Q7P Following what we did in equatio... [FREE SOLUTION]

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Chapter 10: Q7P (page 502)

Following what we did in equations (2.14) to (2.17), show that the direct product of a vector and a $3^{r d}$ -rank tensor is a $4^{r h}$ -rank tensor. Also show that the direct product of two $2^{n d}$ -rank tensors is a $4^{rh}$ -rank tensor. Generalize this to show that the direct product of two tensors of ranks m and n is a tensor of rank m + n .

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Answer

The statement has been proven.

Step by step solution

Given Information

A $2^{n d}$ - rank tensor, $3^{r d}$ -rank tensor and a $4^{t h}$ - rank tensor.

Prove the statement.

The formula that ${(S 鈯� T)}_{i_{1}, i_{2}, j_{1}, j_{2}, j_{m}} = S_{i_{1}, i_{2},, i_{n}} T_{j_{1}, j_{2},, j_{m}}$

Solve the tensor(direct) product mentioned in the formula.

The formula becomes as follows.

${(S 鈯� T)}_{i'_{1}, i'_{2},, i'_{n}, j'_{1}, j'_{2}, j'_{m}} = S_{j'_{1} i'_{2},, j'_{n}} T_{j'_{1} {j^{'}}_{2},, j_{m}} {(S 鈯� T)}_{i'_{1}, i'_{2},, i'_{n}, j'_{1}, j'_{2}, j'_{m}} = \underset{i_{1} - i_{n / 1},, i_{m}}{鈭�} a_{i'_{1} i_{1}} a_{i'_{n} i_{n}} a_{j'_{1} j_{1}} . . a_{j'_{m} j_{m}} S_{i_{1}, i_{2}} . . T_{j_{1}, j_{2} . . j_{m}} {(S 鈯� T)}_{i'_{1}, i'_{2},, i'_{n}, j'_{1}, j'_{2}, j'_{m}} = \underset{i_{1} - i_{n / 1},, i_{m}}{鈭�} a_{i'_{1} i_{1}} a_{i'_{n} i_{n}} a_{j'_{1} j_{1}} . . a_{j'_{m} j_{m}} {(S 鈯� T)}_{i_{1}, i_{2}} . ._{i_{1}, i_{2}, i_{2} . . j_{m}}$

This proves that $(S 鈯� T)$ is a tensor of rank $(n + + m)$ .

Hence, the statement has been proven.

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Given Information

Prove the statement.

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