Question

${\mathit{X}}_{\mathbf{i}\mathbf{j}}{\mathit{A}}_{\mathbf{k}}\mathbf{=}{\mathit{B}}_{\mathbf{i}\mathbf{j}\mathbf{k}}$.

Short answer

Answer

The equation has been proven.

Step-by-step solution

Given Information

The tensor$X_{ij}{A}_k=B_{ijk}$.

Definition of a cartesian tensor.

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

Prove that X is a tensor.

Let $X_{ij}{A}_k=B_{ijk}$where B is a non-zero tensor.

Apply transformation law on A and B.

$$\begin{gathered} 0=X{\text{'}}_{\alpha \beta }A{\text{'}}_{\gamma }-A{\text{'}}_{\alpha \beta \gamma } \\ 0=X{\text{'}}_{\alpha \beta }A{\text{'}}_{\gamma }-a_{\alpha j}{a}_{\beta j}{a}_{\gamma k}{B}_{ijk} \\ 0=X{\text{'}}_{\alpha \beta }A{\text{'}}_{\gamma }-a_{\alpha i}{a}_{\alpha j}{a}_{\beta j}{a}_{\gamma k}{X}_{ij}{A}_n \\ 0=X{\text{'}}_{\alpha \beta }A{\text{'}}_{\gamma }-a_{\alpha i}{a}_{\beta j}{a}_{\gamma k}{a}_{nk}{X}_{ij}A{\text{'}}_n \end{gathered}$$

$$\begin{gathered} 0=X{\text{'}}_{\alpha \beta }A{\text{'}}_{\gamma }-a_{\alpha i}{\delta }_{\gamma k}{a}_{\gamma n}{X}_{ij}A{\text{'}}_n \\ 0=\left(X{\text{'}}_{\alpha \beta }-a_{\alpha i}{a}_{\beta i}{X}_{ij}\right)A{\text{'}}_n \end{gathered}$$

A is an arbitrary vector.

Hence $\left(X{\text{'}}_{\alpha \beta }-a_{\alpha i}{a}_{\beta j}{X}_{ij}\right)=0$

Thus, X is a tensor.