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

Chapter 1: The Groundwork of Crystal Physics

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 $a_{ij}$

$p_{i}'=a_{ij}p_{j}$

$q_{i}'=a_{ij}q_{j}$

${\frac {p_{i}'}{q_{i}'}}={\frac {a_{ij}p_{j}}{a_{ij}q_{j}}}={\frac {p_{j}}{q_{j}}}\neq a_{ij}{\frac {p_{j}}{q_{j}}}$

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 $\sigma _{ij}$:

$\sigma _{ij}'$	$=a_{i1}a_{j1}\overbrace {\sigma _{11}} ^{25}+a_{i1}a_{j2}\overbrace {\sigma _{12}} ^{0}+a_{i1}a_{j3}\overbrace {\sigma _{13}} ^{0}+$
	$+a_{i2}a_{j1}\overbrace {\sigma _{21}} ^{0}+a_{i2}a_{j2}\overbrace {\sigma _{22}} ^{7}+a_{i2}a_{j3}\overbrace {\sigma _{23}} ^{-3{\sqrt {3}}}+$
	$+a_{i3}a_{j1}\overbrace {\sigma _{31}} ^{0}+a_{i3}a_{j2}\overbrace {\sigma _{32}} ^{-3{\sqrt {3}}}+a_{i3}a_{j3}\overbrace {\sigma _{33}} ^{13}$
	$=a_{i1}a_{j1}25+a_{i2}a_{j2}7-a_{i2}a_{j3}3{\sqrt {3}}-a_{i3}a_{j2}3{\sqrt {3}}+a_{i3}a_{j3}13$

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 $Ox_{3}=55^{\circ }9'$.