Question

Show that the sum of two ${\mathbf{3}}^{\mathbf{r}\mathbf{d}}$-rank tensors is a ${\mathbf{3}}^{\mathbf{r}\mathbf{d}}$-rank tensor. Hint: Write the transformation law for each tensor and then add your two equations. Divide out the factors to leave the result $\mathit{T}{\mathbf{\text{'}}}_{\mathbf{\alpha }\mathbf{\beta }\mathbf{\gamma }}\mathbf{+}\mathit{S}{\mathbf{\text{'}}}_{\mathbf{\alpha }\mathbf{\beta }\mathbf{\gamma }}\mathbf{=}{\mathit{a}}_{\mathbf{\alpha }\mathbf{i}}{\mathit{a}}_{\mathbf{\beta }\mathbf{j}}{\mathit{a}}_{\mathbf{\gamma }\mathbf{k}}\left(T_{ijk}+S_{ijk}\right)$.

Short answer

Answer

The equation has been proven.

Step-by-step solution

Given Information

The two$3^{rd}$-rank tensor.

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 the statement.

Let T and S be the two $3^{rd}$-rank tensor.

${\left(T+S\right)}_{ijk}=T_{ijk}+S_{ijk}$

Use transformation law in the above equation.

$$\begin{gathered} \left(T+S\right){\text{'}}_{\alpha \beta \gamma }=a_{\alpha i}{a}_{\beta i}{a}_{\gamma i}{T}_{ijk}+a_{\alpha i}{a}_{\beta i}{a}_{\gamma i}S{}_{ijk} \\ \left(T+S\right){\text{'}}_{\alpha \beta \gamma }=a_{\alpha i}\left(a_{\beta i}{a}_{\gamma i}{T}_{ijk}+a_{\beta i}{a}_{\gamma i}{S}_{ijk}\right) \\ \left(T+S\right){\text{'}}_{\alpha \beta \gamma }=a_{\alpha i}{a}_{\beta i}\left(a_{\gamma i}{T}_{ijk}+a_{\gamma i}{S}_{ijk}\right) \\ \left(T+S\right){\text{'}}_{\alpha \beta \gamma }=a_{\alpha i}{a}_{\beta i}{a}_{\gamma i}\left(T_{ijk}+S_{ijk}\right) \end{gathered}$$

Hence, $\left(T+S\right){\text{'}}_{\alpha \beta \gamma }=a_{\alpha i}{a}_{\beta i}{a}_{\gamma i}{\left(T+S\right)}_{ijk}$and the sum of two $3^{rd}$-rank tensor is a -rank tensor.