Question

Show that the contracted tensor ${\mathit{T}}_{\mathbf{i}\mathbf{j}\mathbf{k}}{\mathit{V}}_{\mathbf{k}}$is a ${\mathbf{2}}^{\mathbf{n}\mathbf{d}}$-rank tensors.

Short answer

Answer

The equation has been proven.

Step-by-step solution

Given Information

The contracted tensor$T_{ijk}{V}_k$.

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.

Prove the statement.

The contracted tensor $T_{ijk}{V}_k$.

$$\begin{gathered} T_{ijk}{V}_k=a_{il}{a}_{jm}{a}_{kn}TImn{a}_{kp}{V}_p \\ {T}_{ijk}{V}_k=a_{iI}{a}_{jm}{a}_{kn}{a}_{kp}{T}_{Imn}{V}_p \\ {T}_{ijk}{V}_k=a_{iI}{a}_{im}{\delta }_{pn}{T}_{Imn}{V}_p \\ {T}_{ijk}{V}_k=a_{ij}{a}_{jm}{T}_{Imn}{V}_n \end{gathered}$$

Hence, $T_{kj}$is a $2^{nd}$rank vector.