2.2 Index Notation for Vector and Tensor Operations

Operations on Cartesian components of vectors and tensors may be expressed very efficiently and clearly using index notation.

2.1. Vector and tensor components.

Let x be a (three dimensional) vector and let S be a second order tensor.   Let $\left\{{e}_{1},{e}_{2},{e}_{3}\right\}$ be a Cartesian basis. Denote the components of x in this basis by $\left({x}_{1},{x}_{2},{x}_{3}\right)$, and denote the components of S by

$\left[\begin{array}{l}{S}_{11}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{12}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{13}\\ {S}_{21}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{22}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{23}\\ {S}_{31}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{32}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{33}\end{array}\right]$

Using index notation, we would express x and S as

$x\equiv {x}_{i\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}S\equiv {S}_{ij}$

#### 2.2. Conventions and special symbols for index notation

Range Convention: Lower case Latin subscripts (i, j, k…) have the range $\left(1,2,3\right)$.  The symbol ${x}_{i}$ denotes three components of a vector ${x}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}_{2}$ and ${x}_{3}$.  The symbol ${S}_{ij}$ denotes nine components of a second order tensor, ${S}_{11},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{12},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{13},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{21}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{33}$

Summation convention (Einstein convention): If an index is repeated in a product of vectors or tensors, summation is implied over the repeated index.  Thus

$\begin{array}{l}\lambda ={a}_{i}{b}_{i}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\lambda =\text{\hspace{0.17em}}\sum _{i=1}^{3}{a}_{i}{b}_{i}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\lambda ={a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3}\\ {c}_{i}={S}_{ik}{x}_{k}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{c}_{i}=\sum _{k=1}^{3}{S}_{ik}{x}_{k}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left\{\begin{array}{l}{c}_{1}={S}_{11}{x}_{1}+{S}_{12}{x}_{2}+{S}_{13}{x}_{3}\\ {c}_{2}={S}_{21}{x}_{1}+{S}_{22}{x}_{2}+{S}_{23}{x}_{3}\\ {c}_{3}={S}_{31}{x}_{1}+{S}_{32}{x}_{2}+{S}_{33}{x}_{3}\end{array}\text{\hspace{0.17em}}\\ \lambda ={S}_{ij}{S}_{ij}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\lambda =\sum _{i=1}^{3}\sum _{j=1}^{3}{S}_{ij}{S}_{ij}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\lambda ={S}_{11}{S}_{11}+{S}_{12}{S}_{12}+\dots +{S}_{31}{S}_{31}+{S}_{32}{S}_{32}+{S}_{33}{S}_{33}\\ {C}_{ij}={A}_{ik}{B}_{kj}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}_{ij}=\sum _{k=1}^{3}{A}_{ik}{B}_{kj}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[C\right]=\left[A\right]\left[B\right]\\ {C}_{ij}={A}_{ki}{B}_{kj}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}_{ij}=\sum _{k=1}^{3}{A}_{ki}{B}_{kj}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[C\right]={\left[A\right]}^{T}\left[B\right]\end{array}$

In the last two equations, $\left[A\right]$, $\left[B\right]$ and $\left[C\right]$ denote the $\left(3×3\right)$ component matrices of A, B and C.

The Kronecker Delta:  The symbol ${\delta }_{ij}$ is known as the Kronecker delta, and has the properties

${\delta }_{ij}=\left\{\begin{array}{l}1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=j\\ 0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i\ne j\end{array}$

thus

${\delta }_{11}={\delta }_{22}={\delta }_{33}=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\delta }_{12}={\delta }_{21}={\delta }_{13}={\delta }_{31}={\delta }_{32}={\delta }_{32}=0$

You can also think of ${\delta }_{ij}$ as the components of the identity tensor, or a $\left(3×3\right)$ identity matrix.  Observe the following useful results

$\begin{array}{l}{\delta }_{ij}={\delta }_{ji}\\ {\delta }_{kk}=3\\ {a}_{i}={\delta }_{ik}{a}_{k}\\ {A}_{ij}={\delta }_{ik}{A}_{kj}\end{array}$

The Permutation Symbol: The symbol ${\in }_{ijk}$ has properties

thus

$\begin{array}{l}{\in }_{123}={\in }_{231}={\in }_{312}=1\\ {\in }_{321}={\in }_{213}={\in }_{132}=-1\\ {\in }_{111}={\in }_{112}={\in }_{113}={\in }_{121}={\in }_{122}={\in }_{131}={\in }_{133}=0\\ {\in }_{211}={\in }_{212}={\in }_{221}={\in }_{222}={\in }_{223}={\in }_{232}={\in }_{233}=0\\ {\in }_{311}={\in }_{313}={\in }_{322}={\in }_{323}={\in }_{321}={\in }_{332}={\in }_{333}=0\end{array}$

Note that

$\begin{array}{l}{\in }_{ijk}={\in }_{kij}={\in }_{jki}=-{\in }_{jik}=-{\in }_{kji}=-{\in }_{kji}\\ {\in }_{kki}=0\\ {\in }_{ijk}{\in }_{imn}={\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{mk}\\ {\in }_{ijk}{\in }_{lmn}={\delta }_{il}\left({\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{km}\right)-{\delta }_{im}\left({\delta }_{jl}{\delta }_{kn}-{\delta }_{jn}{\delta }_{kl}\right)+{\delta }_{in}\left({\delta }_{jl}{\delta }_{km}-{\delta }_{jm}{\delta }_{kl}\right)\end{array}$

2.3. Rules of index notation

1. The same index (subscript) may not appear more than twice in a product of two (or more) vectors or tensors.  Thus

${A}_{ik}{x}_{k},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{ik}{B}_{kj},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{ij}{B}_{ik}{C}_{nk}$

are valid, but

${A}_{kk}{x}_{k},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{ik}{B}_{kk},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{ij}{B}_{ik}{C}_{ik}$

are meaningless

2. Free indices on each term of an equation must agree.  Thus

$\begin{array}{l}{x}_{i}={u}_{i}+{c}_{i}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=u+c\\ {a}_{i}={A}_{ki}{B}_{kj}{x}_{j}+{C}_{ik}{u}_{k}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a={A}^{T}B\text{\hspace{0.17em}}x+Cu\end{array}$

are valid, but

$\begin{array}{l}{x}_{i}={A}_{ij}\\ {x}_{j}={A}_{ik}{u}_{k}\\ {x}_{i}={A}_{ik}{u}_{k}+{c}_{j}\end{array}$

are meaningless.

3.  Free and dummy indices may be changed without altering the meaning of an expression, provided that rules 1 and 2 are not violated. Thus

${x}_{i}={A}_{ik}{x}_{k}⇔{x}_{j}={A}_{jk}{x}_{k}⇔{x}_{j}={A}_{ji}{x}_{i}$

2.4. Vector operations expressed using index notation

Addition.   $c=a+b\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{c}_{i}={a}_{i}+{b}_{i}$

Dot Product  $\lambda =a\cdot b\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\lambda ={a}_{i}{b}_{i}$

Vector Product $c=a×b\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{c}_{i}=\text{\hspace{0.17em}}{\in }_{ijk}{a}_{j}{b}_{k}$

Dyadic Product   $S=a\otimes b\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{S}_{ij}={a}_{i}{b}_{j}$

Change of Basis.  Let a be a vector. Let $\left\{{e}_{1},{e}_{2},{e}_{3}\right\}$ be a Cartesian basis, and denote the components of a in this basis by ${a}_{i}$.  Let $\left\{{m}_{1},{m}_{2},{m}_{3}\right\}$ be a second basis, and denote the components of a in this basis by ${\alpha }_{i}$.  Then, define

${Q}_{ij}={m}_{i}\cdot {e}_{j}=\mathrm{cos}\theta \left({m}_{i},{e}_{j}\right)$

where $\theta \left({m}_{i},{e}_{j}\right)$ denotes the angle between the unit vectors ${m}_{i}$  and ${e}_{j}$.  Then

${\alpha }_{i}={Q}_{ij}{a}_{j}$

2.5. Tensor operations expressed using index notation.

Addition.   $C=A+B\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}_{ij}={A}_{ij}+{B}_{ij}$

Transpose  $A={B}^{T}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{ij}={B}_{ji}$

Scalar Products $\begin{array}{l}\lambda =A:B\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\lambda ={A}_{ij}{B}_{ij}\\ \lambda =A\cdot \cdot \text{\hspace{0.17em}}B\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \lambda ={A}_{ji}{B}_{ij}\end{array}$

Product of a tensor and a vector $\begin{array}{l}c=Ab\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{c}_{i}={A}_{ij}{b}_{j}\\ c={A}^{T}b\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{c}_{i}={A}_{ji}{b}_{j}\end{array}$

Product of two tensors  $\begin{array}{l}C=AB\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}_{ij}={A}_{ik}{B}_{kj}\\ C={A}^{T}B\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}_{ij}={A}_{ki}{B}_{kj}\end{array}$

Determinant $\begin{array}{l}\lambda =\mathrm{det}A\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\lambda =\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{6}{\in }_{ijk}{\in }_{lmn}{A}_{li}{A}_{mj}{A}_{nk}\text{\hspace{0.17em}}={\in }_{ijk}{A}_{i1}{A}_{j2}{A}_{k3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⇔\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\in }_{lmn}\lambda \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\in }_{ijk}{A}_{li}{A}_{mj}{A}_{nk}={\in }_{ijk}{A}_{il}{A}_{jm}{A}_{kn}\end{array}$

Inverse ${S}_{ji}^{-1}=\frac{1}{2\mathrm{det}\left(S\right)}{\in }_{ipq}{\in }_{jkl}{S}_{pk}{S}_{ql}$

Change of Basis.  Let A be a second order tensor. Let $\left\{{e}_{1},{e}_{2},{e}_{3}\right\}$ be a Cartesian basis, and denote the components of A in this basis by ${A}_{ij}$.  Let $\left\{{m}_{1},{m}_{2},{m}_{3}\right\}$ be a second basis, and denote the components of A in this basis by ${\Lambda }_{ij}$.  Then, define

${Q}_{ij}={m}_{i}\cdot {e}_{j}=\mathrm{cos}\theta \left({m}_{i},{e}_{j}\right)$

where $\theta \left({m}_{i},{e}_{j}\right)$ denotes the angle between the unit vectors ${m}_{i}$  and ${e}_{j}$.  Then

${\Lambda }_{ij}={Q}_{ik}{A}_{km}{Q}_{jm}$

2.6. Calculus using index notation

The derivative $\partial {x}_{i}/\partial {x}_{j}$ can be deduced by noting that $\partial {x}_{i}/\partial {x}_{j}=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=j$ and  $\partial {x}_{i}/\partial {x}_{j}=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i\ne j$.  Therefore

$\frac{\partial {x}_{i}}{\partial {x}_{j}}={\delta }_{ij}$

The same argument can be used for higher order tensors

$\frac{\partial {A}_{ij}}{\partial {A}_{kl}}={\delta }_{ik}{\delta }_{jl}$

2.7. Examples of algebraic manipulations using index notation

1. Let a, b, c, d be vectors.  Prove that

$\left(a×b\right)\cdot \left(c×d\right)=\left(a\cdot c\right)\left(b\cdot d\right)-\left(b\cdot c\right)\left(a\cdot d\right)$

Express the left hand side of the equation using index notation (check the rules for cross products and dot products of vectors to see how this is done)

$\left(a×b\right)\cdot \left(c×d\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\in }_{ijk}{a}_{j}{b}_{k}{\in }_{imn}{c}_{m}{d}_{n}$

Recall the identity

${\in }_{ijk}{\in }_{imn}={\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{mk}$

so

$\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\in }_{ijk}{a}_{j}{b}_{k}{\in }_{imn}{c}_{m}{d}_{n}=\left({\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{mk}\right){a}_{j}{b}_{k}{c}_{m}{d}_{n}$

Multiply out, and note that

${\delta }_{jm}{a}_{j}={a}_{m}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\delta }_{kn}{b}_{k}={b}_{n}$

(multiplying by a Kronecker delta has the effect of switching indices…) so

$\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left({\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{mk}\right){a}_{j}{b}_{k}{c}_{m}{d}_{n}={a}_{m}{b}_{n}{c}_{m}{d}_{n}-{a}_{n}{b}_{m}{c}_{m}{d}_{n}$

Finally, note that

${a}_{m}{c}_{m}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\equiv a\cdot c$

and similarly for other products with the same index, so that

${a}_{m}{b}_{n}{c}_{m}{d}_{n}-{a}_{n}{b}_{m}{c}_{m}{d}_{n}={a}_{m}{c}_{m}{b}_{n}{d}_{n}-{b}_{m}{c}_{m}{a}_{n}{d}_{n}\equiv \left(a\cdot c\right)\left(b\cdot d\right)-\left(b\cdot c\right)\left(a\cdot d\right)$

2. The stress$—$strain relation for linear elasticity may be expressed as

${\sigma }_{ij}=\frac{E}{1+\nu }\left({\epsilon }_{ij}+\frac{\nu }{1-2\nu }{\epsilon }_{kk}{\delta }_{ij}\right)$

where ${\sigma }_{ij}$ and ${\epsilon }_{ij}$ are the components of the stress and strain tensor, and $E$ and $\nu$ denote Young’s modulus and Poisson’s ratio.  Find an expression for strain in terms of stress.

Set i=j to see that

${\sigma }_{ii}=\frac{E}{1+\nu }\left({\epsilon }_{ii}+\frac{\nu }{1-2\nu }{\epsilon }_{kk}{\delta }_{ii}\right)$

Recall that ${\delta }_{ii}=3$, and notice that we can replace the remaining ii by kk

$\begin{array}{l}{\sigma }_{kk}=\frac{E}{1+\nu }\left({\epsilon }_{kk}+\frac{\nu }{1-2\nu }3{\epsilon }_{kk}\right)=\frac{E}{1-2\nu }{\epsilon }_{kk}\\ ⇔\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\epsilon }_{kk}=\frac{1-2\nu }{E}{\sigma }_{kk}\end{array}$

Now, substitute for ${\epsilon }_{kk}$ in the given stress$—$strain relation

$\begin{array}{l}{\sigma }_{ij}=\frac{E}{1+\nu }\left({\epsilon }_{ij}+\frac{\nu }{E}{\sigma }_{kk}{\delta }_{ij}\right)\\ ⇔{\epsilon }_{ij}=\frac{1+\nu }{E}\left({\sigma }_{ij}-\frac{\nu }{1+\nu }{\sigma }_{kk}{\delta }_{ij}\right)\end{array}$

3. Solve the equation

$\mu \left\{{\delta }_{kj}{a}_{i}{a}_{i}+\frac{1}{1-2\nu }{a}_{k}{a}_{j}\right\}{U}_{k}={P}_{j}$

for ${U}_{k}$ in terms of ${P}_{i}$ and ${a}_{i}$

Multiply both sides by ${a}_{j}$ to see that

$\begin{array}{l}\mu \left\{{a}_{j}{\delta }_{kj}{a}_{i}{a}_{i}+\frac{1}{1-2\nu }{a}_{k}{a}_{j}{a}_{j}\right\}{U}_{k}={P}_{j}{a}_{j}\\ ⇔\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mu \left\{{a}_{k}{a}_{i}{a}_{i}+\frac{1}{1-2\nu }{a}_{k}{a}_{j}{a}_{j}\right\}{U}_{k}={P}_{j}{a}_{j}\\ ⇔\mu {U}_{k}{a}_{k}\frac{2\left(1-\nu \right)}{1-2\nu }{a}_{i}{a}_{i}={P}_{j}{a}_{j}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⇔{U}_{k}{a}_{k}=\frac{\left(1-2\nu \right){P}_{j}{a}_{j}}{2\mu \left(1-\nu \right){a}_{i}{a}_{i}}\end{array}$

Substitute back into the equation given for ${U}_{k}{a}_{k}$ to see that

$\mu {U}_{j}{a}_{i}{a}_{i}+\frac{{P}_{k}{a}_{k}}{2\left(1-\nu \right){a}_{i}{a}_{i}}{a}_{j}={P}_{j}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⇒{U}_{j}=\frac{1}{\mu {a}_{i}{a}_{i}}\left({P}_{j}-\frac{{P}_{k}{a}_{k}}{2\left(1-\nu \right){a}_{n}{a}_{n}}{a}_{j}\right)$

4. Let $r=\sqrt{{x}_{k}{x}_{k}}$.  Calculate $\frac{\partial r}{\partial {x}_{i}}$

We can just apply the usual chain and product rules of differentiation

$\frac{\partial r}{\partial {x}_{i}}=\frac{1}{2}\frac{1}{\sqrt{{x}_{k}{x}_{k}}}\left({x}_{k}\frac{\partial {x}_{k}}{\partial {x}_{i}}+\frac{\partial {x}_{k}}{\partial {x}_{i}}{x}_{k}\right)=\frac{1}{\sqrt{{x}_{k}{x}_{k}}}{x}_{k}{\delta }_{ik}=\frac{{x}_{i}}{\sqrt{{x}_{k}{x}_{k}}}=\frac{{x}_{i}}{r}$

5. Let $\lambda ={A}_{ij}{A}_{ij}$.  Calculate $\partial \lambda /\partial {A}_{kl}$

Using the product rule