•   Wikis

# Continuum mechanics/Tensor-vector identities: Wikis

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

# Study guide

Up to date as of January 14, 2010

## 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,

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

## Tensor-vector identity 2

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

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

Proof:

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

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

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

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

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

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

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

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

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

Therefore, the final form of the first term is

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

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

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

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

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

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

Therefore, the final form of the second term is

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

Adding the two terms, we get

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

Therefore,

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