Question

As in problem 6, show that the sum of two ${\mathbf{2}}^{\mathbf{n}\mathbf{d}}$-rank tensors is a ${\mathbf{2}}^{\mathbf{n}\mathbf{d}}$-rank tensor; that the sum of two ${\mathbf{4}}^{\mathbf{t}\mathbf{h}}$-rank tensors is a ${\mathbf{4}}^{\mathbf{t}\mathbf{h}}$-rank tensor.

Short answer

Answer

The equation has been proven.

Step-by-step solution

Given Information

The two $2^{nd}$-rank tensor and$4^{th}$- 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 n-rank tensor.

$$\begin{gathered} \left(T+S\right){\text{'}}_{i_1,i_2,..i_n}=T{\text{'}}_{i_1,i_2,...i_n}+S{\text{'}}_{i_1,i_2,..i_n} \\ \left(T+S\right){\text{'}}_{i_1,i_2,..i_n}=a_{i_1{j}_1...}{a}_{j_n{j}_n}{T}_{j_1...j_n}+a_{i_1{j}_1}....a_{i_n{j}_n}S{\text{'}}_{j_1{j}_n,..i_n} \\ \left(T+S\right){\text{'}}_{i_1,i_2,..i_n}=T{\text{'}}_{j_1,i_1....}{a}_{i_n{j}_n}{\left(T+S\right)}_{j_1....j_n} \end{gathered}$$

Hence, sum of the two n-rank tensor is n-rank.