（二十三）特殊的四阶张量 ——四阶单位张量

1. 四阶单位张量

考虑向量函数 $\mathbf{f}(\mathbf{v})=\mathbf{v}$，则
$${\mathbf{f}}^{\prime }(\mathbf{v})[\mathbf{u}]=\underset{h\to 0}{\lim }\frac{\mathbf{f}(\mathbf{v}+h\mathbf{u})-\mathbf{f}(\mathbf{v})}{h}=\mathbf{u}=\mathbf{I}\cdot \mathbf{u}=\mathbf{u}\cdot \mathbf{I}$$故可将二阶单位张量视为：
$${\frac{d\mathbf{v}}{d\mathbf{v}}}_L={\frac{d\mathbf{v}}{d\mathbf{v}}}_R=\mathbf{I}={\mathbf{g}}^k\otimes {\mathbf{g}}_k={\mathbf{g}}_k\otimes {\mathbf{g}}^k$$类比考虑四阶单位张量的定义，对二阶张量函数 $\mathbf{F}(\mathbf{A})=\mathbf{A}$，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
$$\begin{array}{rl}{\mathbf{g}}_k\otimes {\mathbf{g}}_l\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l& =g_{km}{g}_{ln}{g}^{kp}{g}^{lq}{\mathbf{g}}^m\otimes {\mathbf{g}}^n\otimes {\mathbf{g}}_p\otimes {\mathbf{g}}_q\\ & ={\delta }_m^p{\delta }_n^q{\mathbf{g}}^m\otimes {\mathbf{g}}^n\otimes {\mathbf{g}}_p\otimes {\mathbf{g}}_q\\ & ={\mathbf{g}}^m\otimes {\mathbf{g}}^n\otimes {\mathbf{g}}_m\otimes {\mathbf{g}}_n={\mathbf{g}}^k\otimes {\mathbf{g}}^l\otimes {\mathbf{g}}_k\otimes {\mathbf{g}}_l\\ {\mathbf{g}}_k\otimes {\mathbf{g}}_l\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l& =g_{km}{g}^{kn}{\mathbf{g}}^m\otimes {\mathbf{g}}_l\otimes {\mathbf{g}}_n\otimes {\mathbf{g}}^l\\ & ={\delta }_m^n{\mathbf{g}}^m\otimes {\mathbf{g}}_l\otimes {\mathbf{g}}_n\otimes {\mathbf{g}}^l={\mathbf{g}}^k\otimes {\mathbf{g}}_l\otimes {\mathbf{g}}_k\otimes {\mathbf{g}}^l\\ {\mathbf{g}}_k\otimes {\mathbf{g}}_l\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l& =g_{lm}{g}^{ln}{\mathbf{g}}_k\otimes {\mathbf{g}}^m\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}_n\\ & ={\delta }_m^n{\mathbf{g}}_k\otimes {\mathbf{g}}^m\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}_n={\mathbf{g}}_k\otimes {\mathbf{g}}^l\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}_l\end{array}$$故定义四阶单位张量（$1-3,2-4$ 指标为哑指标）：
$$\begin{array}{rl}\mathbb{I}\triangleq {\frac{d\mathbf{A}}{d\mathbf{A}}}_L={\frac{d\mathbf{A}}{d\mathbf{A}}}_R& ={\mathbf{g}}_k\otimes {\mathbf{g}}_l\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l={\mathbf{g}}^k\otimes {\mathbf{g}}^l\otimes {\mathbf{g}}_k\otimes {\mathbf{g}}_l\\ & ={\mathbf{g}}_k\otimes {\mathbf{g}}^l\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}_l={\mathbf{g}}^k\otimes {\mathbf{g}}_l\otimes {\mathbf{g}}_k\otimes {\mathbf{g}}^l\end{array}$$显然，四阶单位张量不等于度量张量的并矢（$1-2,3-4$ 指标为哑指标），即
$$\begin{array}{c}\mathbb{I}\ne \mathbf{I}\otimes \mathbf{I}={\mathbf{g}}_k\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}_l\otimes {\mathbf{g}}^l=\cdots \end{array}$$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：
$$\begin{array}{c}\mathbb{I}:\mathbf{B}=\mathbf{B}:\mathbb{I}=\mathbf{B}(\forall \mathbf{B}\in {\mathcal{T}}^2)\end{array}$$四阶单位张量与其它四阶张量间满足：
$$\begin{array}{c}\{\begin{array}{l}\mathbb{I}:\mathbb{A}=({\mathbf{g}}^k\otimes {\mathbf{g}}^l\otimes {\mathbf{g}}_k\otimes {\mathbf{g}}_l):(A_{mnpq}{\mathbf{g}}^m\otimes {\mathbf{g}}^n\otimes {\mathbf{g}}^p\otimes {\mathbf{g}}^q)=\mathbb{A}\\ \mathbb{A}:\mathbb{I}=(A_{mnpq}{\mathbf{g}}^m\otimes {\mathbf{g}}^n\otimes {\mathbf{g}}^p\otimes {\mathbf{g}}^q):({\mathbf{g}}_k\otimes {\mathbf{g}}_l\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l)=\mathbb{A}(\forall \mathbb{A}\in {\mathcal{T}}^4)\end{array}\end{array}$$

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 $\mathbb{I}={\mathbf{g}}_k\otimes {\mathbf{g}}_l\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l$ 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 $\mathbf{I}\otimes \mathbf{I}$ 与 $\mathbb{I}$ 外的结果称作四阶单位张量的转置（$1-4,2-3$ 指标为哑指标），记为
$$\begin{array}{rl}{\mathbb{I}}^T& ={\mathbf{g}}_l\otimes {\mathbf{g}}_k\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l={\mathbf{g}}^l\otimes {\mathbf{g}}_k\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}_l\\ & ={\mathbf{g}}_l\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}_k\otimes {\mathbf{g}}^l={\mathbf{g}}^l\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}_k\otimes {\mathbf{g}}_l\end{array}$$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$1-2$指标进行。 注意到，四阶单位张量的转置满足如下性质：
$$\begin{array}{c}\{\begin{array}{l}{\mathbb{I}}^T:\mathbf{B}=({\mathbf{g}}_l\otimes {\mathbf{g}}_k\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l):(B^{mn}{\mathbf{g}}_m\otimes {\mathbf{g}}_n)=B^{mn}{\mathbf{g}}_n\otimes {\mathbf{g}}_m={\mathbf{B}}^T\\ \mathbf{B}:{\mathbb{I}}^T=(B_{mn}{\mathbf{g}}^m\otimes {\mathbf{g}}^n):({\mathbf{g}}_l\otimes {\mathbf{g}}_k\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l)=B_{mn}{\mathbf{g}}^n\otimes {\mathbf{g}}^m={\mathbf{B}}^T\end{array}\end{array}$$易知：${\mathbb{I}}^T$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
$$\{\begin{array}{l}{\mathbb{I}}^T:\mathbf{A}=A_{ijkl}({\mathbb{I}}^T:{\mathbf{g}}^i\otimes {\mathbf{g}}^j)\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l=A_{ijkl}{\mathbf{g}}^j\otimes {\mathbf{g}}^i\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l\\ \mathbf{A}:{\mathbb{I}}^T=A_{ijkl}{\mathbf{g}}^i\otimes {\mathbf{g}}^j(\otimes {\mathbf{g}}^k\otimes {\mathbf{g}}^l:{\mathbb{I}}^T)=A_{ijkl}{\mathbf{g}}^i\otimes {\mathbf{g}}^j\otimes {\mathbf{g}}^l\otimes {\mathbf{g}}^k\end{array}$$则
$${\mathbb{I}}^T:{\mathbb{I}}^T=\mathbb{I}$$
对于彷射量$\mathbf{A}$，易证得：
$$\frac{d{\mathbf{A}}^{\mathbf{T}}}{d\mathbf{A}}={\mathbb{I}}^T$$

2.2 对称的四阶单位张量

将四阶单位张量关于$1-2$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：
$$\begin{array}{c}{\stackrel{4s}{\mathbb{I}}}_{ijkl}={\stackrel{4s}{\mathbb{I}}}_{klij},{\stackrel{4s}{\mathbb{I}}}_{ijkl}={\stackrel{4s}{\mathbb{I}}}_{jikl}={\stackrel{4s}{\mathbb{I}}}_{ijlk}\end{array}$$易知：$\stackrel{4s}{\mathbb{I}}$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
$$\stackrel{4s}{\mathbb{I}}:\stackrel{4s}{\mathbb{I}}=\stackrel{4s}{\mathbb{I}}$$
对于彷射量$\mathbf{A}$，易证得：
$$\frac{dsym(\mathbf{A})}{d\mathbf{A}}=\stackrel{4s}{\mathbb{I}}$$

2.3 四阶投影张量

通常，对任意对称仿射量 $\mathbf{B}$ ，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$$\begin{array}{c}\{\begin{array}{l}{\mathbb{I}}_m\triangleq \frac{1}{3}\mathbf{I}\otimes \mathbf{I}\\ {\mathbb{I}}_s\triangleq \stackrel{4s}{\mathbb{I}}-{\mathbb{I}}_m\end{array}⟹\stackrel{4s}{\mathbb{I}}:{\mathbb{I}}_m={\mathbb{I}}_m:\stackrel{4s}{\mathbb{I}}={\mathbb{I}}_m\end{array}$$则对称仿射量的偏量部分与球量部分可分别表示为：
$$\begin{array}{c}\{\begin{array}{l}dev\mathbf{B}={\mathbb{I}}_s:\mathbf{B}=\mathbf{B}:{\mathbb{I}}_s\\ \mathbf{B}-dev\mathbf{B}={\mathbb{I}}_m:\mathbf{B}=\mathbf{B}:{\mathbb{I}}_m\end{array}\end{array}$$
此外，${\mathbb{I}}_m$ 与四阶投影张量 ${\mathbb{I}}_s$ 间还满足：
$$\{\begin{array}{l}{\mathbb{I}}_m:{\mathbb{I}}_m=\frac{1}{9}\mathbf{I}\otimes \mathbf{I}:\mathbf{I}\otimes \mathbf{I}=\frac{1}{9}(tr\mathbf{I})\mathbf{I}\otimes \mathbf{I}={\mathbb{I}}_m\\ \\ {\mathbb{I}}_s:{\mathbb{I}}_s=(\stackrel{4s}{\mathbb{I}}-{\mathbb{I}}_m):(\stackrel{4s}{\mathbb{I}}-{\mathbb{I}}_m)={\mathbb{I}}_s\\ \\ {\mathbb{I}}_m:{\mathbb{I}}_s={\mathbb{I}}_s:{\mathbb{I}}_m=0\end{array}$$