User:Wikieditoroftoday/Solutions To Physics Textbooks/Physical Properties of Crystals (9780198511656)/Chapter 1

Exercise 1.1
p.15

Problem: If $$p_i$$ and $$q_i$$ are vectors, then $$p_1/q_1$$, $$p_2/q_2$$, $$p_3/q_3$$ are three numbers whose values are defined for any set of axes. Are they the components of a vector?

Solution: If $$p_i/q_i$$ are components of a vector they should transform like the components of a vector. See Table 2 p. 13 $$p_i' = a_{ij}p_j$$

A transformation with $$a_{ij}$$

$$p_i' = a_{ij}p_j$$

$$q_i' = a_{ij}q_j$$

$$\frac{p_i'}{q_i'} = \frac{a_{ij}p_j}{a_{ij}q_j} = \frac{p_j}{q_j} \ne a_{ij} \frac{p_j}{q_j}$$

shows that $$p_i/q_i$$ are not the components of a vector.

Answer: No, because they do not transform like vector components.

Exercise 1.2
p.22

Exercise 1.3
p. 31

[1]
Problem: Electrical conductivity tensor with axes $$x_1$$, $$x_2$$, $$x_3$$ $$ \begin{bmatrix} \sigma_{ij} \end{bmatrix} = \begin{bmatrix} 25 & 0 & 0 \\ 0 & 7 & -3\sqrt{3} \\ 0 & -3\sqrt{3} & 13\end{bmatrix} \cdot 10^7 \mathrm{ohm}^{-1}\mathrm{m}^{-1} $$ is transformed to a new set of axes $$x_1$$, $$x_2$$, $$x_3$$ with the following angles

$$x_1'Ox_1=0^{\circ}$$, $$x_2'Ox_2=30^{\circ}$$, $$x_2'Ox_3=60^{\circ}$$, $$x_3'Ox_3=30^{\circ}$$.

Draw up a transformation table similar to (11) on p.9 and check that the sum of the squares of $$a_{ij}$$ for each row and column is 1.

Solution: It follows from the given angles that $$x_3'Ox_2 = 30^{\circ} + 60^{\circ} + 30^{\circ} = 120^{\circ}$$. $$a_{ij}$$ consists of the direction cosines of the angles:

$$ \begin{pmatrix} a_{ij} \end{pmatrix} = \begin{pmatrix} \cos(0^{\circ}) & 0 & 0 \\ 0 & \cos(30^{\circ}) & \cos(60^{\circ}) \\ 0 & \cos(120^{\circ}) & \cos(30^{\circ})\end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \sqrt{3}/2 & 1/2 \\ 0 & -1/2 & \sqrt{3}/2\end{pmatrix} $$

Checking the columns and rows

$$\cos(0^{\circ})^2 = 1^2 = 1$$

$$\cos(30^{\circ})^2 + \cos(60^{\circ})^2 = 3/4 + 1/4 = 1$$

$$\cos(30^{\circ})^2 + \cos(120^{\circ})^2 = 1$$

[2]
Problem: Transform $$\sigma_{ij}$$ to $$\sigma_{ij}'$$ and interpret the result.

Solution: A second-rank tensor transforms according to (22) on p.11: $$\sigma_{ij}' = a_{ik}a_{jl}\sigma_{kl}$$:

This can be reduced because we have some zero components in the electrical conductivity tensor $$\sigma_{ij}$$:

now it is just a matter of calculating the components

$$\sigma_{11}' = \underbrace{ a_{11}a_{11} }_{1} 25 + \underbrace{\ldots}_0 = 25$$

$$\sigma_{12}' = 0$$

$$\sigma_{13}' = 0$$

$$\sigma_{21}' = 0$$

$$\sigma_{22}' = \underbrace{a_{21}a_{21}}_0 25 + \underbrace{a_{22}a_{22}}_{3/4} 7 - \underbrace{a_{22}a_{23}}_{\sqrt{3}/4} 3\sqrt{3} - \underbrace{a_{23}a_{22}}_{\sqrt{3}/4} 3\sqrt{3} + \underbrace{a_{23}a_{23}}_{1/4} 13 = \frac{1}{4} \left( 21 - 9 - 9 + 13 \right) = 4$$

$$\sigma_{23}' = \underbrace{ a_{21} a_{31} }_0 25 + \underbrace{a_{22}a_{32}}_{-\sqrt{3}/2} 7 - \underbrace{a_{22} a_{33}}_{3/4} 3\sqrt{3} - \underbrace{a_{23} a_{32} }_{-1/4} 3\sqrt{3} + \underbrace{a_{23}a_{33}}_{\sqrt{3}/4} 13 = \frac{\sqrt{3}}{4} \left( -7 - 9 + 3 + 13 \right) = 0$$

$$\sigma_{31}' = 0$$

$$\sigma_{32}' = \underbrace{a_{31}a_{21}}_0 25 + \underbrace{a_{32}a_{22}}_{-\sqrt{3}/4} 7 - \underbrace{a_{32}a_{23}}_{-1/4} 3\sqrt{3} - \underbrace{a_{33}a_{22}}_{3/4} 3\sqrt{3} + \underbrace{a_{33}a_{23}}_{\sqrt{3}/4} 13 = \frac{\sqrt{3}}{4}\left( -7 + 3 - 9 + 13\right) = 0$$

$$\sigma_{33}' = \underbrace{a_{31}a_{31}}_0 25 + \underbrace{a_{32}a_{32}}_{1/4} 7 - \underbrace{a_{32}a_{33}}_{-\sqrt{3}/4} 3\sqrt{3} - \underbrace{a_{33}a_{32}}_{-\sqrt{3}/4} 3\sqrt{3} + \underbrace{a_{33}a_{33}}_{3/4} 13 = \frac{1}{4}\left( 7 + 18 + 49\right) = 16$$

or written in array notation

$$\begin{bmatrix} \sigma_{ij}' \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 16 \end{bmatrix} \cdot 10^7 \mathrm{ohm}^{-1}\mathrm{m}^{-1}$$

Interpretation: The new set of axes $$x_i'$$ is the principle axes of the tensor.

[5]
Problem: Radial vector $$OP$$ with direction cosines in $$x_i$$ axes $$\begin{pmatrix}0, & \frac{1}{2}, & \frac{\sqrt{3}}{2}\end{pmatrix}$$. Find the electrical conductivity in that direction with an analytical expression.

Solution: Get the vector components in the principle axes $$x_i'$$: Using the transformation for polar vector components $$l_i' = a_{ij} l_j$$ leads to $$\begin{pmatrix}0, &\frac{\sqrt{3}}{2}, &\frac{1}{2}\end{pmatrix}$$. Using equation (32) on p.25 $$\sigma = l_i^2\sigma_i$$ gives us $$\sigma = \frac{3}{4} 4\mathrm{ohm}^{-1}\mathrm{m}^{-1} + \frac{1}{4} 16\mathrm{ohm}^{-1}\mathrm{m}^{-1} = 7\mathrm{ohm}^{-1}\mathrm{m}^{-1}$$.

[6]
Problem: An electric field $$E = 1\;\mathrm{volt}\mathrm{m}^{-1}$$ is applied in direction $$OP$$. Calculate the components of $$E_i$$ and the current density $$j_i$$ along the $$x_i$$ axes.

Solution: The components of the electric field are $$E_i = l_i E$$:

$$E_1 = 0$$

$$E_2 = \frac{1}{2}\;\mathrm{volt}/\mathrm{m}$$

$$E_3 = \frac{\sqrt{3}}{2}\;\mathrm{volt}/\mathrm{m}$$

The components of the current density are $$j_i = \sigma_{ik} E_k = \sigma_{ik} l_k E$$:

$$j_1 = 0$$

$$j_2 = \left( \frac{7}{2} - \frac{9}{2} \right) \cdot 10^7\mathrm{amps}/\mathrm{m}^2 = -10^7 \mathrm{amps}/\mathrm{m}^2$$

$$j_3 = \left( -\frac{3\sqrt{3}}{2} + \frac{13\sqrt{3}}{2} \right) \mathrm{amps}/\mathrm{m}^2 = 5 \sqrt{3} \cdot 10^7 \mathrm{amps}/\mathrm{m}^2$$

[7]
Problem: determine the magnitude and direction of the current density $$j_i$$.

Solution:

Magnitude: $$j = \sqrt{ j_1^2 + j_2^2 + j_3^2 } = \sqrt{ 76 } \cdot 10^7 \mathrm{amps}\mathrm{m}^{-2}$$

Direction: $$j = \begin{pmatrix}0, & -1/\sqrt{ 76 }, & 5 \sqrt{3/76} \end{pmatrix}$$. It lies in the $$x_2$$, $$x_3$$ plane with angles $$Ox_3 = \cos^{-1}(5 \sqrt{3/76}) \approx 6^{\circ} 35'$$ and $$-Ox_2 = \cos^{-1}(1/\sqrt{ 76 }) \approx 83^{\circ} 25'$$.

[8]
Problem: Repeat [6] and [7] but with the $$x_i'$$ axes.

Solution:

Components: $$j_i = \sigma_i l_i E$$

$$j_1 = 0$$

$$j_2 = 4 \frac{1}{2} 10^7 \mathrm{amps}/\mathrm{m}^2 = 2 \cdot 10^7\mathrm{amps}/\mathrm{m}^2$$

$$j_3 = 16 \frac{\sqrt{3}}{2} 10^7 \mathrm{amps}/\mathrm{m}^2 = 8 \sqrt{3} \cdot 10^7\mathrm{amps}/\mathrm{m}^2$$

Magnitude: $$j = \sqrt{ 2^2 + 8^2 3 } 10^7\mathrm{amps}/\mathrm{m}^2 = 14 \cdot 10^7\mathrm{amps}/\mathrm{m}^2$$

Direction: $$\begin{pmatrix} 0, & 2/7, & 4/7\end{pmatrix}$$ with angles $$Ox_2 = 73^{\circ} 24'$$ and $$Ox_3 = 55^{\circ} 9'$$.