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Chapter 10: Q9P (page 505)

$X_{i} A_{ij} = B_{j}$ .

Short Answer

Expert verified

Answer

The equation has been proven.

Step by step solution

01

Given Information

The tensor $X_{i} A_{i j} = B_{j}$ .

01

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.

02

Prove that X is a tensor.

Let $X_{i} A_{i j} = B_{i}$ where B is a non-zero tensor.

Apply transformation law on A and B.

$0 = X'_{α} A'_{α β} - a_{β i} B'_{β} 0 = X'_{α} A'_{α β} - a_{β i} X_{i} A_{i j} 0 = X'_{α} A'_{α β} - a_{β j} a_{α i} a_{γ i} X_{i} A'_{α γ} 0 = X'_{α} A'_{α β} - δ_{β j} a_{α i} X_{i} A'_{α γ} 0 = (X'_{α} - a_{α i} X_{i}) A'_{α β}$

Let $S_{α} = (X'_{α} - a_{α i} X_{i}) A'_{α β}$

Hence, $S^{T} A = 0$

The equation mentioned above is true for any matrix A.

Thus, X is a tensor.

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Most popular questions from this chapter

$X_{i j k l} A_{k l} = B_{i j}$ .

Evaluate:

$\begin{array}{l} (a) δ_{i j} δ_{j k} δ_{k m} δ_{i m} \\ (c) ϵ_{j k 2} ϵ_{k 2 j} \\ (e) ϵ_{23 i} ϵ_{2 i 3} \end{array}$ $\begin{array}{l} (b) ϵ_{i j k} δ_{j k} \\ (d) ϵ_{3 j k} ϵ_{k j 3} \\ (f) ϵ_{k 31} ϵ_{3 k 1} \end{array}$

Show that (3.9) follows from (3.8) . Hint: Give a proof by contradiction. Let $S_{α β}$ be the parenthesis in ; you may find it useful to think of the components written as a matrix. You want to prove that all 9 components of $S_{αβ}$ are zero. Suppose it is claimed that $S_{12}$ is not zero. Since $V'_{β}$ is an arbitrary vector, take it to be the vector , and observe that $S_{αβ} V'_{β}$ is then not zero in contradiction to .Similarly show that all components of $S_{αβ}$ are zero as claims.

Let $T_{j k m n}$ be the tensor in $(5 .8)$ . This is a $4^{t h}$ -rank tensor and so has $3^{4} = 81$ components. Most of the components are zero. Find the nonzero components and their values. Hint: See discussion after $(5 .8)$ .

Point masses 1 at (1, 1, 1) and at (-1, 1, 1).

See all solutions

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