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Chapter 10: Q10P (page 505)

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

Short Answer

Expert verified

Answer

The equation has been proven.

Step by step solution

01

Given Information

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

02

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.

03

Prove that X is a tensor.

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

Apply transformation law on A and B.

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

A is an arbitrary vector.

Hence $(X'_{α} - a_{α i} X_{i}) = 0$

Thus, X is a tensor.

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Most popular questions from this chapter

Show that the transformation equation for a $2^{n d}$ -rank Cartesian tensor is equivalent to a similarity transformation. Warning hint: Note that the matrix C in Chapter 3 , Section 11 , is the inverse of the matrix A we are using in Chapter 10 (compare $r' = Ar and r = Cr'$ ). Thus a similarity transformation of the matrix T with tensor components $T_{i j}$ is $T' = {ATA}^{- 1}$ . Also, see “Tensors and Matrices” in Section 3 and remember that A is orthogonal.

Write out the sums $P_{i j} e_{j}$ for each value of and compare the discussion of $(1 .1)$ .Hint: For example, if $i = 2$ [or y in $(1 .1)$ ], then the pressure across the face perpendicular to the $x_{2}$ axis is $P_{21} e_{1} + P_{22} e_{22} + P_{23} e_{3}$ , or, in the notation of (1.1), $P_{y x} i + P_{y y} j + P_{y z} k$ .

Parabolic cylinder coordinates $u, v, z .$

$x = \frac{1}{2} (u^{2} - v^{2}) y = u v z = z$

$E = \frac{F}{q}$

Show by the quotient rule (Section 3 ) that $C_{ijkm}$ in $(7.5)$ is a $4^{th}$ -rank tensor.

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