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(二十五)各向同性张量(函数)

(二十五)各向同性张量(函数)

1. 各向同性张量

定义直角坐标系(标准正交基不变)中,若某张量的分量值不随基的正交变换而变化,则称该张量为欧氏空间中的各向同性张量。即,

r r r 阶张量 Φ \bold\Phi Φ:
Φ = Φ i 1 i 2 … i r e ⃗ i 1 ⊗ e ⃗ i 2 ⊗ ⋯ ⊗ e ⃗ i r = Φ i 1 ′ i 2 ′ … i r ′ e ′ ⃗ i 1 ⊗ e ′ ⃗ i 2 ⊗ ⋯ ⊗ e ′ ⃗ i r \bold\Phi =\Phi_{i_1i_2\dots i_r}\vec{e}_{i_1}\otimes\vec{e}_{i_2}\otimes\dots\otimes\vec{e}_{i_r} =\Phi_{i'_1i'_2\dots i'_r}\vec{e'}_{i_1}\otimes\vec{e'}_{i_2}\otimes\dots\otimes\vec{e'}_{i_r} Φ=Φi1i2ire i1e i2e ir=Φi1i2ire i1e i2e ir
其中,
e ′ ⃗ i t = Q ⋅ e ⃗ i t ( t = 1 , 2 , … , r ; i t = 1 , 2 , 3 ) , Q 为任意正交仿射量 \vec{e'}_{i_t}=\bold Q\cdot\vec{e}_{i_t}\quad(t=1,2,\dots,r;i_t=1,2,3),\bold Q为任意正交仿射量 e it=Qe it(t=1,2,,r;it=1,2,3)Q为任意正交仿射量

Φ i 1 i 2 … i r = Φ i 1 ′ i 2 ′ … i r ′ \Phi_{i_1i_2\dots i_r}=\Phi_{i'_1i'_2\dots i'_r} Φi1i2ir=Φi1i2ir
则将 Φ \bold\Phi Φ 称作各向同性张量,记 Φ ∈ i s o \bold\Phi\in iso Φiso

定义一种新的符号记法:
Q ∘ Φ ≜ Φ i 1 i 2 … i r ( Q ⋅ e ⃗ i 1 ) ⊗ ( Q ⋅ e ⃗ i 2 ) ⊗ ⋯ ⊗ ( Q ⋅ e ⃗ i r ) \bold Q\circ\bold\Phi\triangleq\Phi_{i_1i_2\dots i_r}(\bold Q\cdot\vec{e}_{i_1})\otimes(\bold Q\cdot\vec{e}_{i_2})\otimes\dots\otimes(\bold Q\cdot\vec{e}_{i_r}) QΦΦi1i2ir(Qe i1)(Qe i2)(Qe ir)
那么,
∀ Q ∈ O 3 , Q ∘ Φ = Φ ⟺ Φ ∈ i s o \forall \bold Q\in\mathcal{O}_3,\bold Q\circ\bold\Phi=\bold\Phi\Longleftrightarrow\bold\Phi\in iso QO3QΦ=ΦΦiso

2. 各项同性张量的实例

2.1. 标量(零阶张量)

标量都是各向同性的,即:
f o r   ∀ φ ∈ T 0 ( V ) , φ ∈ i s o for\ \forall\varphi\in\mathscr{T}_0(V),\varphi\in iso for φT0(V)φiso

2.2. 向量(一阶张量)


v ⃗ ∈ i s o \vec{v}\in iso v iso
那么,
∀ Q ∈ O 3 , v ⃗ = v i e ⃗ i = v i ( Q ⋅ e ⃗ i ) \forall\bold{Q}\in\mathcal{O}_3,\vec{v}=v_i\vec{e}_i=v_i(\bold{Q}\cdot\vec{e}_i) QO3v =vie i=vi(Qe i)

Q = − G \bold{Q}=-\bold{G} Q=G

v i e ⃗ i = − v i e ⃗ i ⟹ 2 v i e ⃗ i = 2 v ⃗ = 0 ⟹ v ⃗ = 0 v_i\vec{e}_i=-v_i\vec{e}_i\Longrightarrow2v_i\vec{e}_i=2\vec{v}=0\Longrightarrow\vec{v}=0 vie i=vie i2vie i=2v =0v =0
故,向量中只有零向量是各向同性的,即:
f o r   v ⃗ ∈ T 1 ( V ) , v ⃗ ∈ i s o ⟺ v ⃗ = 0 for\ \vec{v}\in\mathscr{T}_1(V),\vec{v}\in iso\Longleftrightarrow\vec{v}=0 for v T1(V)v isov =0

2.3. 仿射量(二阶张量)


T ∈ i s o \bold{T}\in iso Tiso
那么,
∀ Q ∈ O 3 , T = Q ∘ T = T i j ( Q ⋅ e ⃗ i ) ⊗ ( Q ⋅ e ⃗ j ) = T i j ( Q ⋅ e ⃗ i ) ⊗ ( e ⃗ j ⋅ Q T ) = Q ⋅ T ⋅ Q T \forall\bold{Q}\in\mathcal{O}_3, \bold{T} =\bold{Q}\circ\bold{T} =T_{ij}(\bold{Q}\cdot\vec{e}_i)\otimes(\bold{Q}\cdot\vec{e}_j) =T_{ij}(\bold{Q}\cdot\vec{e}_i)\otimes(\vec{e}_j\cdot\bold{Q}^T) =\bold{Q}\cdot\bold{T}\cdot\bold{Q}^T QO3T=QT=Tij(Qe i)(Qe j)=Tij(Qe i)(e jQT)=QTQT
故,
Q ⋅ T = T ⋅ Q \bold{Q}\cdot\bold{T}=\bold{T}\cdot\bold{Q} QT=TQ
λ \lambda λ r ⃗ \vec{r} r 分别为 T \bold{T} T 的右特征值与右特征向量,即
T ⋅ r ⃗ = λ r ⃗ \bold{T}\cdot\vec{r}=\lambda\vec{r} Tr =λr
则有:
T ⋅ Q ⋅ r ⃗ = Q ⋅ ( T ⋅ r ⃗ ) = λ Q ⋅ r ⃗ \bold{T}\cdot\bold{Q}\cdot\vec{r}=\bold{Q}\cdot(\bold{T}\cdot\vec{r})=\lambda\bold{Q}\cdot\vec{r} TQr =Q(Tr )=λQr

( T − λ G ) ⋅ Q ⋅ r ⃗ = 0 (\bold{T}-\lambda\bold{G})\cdot\bold{Q}\cdot\vec{r}=0 (TλG)Qr =0
根据 Q \bold Q Q 的任意性,知:
T = λ G ( λ 为 T 的特征值 ) \bold{T}=\lambda\bold{G}\quad(\lambda为\bold T 的特征值) T=λG(λT的特征值)
换而言之,仿射量只有球形张量是各向同性的

2.4. 三阶张量

考虑特殊的三阶张量——置换张量

现有两组直角坐标系,且二者满足正交变换的关系,即:
∀ Q ∈ O 3 , { e ′ ⃗ 1 , e ′ ⃗ 2 , e ′ ⃗ 3 } = { Q ⋅ e ⃗ 1 , Q ⋅ e ⃗ 2 , Q ⋅ e ⃗ 3 } \forall\bold{Q}\in\mathcal{O}_3,\{\vec{e'}_1,\vec{e'}_2,\vec{e'}_3\} =\{\bold{Q}\cdot\vec{e}_1,\bold{Q}\cdot\vec{e}_2,\bold{Q}\cdot\vec{e}_3\} QO3{e 1,e 2,e 3}={Qe 1,Qe 2,Qe 3}
则,
ϵ = ϵ i j k e ⃗ i ⊗ e ⃗ j ⊗ e ⃗ k = ϵ i ′ j ′ k ′ e ⃗ i ′ ⊗ e ⃗ j ′ ⊗ e ⃗ k ′ \epsilon=\epsilon^{ijk}\vec{e}_i\otimes\vec{e}_j\otimes\vec{e}_k=\epsilon^{i'j'k'}\vec{e}_{i'}\otimes\vec{e}_{j'}\otimes\vec{e}_{k'} ϵ=ϵijke ie je k=ϵijke ie je k
其中
ϵ i j k = e i j k g ϵ i ′ j ′ k ′ = e i ′ j ′ k ′ g ′ = e i ′ j ′ k ′ [ e ⃗ 1 ′ e ⃗ 2 ′ e ⃗ 3 ′ ] = e i ′ j ′ k ′ [ Q ⋅ e ⃗ 1 Q ⋅ e ⃗ 2 Q ⋅ e ⃗ 3 ] = e i ′ j ′ k ′ d e t ( Q ) [ e ⃗ 1 e ⃗ 2 e ⃗ 3 ] = e i ′ j ′ k ′ d e t ( Q ) g = e i j k d e t ( Q ) g = ϵ i j k d e t ( Q )

ϵijk=eijkgϵijk=eijkg=eijk[e1e2e3]=eijk[\boldQe1\boldQe2\boldQe3]=eijkdet(\boldQ)[e1e2e3]=eijkdet(\boldQ)g=eijkdet(\boldQ)g=ϵijkdet(\boldQ)
ϵijk=g eijkϵijk=g eijk=[e 1e 2e 3]eijk=[Qe 1Qe 2Qe 3]eijk=det(Q)[e 1e 2e 3]eijk=det(Q)g eijk=det(Q)g eijk=det(Q)ϵijk
因此 三阶张量 λ ϵ ( λ ∈ R ) \lambda\epsilon(\lambda\in R) λϵ(λR) 在旋转正交变换(无镜面反射)下是各向同性的

2.5. 四阶张量

定理 在直角坐标系中,四阶各向同性张量 Φ \bold\Phi Φ 的分量可写为:
Φ i j k l = λ δ i j δ k l + α δ i k δ j l + β δ i l δ j k ( 具有对称性: Φ i j k l = Φ k l i j ) \Phi_{ijkl}=\lambda\delta_{ij}\delta_{kl}+\alpha\delta_{ik}\delta_{jl}+\beta\delta_{il}\delta_{jk}\quad(具有对称性:\Phi_{ijkl}=\Phi_{klij}) Φijkl=λδijδkl+αδikδjl+βδilδjk(具有对称性:Φijkl=Φklij)
其中, λ , α , β ∈ R \lambda,\alpha,\beta\in R λ,α,βR。令
α = μ + ν ; β = μ − ν \alpha=\mu+\nu;\beta=\mu-\nu α=μ+νβ=μν
则有
Φ i j k l = λ δ i j δ k l + μ ( δ i k δ j l + δ i l δ j k ) + ν ( δ i k δ j l − δ i l δ j k ) \Phi_{ijkl}=\lambda\delta_{ij}\delta_{kl}+\mu(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})+\nu(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}) Φijkl=λδijδkl+μ(δikδjl+δilδjk)+ν(δikδjlδilδjk)
进一步,若四阶张量关于前两个指标或后两个指标还具有对称性:

以前两个指标对称为例,交换 i , j i,j ij 指标:
Φ j i k l = λ δ i j δ k l + μ ( δ j k δ i l + δ j l δ i k ) + ν ( δ j k δ i l − δ j l δ i k ) \Phi_{jikl}=\lambda\delta_{ij}\delta_{kl}+\mu(\delta_{jk}\delta_{il}+\delta_{jl}\delta_{ik})+\nu(\delta_{jk}\delta_{il}-\delta_{jl}\delta_{ik}) Φjikl=λδijδkl+μ(δjkδil+δjlδik)+ν(δjkδilδjlδik)
两式相加得:
Φ i j k l = 1 2 ( Φ i j k l + Φ j i k l ) = λ δ i j δ k l + μ ( δ i k δ j l + δ i l δ j k ) ( ∗ ) \Phi_{ijkl}=\frac{1}{2}(\Phi_{ijkl}+\Phi_{jikl})=\lambda\delta_{ij}\delta_{kl}+\mu(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})(*) Φijkl=21(Φijkl+Φjikl)=λδijδkl+μ(δikδjl+δilδjk)
根据上式可知:

  • 关于前两个指标 i i i j j j 对称 ⟹ Φ i j k l = Φ j i k l = Φ i j l k = Φ k l i j \Longrightarrow \Phi_{ijkl}=\Phi_{jikl}=\Phi_{ijlk}=\Phi_{klij} Φijkl=Φjikl=Φijlk=Φklij
  • 关于后两个指标 k k k l l l 对称 ⟹ Φ i j k l = Φ i j l k = Φ j i k l = Φ k l i j \Longrightarrow \Phi_{ijkl}=\Phi_{ijlk}=\Phi_{jikl}=\Phi_{klij} Φijkl=Φijlk=Φjikl=Φklij

对(*)式采用实体记法:
Φ = λ G ⊗ G + 2 μ I 4 s ( ∗ ∗ ) \bold\Phi=\lambda\bold{G}\otimes\bold{G}+2\mu\stackrel{4s}{\bold I}(**) Φ=λGG+2μI4s
其中,
I 4 s = 1 2 ( δ i k δ j l + δ i l δ j k ) e ⃗ i ⊗ e ⃗ j ⊗ e ⃗ k ⊗ e ⃗ l \stackrel{4s}{\bold I}=\frac{1}{2}(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})\vec{e}_i\otimes\vec{e}_j\otimes\vec{e}_k\otimes\vec{e}_l I4s=21(δikδjl+δilδjk)e ie je ke l
若令
{ I m ≜ 1 3 G ⊗ G I s ≜ I 4 s − I m

{\boldIm13\boldG\boldG\boldIs\boldI4s\boldIm
Im31GGIsI4sIm
二者满足:
{ I m : I m = 1 9 G ⊗ G : G ⊗ G = 1 9 ( g m n g m n ) G ⊗ G = 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
{\boldIm:\boldIm=19\boldG\boldG:\boldG\boldG=19(gmngmn)\boldG\boldG=\boldIm\boldIs:\boldIs=(\boldI4s\boldIm):(\boldI4s\boldIm)=\boldIs\boldIm:\boldIs=\boldIs:\boldIm=0
Im:Im=91GG:GG=91(gmngmn)GG=ImIs:Is=(I4sIm):(I4sIm)=IsIm:Is=Is:Im=0

则有:
Φ = 3 λ I m + 2 μ ( I s + I m ) = ( 3 λ + 2 μ ) I m + 2 μ I s \bold\Phi=3\lambda\bold{I}_m+2\mu(\bold{I}_s+\bold{I}_m) =(3\lambda+2\mu)\bold{I}_m+2\mu\bold{I}_s Φ=3λIm+2μ(Is+Im)=(3λ+2μ)Im+2μIs

k ≜ λ + 2 3 μ k\triangleq\lambda+\dfrac{2}{3}\mu kλ+32μ
则:
Φ = 3 k I m + 2 μ I s ( ∗ ∗ ∗ ) \bold\Phi=3k\bold{I}_m+2\mu\bold{I}_s(***) Φ=3kIm+2μIs
通过上式来表示满足对前两个指标或后两个指标具有对称性的四阶各向同性张量,便于进行两个此类四阶张量的双点积运算。

3. 各向同性张量函数

定义 自变量为张量的函数称作称作张量函数,其函数值可以是标量也可以是高阶张量。

定义 (1)当自变量和函数值在任意正交变换 Q ∈ O 3 \bold Q\in\mathcal{O}_3 QO3 下保持原有函数关系不变,则称该函数为各向同性张量函数;(2)当自变量和函数值在任意正常正交变换 Q ∈ O 3 + \bold Q\in\mathcal{O}_3^+ QO3+ 下保持原有函数关系不变,则称该函数为半各向同性张量函数。即,

对于张量函数:
F ( A , B , … ) \bold F(\bold{A,B,\dots}) F(A,B,)
(1)若满足:
∀ Q ∈ O 3 , Q ∘ F = F ( Q ∘ A , Q ∘ B , …   ) \forall\bold Q\in\mathcal{O}_3,\bold Q\circ\bold F=\bold F(\bold Q\circ\bold A,\bold Q\circ\bold B,\dots) QO3QF=F(QA,QB,)
则称 F \bold F F 为各向同性张量函数。
(2)若满足:
∀ Q ∈ O 3 + , Q ∘ F = F ( Q ∘ A , Q ∘ B , …   ) \forall\bold Q\in\mathcal{O}_3^+,\bold Q\circ\bold F=\bold F(\bold Q\circ\bold A,\bold Q\circ\bold B,\dots) QO3+QF=F(QA,QB,)
则称 F \bold F F 为半各向同性张量函数。

4. 仿射量的主不变量是函数值为标量的各向同性函数

命题 仿射量 B \bold B B 的三个主不变量是标量值的各向同性函数。

证明:第一主不变量:
C 1 ( B ) = t r ( B ) \mathscr{C}_1(\bold B)=tr(\bold B) C1(B)=tr(B)
根据迹的运算性质
∀ Q ∈ O 3 , t r ( Q ∘ B ) = t r ( Q ∘ B ) = t r ( Q ⋅ B ⋅ Q T ) = t r ( B ⋅ Q T ⋅ Q ) = t r ( B ) (证毕) \forall\bold Q\in\mathcal{O}_3, tr(\bold Q\circ\bold B) =tr(\bold Q\circ\bold B) =tr(\bold Q\cdot\bold B\cdot\bold Q^T) =tr(\bold B\cdot\bold Q^T\cdot\bold Q) =tr(\bold B)(证毕) QO3tr(QB)=tr(QB)=tr(QBQT)=tr(BQTQ)=tr(B)(证毕)
第二主不变量:
C 2 ( B ) = 1 2 ( t r 2 ( B ) − t r ( B 2 ) ) \mathscr{C}_2(\bold B)=\frac{1}{2}(tr^2(\bold B)-tr(\bold B^2)) C2(B)=21(tr2(B)tr(B2))
其前一项已证明是各向同性的,现关注第二项,由于
t r [ ( Q ∘ B ) 2 ] = t r [ ( Q ⋅ B ⋅ Q T ) ⋅ ( Q ⋅ B ⋅ Q T ) ] = t r ( Q ⋅ B 2 ⋅ Q T ) = t r ( B 2 ) (证毕) tr[(\bold Q\circ\bold B)^2] =tr[(\bold Q\cdot\bold B\cdot\bold Q^T)\cdot(\bold Q\cdot\bold B\cdot\bold Q^T)] =tr(\bold Q\cdot\bold B^2\cdot\bold Q^T) =tr(\bold B^2)(证毕) tr[(QB)2]=tr[(QBQT)(QBQT)]=tr(QB2QT)=tr(B2)(证毕)
第三主不变量:
C 3 ( B ) = d e t ( B ) \mathscr{C}_3(\bold B)=det(\bold B) C3(B)=det(B)
由于
d e t ( Q ∘ B ) = d e t ( Q ⋅ B ⋅ Q T ) = d e t ( Q ) d e t ( Q T ) d e t ( B ) = d e t 2 ( Q ) d e t ( B ) = d e t ( B ) (证毕) det(\bold Q\circ\bold B)=det(\bold Q\cdot\bold B\cdot\bold Q^T) =det(\bold Q)det(\bold Q^T)det(\bold B) =det^2(\bold Q)det(\bold B)=det(\bold B)(证毕) det(QB)=det(QBQT)=det(Q)det(QT)det(B)=det2(Q)det(B)=det(B)(证毕)

5. 仿射量的 k 阶矩是函数值为标量的各向同性函数

二阶张量 B \bold B B k ∈ N k\in N kN 阶矩
C k ∗ ( B ) = t r ( B k ) \mathscr{C}^*_k(\bold B)=tr(\bold B^k) Ck(B)=tr(Bk)
因为
t r [ ( Q ∘ B ) k ] = t r [ ( Q ⋅ B ⋅ Q T ) k ] = t r ( Q ⋅ B k ⋅ Q T ) = t r ( B k ⋅ Q T ⋅ Q ) = t r ( B k ) tr[(\bold Q\circ\bold B)^k] =tr[(\bold Q\cdot\bold B\cdot\bold Q^T)^k] =tr(\bold Q\cdot\bold B^k\cdot\bold Q^T) =tr(\bold B^k\cdot\bold Q^T\cdot\bold Q)=tr(\bold B^k) tr[(QB)k]=tr[(QBQT)k]=tr(QBkQT)=tr(BkQTQ)=tr(Bk)
二阶张量 B \bold B B k k k 阶矩是各向同性标量值函数

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