  τ3=[Tji]= T11T12T13T21T22T23T31T32T33  τ4=[Tij]= T11T21T31T12T22T32T13T23T33

    注:本文采用前(上)指标为行编号,后(下)指标为列编号的对应规则,而在《张量分析》-黄克智中采用前指标为行编号的对应规则,两种规则的区别仅在于 τ 2 \tau_{2} τ2的不同,两种规则得到的 τ 2 \tau_{2} τ2互为矩阵转置的关系。

    特别地,度量张量也是二阶张量,其矩阵形式为:
    G 1 = [ g i j ] = [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ]   G 2 = G 3 = [ δ j i ] = [ 1 0 0 0 1 0 0 0 1 ] = E   G 4 = [ g i j ] = [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ] = G 1 − 1 \mathscr{G}_{1}= [g_{ij}]= \begin{bmatrix} g_{11} & g_{12} & g_{13} \\\ \\ g_{21} & g_{22} & g_{23} \\\ \\ g_{31} & g_{32} & g_{33} \end{bmatrix}\\\ \\ %%%%%%%%%%%% \mathscr{G}_{2}= \mathscr{G}_{3}= [\delta^i_j]= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =E\\\ \\ %%%%%%%%%%%% \mathscr{G}_{4}= [g^{ij}]= \begin{bmatrix} g^{11} & g^{12} & g^{13} \\\ \\ g^{21} & g^{22} & g^{23} \\\ \\ g^{31} & g^{32} & g^{33} \end{bmatrix} =\mathscr{G}_{1}^{-1} G1=[gij]= g11 g21 g31g12g22g32g13g23g33  G2=G3=[δji]= 100010001 =E G4=[gij]= g11 g21 g31g12g22g32g13g23g33 =G11注意到,在一般坐标系下,二阶张量对应的四个矩阵是不相等的,它们之间的转换通过指标升降关系的矩阵形式来实现:
    [ T 11 T 12 T 13 T 21 T 22 T 23 T 31 T 32 T 33 ] =   [ 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 ] T [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ] =   [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ] [ 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 ] =   [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ] [ T 11 T 12 T 13 T 21 T 22 T 23 T 31 T 32 T 33 ] [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ] \begin{bmatrix} T_{11} & T_{12} & T_{13} \\ \\ T_{21} & T_{22} & T_{23} \\ \\ T_{31} & T_{32} & T_{33} \end{bmatrix} =\ \begin{bmatrix} T_{1}^{\bullet 1} & T_{2}^{\bullet 1} & T_{3}^{\bullet 1} \\ \\ T_{1}^{\bullet 2} & T_{2}^{\bullet 2} & T_{3}^{\bullet 2} \\ \\ T_{1}^{\bullet 3} & T_{2}^{\bullet 3} & T_{3}^{\bullet 3} \end{bmatrix}^T \begin{bmatrix} g_{11} & g_{12} & g_{13} \\\ \\ g_{21} & g_{22} & g_{23} \\\ \\ g_{31} & g_{32} & g_{33} \end{bmatrix} =\ \begin{bmatrix} g_{11} & g_{12} & g_{13} \\\ \\ g_{21} & g_{22} & g_{23} \\\ \\ g_{31} & g_{32} & g_{33} \end{bmatrix} \begin{bmatrix} 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{bmatrix} =\ \begin{bmatrix} g_{11} & g_{12} & g_{13} \\\ \\ g_{21} & g_{22} & g_{23} \\\ \\ g_{31} & g_{32} & g_{33} \end{bmatrix} \begin{bmatrix} T^{11} & T^{12} & T^{13} \\ \\ T^{21} & T^{22} & T^{23} \\ \\ T^{31} & T^{32} & T^{33} \end{bmatrix} \begin{bmatrix} g_{11} & g_{12} & g_{13} \\\ \\ g_{21} & g_{22} & g_{23} \\\ \\ g_{31} & g_{32} & g_{33} \end{bmatrix} T11T21T31T12T22T32T13T23T33 =  T11T12T13T21T22T23T31T32T33 T g11 g21 g31g12g22g32g13g23g33 =  g11 g21 g31g12g22g32g13g23g33 T11T12T13T21T22T23T31T32T33 =  g11 g21 g31g12g22g32g13g23g33 T11T21T31T12T22T32T13T23T33 g11 g21 g31g12g22g32g13g23g33 即:
    τ 1 = τ 2 T G 1 = G 1 τ 3 = G 1 τ 4 G 1 \tau_{1} =\tau_{2}^T\mathscr{G}_{1} =\mathscr{G}_{1}\tau_{3} =\mathscr{G}_{1}\tau_{4}\mathscr{G}_{1} τ1=τ2TG1=G1τ3=G1τ4G1那么:
    G 1 − 1 τ 2 T G 1 = τ 3 ⟹ τ 2 T ∼ τ 3 \mathscr{G}_{1}^{-1}\tau_{2}^T\mathscr{G}_{1}=\tau_3\Longrightarrow \tau_{2}^T\sim\tau_{3} G11τ2TG1=τ3τ2Tτ3在笛卡尔坐标系中,由于 G 1 = E \mathscr{G}_{1}=E G1=E 故在笛卡尔坐标系中有: τ 1 = τ 2 T = τ 3 = τ 4 \tau_{1}=\tau_{2}^T=\tau_{3}=\tau_{4} τ1=τ2T=τ3=τ4通常如不加说明,定义 τ 3 \tau_{3} τ3为张量的矩阵。

    2. 二阶张量转置与矩阵转置

    二阶转置张量的分量形式为:
    ( T T ) i j = T j i ⟹ ( τ T ) 1 = [ ( T T ) i j ] = [ ( T T ) 11 ( T T ) 12 ( T T ) 13 ( T T ) 21 ( T T ) 22 ( T T ) 23 ( T T ) 31 ( T T ) 32 ( T T ) 33 ] = [ T 11 T 21 T 31 T 12 T 22 T 32 T 13 T 23 T 33 ] = ( τ 1 ) T   ( T T ) i ∙ j = T ∙ i j ⟹ ( τ T ) 2 = [ ( T T ) i ∙ j ] = [ ( T T ) 1 ∙ 1 ( T T ) 2 ∙ 1 ( T T ) 3 ∙ 1 ( T T ) 1 ∙ 2 ( T T ) 2 ∙ 2 ( T T ) 3 ∙ 2 ( T T ) 1 ∙ 3 ( T T ) 2 ∙ 3 ( T T ) 3 ∙ 3 ] = [ 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 ] = τ 3   ( T T ) ∙ j i = T j ∙ i ⟹ ( τ T ) 3 = [ ( T T ) ∙ j i ] = [ ( T T ) ∙ 1 1 ( T T ) ∙ 2 1 ( T T ) ∙ 3 1 ( T T ) ∙ 1 2 ( T T ) ∙ 2 2 ( T T ) ∙ 3 2 ( T T ) ∙ 2 3 ( T T ) ∙ 2 3 ( T T ) ∙ 3 3 ] = [ 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 ] = τ 2   ( T T ) i j = T j i ⟹ ( τ T ) 4 = [ ( T T ) i j ] = [ ( T T ) 11 ( T T ) 12 ( T T ) 13 ( T T ) 21 ( T T ) 22 ( T T ) 23 ( T T ) 31 ( T T ) 32 ( T T ) 33 ] = [ T 11 T 21 T 31 T 12 T 22 T 32 T 13 T 23 T 33 ] = ( τ 4 ) T (T^{T})_{ij}=T_{ji} \Longrightarrow (\tau^T)_{1}=[(T^{T})_{ij}] =\begin{bmatrix} (T^{T})_{11} & (T^{T})_{12} & (T^{T})_{13} \\ \\ (T^{T})_{21} & (T^{T})_{22} & (T^{T})_{23} \\ \\ (T^{T})_{31} & (T^{T})_{32} & (T^{T})_{33} \end{bmatrix} =\begin{bmatrix} T_{11} & T_{21} & T_{31} \\ \\ T_{12} & T_{22} & T_{32} \\ \\ T_{13} & T_{23} & T_{33} \end{bmatrix} =(\tau_{1})^T\\\ \\ %%%%%%%%%%%% (T^{T})_{i}^{\bullet j}=T^{j}_{\bullet i} \Longrightarrow (\tau^T)_{2}=[(T^{T})_{i}^{\bullet j}] =\begin{bmatrix} (T^{T})_{1}^{\bullet 1} & (T^{T})_{2}^{\bullet 1} & (T^{T})_{3}^{\bullet 1} \\ \\ (T^{T})_{1}^{\bullet 2} & (T^{T})_{2}^{\bullet 2} & (T^{T})_{3}^{\bullet 2} \\ \\ (T^{T})_{1}^{\bullet 3} & (T^{T})_{2}^{\bullet 3} & (T^{T})_{3}^{\bullet 3} \end{bmatrix} =\begin{bmatrix} 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{bmatrix} =\tau_{3}\\\ \\ %%%%%%%%%%%% (T^{T})^{i}_{\bullet j}=T_{j}^{\bullet i} \Longrightarrow (\tau^T)_{3}=[(T^{T})^{i}_{\bullet j}] =\begin{bmatrix} (T^{T})^{1}_{\bullet 1} & (T^{T})^{1}_{\bullet 2} & (T^{T})^{1}_{\bullet 3} \\ \\ (T^{T})^{2}_{\bullet 1} & (T^{T})^{2}_{\bullet 2} & (T^{T})^{2}_{\bullet 3} \\ \\ (T^{T})^{3}_{\bullet 2} & (T^{T})^{3}_{\bullet 2} & (T^{T})^{3}_{\bullet 3} \end{bmatrix} =\begin{bmatrix} T_{1}^{\bullet 1} & T_{2}^{\bullet 1} & T_{3}^{\bullet 1} \\ \\ T_{1}^{\bullet 2} & T_{2}^{\bullet 2} & T_{3}^{\bullet 2} \\ \\ T_{1}^{\bullet 3} & T_{2}^{\bullet 3} & T_{3}^{\bullet 3} \end{bmatrix} =\tau_{2} \\\ \\ %%%%%%%%%%%% (T^{T})^{ij}=T^{ji} \Longrightarrow (\tau^T)_{4}=[(T^{T})^{ij}] =\begin{bmatrix} (T^{T})^{11} & (T^{T})^{12} & (T^{T})^{13} \\ \\ (T^{T})^{21} & (T^{T})^{22} & (T^{T})^{23} \\ \\ (T^{T})^{31} & (T^{T})^{32} & (T^{T})^{33} \end{bmatrix} =\begin{bmatrix} T^{11} & T^{21} & T^{31} \\ \\ T^{12} & T^{22} & T^{32} \\ \\ T^{13} & T^{23} & T^{33} \end{bmatrix} =(\tau_{4})^T (TT)ij=Tji(τT)1=[(TT)ij]= (TT)11(TT)21(TT)31(TT)12(TT)22(TT)32(TT)13(TT)23(TT)33 = T11T12T13T21T22T23T31T32T33 =(τ1)T (TT)ij=Tij(τT)2=[(TT)ij]= (TT)11(TT)12(TT)13(TT)21(TT)22(TT)23(TT)31(TT)32(TT)33 = T11T12T13T21T22T23T31T32T33 =τ3 (TT)ji=Tji(τT)3=[(TT)ji]= (TT)11(TT)12(TT)23(TT)21(TT)22(TT)23(TT)31(TT)32(TT)33 = T11T12T13T21T22T23T31T32T33 =τ2 (TT)ij=Tji(τT)4=[(TT)ij]= (TT)11(TT)21(TT)31(TT)12(TT)22(TT)32(TT)13(TT)23(TT)33 = T11T12T13T21T22T23T31T32T33 =(τ4)T这说明若两个张量互为转置,则它们的 τ 1 、 τ 4 \tau_1、\tau_4 τ1τ4 矩阵也互为转置,而转置张量的 τ 2 \tau_2 τ2 矩阵与原张量的 τ 3 \tau_3 τ3 矩阵相等,转置张量的 τ 3 \tau_3 τ3 矩阵与原张量的 τ 2 \tau_2 τ2 矩阵相等

    • 更特别地,对于对称张量 N \bold{N} N有:
      ( N T ) i j = N i j ⟹ ( N 1 ) T = ( N T ) 1 = [ ( N T ) i j ] = [ ( N T ) 11 ( N T ) 12 ( N T ) 13 ( N T ) 21 ( N T ) 22 ( N T ) 23 ( N T ) 31 ( N T ) 32 ( N T ) 33 ] = [ N 11 N 12 N 13 N 21 N 22 N 23 N 31 N 32 N 33 ] = N 1   ( N T ) i ∙ j = N i ∙ j ⟹ N 3 = ( N T ) 2 = [ ( N T ) i ∙ j ] = [ ( N T ) 1 ∙ 1 ( N T ) 2 ∙ 1 ( N T ) 3 ∙ 1 ( N T ) 1 ∙ 2 ( N T ) 2 ∙ 2 ( N T ) 3 ∙ 2 ( N T ) 1 ∙ 3 ( N T ) 2 ∙ 3 ( N T ) 3 ∙ 3 ] = [ N 1 ∙ 1 N 2 ∙ 1 N 3 ∙ 1 N 1 ∙ 2 N 2 ∙ 2 N 3 ∙ 2 N 1 ∙ 3 N 2 ∙ 3 N 3 ∙ 3 ] = N 2   ( N T ) ∙ j i = N ∙ j i ⟹ N 2 = ( N T ) 3 = [ ( N T ) ∙ j i ] = [ ( N T ) ∙ 1 1 ( N T ) ∙ 2 1 ( N T ) ∙ 3 1 ( N T ) ∙ 1 2 ( N T ) ∙ 2 2 ( N T ) ∙ 3 2 ( N T ) ∙ 2 3 ( N T ) ∙ 2 3 ( N T ) ∙ 3 3 ] = [ N ∙ 1 1 N ∙ 2 1 N ∙ 3 1 N ∙ 1 2 N ∙ 2 2 N ∙ 3 2 N ∙ 2 3 N ∙ 2 3 N ∙ 3 3 ] = N 3   ( N T ) i j = N i j ⟹ ( N 4 ) T = ( N T ) 4 = [ ( N T ) i j ] = [ ( N T ) 11 ( N T ) 12 ( N T ) 13 ( N T ) 21 ( N T ) 22 ( N T ) 23 ( N T ) 31 ( N T ) 32 ( N T ) 33 ] = [ N 11 N 12 N 13 N 21 N 22 N 23 N 31 N 32 N 33 ] = N 4 (N^T)_{ij}=N_{ij} \Longrightarrow (N_{1})^T=(N^T)_{1}=[(N^{T})_{ij}] =\begin{bmatrix} (N^{T})_{11} & (N^{T})_{12} & (N^{T})_{13} \\ \\ (N^{T})_{21} & (N^{T})_{22} & (N^{T})_{23} \\ \\ (N^{T})_{31} & (N^{T})_{32} & (N^{T})_{33} \end{bmatrix} =\begin{bmatrix} N_{11} & N_{12} & N_{13} \\ \\ N_{21} & N_{22} & N_{23} \\ \\ N_{31} & N_{32} & N_{33} \end{bmatrix} =N_1\\\ \\ %%%%%%%%%%%% (N^{T})_{i}^{\bullet j}=N_{i}^{\bullet j} \Longrightarrow N_{3}=(N^T)_{2}=[(N^{T})_{i}^{\bullet j}] =\begin{bmatrix} (N^{T})_{1}^{\bullet 1} & (N^{T})_{2}^{\bullet 1} & (N^{T})_{3}^{\bullet 1} \\ \\ (N^{T})_{1}^{\bullet 2} & (N^{T})_{2}^{\bullet 2} & (N^{T})_{3}^{\bullet 2} \\ \\ (N^{T})_{1}^{\bullet 3} & (N^{T})_{2}^{\bullet 3} & (N^{T})_{3}^{\bullet 3} \end{bmatrix} =\begin{bmatrix} N_{1}^{\bullet 1} & N_{2}^{\bullet 1} & N_{3}^{\bullet 1} \\ \\ N_{1}^{\bullet 2} & N_{2}^{\bullet 2} & N_{3}^{\bullet 2} \\ \\ N_{1}^{\bullet 3} & N_{2}^{\bullet 3} & N_{3}^{\bullet 3} \end{bmatrix} =N_2 \\\ \\ %%%%%%%%%%%% (N^{T})^{i}_{\bullet j}=N^{i}_{\bullet j} \Longrightarrow N_2=(N^T)_{3}=[(N^{T})^{i}_{\bullet j}] =\begin{bmatrix} (N^{T})^{1}_{\bullet 1} & (N^{T})^{1}_{\bullet 2} & (N^{T})^{1}_{\bullet 3} \\ \\ (N^{T})^{2}_{\bullet 1} & (N^{T})^{2}_{\bullet 2} & (N^{T})^{2}_{\bullet 3} \\ \\ (N^{T})^{3}_{\bullet 2} & (N^{T})^{3}_{\bullet 2} & (N^{T})^{3}_{\bullet 3} \end{bmatrix} =\begin{bmatrix} N^{1}_{\bullet 1} & N^{1}_{\bullet 2} & N^{1}_{\bullet 3} \\ \\ N^{2}_{\bullet 1} & N^{2}_{\bullet 2} & N^{2}_{\bullet 3} \\ \\ N^{3}_{\bullet 2} & N^{3}_{\bullet 2} & N^{3}_{\bullet 3} \end{bmatrix} =N_{3} \\\ \\ %%%%%%%%%%%% (N^{T})^{ij}=N^{ij} \Longrightarrow (N_4)^T=(N^T)_{4}=[(N^{T})^{ij}] =\begin{bmatrix} (N^{T})^{11} & (N^{T})^{12} & (N^{T})^{13} \\ \\ (N^{T})^{21} & (N^{T})^{22} & (N^{T})^{23} \\ \\ (N^{T})^{31} & (N^{T})^{32} & (N^{T})^{33} \end{bmatrix} =\begin{bmatrix} N^{11} & N^{12} & N^{13} \\ \\ N^{21} & N^{22} & N^{23} \\ \\ N^{31} & N^{32} & N^{33} \end{bmatrix} =N_{4} (NT)ij=Nij(N1)T=(NT)1=[(NT)ij]= (NT)11(NT)21(NT)31(NT)12(NT)22(NT)32(NT)13(NT)23(NT)33 = N11N21N31N12N22N32N13N23N33 =N1 (NT)ij=NijN3=(NT)2=[(NT)ij]= (NT)11(NT)12(NT)13(NT)21(NT)22(NT)23(NT)31(NT)32(NT)33 = N11N12N13N21N22N23N31N32N33 =N2 (NT)ji=NjiN2=(NT)3=[(NT)ji]= (NT)11(NT)12(NT)23