Preliminary
Reference
Mechanics Lecture Notes Part III: Foundations of Continuum Mechanics , pa.kelly@auckland.ac.nz .
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Tensors
A tensor of order zero is simply another name for a scalar \(\alpha\) .
A first-order tensor is simply another name for a vector \(\pmb{u}\) .
We use uppercase bold-face Latin letters to denote second order tensor.
A second-order tensor \(\pmb{T}\) may be defined as an operator that acts on a vector \(\pmb{u}\) generating another vector \(\pmb{v}\) , such that
\[
\pmb{T}(\pmb{u})=\pmb{v}
\]
which is a linear operator.
the tensor product of two vectors \(\pmb{u}\) and \(\pmb{v}\)
\[
\pmb{u}\otimes \pmb{v}
\]
is defined by
\[
(\pmb{u}\otimes \pmb{v} ) \pmb{w} = \pmb{u}(\pmb{v}\cdot \pmb{w})
\]
Properties
(i)
\[
(\pmb{u}\otimes \pmb{v})(\pmb{w}\otimes \pmb{x})=(\pmb{v}\cdot\pmb{w})(\pmb{u}\otimes \pmb{x})
\]
cause
\[
\begin{align*}
(\pmb{u}\otimes \pmb{v})(\pmb{w}\otimes \pmb{x})\pmb{y}&=(\pmb{u}\otimes \pmb{v})(\pmb{x}\cdot\pmb{y})\pmb{w}\\
&=(\pmb{x}\cdot\pmb{y})(\pmb{u}\otimes \pmb{v})\pmb{w}\\
&=(\pmb{x}\cdot\pmb{y})(\pmb{v}\cdot\pmb{w})\pmb{u}\\
&=(\pmb{v}\cdot\pmb{w})(\pmb{x}\cdot\pmb{y})\pmb{u}\\
&=(\pmb{v}\cdot\pmb{w})(\pmb{u}\otimes \pmb{x})\pmb{y}
\end{align*}
\]
(ii)
\[
\pmb{u}(\pmb{v}\otimes \pmb{w})=(\pmb{u}\cdot \pmb{v})\pmb{w}
\]
cause
\[
\begin{align*}
(\pmb{y}\otimes \pmb{u})(\pmb{v}\otimes \pmb{w})&=(\pmb{u}\cdot \pmb{v})(\pmb{y}\otimes \pmb{w})\\
&=\pmb{y} \otimes [(\pmb{u}\cdot \pmb{v})\pmb{w}]
\end{align*}
\]
Some example
Projection Tensor \((\pmb{e}\otimes \pmb{e})\)
So
\[
(\pmb{e}\otimes \pmb{e})\pmb{u} = (\pmb{e}\cdot \pmb{u})\pmb{e}
\]
is the vector projection of \(\pmb{u}\) on \(\pmb{e}\) , denoted by \(\pmb{P}\) .
A dyadic is a linear combination of dyads (with scalar coefficients).
In the following discussion, we can treat \(\pmb{T}\) as a matrix.
Cartesian Tensors
A second order tensor and the the vector it operates on can be described in terms of Cartesian components.
Example
Identity tensor /(or unit tensor ).
\[
\pmb{I}=\sum_{i=1}^d\pmb{e}_i\otimes \pmb{e}_i
\]
cause it follows
\[
\begin{align*}
\pmb{I} \pmb{u}&=\sum_{i=1}^d(\pmb{e}_i\otimes \pmb{e}_i) \pmb{u}\\
&=\sum_{i=1}^d(\pmb{e}_i \cdot \pmb{u})\pmb{e}_i \\
&=\sum_{i=1}^d u_i\pmb{e}_i \\
&=\pmb{u}
\end{align*}
\]
Or identity tensor can be written as
\[
\pmb{I} = \sum_{i,j=1}^d\delta_{ij}(\pmb{e}_i\otimes \pmb{e}_j)
\]
Second order tensor as a Dyadic
Every second order tensor can always be written as a dyadic involving the Cartesian base vectors \(\pmb{e}_i\) , that is, if we denote
\[
\pmb{E}_i=\pmb{T}\left(\pmb{e}_i\right)
\]
then
\[
\pmb{T}= \sum_{i=1}^d \left(\pmb{E}_i\otimes \pmb{e}_i\right)
\]
Proof
\[
\begin{align*}
\pmb{b}&=\pmb{T}(\pmb{a})\\
&=\pmb{T}\left(\sum_{i=1}^d a_i\pmb{e}_i\right)\\
&=\sum_{i=1}^d a_i \pmb{T}(\pmb{e}_i)\\
\end{align*}
\]
Denote \(\pmb{T}(\pmb{e}_i)\) to be \(\pmb{E}_i\) , then
\[
\begin{align*}
\pmb{b}&=\sum_{i=1}^d a_i \pmb{E}_i\\
&=\sum_{i=1}^d (\pmb{a}\cdot \pmb{e}_i )\pmb{E}_i\\
&=\sum_{i=1}^d (\pmb{E}_i\otimes \pmb{e}_i) \pmb{a}
\end{align*}
\]
So
\[
\pmb{T}= \sum_{i=1}^d (\pmb{E}_i\otimes \pmb{e}_i)
\]
If we write \(\pmb{E}_i\) with base vectors like
\[
\pmb{E}_i=\sum_{j=1}^d E_{ij}\pmb{e}_j, \quad i=1,\cdots, d
\]
Then
\[
\begin{align*}
(\pmb{E}_i\otimes \pmb{e}_i)&=\left(\sum_{j=1}^d E_{ij}\pmb{e}_j\right)\otimes \pmb{e}_i\\
&=\sum_{j=1}^d E_{ij} (\pmb{e}_j\otimes \pmb{e}_i), \quad i=1,\cdots, d
\end{align*}
\]
Thus
\[
\pmb{T}=\sum_{i=1}^d\sum_{j=1}^d E_{ij} (\pmb{e}_j\otimes \pmb{e}_i)
\]
Introduce 9 scalars \(T_{ij}=E_{ji}\) , then
\[
\begin{align*}
\pmb{T}&=\sum_{i=1}^d\sum_{j=1}^d T_{ji} (\pmb{e}_j\otimes \pmb{e}_i)\\
&=\sum_{j=1}^d\sum_{i=1}^d T_{ji} (\pmb{e}_j\otimes \pmb{e}_i) \quad \text{switch summation turn}\\
&=\sum_{i=1}^d\sum_{j=1}^d T_{ij} (\pmb{e}_i\otimes \pmb{e}_j)\quad \text{switch $i$ and $j$}
\end{align*}
\]
We can see that 9 dyads \(\{\pmb{e}_i\otimes \pmb{e}_j\}_{i,j=1}^3\) forms a basis for the space of second order tensors.
Recall that
\[
\begin{align*}
T_{ij}&=E_{ji}\\
&=\pmb{E}_j \cdot \pmb{e}_i \\
&=T(\pmb{e}_j) \cdot (\pmb{e}_i)\\
&=(\pmb{e}_i) \cdot T(\pmb{e}_j)\\
\end{align*}
\]
So we can get the component of a tensor by the above way.
Cauchy Stress Tensor
The traction vector , the limiting value of the ratio of force over area, that is,
\[
\pmb{t}^{\pmb{n}}=\lim_{\Delta_s\rightarrow 0}\frac{\Delta F}{\Delta S}
\]
where \(\pmb{n}\) denotes normal vector to the surface.
The stress \(\pmb{\sigma}\) , a second order tensor which maps \(\pmb{n}\) onto \(\pmb{t}\)
\[
\pmb{t}=\pmb{\sigma}\pmb{n}
\]
If we consider a coordinate system with base vectors \(\pmb{e}_i\) , then \(\pmb{\sigma}=\sum\limits_{i,j=1}^d(\sigma_{ij}\pmb{e}_i \otimes \pmb{e}_j)\)
\[
\pmb{\sigma}=(\sigma_{ij})
\]
So
\[
t_i \pmb{e_i} = \sum_{j=1}^3\sigma_{ij}n_{j} \pmb{e}_i
\]
For example,
\[
\pmb{\sigma}\pmb{e}_j=\sum_{i=1}^3\sigma_{ij}\pmb{e}_i
\]
which denotes the summation of the \(j\) th column of matrix \(\pmb{\sigma}\) .
So the components \(\sigma_{11}, \sigma_{21}, \sigma_{31}\) of the stress tensor are the three components of the traction vector which acts on the plane with normal \(\pmb{e}_1\) .
Hamilton Operator
First we want to introduce the operator
\[
\nabla=\pmb{i}\frac{\partial }{\partial x}+\pmb{j}\frac{\partial }{\partial y}+\pmb{k}\frac{\partial }{\partial z}
\]
then
\[
\nabla f= \pmb{i}\frac{\partial f}{\partial x}+\pmb{j}\frac{\partial f}{\partial y}+\pmb{k}\frac{\partial f}{\partial z}=\text{grad} f
\]
we also have inner product
\[
\begin{align*}
\nabla\cdot \pmb{a}&=\left(\pmb{i}\frac{\partial }{\partial x}+\pmb{j}\frac{\partial }{\partial y}+\pmb{k}\frac{\partial }{\partial z}\right)\cdot (P\pmb{i}+Q\pmb{j}+R\pmb{k})\\
&=\frac{\partial P}{\partial x}+\frac{\partial Q}
{\partial y}+\frac{\partial R}{\partial z}\\
&=\text{div} \pmb{a}
\end{align*}
\]
and cross product
\[
\begin{align*}
\nabla\times \pmb{a}&=\left(\pmb{i}\frac{\partial }{\partial x}+\pmb{j}\frac{\partial }{\partial y}+\pmb{k}\frac{\partial }{\partial z}\right)\times (P\pmb{i}+Q\pmb{j}+R\pmb{k})\\
&=\left|\begin{array}{ccc}
\pmb{i}&\pmb{j}&\pmb{k}\\
\displaystyle \frac{\partial }{\partial x}&\displaystyle \frac{\partial }{\partial y} & \displaystyle \frac{\partial }{\partial z}\\
P & Q & R\\
\end{array}
\right|\\
&=\left(
\frac{\partial R}{\partial y}-
\frac{\partial Q}{\partial z}\right)\pmb{i}+\left(
\frac{\partial R}{\partial x}-
\frac{\partial P}{\partial z}\right)\pmb{j}+\left(
\frac{\partial Q}{\partial x}-
\frac{\partial P}{\partial y}\right)\pmb{k}\\
&=\text{rot} \pmb{a}
\end{align*}
\]
Then Gauss Formula can be expressed by
\[
\iint_{\partial\Omega}\pmb{a}d\pmb{S}=\iiint_{\Omega}\nabla\cdot \pmb{a}dV
\]
Stokes Formula can be expressed by
\[
\int_{\partial \Sigma}\pmb{a}d\pmb{s}=\iint_{\Sigma}(\nabla\times \pmb{a})\cdot d\pmb{S}
\]
