### 93 Fundamentals -- Tensors, Symbols, and Some Basic Identities

Hello everyone! After several posts without any formulas stated, today we are going to have a post with lots of them. Recently I am preparing an interesting post about material mechanics, and it is difficult to fully appreciate that without the knowledge about tensor. So let's see how far we can go!

Before we talk about tensors, let's first ask ourselves: "what is a vector, and why do we need it?" A vector is a physical quantity that specifies both its magnitude and its direction. For example, velocity is a vector, and you need to specify both the speed and the direction in order to track the motion of a tennis ball. To specify the direction, we will need to define a set of coordinates, so that we can decompose the velocity into different components that are parallel to the axes of our coordinates. The mathematical representation of the process is as simple as this:
$\vec{v}=\sum_{i} v_{i}\hat{e}_{i}=v_{i}\hat{e}_{i}$
Here we use Einstein notation to save ourselves from endless summation symbols.

However, there are many ways to define our set of coordinates. Clearly, our physical quantity -- the velocity of the tennis ball -- should not change if we merely change our set of coordinates. Mathematically speaking, when we change our set of coordinates from {e} to {e'}, the velocity itself should stay the same:
$\vec{v}=v_{i}\hat{e}_{i}=v_{i}'\hat{e}_{i}'$
This implies the decompositions of the velocity vector must follow certain transformation rules. This is a major difference between a vector and a column matrix: a vector is a physical quantity that specifies both its magnitude and its direction, and its decompositions must follow certain transformation rules when we change our coordinates. On the other hand, a column matrix is just a mathematical entity, and does not necessarily follow any transformation rules.

What if there is another physical quantity that must be specified by at least 2 directions? For example, when the tennis ball hit the ground, the ball will deform, and the kinetic energy will be transformed into elastic potential energy. For a unit element within the tennis ball, the forces acting upon it can be quite complex: for all its 6 surfaces, there may be forces acting along x, y, and z directions, so you need 2 directions to specify the force field -- one for the direction of the plane upon which the force is acting, and another one for the direction of the force. This is when we need a tensor: the deformation within an elastic material could be specified by a strain tensor, and the internal force within an elastic material could be specified by a stress tensor. Just like vectors, they also have to follow certain transformation rules when we change our coordinates.

After we know a little bit more about what tensor is for, we can now dig into some practical details. First of all, we need to familiarize ourselves with the index notation of tensors, vectors, and matrices. For example, the following: (Here we use bold lowercase letters to represent vectors, bold uppercase letters to represent tensors, letters in square brackets to represent matrices, and letters in braces to represent column matrices.)
$\textbf{a}$
$\textbf{A}$
$\left [ A \right ]\left \{ x \right \}=\left \{ b \right \}$
$\textbf{a}\cdot\textbf{b}=\phi$
$\left [ C \right ]=\left[A\right]\left[B\right]$
$\left [ D \right ]=\left[A\right]\left[B\right]^{T}$
can also be written in index notation:
$a_{i}\textbf{e}_{i}$
$A_{ij}\textbf{e}_{i}\textbf{e}_{j}$
$A_{ij}x_{j}=b_{i}$
$a_{i}b_{i}=\phi$
$C_{ij}=A_{ik}B_{kj}$
$D_{ij}=A_{ik}B_{jk}$

We then need 3 important functions -- the Kronecker delta function( $\delta_{ij}$), permutation symbol/Levi-Civita symbol( $e_{ijk}$) and the direction cosine matrix( $l_{ij}$), which can be defined as below:
$\delta_{ij} = \textbf{e}_{i}\cdot\textbf{e}_{j}=\begin{matrix}1 \; \text{if} \; i=j\\ 0 \; \text{if} \; i\neq j\end{matrix}$

$e_{ijk}=\textbf{e}_{i}\times\textbf{e}_{j} \cdot\textbf{e}_k=\begin{matrix}1 \; \text{if} \; (i,j,k)=(1,2,3),(2,3,1),(3,1,2)\\ -1 \; \text{if} \; (i,j,k)=(3,2,1),(2,1,3),(1,3,2)\\0\;\text{if otherwise}\end{matrix}$

$l_{ij}= \textbf{e}_{i}\cdot\textbf{e}_{j}'$
We can easily express some basic manipulations of tensors with these symbols, and simplify them with identities of Kronecker delta and permutation symbols. Some really basic ones would be:
$f_{i} = \textbf{f}\cdot\textbf{e}_{i},\;\textbf{f}=f_{i}\textbf{e}_{i},\; f_{i}\delta_{ij}=f_{j}$
$e_{ijk}=e_{jki}=e_{kij}=-e_{ikj}=-e_{jik}=-e_{kji}$
$\textbf{a}\times\textbf{b}=a_{i}b_{j}e_{ijk}\textbf{e}_{k}$
$\text{det}(A) = e_{ijk}A_{i1}A_{j2}A_{k3}=e_{rst}A_{1r}A_{2s}A_{3t}$
$e_{ijk}e_{iqr}=\delta_{jq}\delta_{kr} - \delta_{jr}\delta_{kq}$
$e_{ijk}e_{ijr}=2\delta_{kr}$
$e_{ijk}e_{ijk}=6$
$\frac{\partial x_{i}}{\partial x_{j}}=\delta_{ij}$
$\triangledown=\textbf{e}_{i}\partial_{i}$

These identities can help us prove some strange vector identities. For example:
$\triangledown\times(\textbf{a}\times\textbf{b})=(\textbf{b}\cdot\triangledown)\textbf{a}+\textbf{a}(\triangledown\cdot\textbf{b})-\textbf{b}(\triangledown\cdot\textbf{a})-(\textbf{a}\cdot\triangledown)\textbf{b}$
That looks impossible to prove with vector calculus or Green's theorems or what so ever. But it is actually fairly easy to prove with these tensor identities.
$\triangledown\times(\textbf{a}\times\textbf{b})\\ =\textbf{e}_{i}\partial_{i}\times(a_{j}\textbf{e}_{j}\times b_{k}\textbf{e}_{k})\\ =\textbf{e}_{i}\partial_{i}\times(a_{j}b_{k}e_{jkl}\textbf{e}_{l})\\ =\partial_{i}(a_{j}b_{k})e_{jkl}e_{ilm}\textbf{e}_{m}\\ =\partial_{i}(a_{j}b_{k})\textbf{e}_{m}(\delta_{jm}\delta_{ki}-\delta_{ji}\delta_{km})\\ =\partial_{i}(a_{j}b_{k})\textbf{e}_{m}\delta_{jm}\delta_{ki} - \partial_{i}(a_{j}b_{k})\textbf{e}_{m}\delta_{ji}\delta_{km}\\=\partial_{i}(a_{m}b_{i})\textbf{e}_{m}-\partial_{i}(a_{i}b_{m})\textbf{e}_{m}\\ =b_{i}(\partial_{i}a_{m})\textbf{e}_{m}+a_{m}\textbf{e}_{m}(\partial_{i}b_{i})-(\partial_{i}a_{i})b_{m}\textbf{e}_{m}-a_{i}(\partial_{i}b_{m})\textbf{e}_{m}\\ =(\textbf{b}\cdot\triangledown)\textbf{a}+\textbf{a}(\triangledown\cdot\textbf{b})-(\textbf{a}\cdot\triangledown)\textbf{b}-\textbf{b}(\triangledown\cdot\textbf{a})$

That is a lot information for one episode, and I hope that make sense so far. I will have to save the transformation rules for our next episode, so stay tuned!