Question

Evaluate:

$\begin{array}{l}\mathbf{\left(}\mathbf{a}\mathbf{\right)}{\mathbf{\delta }}_{\mathbf{i}\mathbf{j}}{\mathbf{\delta }}_{\mathbf{j}\mathbf{k}}{\mathbf{\delta }}_{\mathbf{k}\mathbf{m}}{\mathbf{\delta }}_{\mathbf{i}\mathbf{m}}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}{\mathbf{ϵ}}_{\mathbf{j}\mathbf{k}\mathbf{2}}{\mathbf{ϵ}}_{\mathbf{k}\mathbf{2}\mathbf{j}}\\ \mathbf{\left(}\mathbf{e}\mathbf{\right)}{\mathbf{ϵ}}_{\mathbf{23}\mathbf{i}}{\mathbf{ϵ}}_{\mathbf{2}\mathbf{i}\mathbf{3}}\end{array}$$\begin{array}{l}\mathbf{\left(}\mathbf{b}\mathbf{\right)}{\mathbf{ϵ}}_{\mathbf{i}\mathbf{j}\mathbf{k}}{\mathbf{\delta }}_{\mathbf{j}\mathbf{k}}\\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}{\mathbf{ϵ}}_{\mathbf{3}\mathbf{j}\mathbf{k}}{\mathbf{ϵ}}_{\mathbf{k}\mathbf{j}\mathbf{3}}\\ \mathbf{\left(}\mathbf{f}\mathbf{\right)}{\mathbf{ϵ}}_{\mathbf{k}\mathbf{31}}{\mathbf{ϵ}}_{\mathbf{3}\mathbf{k}\mathbf{1}}\end{array}$

Short answer

$\begin{array}{l}\left(a\right){\delta }_{ij}{\delta }_{jk}{\delta }_{km}{\delta }_{im}=3\\ \left(b\right){\delta }_{jk}{\epsilon }_{ijk}=0\\ \left(c\right){ϵ}_{jk2}{ϵ}_{k2j}=2\\ \left(d\right){ϵ}_{3jk}{ϵ}_{kj3}=-2\\ \left(e\right){ϵ}_{23i}{ϵ}_{2i3}=-1\\ \left(f\right){ϵ}_{k31}{ϵ}_{3k1}=-1\end{array}$

Step-by-step solution

Definition of a cartesian tensor

The first rank tensor is just a vector. A tensor of second rank has nine components (in three dimensions) in every rectangular coordinate system.

Find  δijδjkδkmδim

In part (a), there are no free indices.

$\begin{array}{c}i=j\\ =k\\ =m\end{array}$

Hence, the required value becomes as follows.

$\begin{array}{c}{\delta }_{ij}{\delta }_{jk}{\delta }_{km}{\delta }_{mi}=\sum _{i=1}^3{\delta }_{ii}{\delta }_{ii}{\delta }_{ii}{\delta }_{ii}\\ =\sum _{i=1}^{3}1\\ =3\end{array}$

Find  ϵijkδjk

The value is not zero if only if $j=k$

For $j=k$, the value of ${\epsilon }_{ijk}$becomes as follows.

Hence, ${\delta }_{ij}{\epsilon }_{ijk}=0$

$\begin{array}{c}i=j\\ =k\\ =m\end{array}$

Find  ϵjk2ϵk2j

The non-zero parts in the sum is$(j,k)=(1,3)or(3,1)$.

Hence, the required value becomes as follows.

$\begin{array}{c}{ϵ}_{jk2}{ϵ}_{j2k}={ϵ}_{132}{ϵ}_{321}+{ϵ}_{312}{ϵ}_{123}\\ =(-1)\cdot (-1)+1\cdot 1\\ =2\end{array}$

Find  ϵ3jkϵkj3

The required value is mentioned below.

$\begin{array}{c}{ϵ}_{3jk}{ϵ}_{kj3}={ϵ}_{312}{ϵ}_{213}+{ϵ}_{321}{ϵ}_{123}\\ =1\cdot (-1)+(-1)\cdot 1\\ =-2\end{array}$

Find  ϵ23iϵ2i3

The required value is mentioned below.

$\begin{array}{c}{ϵ}_{23i}{ϵ}_{2i3}={ϵ}_{231}{ϵ}_{213}+{ϵ}_{232}{ϵ}_{223}+{ϵ}_{223}{ϵ}_{233}\\ =1\cdot (-1)+0+0\\ =-1\end{array}$

Find  ϵk31ϵ3k1

The required value is mentioned below.

$\begin{array}{c}{ϵ}_{k31}{ϵ}_{3k1}={ϵ}_{131}{ϵ}_{311}+{ϵ}_{231}{ϵ}_{321}+{ϵ}_{331}{ϵ}_{331}\\ =0+1\cdot (-1)+0\\ =-1\end{array}$

Hence, the required values are

$\begin{array}{l}\left(a\right){\delta }_{ij}{\delta }_{jk}{\delta }_{km}{\delta }_{mi}=3\\ \left(b\right){\delta }_{ij}{\epsilon }_{ijk}=0\\ \left(c\right){ϵ}_{jk2}{ϵ}_{j2k}=2\\ \left(d\right){ϵ}_{3jk}{ϵ}_{kj3}=-2\\ \left(e\right){ϵ}_{23i}{ϵ}_{2i3}=-1\\ \left(f\right){ϵ}_{k31}{ϵ}_{3k1}=-1\end{array}$