（二十三）特殊的四阶张量 ——四阶单位张量

张量 ——四阶单位张量"/>

（二十三）特殊的四阶张量 ——四阶单位张量

本文主要内容如下：

• 1. 四阶单位张量
• 2. 由四阶单位张量导出的其它特殊的四阶张量
• • 2.1 四阶单位张量的转置
  • 2.2 对称的四阶单位张量
  • 2.3 四阶投影张量

1. 四阶单位张量

考虑向量函数 f ( v ) = v \bm{f}(\bm{v})=\bm{v} f(v)=v，则
f ′ ( v ) [ u ] = lim ⁡ h → 0 f ( v + h u ) − f ( v ) h = u = I ⋅ u = u ⋅ I \bm{f}'(\bm{v})[\bm{u}]=\lim_{h\rightarrow0}\dfrac{\bm{f}(\bm{v}+h\bm{u})-\bm{f}(\bm{v})}{h}=\bm{u}=\mathbf{I}\cdot\bm{u}=\bm{u}\cdot\mathbf{I} f′(v)[u]=h→0lim​hf(v+hu)−f(v)​=u=I⋅u=u⋅I故可将二阶单位张量视为：
d v d v L = d v d v R = I = g k ⊗ g k = g k ⊗ g k {\dfrac{d\bm{v}}{d\bm{v}}}_{L}={\dfrac{d\bm{v}}{d\bm{v}}}_{R}=\mathbf{I}=\bm{g}^k\otimes\bm{g}_k=\bm{g}_k\otimes\bm{g}^k dvdv​L​=dvdv​R​=I=gk⊗gk​=gk​⊗gk类比考虑四阶单位张量的定义，对二阶张量函数 F ( A ) = A \mathbf{F}(\mathbf{A})=\mathbf{A} F(A)=A，则
F ′ ( A ) [ B ] = lim ⁡ h → 0 F ( A + h B ) − F ( A ) h = B = B i j g i ⊗ g j = B i j δ k i δ l j g k ⊗ g l = ( B i j g i ⊗ g j ) : ( g k ⊗ g l ⊗ g k ⊗ g l ) = B : ( g k ⊗ g l ⊗ g k ⊗ g l ) = ( g k ⊗ g l ⊗ g k ⊗ g l ) : ( B i j g i ⊗ g j ) = ( g k ⊗ g l ⊗ g k ⊗ g l ) : B \begin{align*} \mathbf{F}'(\mathbf{A})[\mathbf{B}] &=\lim_{h\rightarrow0}\dfrac{\mathbf{F}(\mathbf{A}+h\mathbf{B})-\mathbf{F}(\mathbf{A})}{h}\\[4mm] &=\mathbf{B} =B_{ij}\bm{g}^i\otimes\bm{g}^j =B_{ij}\delta_{k}^i\delta^j_{l}\bm{g}^k\otimes\bm{g}^l\\[2mm] &=(B_{ij}\bm{g}^i\otimes\bm{g}^j):(\bm g_{k}\otimes\bm{g}_l\otimes\bm{g}^k\otimes\bm{g}^l) =\mathbf{B}:(\bm g_{k}\otimes\bm{g}_l\otimes\bm{g}^k\otimes\bm{g}^l)\\[2mm] &=(\bm{g}^k\otimes\bm{g}^l\otimes\bm g_{k}\otimes\bm{g}_l):(B_{ij}\bm{g}^i\otimes\bm{g}^j) =(\bm{g}^k\otimes\bm{g}^l\otimes\bm g_{k}\otimes\bm{g}_l):\mathbf{B} \end{align*} F′(A)[B]​=h→0lim​hF(A+hB)−F(A)​=B=Bij​gi⊗gj=Bij​δki​δlj​gk⊗gl=(Bij​gi⊗gj):(gk​⊗gl​⊗gk⊗gl)=B:(gk​⊗gl​⊗gk⊗gl)=(gk⊗gl⊗gk​⊗gl​):(Bij​gi⊗gj)=(gk⊗gl⊗gk​⊗gl​):B​且
g k ⊗ g l ⊗ g k ⊗ g l = g k m g l n g k p g l q g m ⊗ g n ⊗ g p ⊗ g q = δ m p δ n q g m ⊗ g n ⊗ g p ⊗ g q = g m ⊗ g n ⊗ g m ⊗ g n = g k ⊗ g l ⊗ g k ⊗ g l g k ⊗ g l ⊗ g k ⊗ g l = g k m g k n g m ⊗ g l ⊗ g n ⊗ g l = δ m n g m ⊗ g l ⊗ g n ⊗ g l = g k ⊗ g l ⊗ g k ⊗ g l g k ⊗ g l ⊗ g k ⊗ g l = g l m g l n g k ⊗ g m ⊗ g k ⊗ g n = δ m n g k ⊗ g m ⊗ g k ⊗ g n = g k ⊗ g l ⊗ g k ⊗ g l \begin{align*} \bm g_{k}\otimes\bm{g}_l\otimes\bm{g}^k\otimes\bm{g}^l &=g_{km}g_{ln}g^{kp}g^{lq}~\bm{g}^m\otimes\bm{g}^n\otimes\bm g_{p}\otimes\bm{g}_q\\[2mm] &=\delta^p_m\delta^q_n~\bm{g}^m\otimes\bm{g}^n\otimes\bm g_{p}\otimes\bm{g}_q\\[2mm] &=\bm{g}^m\otimes\bm{g}^n\otimes\bm g_{m}\otimes\bm{g}_n =\bm{g}^k\otimes\bm{g}^l\otimes\bm g_{k}\otimes\bm{g}_l\\[2mm] \bm g_{k}\otimes\bm{g}_l\otimes\bm{g}^k\otimes\bm{g}^l &=g_{km}g^{kn}~\bm{g}^m\otimes\bm{g}_l\otimes\bm g_{n}\otimes\bm{g}^l\\[2mm] &=\delta_m^{n}~\bm{g}^m\otimes\bm{g}_l\otimes\bm g_{n}\otimes\bm{g}^l =\bm{g}^k\otimes\bm{g}_l\otimes\bm g_{k}\otimes\bm{g}^l\\[2mm] \bm g_{k}\otimes\bm{g}_l\otimes\bm{g}^k\otimes\bm{g}^l &=g_{lm}g^{ln}~\bm{g}_{k}\otimes\bm{g}^{m}\otimes\bm g^{k}\otimes\bm{g}_n\\[2mm] &=\delta_m^{n}~\bm{g}_{k}\otimes\bm{g}^{m}\otimes\bm g^{k}\otimes\bm{g}_n =\bm{g}_{k}\otimes\bm{g}^{l}\otimes\bm g^{k}\otimes\bm{g}_l\\[2mm] \end{align*} gk​⊗gl​⊗gk⊗glgk​⊗gl​⊗gk⊗glgk​⊗gl​⊗gk⊗gl​=gkm​gln​gkpglq gm⊗gn⊗gp​⊗gq​=δmp​δnq​ gm⊗gn⊗gp​⊗gq​=gm⊗gn⊗gm​⊗gn​=gk⊗gl⊗gk​⊗gl​=gkm​gkn gm⊗gl​⊗gn​⊗gl=δmn​ gm⊗gl​⊗gn​⊗gl=gk⊗gl​⊗gk​⊗gl=glm​gln gk​⊗gm⊗gk⊗gn​=δmn​ gk​⊗gm⊗gk⊗gn​=gk​⊗gl⊗gk⊗gl​​故定义四阶单位张量（ 1 − 3 , 2 − 4 1-3,2-4 1−3,2−4 指标为哑指标）：
I ≜ d A d A L = d A d A R = g k ⊗ g l ⊗ g k ⊗ g l = g k ⊗ g l ⊗ g k ⊗ g l = g k ⊗ g l ⊗ g k ⊗ g l = g k ⊗ g l ⊗ g k ⊗ g l \begin{align} \mathbb{I}\triangleq {\dfrac{d\mathbf{A}}{d\mathbf{A}}}_{L}={\dfrac{d\mathbf{A}}{d\mathbf{A}}}_{R} &=\bm g_{k}\otimes\bm{g}_l\otimes\bm{g}^k\otimes\bm{g}^l =\bm{g}^k\otimes\bm{g}^l\otimes\bm g_{k}\otimes\bm{g}_l\notag\\[4mm] &=\bm{g}_{k}\otimes\bm{g}^{l}\otimes\bm g^{k}\otimes\bm{g}_l =\bm{g}^k\otimes\bm{g}_l\otimes\bm g_{k}\otimes\bm{g}^l \end{align} I≜dAdA​L​=dAdA​R​​=gk​⊗gl​⊗gk⊗gl=gk⊗gl⊗gk​⊗gl​=gk​⊗gl⊗gk⊗gl​=gk⊗gl​⊗gk​⊗gl​​显然，四阶单位张量不等于度量张量的并矢（ 1 − 2 , 3 − 4 1-2,3-4 1−2,3−4 指标为哑指标），即
I ≠ I ⊗ I = g k ⊗ g k ⊗ g l ⊗ g l = ⋯ \begin{equation} \mathbb{I}\ne\mathbf{I}\otimes\mathbf{I}=\bm{g}_k\otimes\bm{g}^k\otimes\bm{g}_l\otimes\bm{g}^l =\cdots \end{equation} I=I⊗I=gk​⊗gk⊗gl​⊗gl=⋯​​从四阶张量的引出过程可知四阶单位张量与任意仿射量间满足：
I : B = B : I = B ( ∀ B ∈ T 2 ) \begin{equation} \mathbb{I}:\mathbf{B}=\mathbf{B}:\mathbb{I}=\mathbf{B}\qquad(\forall~\mathbf{B}\in\mathscr{T}^2) \end{equation} I:B=B:I=B(∀ B∈T2)​​四阶单位张量与其它四阶张量间满足：
{ I : A = ( g k ⊗ g l ⊗ g k ⊗ g l ) : ( A m n p q g m ⊗ g n ⊗ g p ⊗ g q ) = A A : I = ( A m n p q g m ⊗ g n ⊗ g p ⊗ g q ) : ( g k ⊗ g l ⊗ g k ⊗ g l ) = A ( ∀ A ∈ T 4 ) \begin{equation} \begin{cases} \mathbb{I}:\mathbb{A} =(\bm g^{k}\otimes\bm{g}^l\otimes\bm{g}_k\otimes\bm{g}_l):(A_{mnpq}\bm g^{m}\otimes\bm{g}^n\otimes\bm{g}^p\otimes\bm{g}^q) =\mathbb{A}\\[4mm] \mathbb{A}:\mathbb{I} =(A_{mnpq}\bm g^{m}\otimes\bm{g}^n\otimes\bm{g}^p\otimes\bm{g}^q):(\bm g_{k}\otimes\bm{g}_l\otimes\bm{g}^k\otimes\bm{g}^l) =\mathbb{A}\qquad(\forall~\mathbb{A}\in\mathscr{T}^4) \end{cases} \end{equation} ⎩ ⎨ ⎧​I:A=(gk⊗gl⊗gk​⊗gl​):(Amnpq​gm⊗gn⊗gp⊗gq)=AA:I=(Amnpq​gm⊗gn⊗gp⊗gq):(gk​⊗gl​⊗gk⊗gl)=A(∀ A∈T4)​​​

2. 由四阶单位张量导出的其它特殊的四阶张量

2.1 四阶单位张量的转置

对四阶单位张量 I = g k ⊗ g l ⊗ g k ⊗ g l \mathbb{I}=\bm g_{k}\otimes\bm{g}_l\otimes\bm{g}^k\otimes\bm{g}^l I=gk​⊗gl​⊗gk⊗gl 进行转置得到结果：
{ 交换  1 − 4 或  2 − 3 指标： I ⊗ I = g l ⊗ g l ⊗ g k ⊗ g k = g k ⊗ g k ⊗ g l ⊗ g l 交换  1 − 3 或  2 − 4 指标： I = g k ⊗ g l ⊗ g k ⊗ g l = g k ⊗ g l ⊗ g k ⊗ g l 交换  1 − 2 或  3 − 4 指标： g l ⊗ g k ⊗ g k ⊗ g l = g k ⊗ g l ⊗ g l ⊗ g k \begin{cases} \text{交换 $1-4$ 或 $2-3$ 指标：} & \mathbf{I}\otimes\mathbf{I} =\bm g^{l}\otimes\bm{g}_l\otimes\bm{g}^k\otimes\bm{g}_k =\bm g_{k}\otimes\bm{g}^k\otimes\bm{g}_l\otimes\bm{g}^l \\[3mm] \text{交换 $1-3$ 或 $2-4$ 指标：} & \mathbb{I} =\bm{g}^k\otimes\bm{g}_l\otimes\bm g_{k}\otimes\bm{g}^l =\bm{g}_{k}\otimes\bm{g}^{l}\otimes\bm g^{k}\otimes\bm{g}_l \\[3mm] \text{交换 $1-2$ 或 $3-4$ 指标：} & \bm{g}_l\otimes\bm{g}_k\otimes\bm g^{k}\otimes\bm{g}^l =\bm{g}_{k}\otimes\bm{g}_{l}\otimes\bm g^{l}\otimes\bm{g}^k \end{cases} ⎩ ⎨ ⎧​交换 1−4 或 2−3 指标：交换 1−3 或 2−4 指标：交换 1−2 或 3−4 指标：​I⊗I=gl⊗gl​⊗gk⊗gk​=gk​⊗gk⊗gl​⊗glI=gk⊗gl​⊗gk​⊗gl=gk​⊗gl⊗gk⊗gl​gl​⊗gk​⊗gk⊗gl=gk​⊗gl​⊗gl⊗gk​将上述结果中除 I ⊗ I \mathbf{I}\otimes\mathbf{I} I⊗I 与 I \mathbb{I} I 外的结果称作四阶单位张量的转置（ 1 − 4 , 2 − 3 1-4,2-3 1−4,2−3 指标为哑指标），记为
I T = g l ⊗ g k ⊗ g k ⊗ g l = g l ⊗ g k ⊗ g k ⊗ g l = g l ⊗ g k ⊗ g k ⊗ g l = g l ⊗ g k ⊗ g k ⊗ g l \begin{align} \mathbb{I}^T&=\bm{g}_l\otimes\bm{g}_k\otimes\bm g^{k}\otimes\bm{g}^l =\bm{g}^l\otimes\bm{g}_k\otimes\bm g^{k}\otimes\bm{g}_l\notag \\[3mm] &=\bm{g}_l\otimes\bm{g}^k\otimes\bm g_{k}\otimes\bm{g}^l =\bm{g}^l\otimes\bm{g}^k\otimes\bm g_{k}\otimes\bm{g}_l \end{align} IT​=gl​⊗gk​⊗gk⊗gl=gl⊗gk​⊗gk⊗gl​=gl​⊗gk⊗gk​⊗gl=gl⊗gk⊗gk​⊗gl​​​上述定义意味着：未特别声明时，四阶单位张量的转置是针对 1 − 2 1-2 1−2指标进行。 注意到，四阶单位张量的转置满足如下性质：
{ I T : B = ( g l ⊗ g k ⊗ g k ⊗ g l ) : ( B m n g m ⊗ g n ) = B m n g n ⊗ g m = B T B : I T = ( B m n g m ⊗ g n ) : ( g l ⊗ g k ⊗ g k ⊗ g l ) = B m n g n ⊗ g m = B T \begin{equation} \begin{cases} \mathbb{I}^T:\mathbf{B} =(\bm{g}_l\otimes\bm{g}_k\otimes\bm g^{k}\otimes\bm{g}^l ):(B^{mn}\bm{g}_m\otimes\bm{g}_n) =B^{mn}\bm{g}_n\otimes\bm{g}_m =\mathbf{B}^T \\[4mm] \mathbf{B}:\mathbb{I}^T =(B_{mn}\bm{g}^m\otimes\bm{g}^n):(\bm{g}_l\otimes\bm{g}_k\otimes\bm g^{k}\otimes\bm{g}^l ) =B_{mn}\bm{g}^n\otimes\bm{g}^m =\mathbf{B}^T \end{cases} \end{equation} ⎩ ⎨ ⎧​IT:B=(gl​⊗gk​⊗gk⊗gl):(Bmngm​⊗gn​)=Bmngn​⊗gm​=BTB:IT=(Bmn​gm⊗gn):(gl​⊗gk​⊗gk⊗gl)=Bmn​gn⊗gm=BT​​​易知： I T \mathbb{I}^T IT与任意四阶张量双点积的结果等于四阶张量对参与双点积的指标进行转置，因为
{ I T : A = A i j k l ( I T : g i ⊗ g j ) ⊗ g k ⊗ g l = A i j k l g j ⊗ g i ⊗ g k ⊗ g l A : I T = A i j k l g i ⊗ g j ( ⊗ g k ⊗ g l : I T ) = A i j k l g i ⊗ g j ⊗ g l ⊗ g k \begin{cases} \mathbb{I}^T:\mathbf{A}=A_{ijkl}(\mathbb{I}^T:\bm{g}^i\otimes\bm{g}^j)\otimes\bm{g}^k\otimes\bm{g}^l =A_{ijkl}\bm{g}^j\otimes\bm{g}^i\otimes\bm{g}^k\otimes\bm{g}^l\\[4mm] \mathbf{A}:\mathbb{I}^T=A_{ijkl}\bm{g}^i\otimes\bm{g}^j(\otimes\bm{g}^k\otimes\bm{g}^l:\mathbb{I}^T) =A_{ijkl}\bm{g}^i\otimes\bm{g}^j\otimes\bm{g}^l\otimes\bm{g}^k \end{cases} ⎩ ⎨ ⎧​IT:A=Aijkl​(IT:gi⊗gj)⊗gk⊗gl=Aijkl​gj⊗gi⊗gk⊗glA:IT=Aijkl​gi⊗gj(⊗gk⊗gl:IT)=Aijkl​gi⊗gj⊗gl⊗gk​则
I T : I T = I \mathbb{I}^T:\mathbb{I}^T=\mathbb{I} IT:IT=I
对于彷射量 A \bm{A} A，易证得：
d A T d A = I T \dfrac{d\bm{A^T}}{d\bm A}=\mathbb{I}^T dAdAT​=IT

2.2 对称的四阶单位张量

将四阶单位张量关于 1 − 2 1-2 1−2指标对称化，则可得到对称的四阶单位张量：
I 4 s ≜ 1 2 ( I + I T ) = 1 2 ( g k ⊗ g l ⊗ g k ⊗ g l + g l ⊗ g k ⊗ g k ⊗ g l ) = 1 2 ( δ k i δ l j + δ k j δ l i ) g i ⊗ g j ⊗ g k ⊗ g l \begin{align} \stackrel{4s}{\mathbb I}\triangleq\dfrac{1}{2}(\mathbb{I}+\mathbb{I}^T) &=\dfrac{1}{2}(\bm g_{k}\otimes\bm{g}_l\otimes\bm{g}^k\otimes\bm{g}^l+\bm g_{l}\otimes\bm{g}_k\otimes\bm{g}^k\otimes\bm{g}^l)\notag\\[4mm] &=\dfrac{1}{2}(\delta^i_{k}\delta^j_{l}+\delta^j_{k}\delta^i_{l})\bm g_{i}\otimes\bm{g}_j\otimes\bm{g}^k\otimes\bm{g}^l \end{align} I4s​≜21​(I+IT)​=21​(gk​⊗gl​⊗gk⊗gl+gl​⊗gk​⊗gk⊗gl)=21​(δki​δlj​+δkj​δli​)gi​⊗gj​⊗gk⊗gl​​显然，对称的四阶单位张量满足 Vogit 对称性：
I 4 s i j k l = I 4 s k l i j , I 4 s i j k l = I 4 s j i k l = I 4 s i j l k \begin{equation} \stackrel{4s}{\mathbb I}_{ijkl}=\stackrel{4s}{\mathbb I}_{klij}, \quad \stackrel{4s}{\mathbb I}_{ijkl}=\stackrel{4s}{\mathbb I}_{jikl}=\stackrel{4s}{\mathbb I}_{ijlk} \end{equation} I4s​ijkl​=I4s​klij​,I4s​ijkl​=I4s​jikl​=I4s​ijlk​​​易知： I 4 s \stackrel{4s}{\mathbb I} I4s​与任意四阶张量双点积的结果等于四阶张量对参与双点积的指标进行对称化。则
I 4 s : I 4 s = I 4 s \stackrel{4s}{\mathbb I}:\stackrel{4s}{\mathbb I}=\stackrel{4s}{\mathbb I} I4s​:I4s​=I4s​
对于彷射量 A \bm{A} A，易证得：
d s y m ( A ) d A = I 4 s \dfrac{d~sym(\bm{A})}{d\bm A}=\stackrel{4s}{\mathbb I} dAd sym(A)​=I4s​

2.3 四阶投影张量

通常，对任意对称仿射量 B \mathbf{B} B ，我们会对它进行如下分解：
B = 1 3 t r ( B ) I + d e v B = 1 3 B : ( I ⊗ I ) + d e v B ⟹ d e v B = B : ( I 4 s − 1 3 ( I ⊗ I ) ) = ( I 4 s − 1 3 ( I ⊗ I ) ) : B \begin{align*} & \mathbf{B}=\dfrac{1}{3}tr(\mathbf{B})\mathbf{I}+dev~\mathbf{B}=\dfrac{1}{3}\mathbf{B}:(\mathbf{I}\otimes\mathbf{I})+dev~\mathbf{B}\\[5mm] ~\Longrightarrow~ &dev~\mathbf{B}=\mathbf{B}:(\stackrel{4s}{\mathbb{I}}-\dfrac{1}{3}(\mathbf{I}\otimes\mathbf{I})) =(\stackrel{4s}{\mathbb{I}}-\dfrac{1}{3}(\mathbf{I}\otimes\mathbf{I})):\mathbf{B} \end{align*}  ⟹ ​B=31​tr(B)I+dev B=31​B:(I⊗I)+dev Bdev B=B:(I4s​−31​(I⊗I))=(I4s​−31​(I⊗I)):B​记 { I m ≜ 1 3 I ⊗ I I s ≜ I 4 s − I m ⟹ I 4 s : I m = I m : I 4 s = I m \begin{equation} \begin{cases} \mathbb{I}_m\triangleq\dfrac{1}{3}\bold{I}\otimes\bold{I}\\[4mm] \mathbb{I}_s\triangleq\stackrel{4s}{\mathbb{I}}-\mathbb{I}_m \end{cases} ~\Longrightarrow~ \stackrel{4s}{\mathbb I}:{\mathbb I}_m={\mathbb I}_m:\stackrel{4s}{\mathbb I}={\mathbb I}_m \end{equation} ⎩ ⎨ ⎧​Im​≜31​I⊗IIs​≜I4s​−Im​​ ⟹ I4s​:Im​=Im​:I4s​=Im​​​则对称仿射量的偏量部分与球量部分可分别表示为：
{ d e v B = I s : B = B : I s B − d e v B = I m : B = B : I m \begin{equation} \begin{cases} dev~\mathbf{B}= \mathbb{I}_s:\mathbf{B}=\mathbf{B}:\mathbb{I}_s\\[4mm] \mathbf{B}-dev~\mathbf{B}=\mathbb{I}_m:\mathbf{B}=\mathbf{B}:\mathbb{I}_m \end{cases} \end{equation} ⎩ ⎨ ⎧​dev B=Is​:B=B:Is​B−dev B=Im​:B=B:Im​​​​
此外， I m {\mathbb I}_m Im​ 与四阶投影张量 I s {\mathbb I}_s Is​ 间还满足：
{ I m : I m = 1 9 I ⊗ I : I ⊗ I = 1 9 ( t r I ) I ⊗ I = I m I s : I s = ( I 4 s − I m ) : ( I 4 s − I m ) = I s I m : I s = I s : I m = 0 \begin{cases} \mathbb{I}_m:\mathbb{I}_m=\dfrac{1}{9}\bold{I}\otimes\bold{I}:\bold{I}\otimes\bold{I} =\dfrac{1}{9}(tr~\bold{I})\bold{I}\otimes\bold{I}=\mathbb{I}_m\\\\ \mathbb{I}_s:\mathbb{I}_s=(\stackrel{4s}{\mathbb{I}}-\mathbb{I}_m):(\stackrel{4s}{\mathbb{I}}-\mathbb{I}_m) =\mathbb{I}_s\\\\ \mathbb{I}_m:\mathbb{I}_s=\mathbb{I}_s:\mathbb{I}_m=0 \end{cases} ⎩ ⎨ ⎧​Im​:Im​=91​I⊗I:I⊗I=91​(tr I)I⊗I=Im​Is​:Is​=(I4s​−Im​):(I4s​−Im​)=Is​Im​:Is​=Is​:Im​=0​