# Continuum mechanics/Tensor-vector identities

## Tensor-vector identity - 1

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

Proof:

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

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

Also, using the definition ${\displaystyle (\mathbf {u} \otimes \mathbf {v} )\cdot \mathbf {a} =(\mathbf {a} \cdot \mathbf {v} )\mathbf {u} }$  we have

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

Therefore,

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

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

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

Finally, using the relation ${\displaystyle (\mathbf {u} \otimes \mathbf {v} )^{T}=\mathbf {v} \otimes \mathbf {u} }$ , we get

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

Hence,

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

## Tensor-vector identity 2

Let ${\displaystyle \mathbf {v} }$  be a vector field and let ${\displaystyle {\boldsymbol {S}}}$  be a second-order tensor field. Let ${\displaystyle \mathbf {a} }$  and ${\displaystyle \mathbf {b} }$  be two arbitrary vectors. Show that

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

Proof:

Using the identity ${\displaystyle {\boldsymbol {\nabla }}\bullet (\varphi ~\mathbf {u} )=\mathbf {u} \cdot {\boldsymbol {\nabla }}\varphi +\varphi ~{\boldsymbol {\nabla }}\bullet \mathbf {u} }$  we have

${\displaystyle {\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]=({\boldsymbol {S}}\cdot \mathbf {b} )\cdot {\boldsymbol {\nabla }}(\mathbf {v} \cdot \mathbf {a} )+(\mathbf {v} \cdot \mathbf {a} )~{\boldsymbol {\nabla }}\bullet ({\boldsymbol {S}}\cdot \mathbf {b} )~.}$

From the identity ${\displaystyle {\boldsymbol {\nabla }}(\mathbf {u} \cdot \mathbf {v} )={\boldsymbol {\nabla }}\mathbf {u} ^{T}\cdot {\boldsymbol {\nabla }}\mathbf {v} +{\boldsymbol {\nabla }}\mathbf {v} ^{T}\cdot \mathbf {u} }$ , we have ${\displaystyle {\boldsymbol {\nabla }}(\mathbf {v} \cdot \mathbf {a} )={\boldsymbol {\nabla }}\mathbf {v} ^{T}\cdot \mathbf {a} +{\boldsymbol {\nabla }}\mathbf {a} ^{T}\cdot \mathbf {v} }$ .

Since ${\displaystyle \mathbf {a} }$  is constant, ${\displaystyle {\boldsymbol {\nabla }}\mathbf {a} =0}$ , and we have

${\displaystyle ({\boldsymbol {S}}\cdot \mathbf {b} )\cdot {\boldsymbol {\nabla }}(\mathbf {v} \cdot \mathbf {a} )=({\boldsymbol {S}}\cdot \mathbf {b} )\cdot ({\boldsymbol {\nabla }}\mathbf {v} ^{T}\cdot \mathbf {a} )~.}$

From the relation ${\displaystyle \mathbf {a} \cdot ({\boldsymbol {A}}^{T}\cdot \mathbf {b} )=\mathbf {b} \cdot ({\boldsymbol {A}}\cdot \mathbf {a} )}$  we have

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

Using the relation ${\displaystyle {\boldsymbol {A}}\cdot ({\boldsymbol {B}}\cdot \mathbf {b} )=({\boldsymbol {A}}\cdot {\boldsymbol {B}})\cdot \mathbf {b} }$ , we get

${\displaystyle {\boldsymbol {\nabla }}\mathbf {v} \cdot ({\boldsymbol {S}}\cdot \mathbf {b} )=({\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}})\cdot \mathbf {b} ~.}$

Therefore, the final form of the first term is

${\displaystyle ({\boldsymbol {S}}\cdot \mathbf {b} )\cdot {\boldsymbol {\nabla }}(\mathbf {v} \cdot \mathbf {a} )=\mathbf {a} \cdot [({\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}})\cdot \mathbf {b} ]~.}$

For the second term, from the identity ${\displaystyle {\boldsymbol {\nabla }}\bullet ({\boldsymbol {S}}^{T}\cdot \mathbf {v} )={\boldsymbol {S}}:{\boldsymbol {\nabla }}\mathbf {v} +\mathbf {v} \cdot ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}})}$  we get, ${\displaystyle {\boldsymbol {\nabla }}\bullet ({\boldsymbol {S}}\cdot \mathbf {b} )={\boldsymbol {S}}^{T}:{\boldsymbol {\nabla }}\mathbf {b} +\mathbf {b} \cdot ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})}$ .

Since ${\displaystyle \mathbf {b} }$  is constant, ${\displaystyle {\boldsymbol {\nabla }}\mathbf {b} =0}$ , and we have

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

From the definition ${\displaystyle (\mathbf {u} \otimes \mathbf {v} )\cdot \mathbf {a} =(\mathbf {a} \cdot \mathbf {v} )\mathbf {u} }$ , we get

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

Therefore, the final form of the second term is

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

Adding the two terms, we get

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

Therefore,

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