在多体动力学里面有并矢的概念,其实就是一个二阶张量的表达。

矢量 a\mathbf{a}a 和矢量 b\mathbf{b}b的并矢积 ab\mathbf{a b}ab定义为按下列规则变换任意矢量的变换
(ab)⋅c=a(b⋅c)(\mathbf{a b}) \cdot \mathbf{c}=\mathbf{a}(\mathbf{b} \cdot \mathbf{c})(ab)c=a(bc)

我们可以说明并矢满足张量的定义(加法和数乘):

(ab)⋅(αc+βd)=a[b⋅(αc+βd)]=a[α(b⋅c)+β(b⋅d)]=αa(b⋅c)+βa(b⋅d)=α(ab)⋅c+β(ab)⋅d\begin{aligned} & (\mathbf{a b}) \cdot(\alpha \mathbf{c}+\beta \mathbf{d})=\mathbf{a}[\mathbf{b} \cdot(\alpha \mathbf{c}+\beta \mathbf{d})] \\ = & \mathbf{a}[\alpha(\mathbf{b} \cdot \mathbf{c})+\beta(\mathbf{b} \cdot \mathbf{d})]=\alpha \mathbf{a}(\mathbf{b} \cdot \mathbf{c})+\beta \mathbf{a}(\mathbf{b} \cdot \mathbf{d}) \\ = & \alpha(\mathbf{a b}) \cdot \mathbf{c}+\beta(\mathbf{a b}) \cdot \mathbf{d} \end{aligned}==(ab)(αc+βd)=a[b(αc+βd)]a[α(bc)+β(bd)]=αa(bc)+βa(bd)α(ab)c+β(ab)d

为什么说它是一个二阶张量呢。我们可以得到它的分量:

(ab)ij=ei⋅(ab)⋅ej=ei⋅[a(b⋅ej)]=ei⋅(abj)=(ei⋅a)bj=aibj\begin{aligned} (\mathbf{a b})_{\mathrm{ij}} & =\mathbf{e}_{\mathrm{i}} \cdot(\mathbf{a b}) \cdot \mathbf{e}_{\mathrm{j}}=\mathbf{e}_{\mathrm{i}} \cdot\left[\mathbf{a}\left(\mathbf{b} \cdot \mathbf{e}_{\mathrm{j}}\right)\right]=\mathbf{e}_{\mathrm{i}} \cdot\left(\mathbf{a} b_{\mathrm{j}}\right) \\ & =\left(\mathbf{e}_{\mathrm{i}} \cdot \mathbf{a}\right) b_{\mathrm{j}}=a_{\mathrm{i}} b_{\mathrm{j}} \end{aligned}(ab)ij=ei(ab)ej=ei[a(bej)]=ei(abj)=(eia)bj=aibj

有时我们也写作:

(ab)ij=(a⊗b)ij=aibj(\mathbf{a b})_{\mathrm{ij}}=(\mathbf{a} \otimes \mathbf{b})_{\mathrm{ij}}=a_{\mathrm{i}} b_{\mathrm{j}}(ab)ij=(ab)ij=aibj

它的分量的总表达式可以写作:

[ab]=[a1b1a1b2a1b3a2b1a2b2a2b3a3b1a3b2a3b3]=[a1a2a3][b1b2b3][\mathbf{a b}]=\left[\begin{array}{lll} a_{1} b_{1} & a_{1} b_{2} & a_{1} b_{3} \\ a_{2} b_{1} & a_{2} b_{2} & a_{2} b_{3} \\ a_{3} b_{1} & a_{3} b_{2} & a_{3} b_{3} \end{array}\right]=\left[\begin{array}{l} a_{1} \\ a_{2} \\ a_{3} \end{array}\right]\left[\begin{array}{lll} b_{1} & b_{2} & b_{3} \end{array}\right][ab]= a1b1a2b1a3b1a1b2a2b2a3b2a1b3a2b3a3b3 = a1a2a3 [b1b2b3]

它的标架可以表示为:eiej,i,j=1,2,3\mathbf{e}_i\mathbf{e}_j,i,j=1,2,3eieji,j=1,2,3

列举两个:

e1e1=[100][100]=[100000000]e1e2=[100][010]=[010000000]⋯\begin{array}{l} {\mathbf{e}_{1} \mathbf{e}_{1}=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 0 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]} \\ {\mathbf{e}_{1} \mathbf{e}_{2}=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 0 \end{array}\right]=\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]} \end{array}\\\cdotse1e1= 100 [100]= 100000000 e1e2= 100 [010]= 000100000

于是可以并矢(二阶张量)可以写作:

T=T11(e1e1)+T12(e1e2)+⋯+T33(e3e3) 即 T=Tijeiej\begin{array}{c} \mathbf{T}=T_{11}\left(\mathbf{e}_{1} \mathbf{e}_{1}\right)+T_{12}\left(\mathbf{e}_{1} \mathbf{e}_{2}\right)+\cdots+T_{33}\left(\mathbf{e}_{3} \mathbf{e}_{3}\right) \\ \quad \text { 即 } \quad \mathbf{T}=T_{{ij}} \mathbf{e}_{{i}} \mathbf{e}_{{j}} \end{array}T=T11(e1e1)+T12(e1e2)++T33(e3e3)  T=Tijeiej

张量的本质我自我感觉就是标架(一阶张量为ei\mathbf{e}_iei,二阶张量为eiej\mathbf{e}_i\mathbf{e}_jeiej,高阶以此类推)和坐标(分量)。

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