Micromechanics of composites/Proof 1

Tensor-vector identity - 1


[(\mathbf{v}\bullet\mathbf{a})(\boldsymbol{S}\bullet\mathbf{b})]\cdot\mathbf{n} = \mathbf{a}\cdot[\{\mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})\}\cdot\mathbf{b}] ~. $$

Proof:

Using the identity $$ \mathbf{a}\cdot(\boldsymbol{A}^T\cdot\mathbf{b}) = \mathbf{b}\cdot(\boldsymbol{A}\cdot\mathbf{a})$$ we have

\mathbf{n}\cdot[(\mathbf{v}\bullet\mathbf{a})(\boldsymbol{S}\bullet\mathbf{b})] = \mathbf{b}\cdot[(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}^T\cdot\mathbf{n})] ~. $$ Also, using the definition $$ (\mathbf{u}\otimes\mathbf{v})\cdot\mathbf{a} = (\mathbf{a}\cdot\mathbf{v})\mathbf{u} $$ we have

(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}^T\cdot\mathbf{n}) = [(\boldsymbol{S}^T\cdot\mathbf{n})\otimes\mathbf{v}]\cdot\mathbf{a}~. $$ Therefore,

\mathbf{n}\cdot[(\mathbf{v}\bullet\mathbf{a})(\boldsymbol{S}\bullet\mathbf{b})] = \mathbf{b}\cdot[\{(\boldsymbol{S}^T\cdot\mathbf{n})\otimes\mathbf{v}\}\cdot\mathbf{a}]~. $$ Using the identity $$ \mathbf{a}\cdot(\boldsymbol{A}^T\cdot\mathbf{b}) = \mathbf{b}\cdot(\boldsymbol{A}\cdot\mathbf{a})$$ we have

\mathbf{b}\cdot[\{(\boldsymbol{S}^T\cdot\mathbf{n})\otimes\mathbf{v}\}\cdot\mathbf{a}] = \mathbf{a}\cdot[\{(\boldsymbol{S}^T\cdot\mathbf{n})\otimes\mathbf{v}\}^T\cdot\mathbf{b}]~. $$ Finally, using the relation $$(\mathbf{u}\otimes\mathbf{v})^T = \mathbf{v}\otimes\mathbf{u}$$, we get

\mathbf{a}\cdot[\{(\boldsymbol{S}^T\cdot\mathbf{n})\otimes\mathbf{v}\}^T\cdot\mathbf{b}] = \mathbf{a}\cdot[\{\mathbf{v}\otimes(\boldsymbol{S}^T\cdot\mathbf{n})\}\cdot\mathbf{b}]~. $$ Hence,

{ [(\mathbf{v}\bullet\mathbf{a})(\boldsymbol{S}\bullet\mathbf{b})]\cdot\mathbf{n} = \mathbf{a}\cdot[\{\mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})\}\cdot\mathbf{b}]} \qquad \qquad \qquad \square $$