Question

Write in terms of$\mathit{\delta }\mathbf{\text{'}}\mathit{s}$as in$\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{)}$and$\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{9}\mathbf{)}$:

(a)${\mathit{\epsilon }}_{\mathbf{i}\mathbf{j}\mathbf{k}}{\mathit{\epsilon }}_{\mathbf{p}\mathbf{j}\mathbf{q}}$(b)${\mathit{\epsilon }}_{\mathbf{a}\mathbf{b}\mathbf{c}}{\mathit{\epsilon }}_{\mathbf{p}\mathbf{q}\mathbf{c}}$

Short answer

The terms of part $\left(a\right)$and$\left(b\right)$ are${\delta }_{qk}{\delta }_{pi}-{\delta }_{qi}{\delta }_{pk}$and${\delta }_{ap}{\delta }_{bq}-{\delta }_{aq}{\delta }_{bp}$

Step-by-step solution

Given Information

The equation, i.e.,${ϵ}_{ijk}{ϵ}_{imn}={\delta }_{jm}{\delta }_{kn}-{\delta }_{jn}{\delta }_{km}$.

The equation, i.e., ${ϵ}_{abc}{ϵ}_{pqb}={ϵ}_{bca}{ϵ}_{bpq}={\delta }_{cp}{\delta }_{aq}-{\delta }_{cq}{\delta }_{ap}$.

Definition ofIsotropic tensor.

The term “isotropic tensor” refers to a tensor whose components remain intact when the coordinate system is rotated arbitrarily.

Solve for part (a)

Use the equation.$\left(5.8\right)$

$\begin{array}{c}{ϵ}_{ijk}{ϵ}_{pjq}={ϵ}_{jki}{ϵ}_{jqp}\\ ={\delta }_{qk}{\delta }_{pi}-{\delta }_{qi}{\delta }_{pk}\end{array}$

Hence, the terms $\delta \text{'}s$are${\delta }_{qk}{\delta }_{pi}-{\delta }_{qi}{\delta }_{pk}$

Solve for part (b)

Use the equation.$\left(5.9\right)$

$\begin{array}{c}{ϵ}_{abc}{ϵ}_{pqc}={ϵ}_{cab}{ϵ}_{cpq}\\ ={\delta }_{ap}{\delta }_{bq}-{\delta }_{aq}{\delta }_{bp}\end{array}$

Hence,the terms $\delta \text{'}s$are.${\delta }_{ap}{\delta }_{bq}-{\delta }_{aq}{\delta }_{bp}$