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Chapter 10: Q3P (page 505)

As we did in (3.3) , show that the contracted tensor $T_{i i j}$ is a first-rank tensor, that is, a vector.

Short Answer

Expert verified

Answer

The equation has been proven.

Step by step solution

01

Given Information

The contracted tensor $T_{i i j}$ .

02

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.

03

Prove the statement.

The contracted tensor $T_{i i j}$ .

A is an orthogonal matrix.

$A A^{T} = δ_{i j}$

$T'_{i i j} = a_{i k} a_{i j} a_{j r} T_{k l r} T'_{i i j} = δ_{k l} a_{j r} T_{k l r} T'_{i i j} = a_{j r} T_{l l r} T'_{i i j} = a_{j r} T_{i i r}$

Hence, $T_{i i j}$ is a vector.

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Most popular questions from this chapter

In equation (5.16), show that if $T_{jk}$ is a tensor (that is, not a pseudotensor), then $V_{i}$ is a pseudovector (axial vector). Also show that if $V_{i}$ is a pseudotensor, then $T_{jk}$ is a vector (true or polar vector). You know that if role="math" localid="1659251751142" $V_{i}$ is a cross product of polar vectors, then it is a pseudovector. Is its dual $T_{jk}$ a tensor or a pseudotensor?

Show that the fourth expression in (3.1) is equal to $\frac{\partial u}{\partial x'_{i}}$ . By equations (2.6) and (2.10) , show that $\frac{\partial x_{j}}{\partial x'_{i}} = a$ , so $\frac{\partial u}{\partial x'_{i}} = \frac{\partial u}{\partial x_{j}} \frac{\partial x_{j}}{\partial x'_{i}} = a_{i j} \frac{\partial u}{\partial x_{j}}$

Compare this with equation (2.12) to show that $\nabla u$ is a Cartesian vector. Hint: Watch the summation indices carefully and if it helps, put back the summation signs or write sums out in detail as in (3.1) until you get used to summation convention.

Use equations (9.2), (9.8), and (9.11) to evaluate the following expressions. In spherical coordinates $\nabla \times (r e_{θ}), \nabla (r c o s θ), \nabla . r$ .

Show that the contracted tensor $T_{i j k} V_{k}$ is a $2^{n d}$ -rank tensors.

$V_{j}$ If role="math" localid="1659267226224" $V_{i} = g_{i j}, V^{j}$ is a contravariant vector and is a covariant vector, show that $U^{i} V_{j}$ is a $2^{n d}$ -rank mixed tensor. Hint:Write the transformation equations for U and V and multiply them.

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