Micromechanics of composites/Proof 4

Question

Let $\partial {\Omega }$ be a surface. Let $\mathbf {x}$ be the position vector of points on the surface and let $\mathbf {t}$ be a vector field that are defined on $\partial {\Omega }$. If

$\int _{\partial {\Omega }}\mathbf {x} \times \mathbf {t} ~{\text{dA}}=\mathbf {0}$

show that

$\int _{\partial {\Omega }}\mathbf {x} \otimes \mathbf {t} ~{\text{dA}}=\int _{\partial {\Omega }}\mathbf {t} \otimes \mathbf {x} ~{\text{dA}}~.$

Proof

If we assume a Cartesian basis, we can write the given relation in index notation as

$\int _{\partial {\Omega }}e_{ijk}~x_{j}~t_{k}~{\text{dA}}=0$

where $e_{ijk}$ is the Levi-Civita (permutation) symbol. Since $e_{ijk}$ is does not depend upon the position we can write

$e_{ijk}\int _{\partial {\Omega }}~x_{j}~t_{k}~{\text{dA}}=0~.$

Define

$A_{jk}:=\int _{\partial {\Omega }}x_{j}~t_{k}~{\text{dA}}~.$

Then,

$e_{ijk}A_{jk}=0~.$

Expanding, we get

$e_{i11}~~A_{11}+e_{i12}A_{12}+e_{i13}A_{13}+e_{i21}A_{21}+e_{i22}~~A_{22}+e_{i23}A_{23}+e_{i31}A_{31}+e_{i32}A_{32}+e_{i33}~~A_{33}=0\qquad i=1,2,3~.$

Expanding further, we get three equations

${\begin{aligned}e_{112}A_{12}+e_{113}A_{13}+e_{121}A_{21}+e_{123}A_{23}+e_{131}A_{31}+e_{132}A_{32}&=0\\e_{212}A_{12}+e_{213}A_{13}+e_{221}A_{21}+e_{223}A_{23}+e_{231}A_{31}+e_{232}A_{32}&=0\\e_{312}A_{12}+e_{313}A_{13}+e_{321}A_{21}+e_{323}A_{23}+e_{331}A_{31}+e_{332}A_{32}&=0\end{aligned}}$

or,

${\begin{aligned}e_{123}A_{23}+e_{132}A_{32}&=0\\e_{213}A_{13}+e_{231}A_{31}&=0\\e_{312}A_{12}+e_{321}A_{21}&=0\end{aligned}}$

or,

$A_{23}=A_{32}~;~~A_{13}=A_{31}~;~~A_{12}=A_{21}~.$

Therefore,

$\int _{\partial {\Omega }}x_{2}~t_{3}~{\text{dA}}=\int _{\partial {\Omega }}x_{3}~t_{2}~{\text{dA}}=\int _{\partial {\Omega }}t_{2}~x_{3}~{\text{dA}}~;~~\int _{\partial {\Omega }}x_{1}~t_{3}~{\text{dA}}=\int _{\partial {\Omega }}x_{3}~t_{1}~{\text{dA}}=\int _{\partial {\Omega }}t_{1}~x_{3}~{\text{dA}}~;~~\int _{\partial {\Omega }}x_{1}~t_{2}~{\text{dA}}=\int _{\partial {\Omega }}x_{2}~t_{1}~{\text{dA}}=\int _{\partial {\Omega }}t_{1}~x_{2}~{\text{dA}}~.$

Also, by symmetry,

$\int _{\partial {\Omega }}x_{1}~t_{1}~{\text{dA}}=\int _{\partial {\Omega }}t_{1}~x_{1}~{\text{dA}}~;~~\int _{\partial {\Omega }}x_{2}~t_{2}~{\text{dA}}=\int _{\partial {\Omega }}t_{2}~x_{2}~{\text{dA}}~;~~\int _{\partial {\Omega }}x_{3}~t_{3}~{\text{dA}}=\int _{\partial {\Omega }}t_{3}~x_{3}~{\text{dA}}~.$

Therefore we may write,

$\int _{\partial {\Omega }}x_{j}~t_{k}~{\text{dA}}=\int _{\partial {\Omega }}t_{j}~x_{k}~{\text{dA}}~.$

Reverting back to direct tensor notation, we get

${\int _{\partial {\Omega }}\mathbf {x} \otimes \mathbf {t} ~{\text{dA}}=\int _{\partial {\Omega }}\mathbf {t} \otimes \mathbf {x} ~{\text{dA}}~.}$

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Micromechanics of composites

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