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Hello everyone! Just a brief disclaimer: this post is never intended to be very comprehensive, but just to familiarize our readers with these symbols and basic operation rules of tensors that you may encounter in the future. I hope you guys enjoy so far, and we will continue our journey today!

Last time we have briefly mentioned about the "transformation rules" that both tensors and vectors must follow. Both tensors and vectors are physical properties that should not change by our choices of coordinate systems (here we always assume Cartesian coordinate systems). When we change our coordinate systems from { $\textbf{e}_{i}$} to { $\textbf{e}_{i}'$}, the components of direction cosine matrix [L] would be:

$l_{ij} = \textbf{e}_{i}\cdot\textbf{e}_{j}'$

For any vector $\textbf{u}$, the following identity must hold when we change our coordinate systems:

$\textbf{u}=u_{i}\textbf{e}_{i}=u_{j}'\textbf{e}_{j}'$

We can do inner product to $\textbf{e}_{i}$ for the both side and we will get:

$u_{i}=u_{j}'\textbf{e}_{j}'\cdot\textbf{e}_{i}=l_{ij}u_{j}'$

Or $\{u\}=[L]\{u'\}$. As for the basis, it is just quite similar:

$\textbf{e}_{i}=l_{ij}\textbf{e}_{j}'$

Now for the tensor part. First of all, we can write down the components of a second-order tensor by $\textbf{T}=T_{ij}\textbf{e}_{i}\textbf{e}_{j}$. Note that the relationship between $\textbf{e}_{i}$ and $\textbf{e}_{j}$ is not just normal "multiplication" or "dot product" or "cross product" or whatsoever. It is a "tensor product," which means:

$\textbf{a}\textbf{b}=\textbf{a}\otimes\textbf{b}=\{a\}\{b\}^{T}=a_{i}b_{j}\textbf{e}_{i}\textbf{e}_{j}=\begin{bmatrix} a_1b_1 & a_1b_2 & a_1b_3\\ a_2b_1 & a_2b_2 & a_2b_3\\ a_3b_1 & a_3b_2 & a_3b_3\end{bmatrix}$

When a tensor product meets another dot product or a cross product, it should follow the following rules:

$\textbf{a}\textbf{b}\cdot\textbf{c}=\textbf{a}(\textbf{b}\cdot\textbf{c})$

$\textbf{a}\textbf{b}\times\textbf{c}=\textbf{a}(\textbf{b}\times\textbf{c})$

These properties define the basic operation rules of a tensor, and is sometimes called "the dyadic definition of a second-order tensor."

We also need a double dot product for a tensor, which is defined like this:

$(\textbf{a}\textbf{b}):(\textbf{c}\textbf{d})=(\textbf{a}\cdot\textbf{c})(\textbf{b}\cdot\textbf{d})$

A double dot product is very useful, as it can help us extract the components, the trace, and the norm of a second-order tensor: ( $\textbf{I}$ is the unit tensor)

$\textbf{T}:\textbf{e}_{i}\textbf{e}_{j}=T_{ab}\textbf{e}_{a}\textbf{e}_{b}:\textbf{e}_{i}\textbf{e}_{j}=T_{ab}\delta_{ai}\delta_{bj}=T_{ij}$

$\text{tr}(\textbf{T})=\textbf{I}:\textbf{T} = T_{ij}\delta_{ij}=T_{ii}$

$\parallel \textbf{T}\parallel =\sqrt{\textbf{T}:\textbf{T}} = \sqrt{T_{ij}T_{ij}} $

Here we have talked about the dyadic definition of a second-order tensor, but there are still other ways to define a tensor. For example, a tensor can be defined by the transformation rules just like a vector:

$\textbf{T}=T_{ij}\textbf{e}_i\textbf{e}_j=T_{mn}'\textbf{e}_m'\textbf{e}_n'$

That implies:

$T_{ij}=T_{mn}'\textbf{e}_m'\textbf{e}_n':\textbf{e}_i\textbf{e}_j=T_{mn}'l_{im}l_{jn}$

Or:

$T=LT'L^{T}$

A second-order tensor can also be defined as a linear mapping. When we say an operation is linear, it means that addition and multiplication can happen before or after the operation without changing the results:

$\textbf{T}(\alpha\textbf{u}+\beta\textbf{v})=\alpha\textbf{T}(\textbf{u})+\beta\textbf{T}(\textbf{v})$

And the mapping process goes like this:

$\textbf{T}(\textbf{u})=\textbf{T}\cdot\textbf{u}=T_{ij}\textbf{e}_i\textbf{e}_j\cdot u_m\textbf{e}_m=T_{ij}u_m\textbf{e}_i\delta_{jm}=T_{ij}u_{j}\textbf{e}_{i}$

Here we mentioned some really basic properties of tensors. But to conclude this episode, we still need to briefly mention a very special group of tensors -- the isotropic tensors. A tensor is isotropic if its decomposition is invariant to coordination changes. That means:

$A_{ijk...}=l_{ip}l_{jq}l_{kr}...A_{pqr...}'=l_{ip}l_{jq}l_{kr}...A_{pqr...}$

Not all tensors are isotropic. For a second-order tensor, it must follow:

$\textbf{A}=k\textbf{I}^{(2)}=k\delta_{ij}\textbf{e}_i\textbf{e}_j$

And for a fourth-order tensor, it must follow:

$\textbf{C}=(\alpha\delta_{ij}\delta_{kl}+\beta\delta_{ik}\delta_{jl}+\gamma\delta_{il}\delta_{jk})\textbf{e}_i\textbf{e}_j\textbf{e}_k\textbf{e}_l$

in order to be an isotropic tensor. This can be proved by imposing a small rotation about $x_{k}$ axis and see how it will affect the tensor decompositions. (The proof can be found here.) Isotropic tensors are important because they can make calculation a lot more easier. To fully describe an anisotropic fourth-order tensor, you need 3x3x3x3 = 81 parameters, while you only need 3 parameters to fully describe an isotropic fourth-order tensor.

And that's a lot for today! Please scroll up to the figure in the beginning and make sure everything make sense to you. In our next episode, we will proceed to some basic concepts in material mechanics, so stay tuned!

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