# Micromechanics of composites/Proof 12

## Question

Let ${\displaystyle {\boldsymbol {A}}}$  and ${\displaystyle {\boldsymbol {B}}}$  be two second-order tensor fields. Let the average of any second-order tensor field (${\displaystyle {\boldsymbol {S}}}$ ) over the region ${\displaystyle \Omega }$  (of volume ${\displaystyle V}$ ) be defined as

${\displaystyle \langle {\boldsymbol {S}}\rangle :={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {S}}~{\text{dV}}~.}$

Show that

${\displaystyle \langle {\boldsymbol {A}}\cdot {\boldsymbol {B}}\rangle -\langle {\boldsymbol {A}}\rangle \cdot \langle {\boldsymbol {B}}\rangle =\langle [{\boldsymbol {A}}-\langle {\boldsymbol {A}}\rangle ]\cdot [{\boldsymbol {B}}-\langle {\boldsymbol {B}}\rangle ]\rangle ~.}$

## Proof

Expanding out the right hand side, we have

{\displaystyle {\begin{aligned}\langle [{\boldsymbol {A}}-\langle {\boldsymbol {A}}\rangle ]\cdot [{\boldsymbol {B}}-\langle {\boldsymbol {B}}\rangle ]\rangle &={\cfrac {1}{V}}\int _{\Omega }[{\boldsymbol {A}}-\langle {\boldsymbol {A}}\rangle ]\cdot [{\boldsymbol {B}}-\langle {\boldsymbol {B}}\rangle ]~{\text{dV}}\\&={\cfrac {1}{V}}\int _{\Omega }[{\boldsymbol {A}}\cdot {\boldsymbol {B}}-\langle {\boldsymbol {A}}\rangle \cdot {\boldsymbol {B}}-{\boldsymbol {A}}\cdot \langle {\boldsymbol {B}}\rangle +\langle {\boldsymbol {A}}\rangle \cdot \langle {\boldsymbol {B}}\rangle ]~{\text{dV}}\\&={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {A}}\cdot {\boldsymbol {B}}~{\text{dV}}-{\cfrac {1}{V}}\int _{\Omega }\langle {\boldsymbol {A}}\rangle \cdot {\boldsymbol {B}}~{\text{dV}}-{\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {A}}\cdot \langle {\boldsymbol {B}}\rangle ~{\text{dV}}+{\cfrac {1}{V}}\int _{\Omega }\langle {\boldsymbol {A}}\rangle \cdot \langle {\boldsymbol {B}}\rangle ~{\text{dV}}~.\end{aligned}}}

Now ${\displaystyle \langle {\boldsymbol {A}}\rangle }$  and ${\displaystyle \langle {\boldsymbol {B}}\rangle }$  are constants with respect to the integration. Hence,

{\displaystyle {\begin{aligned}\langle [{\boldsymbol {A}}-\langle {\boldsymbol {A}}\rangle ]\cdot [{\boldsymbol {B}}-\langle {\boldsymbol {B}}\rangle ]\rangle &={\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {A}}\cdot {\boldsymbol {B}}~{\text{dV}}-\langle {\boldsymbol {A}}\rangle \cdot \left({\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {B}}~{\text{dV}}\right)-\left({\cfrac {1}{V}}\int _{\Omega }{\boldsymbol {A}}~{\text{dV}}\right)\cdot \langle {\boldsymbol {B}}\rangle +\langle {\boldsymbol {A}}\rangle \cdot \langle {\boldsymbol {B}}\rangle \left({\cfrac {1}{V}}\int _{\Omega }~{\text{dV}}\right)\\&=\langle {\boldsymbol {A}}\cdot {\boldsymbol {B}}\rangle -\langle {\boldsymbol {A}}\rangle \cdot \langle {\boldsymbol {B}}\rangle -\langle {\boldsymbol {A}}\rangle \cdot \langle {\boldsymbol {B}}\rangle +\langle {\boldsymbol {A}}\rangle \cdot \langle {\boldsymbol {B}}\rangle \\&=\langle {\boldsymbol {A}}\cdot {\boldsymbol {B}}\rangle -\langle {\boldsymbol {A}}\rangle \cdot \langle {\boldsymbol {B}}\rangle ~.\end{aligned}}}

Therefore,

${\displaystyle {\langle {\boldsymbol {A}}\cdot {\boldsymbol {B}}\rangle -\langle {\boldsymbol {A}}\rangle \cdot \langle {\boldsymbol {B}}\rangle =\langle [{\boldsymbol {A}}-\langle {\boldsymbol {A}}\rangle ]\cdot [{\boldsymbol {B}}-\langle {\boldsymbol {B}}\rangle ]\rangle \qquad \square }}$