Appendix C
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.

 

 

C.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

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

Using index notation, we would express x and S as

$x\equiv x_{i}S\equiv S_{ij}$

 

 

 

C.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 $x_1,x_2$ and $x_3$.  The symbol $S_{ij}$ denotes nine components of a second order tensor, $S_{11},S_{12},S_{13},S_{21}\dots S_{33}$

 

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

$\begin{array}{l}\lambda =a_i{b}_i\equiv \lambda ={\displaystyle \sum _{i=1}^3{a}_i{b}_i\equiv \lambda =a_1{b}_1+a_2{b}_2+a_3{b}_3}\\ c_i=S_{ik}{x}_k\equiv c_i={\displaystyle \sum _{k=1}^3{S}_{ik}{x}_k}\equiv \left\{\begin{array}{l}{c}_1=S_{11}{x}_1+S_{12}{x}_2+S_{13}{x}_3\\ c_2=S_{21}{x}_1+S_{22}{x}_2+S_{23}{x}_3\\ c_3=S_{31}{x}_1+S_{32}{x}_2+S_{33}{x}_3\end{array}\right.\\ \lambda =S_{ij}{S}_{ij}\equiv \lambda ={\displaystyle \sum _{i=1}^3{\displaystyle \sum _{j=1}^3{S}_{ij}{S}_{ij}}}\\ \equiv \lambda =S_{11}{S}_{11}+S_{12}{S}_{12}+\dots +S_{31}{S}_{31}+S_{32}{S}_{32}+S_{33}{S}_{33}\\ C_{ij}=A_{ik}{B}_{kj}\equiv C_{ij}={\displaystyle \sum _{k=1}^3{A}_{ik}{B}_{kj}}\equiv \left[C\right]=\left[A\right]\left[B\right]\\ C_{ij}=A_{ki}{B}_{kj}\equiv C_{ij}={\displaystyle \sum _{k=1}^3{A}_{ki}{B}_{kj}}\equiv \left[C\right]={\left[A\right]}^T\left[B\right]\end{array}$

 

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

 

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

${\delta }_{ij}=\left\{\begin{array}{l}1i=j\\ 0i\ne j\end{array}\right.$

thus

${\delta }_{11}={\delta }_{22}={\delta }_{33}=1{\delta }_{12}={\delta }_{21}={\delta }_{13}={\delta }_{31}={\delta }_{23}={\delta }_{32}=0$

You can also think of ${\delta }_{ij}$ as the components of the identity tensor, or a $\left(3\times 3\right)$ 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

${\in }_{ijk}=\left\{\begin{array}{l}1i,j,k=1,2,3;2,3,1\text{or}3,1,2\\ -1i,j,k=3,2,1;2,1,3\text{or }\text{1,3,2}\\ \text{0}\text{otherwise}\end{array}\right.$

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

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

 

 

C.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

$A_{ik}{x}_k,A_{ik}{B}_{kj},A_{ij}{B}_{ik}{C}_{nk}$

are valid, but

$A_{kk}{x}_k,A_{ik}{B}_{kk},A_{ij}{B}_{ik}{C}_{ik}$

are meaningless

 

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

$\begin{array}{l}{x}_i=u_i+c_i\equiv x=u+c\\ a_i=A_{ki}{B}_{kj}{x}_j+C_{ik}{u}_k\equiv a=A^{T}Bx+Cu\end{array}$

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

$x_i=A_{ik}{x}_k\iff x_j=A_{jk}{x}_k\iff x_j=A_{ji}{x}_i$

 

 

 

C.4. Vector operations expressed using index notation

 

· Addition.   $c=a+b\equiv c_i=a_i+b_i$

 

· Dot Product  $\lambda =a\cdot b\equiv \lambda =a_i{b}_i$

 

· Vector Product $c=a\times b\equiv c_i={\in }_{ijk}{a}_j{b}_k$

 

· Dyadic Product   $S=a\otimes b\equiv S_{ij}=a_i{b}_j$

 

· 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

$Q_{ij}=m_i\cdot e_j=\cos \theta (m_i,e_j)$

where $\theta (m_i,e_j)$ denotes the angle between the unit vectors $m_i$  and $e_j$.  Then

${\alpha }_i=Q_{ij}{a}_j$

 

 

 

C.5. Tensor operations expressed using index notation

 

· Addition.   $C=A+B\equiv C_{ij}=A_{ij}+B_{ij}$

 

· Transpose  $A=B^T\equiv A_{ij}=B_{ji}$

 

· Scalar Products

 

$\begin{array}{l}\lambda =A:B\equiv \lambda =A_{ij}{B}_{ij}\\ \lambda =A\cdot \cdot B\equiv \lambda =A_{ji}{B}_{ij}\end{array}$

 

· Product of a tensor and a vector

$\begin{array}{l}c=Ab\equiv c_i=A_{ij}{b}_j\\ c=A^{T}b\equiv c_i=A_{ji}{b}_j\end{array}$

 

· Product of two tensors

$\begin{array}{l}C=AB\equiv C_{ij}=A_{ik}{B}_{kj}\\ C=A^{T}B\equiv C_{ij}=A_{ki}{B}_{kj}\end{array}$

· Determinant

$\begin{array}{l}\lambda =\det A\equiv \lambda ={\displaystyle \frac{1}{6}{\in }_{ijk}{\in }_{lmn}{A}_{li}{A}_{mj}{A}_{nk}}\\ \iff {\in }_{lmn}\lambda ={\in }_{ijk}{A}_{li}{A}_{mj}{A}_{nk}\end{array}$

 

· 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

$Q_{ij}=m_i\cdot e_j=\cos \theta (m_i,e_j)$

where $\theta (m_i,e_j)$ denotes the angle between the unit vectors $m_i$  and $e_j$.  Then

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

 

 

 

C.6. Calculus using index notation

 

The derivative $\partial x_i/\partial x_j$ can be deduced by noting that $\partial x_i/\partial x_j=1i=j$ and  $\partial x_i/\partial x_j=0i\ne j$.  Therefore

                                                                  ${\displaystyle \frac{\partial x_i}{\partial x_j}={\delta }_{ij}}$

The same argument can be used for higher order tensors

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

 

 

 

C.7. Examples of algebraic manipulations using index notation

 

 

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

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

 

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)

$\left(a\times b\right)\cdot \left(c\times d\right)\equiv {\in }_{ijk}{a}_j{b}_k{\in }_{imn}{c}_m{d}_n$

Recall the identity

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

so

${\in }_{ijk}{a}_j{b}_k{\in }_{imn}{c}_m{d}_n=\left({\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{mk}\right)a_j{b}_k{c}_m{d}_n$

Multiply out, and note that

${\delta }_{jm}{a}_j=a_m{\delta }_{kn}{b}_k=b_n$

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

$\left({\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{mk}\right)a_j{b}_k{c}_m{d}_n=a_m{b}_n{c}_m{d}_n-a_n{b}_m{c}_m{d}_n$

Finally, note that

$a_m{c}_m\equiv a\cdot c$

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

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

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

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

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

$\begin{array}{l}{\sigma }_{kk}={\displaystyle \frac{E}{1+\nu }\left({\epsilon }_{kk}+{\displaystyle \frac{\nu }{1-2\nu }3{\epsilon }_{kk}}\right)={\displaystyle \frac{E}{1-2\nu }{\epsilon }_{kk}}}\\ \iff {\epsilon }_{kk}={\displaystyle \frac{1-2\nu }{E}{\sigma }_{kk}}\end{array}$

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

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

 

 

3. Solve the equation

 

$\mu \left\{{\delta }_{kj}{a}_i{a}_i+{\displaystyle \frac{1}{1-2\nu }{a}_k{a}_j}\right\}{U}_k=P_j$

for $U_k$ in terms of $P_i$ and $a_i$

 

Multiply both sides by $a_j$ to see that

$\begin{array}{l}\mu \left\{a_j{\delta }_{kj}{a}_i{a}_i+{\displaystyle \frac{1}{1-2\nu }{a}_k{a}_j{a}_j}\right\}{U}_k=P_j{a}_j\\ \iff \mu \left\{a_k{a}_i{a}_i+{\displaystyle \frac{1}{1-2\nu }{a}_k{a}_j{a}_j}\right\}{U}_k=P_j{a}_j\\ \iff \mu U_k{a}_k{\displaystyle \frac{2\left(1-\nu \right)}{1-2\nu }{a}_i{a}_i=P_j{a}_j\iff U_k{a}_k={\displaystyle \frac{(1-2\nu )P_j{a}_j}{2\mu \left(1-\nu \right)a_i{a}_i}}}\end{array}$

Substitute back into the equation given for $U_k{a}_k$ to see that

$\mu U_j{a}_i{a}_i+{\displaystyle \frac{P_k{a}_k}{2(1-\nu )a_i{a}_i}{a}_j=P_j\Rightarrow U_j={\displaystyle \frac{1}{\mu a_i{a}_i}\left(P_j-{\displaystyle \frac{P_k{a}_k}{2(1-\nu )a_n{a}_n}{a}_j}\right)}}$

 

 

4. Let $r=\sqrt{x_k{x}_k}$.  Calculate ${\displaystyle \frac{\partial r}{\partial x_i}}$

 

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

${\displaystyle \frac{\partial r}{\partial x_i}={\displaystyle \frac{1}{2\sqrt{x_k{x}_k}}\left(x_k{\displaystyle \frac{\partial x_k}{\partial x_i}+{\displaystyle \frac{\partial x_k}{\partial x_i}{x}_k}}\right)={\displaystyle \frac{1}{\sqrt{x_k{x}_k}}{x}_k{\delta }_{ik}={\displaystyle \frac{x_i}{\sqrt{x_k{x}_k}}={\displaystyle \frac{x_i}{r}}}}}}$

 

 

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

 

Using the product rule

${\displaystyle \frac{\partial \lambda }{\partial A_{kl}}=A_{ij}{\delta }_{ik}{\delta }_{jl}+{\delta }_{ik}{\delta }_{jl}{A}_{ij}=2A_{kl}}$