Mathematical Crystallography [Revised Edition] 093995026X, 0939950146, 0939950138, 0939950065, 0939950081, 0939950200, 0939950219, 0939950103

Table of contents :
Page 1
Titles
MATHEMATICAL
Page 1
Titles
FIRST EDmON 1985
Printed by BookCrafters, Inc., Chelsea, Michigan 48118
REVIEWS in MINERALOGY
------------------------------------------------------------------------------------------------------------
~~:n~r:%~~~~N~;~~~I.SyntresiS, phase
~rr:,~':'.~s~ 7~~r:sr:n,::~:~'r~~~~.e~~n':~~:'::~~1 :..~~'M'~:~
~ct~~:.'~.=~~arxl Rietveld rafinement of aystal
Page 1
Titles
MA THEMATICAL
CRYST ALLOGRAPHY
PREFACE
Page 2
Titles
ACKNOWLEDGEMENTS
Page 1
Titles
REVIEWS IN MINERALOGY
Preface to the Revised Edition
Page 2
Page 1
Titles
EXPLANATION OF SYMBOLS
DESCRIPTION
Tables
Table 1
Page 1
Titles
MATHEMATICAL
CONTENTS
Chapter 1. MODELING SYMMETRICAL PATTERNS AND GEOMETRIES
Chapter 2. SOME GEOMETRICAL ASPECTS OF CRYSTALS
Page 2
Titles
Chapter 3. POINT ISOMETRIES - VEHICLES FOR
Chapter 4. THE MONAXIAL CRYSTALLOGRAPHIC POINT GROUPS
Page 3
Titles
Chapter 5. THE POLYAXIAL CRYSTALLOGRAPHIC POINT GROUPS
Tables
Table 1
Page 4
Titles
Chapter 7. THE CRYSTALLOGRAPH IC SPACE GROUPS
Appendix 1.
Appendix 2.
MAPPINGS
MA TR IX METHODS
Appendix 3.
CONSTRUCTION AND INTERPRETATION OF
Page 5
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NEW to the REVISED EDITION
SOLUTIONS TO PROBLEMS
INDEX
456
Page 1
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The minimum energy
Page 2
Titles
s.
Page 3
Page 4
Titles
a
4
Page 5
Titles
5
Page 6
Page 7
Titles
(P1.l) Problem:
Page 8
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A MATHEMATICAL DESCRIPTION OF THE GEOMETRIES
Page 9
Page 10
Titles
o
-./2v
o
!V
Vector addition and scalar multiplication:
Page 11
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S = {xa + yb + zc 1 x,y,z E R} .
Page 12
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( [;]1 ',Y,' , R) .
Space lattice:
LD == {ua + vb + wc I u,v,w E Z}
Page 13
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• •
•
•
13
Page 14
Titles
"-
Page 15
Titles
15
Page 16
Titles
(Pl.2) Problem:
(b) [r210
Hl·
Page 17
Titles
[Xl] [XX1]
(Pl.3) Problem:
Page 18
Titles
v
Page 19
Titles
x
y
x + y
z = v,
o
Hl
Page 20
Titles
(Pl.5) Problem: Show that
(Tl.8) Theorem:
[:] , [=;] , [=;]
Page 21
Titles
[::] ,
w,a + w,b + w,e -[;:1
Page 22
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(2) [xU]D~ x[U]D for all x t Rand U t S.
Hence [u + V]D = [U]D + [V]O'
(a) [-r]O
(b) [6r + 2U]D
Page 23
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[:], [biD
m' [eiD
[xalD
[:], [yblD
[~l ' [,eID
Page 24
Titles
(a)
24
Page 25
Titles
x = {[':]I r , e R}
Y
LENGTHS AND ANGLES
Page 26
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oL
~_j_
~
u • (v + w)
v
w
+
v • w
Tables
Table 1
Page 27
Titles
+
a • b
[; : :
k • i
k • j
Page 28
Titles
v • w
Tables
Table 1
Page 29
Tables
Table 1
Page 30
Tables
Table 1
Page 31
Titles
(Pl.l0) Problem:
(El.13) Example - Calculation of bond distances and angles for a-quartz:
Page 32
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Table 1.3: The coordinates of the atoms in a unit cell of ex-quartz.
b • a
a • a
b • b
Tables
Table 1
Page 33
Titles
o 0
e
Page 34
Titles
(Pl.13) Problem:
Page 35
Titles
v
w
v x w
+
k ,
Tables
Table 1
Page 36
Titles
v x w
Tables
Table 1
Page 37
Titles
o
Table 1.5: Coordinates of the oxygen atoms
x
y
z
Page 38
Titles
b
Triple scalar product:
u • (v x w)
(u,; + u,j + u,k)' (
i -
+
u • (v x w)
Page 39
Titles
v
a • (b x c)
a , b , c ,
Tables
Table 1
Page 40
Titles
c = ck
v = a > (b x c)
o b 0
c = ck
Page 1
Titles
CHAPTER 2
SOME GEOMETRICAL ASPECTS OF CRYSTALS
INTRODUCTION
Page 2
Titles
EQUATION OF PLANES AND LATTICE PLANES
p
r
P
{p + ns + mt I n,m E l}
u = {u + ns + mt I n,m E l}
s • (x - p) = 0
s • x
s • p
s • p ,
Page 3
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o
.~- '''''''''E:=:::'±::':L, f.",
w
(E2.1) Example - Verification that the termini of a set of vectors lie on
(3/2)x + (4/3)y - (2/5)z = 3/2
Page 4
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u = a
m,
Page 5
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[s]~ = [hk£]G-1
Page 6
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y
z
hx + ky + tz
Page 7
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(E2.3) Example - Calculation of the d-spacing for a plane in a crystal:
Solution:
RECIPROCAL BASIS VECTORS
Page 8
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hx + ky + ~z
s
Page 9
Titles
~
reb X c) • a = 1
r = 1/[a· b x c]
a
b'~
,~
(b xc) /[ a • b xc]
(c x a)/[a • b X c]
(P2.1) Problem:
b"
c
(a X b) /[ a • b xc]
a'~. a = b'~·
c = 1
a • b
c"· b
o
,': ,,: * * *
Page 10
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['10* = [:]
hx + ky + t z = 1
s
s •
• a
s
so
Tables
Table 1
Page 11
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(P2.4) Problem: In the h
(P2.5) Problem: In the h 1 0, k
['10* = [:]
called the direct basis.
=o" = {ha"
+ kb"
I h,k,~ E l}
* *
Page 12
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r • a
~
(P2.6) Problem: Show that r • b
(T2.5) Theorem:
r • a
r • b'"
r • c
Tables
Table 1
Page 13
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a • b
b • b
c • b
a • c
b • c
c • c
Proof:
0* be
~ ~
Tables
Table 1
Page 14
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(0 )
b"· (a")" = c"· (a")" = 0
a.
{a,b,c} .
Page 15
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(T2.8) Theorem:
...
a·(bxc).
Since lib xcii
(lib X cll)/(a· (b x e))
(x x y) • (z x w)
x • WI.
y • W
a • b
Tables
Table 1
Page 16
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*
* * ,~
CHANGE OF BASIS
*
G[rlD = [rlD*
Page 17
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By T1.9, we have [r]D = rdazjD + r,[bdD + rdcdD. But then
Tables
Table 1
Page 18
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[ [adO,
[bdO, I [']0,
Page 19
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D~ = {a~,b~,c~}. By T2.5,
[ r] ,~
[ r] ,~
* ... "
Tables
Table 1
Table 2
Page 20
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According to this diagram,
-1
for all rES. Hence, SGIT = G2• Consequently we need to calculate
S. By T2.6, D. is the reciprocal basis of 0:. Hence,
la:: · a2j
According to the analogous statement to (2.21), st is the change of basis
-1 t t -1 -t
(T ) = (T) , we write T . Hence we have the following theorem.
(T2.12) Theorem: Let 01 and O2 denote bases, T the change of basis
60
Tables
Table 1
Table 2
Page 21
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M = [(ale
(ele 1
Tables
Table 1
Page 22
Titles
a
r X s
b
c
r x s
a
,"
Tables
Table 1
Table 2
Page 23
Titles
v
v
1/v'~ .
Zones:
Tables
Table 1
Page 24
Tables
Table 1
Table 2
Page 25
Titles
(P2.9) Problem:
APPLICATIONS
Tables
Table 1
Page 26
Titles
XI
Page 27
Titles
o 1
Tables
Table 1
Page 28
Titles
a1 = 2a2 + 2b2
o
Tables
Table 1
Page 29
Titles
(E2.18) Example - Transforming indices of planes with change of basis:
, *
-h2 + £2
Page 30
Page 31
Titles
(P2.13) Problem:
Tables
Table 1
Page 32
Titles
z*
* • * •
* ""
"-......y
A DESCRIPTION OF THE GEOMETRY OF A CRYSTAL IN TERMS OF
Page 33
Titles
A
[ ['Ie
Tables
Table 1
Table 2
Page 34
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*
Tables
Table 1
Table 2
Page 35
Titles
* * * * *
* *
b = b j
* * * * * * ,~
A .
Calculation of angular coordinates from crystallographic data: When the
*
Tables
Table 1
Page 36
Titles
x
z
*
*
R = (R • i)i + (R • j)j + (R • k jk
(E2.20) Example - Calculation of angles between zones and face poles:
Page 37
Titles
o
Tables
Table 1
Page 38
Titles
""
Tables
Table 1
Page 39
Titles
S • R
Page 40
Titles
[l
Tables
Table 1
Page 41
Titles
e[R[Ioo]]c = e
Page 42
Titles
* * * * * *
c /b
o
Page 43
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A
DRAWING CRYSTAL STRUCTURES
Page 44
Titles
[O'~: 10]
Tables
Table 1
Page 45
Titles
r
a
b
c
k
(1/ r) r
Page 46
Titles
b
c
Page 47
Titles
Table 2.3: Atomic coordinates of the SiO. tetrahedron in jadeite.
Tables
Table 1
Page 48
Tables
Table 1
Page 49
Titles
(P2.18) Problem:
Table 2.4: Cartesian coordinates of an Si04 tetrahedron in pyroxferroite.
Tables
Table 1
Page 50
Tables
Table 1
Page 1
Titles
CHAPTER 3
POINT ISOMETRIES - VEHICLES FOR DESCRIBING SYMMETRY
"In asking what operations will turn a pattern into itself, we are dis­
INTRODUCTION
ISOMETRIES
Page 2
Titles
r r
\1/
jo
(c)
Tables
Table 1
Page 3
Titles
1
·0
I
Page 4
Titles
94
Tables
Table 1
Table 2
Table 3
Table 4
Page 5
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-1 -1
r = ua + vb + we
[100] [110] [010]
Page 6
Titles
Proof:
Tables
Table 1
Page 7
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1
i2(r) = 2(r)
Page 8
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1
c
1(0)
leb)
---
I( 0)
I(c)
(a)
o
r
u
Page 9
Titles
i(-r)
Page 10
Titles
DEFINING SYMMETRY
Page 11
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(P3.2)
Problem:
LINEAR MAPPINGS
Page 12
Page 13
Titles
z
(a)
z
~
z
~
o
a(r+s) = a(r) + a(S)
(b)
(c)
Page 14
Titles
(E3.8) Example - The inversion operation is a linear mapping: Show that
Page 15
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Solution:
Since i(r + s) = -(r + s) = -r - s = i(r) + i(s) and i(xr) =
r~
Page 16
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, , , , I
, I , I , I
I I I
Page 17
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[a(b)]o
[a(e)]o
[a(a)]o
[a(b)]o
[a(e)]o
[[2[')10 : [2(b)IO : [2(c)[0 1
Page 18
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2(r)
2(xa + yb + ze) = -xa - yb + (x + y + z)e .
(P3.5) Problem:
Page 19
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[1 1 1]
Page 20
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THE CONSTRUCTION OF A SET OF MATRICES DEFINING THE RO­
Page 21
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(a)
(b)
Page 22
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c
j
-c
b
[ [3(0)10
[3(b) 10
-1 -1 -1
[ [3-1(0)10
[-1 1 0]
o 0 n-
Page 23
Tables
Table 1
Page 24
Titles
Construction of MD (1 ):
-1 -1 -1
Page 25
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. [100] [100] [100].
h If .. . . [100]2 [100]2-1 Th
. [100] [100J [100]
[0 1 0]
Tables
Table 1
Page 26
Titles
z
X
. [110J [110]
.[-1 0 0]
Page 27
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322 = {1 3 3-1 [100]2 [110]2 [010]2
'" , , }
[1 -1 0] [0 1 0]
° -1 ° 1 ° °
Page 28
Titles
[100] [110] -1
-1
[ 100]
Tables
Table 1
Page 29
Titles
h· f' R -1
Tables
Table 1
Page 30
Titles
-1 [100] [100] [110J
Page 31
Titles
[100]
c6
[110] 2(c)
(P3.15) Problem:
Page 32
Titles
- -1 --1 .
Page 1
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CHAPTER FOUR
THE MONAXIAL CRYSTALLOGRAPHIC POINT GROUPS
INTRODUCTION
ALGEBRAIC CONCEPTS
Page 2
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322
{1,[100]2,[110]2}
T
[100]2[110]2 = 3-1 is
(P4.1) Problem:
Page 3
Titles
(E4.3) Example - (Z ,+) is a group: The integers under addition (l,+)
(E4.4) Example - 322 is a group under composition: The set 322 =
-1 [100] [110] [010] ..
Page 4
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(E4.6) Example - (Z,-) is not a group: The set of integers Z under
Page 5
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-1 -1
The set ~IC(6) = {MC(1),MC(6),MC(3),~IC(2),MC(3 ),MC(6 )}
[1 0 01 [0 -1 0] [0 0 1] [1 1 1]
[1 0 0] [0 1 0] [0 1 -1] [1 0 -1]
Tables
Table 1
Page 6
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(P4.3) Problem:
point symmetry group of 8.
Page 7
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(D4.12) Defi nition: A crystallographic group is a group of isometries
CRYSTALLOGRAPHIC RESTRICTIONS
(T4.14) Theorem:
Page 8
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130
Page 9
Page 10
Titles
(T4.15) Theorem:
Page 11
Titles
Proof:
" "
Table 4.1: Solutions sets of the turn angles p for any
N 1 cp = (N - 1)/21 pO
~
1 N 1 cp = -(N + 1)/21 pO
~
Tables
Table 1
Page 12
Titles
MONAXIAL ROTATION GROUPS
Page 13
Tables
Table 1
Page 14
Titles
n m n + m
(E4.23) Example:
H = {gn I n £ Z} ,
xy
Page 15
Titles
1 < . < k F 1 -1 _ k
Page 16
Titles
-1
4 = = {1,4,2,4 } .
(D4.27) Definition:
axis is called a proper monaxial group.
(T4.28) Theorem:
(The improper point group generating theorem.) If
Tables
Table 1
Page 17
Titles
#(C)/fI(H)
C U ct
4 U 4i
-1 -1
{1,4,2,4 } U {1,4,2,4 }i
{ 1 ,4,2,4 -1 , 1 i ,4i .u ,4 -1 i}
-1 - --1
{1,4,2,4 ,i,4,m,4 }.
-1
H U (C \ H)i
2 U (4 \ 2)i = {1 ,2} U ({1,4,2,4-1} \ {1,2})i
-1
- --1
Page 18
Titles
3 U 3i
-1 - --1
{1,3,3 }U{i,3,3}
-1 - --1
Page 19
Titles
2 U 2i
{1,2,i,m}
1 U (2 \ 1) i
1 U 1i
. -1 -1. - - - -1 --1
. -1 - --1
Matrix representations and basis vectors:
Page 20
Titles
Table 4.2: The 13 monaxial crystallographic point groups and their or-der-s
C U Ci I It(C U Ci) I H U (C ~_~)i I It(H U (C '
H
It(C)
1 (identity): Since
Tables
Table 1
Page 21
Tables
Table 1
Page 22
Titles
(4.3)
(4.3). 0
Equivalent Points and Planes:
~(X)=y.
Tables
Table 1
Page 23
Titles
(T4.33) Theorem:
Proof:
x
Page 24
Titles
[x]
{Y E S
x - Y}
{g(x) I g E C}.
[1 0 0] [1. 281]
[1. 281]
Page 25
Titles
o -1
o
1
o
o
o
o -1
Tables
Table 1
Table 2
Page 26
Titles
,"
o
0113
Mp*(6) [Slp'"
01 12
o
o -1111
Page 27
Titles
--1
Mp*(6 ) [Slp*
31 -51
1
Page 28
Tables
Table 1
Page 29
Titles
Figure 4.5 (on this and the following pages):
1
2
Page 30
Titles
3
4
6
Page 31
Page 32
Page 33
Page 34
Page 1
Titles
CHAPTER S
THE POLYAXIAL CRYSTALLOGRAPHIC POINT GROUPS
INTRODUCTION
PROPER POLYAXIAL POINT GROUPS
Page 2
Titles
(322)33= {I,(OIO]2}
r
(322)22= {I,(OIO]2}
P(322)
Page 3
Titles
(P5.1) Problem: Verify that C2(322)
Page 4
Titles
p
P
Tables
Table 1
Page 5
Titles
-1 -1
-1
{1,[100]2}
{1,[010]2}
Page 6
Titles

conjugate to H 2 •
P - q
t
r n.t», - 1)
Page 7
Titles
t
L n.t», - 1)
Tables
Table 1
Page 8
Titles
(P5.4) Problem:
P
-1
-1
9 (q)
p
-1
P P
p
P
p p
#(C)
N
Page 9
Titles
(P5.5) Problem:
flCC)
E n.(v. - 1)
i=l I
E (1 - l/v.)
i=l I
Tables
Table 1
Page 10
Titles
2 - 2/N
i=l I
2 - 2/N
1 + 2/N
Page 11
Titles
N
Tables
Table 1
Page 12
Titles
~
CONSTRUCTION OF THE DIHEDRAL GROUPS
Tables
Table 1
Page 13
Titles
o
Page 14
Page 15
Titles
* * * *
"-,J :;; -1
..!&
171
Tables
Table 1
Table 2
Page 16
Titles
a
•
-e If
622
Page 17
Titles
CONSTRUCTION OF THE CUBIC AXIAL GROUPS
(E5.11) Example:
Solution:
Page 18
Titles
[a]
{Mp(g)[a]p I 9 E 222}
Construction of 332(=23):
Page 19
Titles
a
(P5.10) Problem: Show that
Page 20
Page 21
Titles
)\
[Iii]
[ill]
[IIi]
{1,[100]2}
{1, [010]2}
2
{1,2}
23 = 332 = {1,2,[100]2,[010]2,[111]3,[111]3-1
177
Page 22
Tables
Table 1
Page 23
Titles
" ........
f I l\ [all]
Page 24
Titles
CONSTRUCTION OF THE IMPROPER CRYSTALLOGRAPHIC POINT
Page 25
Titles
[100] [110] .
- -1 --1 -
({1,i,[110]2,[110]m} which is the monaxial group 21m)
322.
{1 ,3,3-1, [100]m, [110]m, [010]m}
Page 26
Titles
622 U 622i = 61mmm
{1,6,3,2,3-1 ,6-1 ,[100]2,[210]2,[110]2,[120]2,[010]2,[110]2,
I, , " , , m, m, m, m , m, m
q32 U q32i = (4Im)3(2im)
Page 27
Titles
i ,4,m,'4-1, [100]4, [100]m, [100]'4-1, [010]'4, [010]m, [010]'4-1,
THE CRYSTAL SYSTEMS
Page 28
Titles
222
-~/
32
Page 29
Titles
\~l?\~~
Page 30
Titles
~! ~v
~v ~\V~
Page 31
Page 32
Titles
-
~/ ~I
\' \
Page 33
Page 34
Page 35
Titles
9"j
- -
o
o
glJ]
o .
gJ J
Page 36
Titles
2mm(=mm2) ,
The icosahedral point groups:
Tables
Table 1
Page 37
Page 38
Titles
k
(P5.28) Problem:
Page 39
Titles
(P5.29) Problem: Confirm that
t 1
(P5.30) Problem: Show that
Page 40
Titles
(P5.31) Problem:
(P5.32) Problem:
Tables
Table 1
Page 41
Page 42
Page 1
Titles
CHAPTER 6
THE BRAVAIS LATTICE TYPES
INTRODUCTION
LATTICES
Page 2
Titles
LO = {ua + vb + we I u, v ,W 1: Z}
L02 .
(D6.2) Definition:
an integral matrix.
Page 3
Titles
-1 . -1
-1
(E6.4) Example - When two bases generate the same lattice:
Page 4
Titles
T
[:
2/3]
[ 1 0 1]
T
LO = {ua I u 1: l}
Page 5
Titles
LO = {ua + vb I u,v 1: l}
(DG.G) Definition:
Page 6
Titles
(P6.2) Problem: Show that if [VlO
Page 7
Titles
L 0 + 0 and L 0 + (ta + tb + tc)
Page 8
Titles
geL)
L
[[.(',ID
(E6.14) Example - A lattice left invariant under a point group C:
Page 9
Titles
[ 0 -1 0]
(PG.3) Problem:
A DERIVATION OF THE 14 BRAVAIS LATTICE TYPES
Page 10
Titles
208
Page 11
Titles
[2(f) lp
[-1 0 0] [f 1] [-f 1]
o -1 0 f2 = -f2
[elp
PIP
AlP I [:] [:]
liP
[:] [:]
[~ .l]t [.l1]t_[.l.l]t
Page 12
Titles
Table 6.1: Combinations of fractional coordinates for 2.
o 0
Tables
Table 1
Page 13
Titles
a + b, c2 - c.
A = (P + 0) U (P + tb + tc)
{ua + v(tb + tc) + we I u,v,w 1: Z}
Page 14
Titles
(P6.G) Problem:
Page 15
Titles
[f-3(f)]p
[fl+f2]
o
Page 16
Titles
Table 6.2: Combinations of fractional coordinates for 3.
Tables
Table 1
Table 2
Page 17
Titles
PIP
R(rev)IP
2/3]
R(obv)IP
o 1/3 2/3
CPG.9) Problem: Show that
Page 18
Titles
°R(obv)
Page 19
Titles
Lattices invariant under 'I:
(ib + ~c) - (-ia + ic)
ia + ib
(P + 0) U (P + (ta + tb + ic))
CP6.10) Problem: Confirm that 01
{a, b, ia + ib + ~c} is a basis for
Page 20
Tables
Table 1
Table 2
Page 21
Tables
Table 1
Table 2
Page 22
Titles
6(P)
{a + b, -a, c}
Page 23
Titles
[:] [:] [;]
[:] [;] [:]
F/P [:] [:] r:J m
[:] [:] [:] m [:]
[:] m [:] [;] m
r:J m m [:] [:]
m m m [:] [:]
Page 24
Titles
THE LATTICES INVARIANT UNDER '122 AND UNDER 622
Page 25
Titles
(P6.21) Problem:
(P6.22) Problem:
THE 14 BRAVAIS LATTICE TYPES
Page 26
Tables
Table 1
Page 27
Titles
MATRIX GROUPS REPRESENTING THE CRYSTALLOGRAPHIC
. [100] [110] [010] .
[210] [120] [110]
Page 28
Titles
Mp(3)TMp( 2)T Mp(3 ) = Mp(3)Mp( 2)Mp(3) = Mp( 2).
Page 29
Tables
Table 1
Page 30
Titles
N ([100]2) H ([110] )
p 'p m ,
Mp([100]m), Mp([110]2),
(P6.25) Problem:
Page 1
Titles
CHAPTER 7
THE CRYSTALLOGRAPH IC SPACE GROUPS
INTRODUCTION
TRANSLATIONS
229
Page 2
Titles
1----ly=T(X)
p _---- x
TI"J,\ :\
u v
230
Page 3
Titles
(T7.2) Theorem:
x + t
Page 4
Titles
(E7.3) Example - The composition of two translational isometries:
(P7.4) Problem:
(D7.4) Definition:
ab = ba
(P7.5) Problem:
(P7.G) Problem:
Page 5
Titles
-1
Page 6
Titles
(E7.7) Example: Show that
x + ra + sb + tc
x y z
Page 7
Titles
x y z
T = [[,,(0.)10,(0,) i ['y(o.)IO,(o,)
Page 8
Titles
(T7.8) Theorem:
Proof:
y
[Y]0(02) = [pqr]t
Page 9
Titles
ISOMETRIES
Tables
Table 1
Page 10
Page 11
Titles
[O('lIO]
Solution: Let
z
x
o
o
o
Tables
Table 1
Table 2
Page 12
Titles
[~(r)lO
+
o
o
[~(r) + t10
(D7.13) Definition:
Page 13
Page 14
Titles
-1 -1
W I t}{N I r}
(P7.13) Problem: Show that
Tables
Table 1
Page 15
Page 16
Page 17
Titles
p
Page 18
Titles
{T I -Tp}
Page 19
Titles
247
Tables
Table 1
Page 20
Titles
c2 = -b,
1 0 o-t
o
Tables
Table 1
Page 21
Titles
CRYSTALLOGRAPH IC SPACE GROUPS
(D7.20) Definition:
x y . z
T
(D7.21) Definition:
Page 22
Titles
(T7.23) Theorem:
(P7.18) Problem: Prove T7.23.
Page 23
Titles
(P7.19) Problem:
(D7.24) Definition:
(T7.25) Theorem:
Proof:
Page 24
Page 25
Tables
Table 1
Page 26
Page 27
Titles
n n
Page 28
Titles
256
Page 29
Page 30
Titles
CRYSTALLOGRAPHIC SPACE GROUP OPERATIONS
Page 31
Titles
1, t
represent quarter-turn screw operations.
[xyz) to [-y,x,t + z) .
. 1 . h [UVW] th b h
m
Page 32
Page 33
Titles
[1 0 0]
o
o
Tables
Table 1
Page 34
Titles
THE CRYSTALLOGRAPHIC SPACE GROUP TYPES DERIVED FROM THE
Page 35
Titles
i i
r}i + i
Ulld+i
Tables
Table 1
Page 36
Titles
(P7.32)
Problem:
Page 37
Titles
M + M2 + M3 + M" + M5 + M6, we have
Page 38
Titles
[J .
denoted P6, P6" P62, P63, P6. and P6s.
266
Page 39
Titles
_u Rp(T p){M I r}
RpCT p){H I r} U Rp(T p){~! I r}2 U Rp(T p){M I r}3
U Rp(T p){H I r}4 U Rp(T p){M I r}5 U Rp(T p){M I r}6 ,
o 0 u
(1/3)(2u - v)a + (1/3)(u + v)b
Page 40
Titles
(P7.34) Problem:
Tables
Table 1
Page 41
Titles
b
.F
., '-\ a
269
Page 42
Page 43
Titles
(P7.39) Problem:
v
x
z
x
y
z
Page 44
Titles
[x - y,x,(l/6) + z]t, [-y,x - y.(l/3) + Z]t, [-x,-y,t + z]t,
(P7.40) Problem:
Page 45
Page 46
Titles
~I3 I s} I s = m + u[i,i,O]t where m E Z3 and u
[0,2r2,0]t = m + u[t,i,O]t
REV015C007_p275-320.pdf
Page 1
Titles
C2.
RpCTC) = RpCTp){I3 I [OOD)t} U RpCTp){I3 I [t,t,D)t}
- RpCT p) U RpCT p) {I3 I [t,t,D) }
[010]
RpCC2) = [RpCTp) U RpCTp){I3 I [t,t,O)t}]{M I [DOD)t}
[-x,y,-z) , [! - x,t + y,-z) ,[x,y,z) and [t + x,t + y,z) .
Page 2
Titles
THE CRYSTALLOGRAPHIC SPACE GROUP TYPES DERIVED
(P7.43) Problem:
(P7.44) Problem: Show that
Tables
Table 1
Page 3
Titles
1.
Solution: Suppose an~m
~m
m -n
But S £ and a E
~m e n .
(P7.45) Problem:
Tables
Table 1
Page 4
Titles
K
i ,i £ Z
-1 3
-1
i .
Page 5
Titles
K
[100] [100] t
Page 6
Titles
1, ~2
Sol ution : Since 0 (c )
Tables
Table 1
Page 7
Titles
Nl
t
[r 1
+ r, - 351j
+ r, - 351
rl + r, - 351 £ Z
Page 8
Titles
--r 1 and
Page 9
Titles
(E7.50) Example:
[0 -1 0]
Page 10
Titles
[-1 0 0]
[x,y,z) , [-y,x - y, (2/3) + z) , [y - x, -x, (113) + z) ,
[x - y, -y, -z) , [y, x, (2/3) - z) , [-x, y - x, (1/3) - z) 0
(P7.46) Problem:
[x, y, z) , [-y, x - y, (2/3) + z) , [y - x, -x, (1/3) + z) ,
[y,x,-z) , [x - y, -y, (1/3) - z) , [-x, y - x, C2/3) - z)
[y, -x, (3/4) + Z)t, [i + x, t - y, _z)t, [y - I, t + x, i - z)t,
[t + x, t - y, (3/4) - z)\ [y, x, _z)t, [-y, -x, t - z]t
284
Page 11
Titles
d
Tables
Table 1
Page 12
Titles
(E7.52) Example:
o
o 0
o
Page 13
Titles
I \.
287
Tables
Table 1
Page 14
Titles
o
o
o
1
o
o 1
o
o
o 1
o 0
o
o
o
o 0
o
o
o
Page 15
Titles
[-1 0 0]
[-1 0 0]
Tables
Table 1
Page 16
Titles
{Ml I r}{M2 I S}{MI I r}
{I3 I V}{M2 I s} .
290
Page 17
Titles
{H, I s}
{H2 I s} where r = [rl .r « ,m12) and S = [nI2, r2 + k12, 53)
t
Page 18
Page 19
Tables
Table 1
Page 20
Titles
-
[x,y,z)t, [t + x,y,t + z)t, [-x,-y,-z)t,
[t-x,-y,t-z), [-x,t-y,z), [t-x,t-y,±+z),
[x,± + y,_z)t, [± + x,± + y,t - z)t.
[!l
Tables
Table 1
Page 21
Titles
o 0
[P)Dl
o
[1 0 0][-1 0 OJ [1 0 0]-1
1 -1 0 0 0 -1 1 -1 0
o 0-1
o
o
o
o 1
(P7.55) Problem:
Page 22
Titles
THE CRYSTALLOGRAPH IC SPACE GROUPS BASED ON THE
THREE-GENERATOR POINT GROUPS
296
Tables
Table 1
Page 23
Tables
Table 1
Table 2
Page 24
Titles
-1 -1
Tables
Table 1
Page 25
Page 26
Page 27
Titles
r
[rl, -k14, k14,) and s
[(3kI4), -3kI4,sJ) ,
if we set PI = P2 = PJ = 53/2, {H2 I s} becomes {H, I [3kI4,-3kI4,O)t}.
Table 7.5: The space group types based on 1132 and P.
Tables
Table 1
Page 28
Page 29
Titles
APPENDIX 1
MAPPINGS
Page 30
Tables
Table 1
Table 2
Page 31
Titles
(TA1.4) Theorem: Let a:A-+8, ~:8-+C and ~:C-+D, then n~a) = (ra)a.
rS(a(a))
(DA1.5) Definition: Let a:A-+8 denote a mapping.
305
Page 32
Titles
fI(8), i.e., fI(A) = fl(8).
(DA1.5) Definition:
Page 33
Page 34
Page 35
Titles
APPENDIX 2
MATRIX METHODS
(DA2.1) Definition:
nth row
a
'I
'I
Tables
Table 1
Page 36
Titles
OPERATIONS
. . .
c
4 ] + [-5
Tables
Table 1
Table 2
Page 37
Titles
(EA2.4) Example: If C
1
Page 38
Titles
SOLVING SYSTEMS OF LINEAR EQUATIONS
aUxl + a12x2 +
b
augmented matrix
Tables
Table 1
Page 39
Titles
d
d
Page 40
Titles
o
(EA2.6) Example - Determining whether a set of vectors is a basis: Three
Solution:
Page 41
Titles
[ : -1
1
[ 1 0 11 9/5]
Tables
Table 1
Page 42
Titles
[ :
1
[ :
1
[ :
o
o
Page 43
Titles
[ :
o
(DA2.7) Definition: A matrix is in reduced row echelon form if
Page 44
Tables
Table 1
Table 2
Page 45
Titles
1 '
then there is exactly one solution:
-114.
1
l :
1
x 3 = -3 -4m - 2n
Page 46
Titles
m
n.
REV015C007_p275-320.pdf
Page 29
Titles
APPENDIX 1
MAPPINGS
Page 30
Tables
Table 1
Table 2
Page 31
Titles
(TA1.4) Theorem: Let a:A-+8, ~:8-+C and ~:C-+D, then n~a) = (ra)a.
rS(a(a))
(DA1.5) Definition: Let a:A-+8 denote a mapping.
305
Page 32
Titles
fI(8), i.e., fI(A) = fl(8).
(DA1.5) Definition:
Page 33
Page 34
REV015C007_p275-320.pdf
Page 35
Titles
APPENDIX 2
MATRIX METHODS
(DA2.1) Definition:
nth row
a
'I
'I
Tables
Table 1
Page 36
Titles
OPERATIONS
. . .
c
4 ] + [-5
Tables
Table 1
Table 2
Page 37
Titles
(EA2.4) Example: If C
1
Page 38
Titles
SOLVING SYSTEMS OF LINEAR EQUATIONS
aUxl + a12x2 +
b
augmented matrix
Tables
Table 1
Page 39
Titles
d
d
Page 40
Titles
o
(EA2.6) Example - Determining whether a set of vectors is a basis: Three
Solution:
Page 41
Titles
[ : -1
1
[ 1 0 11 9/5]
Tables
Table 1
Page 42
Titles
[ :
1
[ :
1
[ :
o
o
Page 43
Titles
[ :
o
(DA2.7) Definition: A matrix is in reduced row echelon form if
Page 44
Tables
Table 1
Table 2
Page 45
Titles
1 '
then there is exactly one solution:
-114.
1
l :
1
x 3 = -3 -4m - 2n
Page 46
Titles
m
n.
REV015C007_p321-366.pdf
Page 1
Titles
Q = HA + KB + LC
HIA + KIB + L1C + El
HA+KB+LC+E
n n n n
321
Tables
Table 1
Table 2
Page 2
Titles
Q
n
v
V
Tables
Table 1
Page 3
Titles
1
1
1
Tables
Table 1
Page 4
Titles
l/b,r and c
Page 5
Titles
Table A2.2. Selected interplanar spacings
[h k I]
o
o
c,,.2
[ : 1
HA + LC = Q
Dtn =
Page 6
Titles
nto =
DETERMINANTS
(EA2.13) Example: Suppose that A
Solution: det(A) = 5 4
Page 7
Titles
(EA2.14) Example: If
A
[ -~
IAI
IAI
81 + (-1)1+\5)13
Tables
Table 1
Table 2
Table 3
Page 8
Titles
IAI
IAI
n ..
i= 1 'I 'I
Page 9
Titles
AB
Page 10
Titles
INVERSES
: I
Page 11
Page 12
Titles
(EA2.22) Example:
Page 13
Titles
[
Page 14
Titles
[
o
[
Tables
Table 1
Table 2
Page 15
Page 16
Titles
336
Tables
Table 1
Page 17
Titles
Proof:
(PA2.S) Problem: Given that
I' I'
Page 18
Tables
Table 1
REV015C007_p321-366.pdf
Page 19
Titles
APPENDIX 3
CONSTRUCTION AND INTERPRETATION OF MATRICES REPRESENTING
INTRODUCTION
CONSTRUCTION OF MATRIX REPRESENTATIONS
Cartesian Bases: The general cartesian rotation matrix
Tables
Table 1
Page 20
Page 21
Titles
A
o
A
"0(·) ~
A
1
Page 22
Titles
] ,
INTERPRETATION OF MATRICES REPRESENTING POINT ISOMETRIES
Tables
Table 1
Page 23
Titles
(RA3.2) Rules for interpreting a general cartesian matrix M representing
some point isometry, w:
M
1
(A3.7)
343
Tables
Table 1
Page 24
Titles
o
Tables
Table 1
Page 25
Titles
General bases:
Tables
Table 1
Table 2
Page 26
Page 27
Titles
(PA3.10) Problem: Prove CA3.8.
Tables
Table 1
Page 28
Page 29
Titles
(EA3.10) Example - Determination of the properties of a point isometry:
Tables
Table 1
Page 30
Titles
{-xa + xb + xcix E R}
{x(-a + b + C)lx E R}
{xrlx E R}
Page 31
Titles
is [111]3.
Proof: Since
n
Tables
Table 1
Page 32
Titles
PROOFS OF (A3.1), (A3.2) AND TA3 .9
Tables
Table 1
Page 33
Titles
(a)
(b)
Page 34
Titles
Discussion and proof of theorem TA3.9:
Tables
Table 1
Page 35
Titles
[~(t)]o
~ 1
w J
Page 36
REV015C007_p321-366.pdf
Page 37
Titles
APPENDIX 4
POTPOURRI
HANDEDNESS OF BASES
w 0 (u x V)
u~v
(0)
u
w
(b)
v
u
o
w
(e)
v
Page 38
Titles
W
(0)
v
W
(b)
DISCUSSION AND PROOF OF T6.15
(TA4. 1) Theorem: Let V E S and let
Page 39
Page 40
Page 41
Titles
APPENDIX 5
SOME PROPERTIES OF LATTICE PLANES
w
Page 42
Titles
hx + ky + z z
m
[-~
1
Page 43
Titles
(TA5.3) Theorem: The equation
o
Proof:
Page 44
Titles
m
p + q
Page 45
Titles
(TA5.5) Theorem: The equation
hx + ky + II
Proof:
365
Page 46
Titles
(TA5.6) Theorem: The equation
t(nx + my) = 1
nx + my
REV015C007_p321-366.pdf
Page 41
Titles
APPENDIX 5
SOME PROPERTIES OF LATTICE PLANES
w
Page 42
Titles
hx + ky + z z
m
[-~
1
Page 43
Titles
(TA5.3) Theorem: The equation
o
Proof:
Page 44
Titles
m
p + q
Page 45
Titles
(TA5.5) Theorem: The equation
hx + ky + II
Proof:
365
Page 46
Titles
(TA5.6) Theorem: The equation
t(nx + my) = 1
nx + my
Page 1
Page 2
Page 3
Titles
Ilr.llllr,11 1.621'
Page 4
Page 5
Page 6
Titles
o c
Page 7
Page 8
Titles
-1]
Page 9
Titles
[1 0 -1] [-2] [-3]
[bll
B
Page 10
Tables
Table 1
Page 11
Page 12
Titles
[1 -1 0]
[ 1 -1 0] [1'1] [rl - 1'2]
Tables
Table 1
Page 13
Tables
Table 1
Table 2
Page 14
Tables
Table 1
Table 2
Page 15
Tables
Table 1
Page 16
Page 17
Titles
= { [~:] , [=~~] , [=~:] , [~n }
Page 18
Page 19
Page 20
Tables
Table 1
Page 21
Titles
{[a/alp, [b/blp, [-~ - ~L, [~+ ~L ,[-a/alp, [-b/blp, [c/clp, [-c/clp}
[al = {Mp(g)[alp) I g E 322} = {[alp, [blp, [-a - blp}
[a + b] = {Mp(g)[a + blp) I g E 322} = {[-alp, [-blp, [a + b]»]
Tables
Table 1
Table 2
Page 22
Tables
Table 1
Page 23
Titles
" " , , , , , , , , , , ,
., "" , " """,
" " , , , , " " , , ,
{MP(g) [~] I g E 422U 422i} = { [~] , [!] , [-~] , [ -n}
Page 24
Titles
[1 _1 0]
~ 0]
:Jio
---"> ---t ---t {--> --> -->}
---t ---t ---t {---t ---t ---t}
---t {---t ---t ---> ---t ---t ---t}
Tables
Table 1
Table 2
Page 25
Titles
[1- T -T
Page 26
Page 27
Titles
J]
Page 28
Titles
0] [11] [- h]
o 12 = h·
-~]
-1
~]
-1] [0] [!] [1
J]
Tables
Table 1
Page 29
Titles
[0 -1 -1]
~]
~] .
Tables
Table 1
REV015C007_p431-460.pdf
Page 1
Titles
-1]
1 .
[i
~] .
~] .
Page 2
Titles
-1]
o .
Page 3
Titles
-t23]
-t23
[ -2t12 t12 0]
T = -tl2 -tl2 0
o 0 t33
which gives det(T) = t33(2t~2 + t~2) = t33(3t~2)'
TMp(3)T-1 = Mp(3-1). Then
0] [ t12
t23 = 2tl2
t33 0
-t12 + t22 t23]
-t12 - tn 0
t32 t33
[ -2t12 t12 0]
T = -tl2 2tl2 0
o 0 t33
which gives det(T) = t33( -4t~2 + t~2) = t33( -3t~2)'
Page 4
Titles
[! 0 1]
MD.(,noi2) ~ =! -; =1
Page 5
Tables
Table 1
Page 6
Tables
Table 1
Page 7
Titles
O(a)O(.8) = Rn(T(G»){M",lt",}Rn(T(G»){M{3lt{3}
Page 8
Titles
{MI[i,i,H}
[1 0 0] [2tl]
[1 0 0]
Tables
Table 1
Page 9
Titles
= {hlsl - Ia(S + sl)}{hls} by (7.23)
= {laIO}.
Thus A-I = {lal- Mo(a)-i(s + sl)}{Mlr}o(a)-i.
Since {I31 - Mo(a)-i(s + Sl)} E Rp(TL) and {Mlry(a)-i E {{Mlr}ili = 1, ... ,o(a)}, A-I E j,
P?32 (page 264) Suppose {I3It} = {I3IsI}{Mlr}i where 1 :::; i :::; o(a). {Mlr}i = {MilMir
M,-lr+ ... +Mr+r}. So .
{I3IsI}{Mlr}i = {laMilMir + ... + r + sI}
1 6 . [0] [1 -1 0] [ u ]
i=l "6 + wOO 1 "6 + W
t - d = (la - M)e
(~u)a + (~v)b.
(u - v)a + (u)b.
Page 10
Titles
o 1 0 3
[1 -1 0]
[!],[i],[l]
Page 11
Page 12
Titles
1-1 -11
Page 13
Titles
p = [~,O,-~lt.
Tables
Table 1
Page 14
Titles
[-1 0 0]
Tables
Table 1
Page 15
Titles
[0 0 1]
[0 0 1]
Page 16
Titles
(9)
Page 17
Titles
1 = ,8-ia-ial,8m-yn-y-"
= ai-I,8-i,8m-yn-" (a,8 = ,8a-l)
Page 18
Titles
1~ (47U] _9[~1] -14[JJ) = [~] =[t1n
Tables
Table 1
Table 2
Page 19
Titles
-! I = ~
1 0 0]
[ t(1+~)-~ i(i)-~ iW+4(4-)] [0 0 1]
Mc(a)= ~(1+~)+4(4-) Hi)-~ tW-4(4-) = 1 0 0
1(1 + 1) _ fl(fl) 1(!!) + 1 1(!!) _ 1 0 1 0
Tables
Table 1
Page 20
Page 21
Titles
[-2 -1 -1 I 0]
Page 22
Titles
a • = - Va - Va Va Va Va Va = - Va'
• = Va Va Va Va Va Va = 1
Page 23
Page 24
Titles
[ 1 0 -1]
[0 1 0]
Page 25
Titles
, " , , , ",
Tables
Table 1
Table 2
REV015C007_p431-460.pdf
Page 26
Titles
INDEX
A
c
D
8
Page 27
Titles
E
H
F
GL(3,R) 126
G
Page 28
Titles
J
L
M
m
N
p
o
K
Page 29
Titles
Q
R
Page 30
Titles
s
v
u
Z
T

Citation preview

MATHEMATICAL CRYSTALLOGRAPHY AN INTRODUCTION TO THE MA THEMATICAL FOUNDATIONS OF CRYSTALLOGRAPHY THE REVISED EDITION OF VOLUME REVIEWS

15

IN MINERALOGY

AUTHORS:

M. B. BOISEN, JR. DEPARTMENT

OF MATHEMATICS

G. V. GIBBS DEPARTMENT

OF GEOLOGICAL

SCIENCES

Virginia Polytechnic Institute & State University Blacksburg, Virginia 24061

SERIES EDITOR:

PAUL H. RIBBE DEPARTMENT OF GEOLOGICAL SCIENCES

Virginia Polytechnic Institute & State University Blacksburg, Virginia 24061

MINERALOGICAL SOCIETY OF AMERICA WASHINGTON, D.C.

COPYRIGHT: FIRST EDmON 1985 REVISED EDmON 1990

MINERALOGICAL

SOCIETY

Printed by BookCrafters,

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in MINERALOGY SHORT

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NOTES)

ISSN 0275-0279 VOLUME

15, Revised

MATHEMATICAL

Edition

CRYSTALLOGRAPHY

ISBN 0-939950-26-X

-----------------------------------------------------------------------------------------------------------ADDmONAL COPIES of this volume as well as those listed below may be obtained from the MlNERALoGICAL SOCIETYOF AMERICA 1625 I Street N.W., Suite 414, Washington, D.C. 20006 U.S.A. Volume 1: Sulfide Mineralogy, 1974; P. H. Rlbbe, Ed. 284 pp. Six chapters on the structures of sulfk:les and sultcsafts: the

~~:n~r:%~~~~N~;~~~I.SyntresiS,phase Volume 2: Feldspar Mineralogy, 2nd Edttion, 1983; P. H. Ribbe. Ed. 382 pp. $17. Thirteen chapters on feldspar chemistry. structure and nomenclature; AI,S; order/disorder in relation to domain textures. diffraction patterns, lattice parameters and optical properties; determinative methods; subsolidus phase relations, microstructures, kinetics and mechanisms of exsoluton. and diffusion; color and Interference colors; chemical properties; deformation.

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X:.

H ::::~:,

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~~:':~:.: 1~~:1~. ~~~:::y:':i.H~~;a~r=;;~::.: Ed •. 570 pp. Starting with the theoretical, kinetic and experimental aspects of isotopic fractionation, 14 chapters deal with stable isotopes in the early solar system, in the mantle, and in the igneous and metamorphic rocks and ore deposits, as well as in magmatic volatiles, natural water, seawater, and in meteoric-hydrothermal systems. ISBN #0-939950-20-0.

~rr:,~':'.~s~ 7~~r:sr:n,::~:~'r~~~~.e~~n':~~:'::~~1 :..~~'M'~:~

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Volume 18: Spectroscopic Methods In Mineralogy and Geology, 1988; F. C. Hawthorne, Ed. 898 pp. Detailed explanations and encyclopedic discussion of applications of spectroscopies of major importance to earth sciences. Included are lA, optical. Raman, Mossbauer, MAS NMR, EXAFS, XANES, EPA.

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~_~~9:tf~2~~~lnescence,

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Volume 19: Hydrou. Phyllosllicatel (exclutlve of mlc8l), 1988; S. W. BaUey, Ed. 898 pp. Seventeen chapters covering the crystal structures, crystal chemistry, serpentine, kaolin, talc, pyrophyUite, chlorite, vermiculite, smectite, mixed-layer, sepiolite, palygorskite, and modulated type hydrous phyllosilicate minerals.

~neous

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~ct~~:.'~.=~~arxl Volume

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of RI,. Eds.

of

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aystal EJe..

MA THEMATICAL

CRYST ALLOGRAPHY PREFACE

This book is written with two goals in mind.

The first is to derive

the 32 crystallographic

point groups, the 14 Bravais

the 230 crystallographic

space group types.

mathematical

tools necessary

to lay the mathematical in crystallography

for these derivations

in such a manner as

foundation needed to solve numerous basic problems

and to avoid extraneous

how these tools can be employed, and problems are given. particular,

lattice types and

The second is to develop the

discourses.

To

a large number of examples are solved

The book is, by and large, self-contained.

topics usually omitted from the traditional

example,

~he techniques

needed to work

cartesian bases are developed. approach,

in vector

Unlike the traditional

isomorphism is not the essential ingredient

classification

schemes.

Because alternative

In

courses in math-

ematics that are essential to the study of crystallography For

demonstrate

are discussed.

spaces with nongroup-theoretical

in crystallographic

classification

schemes must

be used, the notions of equivalence relations and classes which are fundamental

to such schemes

example,

we

will

are defined, discussed

find that the classification

space groups into the traditional matrix representations.

and illustrated.

For

of the crystallographic

230 types is defined in terms of their

Therefore, the derivation

of these groups

the point groups will be conducted using the 37 distinct

from

matrix groups

rather than the 32 point groups they represent. We have been greatly

influenced by two beautiful

\~eyl's book entitled Symmetry versity gives a wonderful elegant

exposition

Zachariasen

I

s book

of

Hermann

based on his lectures at Princeton Uni-

development symmetry

entitled

books.

of the point groups as well as an

in

art

Theory

of

and X-ray

nature.

Fredrik

Diffraction

in

W.

H.

Crystals

presents important insights on the derivation of the Bravais lattice types and the crystallographic

space groups.

for many of the ideas developed The

theorems,

sequentially

examples,

These two books provided the basis

in this book. definitions

and corollaries

as a group whereas the problems

a group as are the equations. self-explanatory.

are labelled

are labelled separately

as

The manner in which these are labelled is

For example, T4.15 refers to Theorem (T) 15 in Chapter

4 while DA1.1 refers to Definition

(D) 1 in Appendix

v

(A) 1.

We have strived to write this book so that it is self-teaching. reader is encouraged

to attempt

to the solution presented

to solve the examples

before

The

appealing

and to work all of the problems.

ACKNOWLEDGEMENTS We wish to thank Virginia Chapman, ingenuity, project.

for

her

enthusiastic

In particular,

this book is greatly

book.

Margie

Strickler

L. Geisinger

preparation

C.

of the many

this

illustrations

acknowledged

in this

for her preparation

It is also a pleasure to thank Karen

preparing

ellipsoids

to

of the GML files used to produce

is gratefully

for painstakingly

talent and

contribution

We thank John C. Groen for his dedi-

of the GML files for the appendices.

of the C-equivalent

tireless

her preparation

appreciated.

cation and his meticulous

a lady of exceptional

and

the stereoscopic

pair diagrams

for the 32 crystallographic

lIe thank Don Bloss and Hans Wondratschek

point groups

for beneficial discussions,

Sharon Chang for important technical advice on the drafting of the figures and her assistant,

Melody L. Watson,

Bryan C. Chakoumakos, Albuquerque, eralogy,

Columbus,

Johnson,

Virginia

of Geology,

Ohio; Karen

L. Geisinger,

State University,

Department

of

the earlier

drafts

University, and useful

of Geology

Department

Geological

Sciences,

Mexico, and Min-

of Geoscience,

University Park, Pennsylvania

and David R. Veblen, Department

The Johns Hopkins

We also thank

Univers ity of New

New Mexico; James W. Downs, Department

The Pennsylvania E.

Department

for her draftwork.

VPI&SU,

and Neil

Blacksburg,

of Earth and Planetary Sciences,

Baltimore,

Maryland

for their reading

remarks.

However,

they are neither

of re-

sponsible for any errors that may be present in the book nor for the point of view we have taken in this project. the Series Editor

Finally, we gratefully

Paul H. Ribbe for his helpful comments

and the National Science Foundation

Grant EAR-8218743

of this project.

vi

acknowledge

and criticisms

for partial support

REVIEWS IN MINERALOGY Foreword to Volume 15 MATHEMATICAL CRYSTALLOGRAPHY represents a new direction for the Reviews in Mineralogy series. This text book is not a review volume in any sense of the term, but in fact it is, as its subtitle suggests, "An Introduction to the Mathematical Foundations of Crystallography." Written by a mathematician, M. B. Boisen, Jr., and a mineralogist, G. V. Gibbs, Volume 15 was carefully prepared and illustrated over a period of several years. It contains numerous worked examples, in addition to problem sets (many with answers) for the reader to solve. The book was first introduced at a Short Course of the same title in conjunction with the annual meetings of the Mineralogical Society of America and the Geological Society of America at Orlando, Florida, October 24-27,1985. Boisen and Gibbs instructed 35 participants with the assistance ofE. Patrick Meagher, (University of British Columbia), James W. Downs (Ohio State University), and Bryan C. Chakoumakos (University of New Mexico), who led the computer-based laboratory sessions.

Paul H. Ribbe Series Editor Blacksburg, VA 9/13/85 - a Friday

Preface to the Revised Edition of Mathematical Crystallography In the Revised Edition we have corrected the errors, misprints and omissions that we have found and our students and other users have kindly pointed out to us. In particular, we are pleased to thank our bright and talented students K.M. Whalen, K.L. Bartelmehs, L.A. Buterakos, V.K Chapman, B.C. Chakoumakos, J.W Downs and R.T. Downs for bringing some of these errors to our attention. The Revised Edition also includes a more comprehensive index and a set of solutions for all of the problems presented in the book. We are especially pleased to thank V.K. Chapman and K.M Whalen for their care and efforts in working out the solutions to these problems in detail. We also thank Sharon Chiang for providing us with her drafting skills. M. B. Boisen, Jr. and G. V. Gibbs Blacksburg, VA 2/28/90

to our wives

HELEN and NANCY

whose patience and support sincerely appreciated

is

EXPLANATION SYMBOL

DESCRIPTION

S

Geometric

P

A primitive

[rlO

The triple representation

LO

The lattice generated

* = {a'" ,b* ,e"'} 0

The reciprocal

OF SYMBOLS

three-dimensional

space.

lattice or the basis for a primitive

lattice.

of r with respect to the basis O.

by O.

lattice of O.

HO(a)

The 3x3 matrix representation with respect to the basis 0 of a when a is a point isometry and of the linear component of a when a is an isometry.

~IO(C)

The set of all ~10(a)

I

The set of all isometries.

r

The set of all translations.

0(0)

The basis {tx(O),ty(O),tz(O)}

RO(O)(C)

The set of all RO(O) (a)

HI

I

The Seitz notation

t}

where a

Ao(a)

The linear component

Ao(C)

The set of Ao (a)

T(C)

The set of all translations

of an isometry.

of T.

of a.

where a

E

C. in C.

associated

with the lattice

L.

The trace of the matrix ~l. =

IMI

The determinant

of the matrix M.

The number

o(a)

The order of the isometry

[uVWln

An nth-turn whose given basis.

m

elements

in

H.

II(H)

[uvwln

C.

under the translations

The set of translations

tr(M) de t Ctl)

E

for the 4x4 matrix representation

The orbit of

TL

of S where 0

where a

orbT(o)

0

C.

E

a.

axis is along ua + vb + we where

0

{a,b ,e} is a

An nth-turn screw about the vector ua + vb + we with a tran$lation of mr / n where r is the shortest nonzero vector in the lattice in the [uvw]

direction.

The inversion. Basic conventions

Points in S are denoted by lower case letters, vectors and their endpoints by bold-faced lower case letters, lengths of vectors by italics, sets by capital italics and matrices by capital letters. vii

MATHEMATICAL CRYSTALLOGRAPHY CONTENTS COPYRIGHT;

Page

LIST OF PUBLICATIONS

ii

DEDICATION

iii

FOREWORD

iv

PREFACE.

v

ACKNOWLEDG~IENTS EXPLANATION

Chapter

1.

vi

OF SnlBOLS

vii

MODELING SYMMETRICAL PATTERNS OF MOLECULES AND CRYSTALS

AND GEOMETRIES

INTRODUCTION Symmetrical Symmetrical

patterns patterns

A HATHENATICAL DESCRIPTION AND CRYSTALS

in molecular in crystals

structures

OF THE GEOMETRIES

OF MOLECULES 8

Geometric three-dimensional space Vector addition and scalar mUltiplication Triples Space lattices Vector spaces Vector space bases The one-to-one correspondence between Sand Coordinate axes LENGTHS

1 3

9

10 11

12 16

R3

AND ANGLES

25

Inner product ~letrical matrix Cross product Triple scalar product

Chapter

2.

25

26 34 38

SOME GEOMETRICAL

ASPECTS

INTRODOCTION EQUATION

OF PLANES

18 21 23

OF CRYSTALS 41

AND LATTICE

PLANES

Lattice planes The equation of a plane Niller indices d-spacings

42 42 42

44 46 viii

RECIPROCAL

47

BASIS VECTORS

Direct and reciprocal o and 0'" compared

51 51

lattices

56

CHANGE OF BASIS Zones

63

APPLICATIONS

65

A DESCRIPTION OF THE GEONETRY OF A CRYSTAL IN TERNS OF A CARTESIAN BASIS . . . . . . . . . . . .

72

Calculation DRAWING CRYSTAL

Chapter

3.

of angular coordinates

from crystallographic

data

83

STRUCTURES

POINT ISOMETRIES - VEHICLES DESCRIBING SYMMETRY

INTRODUCTION ISmlETRIES

FOR

. . . . . . . . . . . . . . . . . . ..

91

. . . . . . . . . . . . . . . . . . . . ..

91

Rotations Orientation symbols Compositions of isometries Rotoinversions SnlHETRY ELEMENTS DEFINING

92 95 95 96

. . . . . . . . . . . . . . . . . . . ..

SY~IMETRY

LINEAR MAPPINGS

.

Chapter

4.

. . . . . . . . . . . . . . . . . . . . .

OF A SET OF ~IATRICES DEFINING OF 322 . . . . . . . . . . . .

THE MONAXIAL

INTRODUCTION

99 100

Natrix representations of linear mappings Matrix representations of compositions of linear mappings Algebraic properties of 322 THE CONSTRUCTION THE ROTATIONS

75

CRYSTALLOGRAPHIC

101 105 108 117

110

POINT

GROUPS

. . . . . . . . . . . . . . . . . . . . . . .

123

CONCEPTS

123

Binary operations Groups Symmetry groups

123 125 127

ALGEBRAIC

ix

CRYSTALLOGRAPHIC RESTRICTIONS

129

MONAXIAL ROTATION GROUPS

134

Natrix representations and basis vectors Equivalent points and planes

Chapter

5.

THE

POLYAXIAL

CRYSTALLOGRAPHIC

141 144

POINT

INTRODUCTION

GROUPS 157

PROPER POLYAXIAL POINT GROUPS

157

CONSTRUCTION OF THE DIHEDRAL GROUPS

168

CONSTRUCTION OF THE CUBIC AXIAL GROUPS

173

CONSTRUCTION OF THE I~IPROPER CRYSTALLOGRAPHIC POINT GROUPS

180

THE CRYSTAL SYSTEHS

183

SCHOENFLIES SnlBOLS

191

THE ICOSAHEDRAL POINT GROUPS

Chapter

6.

THE

BRAVAIS

192

LATTICE

TYPES

INTRODUCTION

199

LATTICES

199

A DERIVATION OF THE 14 BRAVAIS LATTICE TYPES Lattices Lattices Lattices Lattices Lattices Lattices Lattices Lattices Lattices Lattices Lattices

invariant invariant invariant invariant invariant invariant invariant invariant invariant invariant invariant

under under under under under under under under under under under

1 2 3 4 6 222 322 422 622 23 432

207 208 208 213 217 220 220 222 222 222 222 222

THE 14 BRAVAIS LATTICE TYPES

223

~IATRIX GROUPS REPRESENTING THE CRYSTALLOGRAPHIC POINT GROUPS

x

225

Chapter

7.

THE CRYSTALLOGRAPH

IC SPACE GROUPS

INTRODUCTION

. . . . . . . . . . . . . . . . . 229

TRANSLATIONS

. . . . . . . . . . . . . . . . 229

ISONETRIES

. . . . . . . . . . . . . . . . . 237

CRYSTALLOGRAPHIC

SPACE GROUPS

249

CRYSTALLOGRAPHIC

SPACE GROUP OPERATIONS

258

THE CRYSTALLOGRAPHIC SPACE GROUP TYPES DERIVED FROH THE ONE-GENERATOR POINT GROUPS . . . . . . . . . . . . .

262

THE CRYSTALLOGRAPHIC SPACE GROUP TYPES DERIVED FRO~1 THE TWO-GENERATOR POINT GROUPS . . . . . . . . . . . . .

276

THE CRYSTALLOGRAPHIC SPACE GROUP TYPES DERIVED FROH THE THREE-GENERATOR POINT GROUPS . . . . . . . . . . . .

295

Appendix

1.

MAPPINGS

303

Appendix

2.

MATR IX METHODS

309

OPERATIONS

310

SOLVING SYSTHIS OF LINEAR EQUATIONS

312

Reduced

317

row echelon matrices

DETERmNANTS

326

INVERSES

330

Appendix

3.

CONSTRUCTION AND INTERPRETATION OF MATRICES REPRESENTING POINT ISOMETRIES 339

INTRODUCTION

339 340

Cartesian bases General bases

INTERPRETATION OF NATRICES REPRESENTING

POINT ISmlETRIES

342 342 345

Cartesian bases General bases

352

PROOFS OF MAIN RESULTS

xi

Appendix

4.

POTPOURRI

HANDEDNESS OF BASES . . . .

357

DISCUSSION AND PROOF OF T6.15

358

Appendix

5.

SOME PROPERTIES

Appendix

6.

INTERSECTION

OF LATTICE

ANGLES

PLANES

BETWEEN

DIHEDRAL GROUPS . .

7.

373

EQUIVALENCE AND FACTOR

RELATIONS, GROUPS

COSETS

EQUIVALENCE RELATIONS

379

EQUIVALENCE CLASSES COSETS

382

...

385

FACTOR GROUPS

Appendix

8.

AXES 371

CUBIC AXIAL GROUPS

Appendix

ROTATION

361

389

ISOMORPHISMS

395

REFERENCES

399 NEW to the REVISED EDITION

SOLUTIONS TO PROBLEMS

402

INDEX

456

xii

CHAPTER MODELING

SYMMETRICAL

PATTERNS AND

1 AND

GEOMETRIES

OF MOLECULES

CRYSTALS

•All things are made of atoms - little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repeling upon being squeezed into one another'. -RichardFeynman

INTRODUCTION Symmetrical principle

Patterns in Molecular Structures:

The

minimum

energy

states that the atoms in an aggregate of matter strive to adopt

an arrangement wherein the total energy of the resulting configuration When such a condition is realized, the atoms in the agis minimized. gregate, whether large or small in number, are characteristically repeated at regular intervals in a symmetrical pattern.

In recent years,

molecular orbital methods have had great success in finding minimum total energy structures for small aggregates (molecules), using various algo1986 and rithms for optimizing molecular geometry (cf. Hehre et al., references therein).

Not only do these calculations reproduce known mo-

lecular structures within the experimental error, but they also show that when the total energy is minimized certain atoms in the molecule are repeated at regular intervals in a symmetrical pattern about a point, line or a plane. (El. 1) Example - Repetition of a pattern at regular intervals about a point and a line: Monosilicic acid, H4Si04,

is a small molecule whose atoms

are repeated at regular intervals in a symmetrical pattern about a point and a line (Figure 1.1). An optimization of the geometry of this molecule using molecular orbital methods shows that its total energy is minimized 0

when an OR group of the molecule is repeated at regular intervals of 90

about a point at the center of the Si atom and at regular intervals of 1800 about a line to give the molecular structure displayed in Figure 1.1. The repetition of the group at regular intervals about a line is called rotational symmetry, whereas that about a point is called rotoinversion symmetry.

Rotations and rotoinversions and how they can be used to define

the symmetry of this molecule will be examined in Chapters 3 and 4.

Figure 1.1 (to the left): as

determined

a 'Keeffe Si, of the

four

actual

sizes

of

of

large

spheres

structure bital

three

of

represent

drawing atoms

in

Figure 1.2 (to the right): posed

molecules

A drawing

the

0,

Si(OH)2 the

molecule

0.

or

of

calculations

sphere

centering

smallest

drawing

the

spheres

in this

book

H~SiOIo'

acid,

(Gibbs et al., molecule

represent are not

1981;

represents

H.

The sizes

intended

to mimic

crystals.

the

groups

structure

of

bonded

together

intermediate-sized

was determined

of monosilicic

structure

orbital

and the

or in any other

"tetrahedral" represent

molecular

The intermediate-sized

spheres

in this

of the molecular

Har t r ee -Fock

1984).

largest

spheres

A drawing

near

and Gibbs,

the the

in

spheres

by O'Keeffe

a tricyclosiloxane into Si,

and Gibbs

molecule

a 6-membered

and

the

(1984),

com-

ring.

The

small

ones

H.

using

molecular

The or-

methods.

(El.2)

Example - Repetition of a pattern

and across planes:

Tricyclosiloxane,

at regular

HsSi303,

intervals

about lines

is an example

of a mole-

cule whose atoms are repeated at regular intervals in a symmetric pattern about

lines and across planes

molecule

is minimized

regular intervals

are repeated

intervals of 120°. across

The total energy of this

Si02H2

group

is repeated

about a line to form a planar 6-membered Imagine a line drawn perpendicular

of the ring and passing

repeated

1.2).

when a "tetrahedral"

Si and three 0 atoms.

molecule

(Figure

through its center.

ring of three to the plane

Note that the atoms of the

in a symmetrical pattern

Also note that the hydrogen

the plane of the ring.

at

about this line at regular atoms of the molecule are

The regular

repetition

of the

structure across a plane is called reflection

symmetry.

lines and planes

the atoms of the molecule

are repeated,

in the molecule

about which

There are other

but a study of these elements will be deferred to Chapter

s. 2

Symmetrical

Patterns

24)

in Crystals:

When a large aggregate

ically _10

strives to adopt a configuration

is minimized,

we may again find that some atomic pattern of the aggregate

is

repeated

similarly

across planes. at regular

at

regular

But, unlike a molecule,

intervals

along straight

of atoms in three dimensions. is

said to be a crystal

arrangement

intervals

wherein

of atoms (typ-

about points

envisaged

or a crystalline

solid are envisaged

respect.

to repeat indefinitely the positions

translational

(El.3)

regular symmetry,

Example

is that

along

peated

- Repetition of a pattern

at regular

of silicate of

tetrahedra

a-quartz,

percent

in Figure

lines

at regular

intervals

6 and 7.

intervals

in the string)

is repeated

Consider

the string

from the crystal structure

composition

comprising

in the direction

in a crystal

about

along some

line at regular

intervals,

parallel

to the string and whose

Such a vector

Example crystal

- The

it is

line segment

length equals the repeat unit

(labelled a in this case), drawn next to

in Figure

1.3, is called a translation

crystal structure

structure

is re-

of the string.

(a vector)

the string of tetrahedra

12

(like the pair of tetrahedra

to represent such a periodic pattern by a directed

of the pattern.

a

of atoms is re-

convenient

a-quartz

along

Fix your attention on any two adjacent

indefinitely

some atomic pattern

of a called

in the string and imagine that this pair of tetrahedra

peated at regular

(El.4)

is

feature of a crystalline

lines.

1.3 isolated

crust.

The repetition

in a large aggregate

of Si02

in three

in such a crystal

in Chapters

along straight

a common mineral

of the continental

tetrahedra

Whenever

intervals

intervals

straight

a subject discussed

some atomic pattern

in such a

exactly by a set of atomic

straight line: As stated above, a characteristic solid

are

to a crystal or a crystalline

to an ideal crystal.

intervals

ideal crystals

at regular

to be static and to be specified

atom

In this

Unlike real crystals

of the atoms

Thus, when we make reference

solid, we shall be referring

the actual

Also the atoms

are envisaged

at

of atoms

solid, whereas

coordinates.

structure

aggregate

to as its crystal structure.

in every

In addition,

dimensions.

pattern

of flaws and irregularities,

as being perfect

and lines or

a periodic

Such a three-dimensional

book, we shall only be concerned with ideal crystals. a variety

energy

the pattern may also be repeated

lines to produce

of the solid is referred

which contain

the total

is presented

3

of a-quartz: in Figure

1.4.

vector.

A drawing

of the

As observed

for

a Figure 1.3: A string of silicate tetrahedra isolated from the crystal structure of aquartz. Each silicon atom (small sphere) in the string is honded to four oxygen atoms (large spheres) disposed at the corners of a SiO. silicate tetrahedron. The lines connecting the atoms in the string represent the bonds between 5i and O. Direct your attention on any two adjacent tetrahedra in the string and note that this pair of tetrahedra is repeated at regular intervals so that the repeat unit is given, for example, by the distance between S1 atoms in alternate silicate tetrahedra. The repeat along the string is represented by 8 parallel vector a whose magnitude, 0, equals the repeat unit along the string. The ellipses, ( ... ), at both ends of the string indicates that this sequence of silicate tetrahedra repeats indefinitely in the direction of the string, even though the sequence is terminated in the figure.

Figure 1.4: The crystal structure of a-quartz projected down a direction in the crystal along which the pattern of atoms displayed in the drawing is repeated. The SiD.. groups in the structure share corners and form spirals of tetrahedra that advance toward the reader and that are linked laterally to form a continuous framework of corner silicate tetrahedra. The vector a represents the repeat unit of the structure in the direction of a; b represents the repeat unit in the direction of b.

4

Figure 1.5: A view of the a-quartz structure tilted about 20 off the viewing direction in Figure 1.4. The repeat unit along the viewing direction is represented by the vector c whose magnitude equals the separation between equivalent atoms in the lines of atoms paralleling c. Because crystals consist of periodic three-dimensional patterns of atoms disposed along well-defined lines, whenever a crystal like a-quartz is viewed along one of these directions, the arrangementwill be simplified by the fact that only the atoms in the repeating unit will be seen as in Figure 1.4. Also, the shorter the repeat unit along such a viewing direction, the simplier the arrangementbecause the repeat unit must involve fewer atoms. Whenthe crystal is tilted off this direction, then the view becomes muchmore complicated with the uncovering of the manyatoms that lie beneath the atoms in the repeating unit. Although the repeat pattern of the structure displayed in Figure 1.4 and 1.5 is finite, the pattern is assumed to continue uninterrupted in the direction of a, band c indefinitely in a ideal crystal.

monosilicic

acid, each Si atom in the structure

neighbor 0 atoms disposed

at the corners of a tetrahedron.

as each 0 atom is bonded to two nearest neighbor can

be

viewed

as

is bonded to four nearest

a

framework

As the structure

in Figure

structure

of

In addition,

Si atoms, the structure corner

sharing

silicate

tetrahedra.

which

an atomic

pattern

is

1.4 is viewed down one of the lines along

repeated,

the atoms along this line are one

on top of the other so that only the atoms in one repeat unit along the line

are visible.

Thus beneath

each Si atom displayed

in the figure,

there exists a line of Si atoms equally spaced at regular intervals that extends indefinitely.

Likewise beneath

each oxygen

atom in the figure,

there exists a comparable

line of oxygen atoms also equally spaced at the

same

However,

regular

intervals.

tilted off the viewing

direction

atoms along this direction

when the structure

in Figure

by about 2°, the repeating

is exposed as in Figure

5

1.5.

1.4 is

pattern

of

By convention,

the repeat unit along each of these vector c whose between

lines is represented

length, c, is equal to the separation

adjacent

atoms in the lines of equivalent

by a parallel

(the repeat unit)

Si atoms and 0 atoms

that parallel the viewing direction. Returning to Figure

1.4, note that the tetrahedra running across the

figure in strings from left to right are exact replicas of the ones comprising the string in Figure 1.3.

In fact, several such parallel strings

are displayed in the figure running

left to right.

another

such

diagonally

set

of

strings

of

tetrahedra

In addition, note that

runs

across

the lower left to the upper right, both intersecting strings

0

at exactly

the left-to-right

120

trending

lation vector a, whereas

b.

tetrahedra, regardless

set of strings

the first set

In addition

and

b

on

of

is represented

by the trans-

that along the diagonal set of strings running

to defining

the

repeat

by the translation

unit

along

strings

of

the pattern at both ends of these vectors is exactly the same of their

location

in the structure provided

not been rotated from their original illustrated

figure

As shown in Figure 1.4, the unit repeat along

•

from the lower right to the upper left is represented vector

the

from the lower right to the upper left and another runs from

orientations.

the vectors have

This property can be

by laying a sheet of tracing paper on Figure it and then sliding

making sure that the vectors on the drawing.

(translating)

1.4, tracing a

the sheet over the figure

on the sheet are kept parallel

Note that the pattern

with those

at the end points of both a and

b is an exact copy of that at their origin regardless

of the placement

of the vectors on the drawing, provided that the vectors on the sheet are kept parallel with those on the figure. The lengths {a,b,c} «c:a) when

(magnitudes) a, band

and the interaxial and ~ = «a:b)

a crystal

diffraction

or

angles between

can be determined crystal

powder

them denoted

a

=

«b:c),

in an X-ray diffraction

is bathed

By measuring

of the spots on the film or the peaks on the chart,

lengths and the interaxial

~

experiment

in an X-ray beam and its

pattern is rec~rded on a film or strip chart.

the positions

vectors 0

c of the translation

angles of the translation

vectors

the

can be de-

termined. A set of X-ray data measured for a powdered a-quartz crystal is given in Table A2. 2.

A least-squares

PA2.4) shows that a -8 where 1A = 10 cm.

=

b

=

4.914A,

refinement c

=

6

of the data

S.409A, a

= ~=

(see EA2.12 and

900 and ~

=

1200

Figure

1.6:

A string

Figure

1.7:

A projection

of the structure silicate

of silicate

isolated

of the a-cristobalite

perpendicular

tetrahedra

tetrahedra

structure

to the drawing.

that form spirals

that

from the a-cristobalite

viewed down c , the repeat

The structure

are linked

structure.

consists

laterally

into

unit

of corner sharing

a framework of SiOl

composit ion.

(P1.l)

Problem:

isolated

Study the string

from the structure

found in highly silicic volcanic

(1)

Determine

(2)

Draw a translation

of silicate tetrahedra

of a-cristobalite,

of Si02

rocks such as rhyolite.

the number of silicate tetrahedra

structure

in Figure 1.6

a rare polymorph

in the repeat unit.

vector a alongside the string.

of a-cristobalite

displayed

in Figure

Examine the 1.7 where

it

is viewed down an important repeat direction and find two repeat directions the vectors directions

in the plane that match that in Figure

1.6.

Draw

a and b to represent the repeat units along these as was done for the a-quartz structure in Figure 1.4 angle I = «~:b).

and measure the interaxial 7

As done in E1.4,

Figure 1.8 A view of the a-cristobalite in Figure 1.7.

structure

about 2° off the viewing direction

tilted

trace a and b on a sheet of paper and note, as observed the

a-quartz

structure,

cristobalite

that

the

atoms

for

in a-

over the drawing. study by Pluth et al.

An X-ray diffraction

that a = b = 4.971A and 0 = 90°, viewing

structure

of

is the same at both ends of a and b as the sheet

is translated

the

pattern

direction

has

been

is

The repeat unit,

exposed

tilted

2°,

(1985) shows

in Figure

The

study

c,

along

1.8 where also

shows

the that

e = 6.928A and a = B = 90°,

A MATHEMATICAL

DESCRIPTION

OF MOLECULES

It

is clear

above that

there

from the discussion, are

numerous

crystals that mu~t be described terns and properties, alone, because

OF THE GEOMETRIES

AND CRYSTALS

examples

geometrical

and problems

features

in order to understand

of

presented

molecules

and

their atomic pat-

This cannot be done with pictures or illustrations

accurate

calculations

are necessary

for a description

of

their symmetry and structure together with their bond lengths and angles. To accomplish this we shall create a mathematical

model of the real world

that will enable us to describe all of the geometrical features of a given crystal with respect to its natural frame of reference and translation 8

vectors.

For example,

polymorphs

a-quartz

directions

along which atoms are repeated over and over again indefinitely

in periodic

we have seen in our

and a-cristobalite

patterns.

The mathematical

sense that its principle special directions collection

directions

discussion

of

the

model will be "adjustable"

of the object under study.

We will also see that the into self-coincidence

will be used to describe

its symmetry.

very complicated

can be easily described using matrices.

begin the development

in the

can be chosen so that they are along

of all the motions that send a crystal

motions

silica

that ideal crystals have special

The model will be such that even

of the model by examining the geometry

We will

of the real

world.

Geometric three-dimensional which molecules

three-dimensional set of points culated,

space:

and crystals

The three-dimensional

actually

space and will be denoted by S.

such that the distance

the angle between

real world in

occur will be called the geometric We will view S as a

between any two points can be cal-

any two intersecting

lines can be measured and

such that we can tell when two lines are parallel.

All that we will add

to S as we go along is a frame of reference that will enable us to model

S algebraically.

The first step in imposing

is to establish as

a point

the origin

0

but, as we shall see later, there

cations that arise when the origin

origin

is determined,

nating

from

bold-faced

0

letters.

or a crystal

in S.

Such

The absolute

vectors

The magnitude

value of a scalar

the vector emanating vector and is denoted

will

otherwise

Once the

be

denoted

from

O.

0

to

0

of r will usually

when an ambiguity

by

is possible,

x will be denoted by

Ix I .

be by

Since

is of length zero, it is called the zero

For the most part we shall not distinguish

tween a vector emanating from the origin and its end point. unless

line or plane

is repeated.

The magnitude or length of a vector r is the distance

to its end point.

denoted by an italics r or, sometimes,

IIr II.

simplifi-

S can be described by the set of all vectors ema-

to each point

from the origin

are certain

is chosen on a point,

about which the pattern of a molecule

on S

a frame of reference

Any point in S can be chosen

to be the origin.

stated, we shall

be-

Consequently,

let r denote both the vector from the

origin to the point r and the point as well.

To impose some organization of three noncoplanar,

on this set of vectors,

nonzero vectors

9

D

=

{a,b,c}.

we choose a set

Using the notions

!V o

o

-./2v

Figure

1.9

(to

the

from the origin, their

sum r1

parallel by

left);

vector

Note that

that

at

the vectors

incident

with

v

of rzo

from a to

Figure 1.10 (to the right): generates

the

occurs

at

and -

end point

the

end

nv.

sides

of

the

parallel

of a vector

of

the

r a [; S radiating

,

r i + r2

way, the vector

copy

every point vectors

and that

is to place

a

sum r i + r a is given of

V by scalars

We note that

points

rr

of a parallelogram

way of obtaining

In this

The multiplication

zv , (l/3)v

of any two vectors

An alternate

the end point

radiates

addition

r i and rz form the

+ rz forms its diagonal.

copy of r1

the

The geometric

0 of S.

ra.

Z, 1/3 and - 12

on a line

contained

in

in S cothe

set

(xvlxtR).

of vector vector c.

addition

and scalar multiplication,

in 5 can be expressed

in a compact

we

manner

shall

Hence, the placement of the origin and the choice of

establish

the frame of reference

addition

is defined

by geometric

addition.

the

sum

of

with sides r1 and r2 (Figure 1.9).

The addition

By geometric

addition

two

r1

vectors

In addition

there is another important operation

bers and vectors numbers

To see how

and

operation

R

and consists

10

on 5

(or vector adr2,

denoted

by

constructed

to the operation

of vector

defined between real num-

in 5 called scalar multiplication.

(scalars) is denoted by

of vector

defined on S.

r1 + r2, is the vector that is the diagonal of a parallelogram

addition,

each

{a,b,c} will

we shall explore the important operations

and scalar multiplication:

dition), we mean that

=

D

of the crystal under study.

and scalar multiplication

Vector addition

that

for the model and therefore will be se-

lected to reflect the properties this is accomplished,

see

in terms of a, band

The set of all real

of the set

Z

of all in-

tegers, the set Q of all quotients alb of integers the set of rational

12,

i =!=T. xr

If rES

is defined

Ixl

length

x

(read "r

however,

numbers

r

involving

and x E R,

and with direction

r

the same as that of

The multiplication

when

of a scalar

x times the vector r is what is meant by scalar multiplication. ample, if the vector V is multiplied

then

along the same line as r with a

of r when x < O.

> 0 and the opposite

contain

numbers such as

is an element of the set SIt)

to be a ve~For pointing

times that of

(b # 0) (Q is called

and the set of irrational

The set R does not,

etc.

11,

numbers)

by 2, then the resulting

For exvector

2V

is a vector in the direction of V that has a length twice that of V while

(1/3)v is a vector in the direction of V (Figure 1.10). negative

from r with vector

1- 121

12

rES,

in

E

with a length one third that

O.

Or to be the zero vector

times that of r.

the

opposite

If x

is a

in the opposite direction For example,

direction

as

V

(- (2)v

with

a

is a length

Thus, given any real number x and any vector

times that of v.

scalar multiplication

vector xr

V

is a vector pointing

Ixl

a length

pointing =

We define

then xr

number,

of

to x and r a uniquely

assigns

determined

S.

Recall that S can be viewed as the set of all vectors emanating from If 0 = {a,b,c} is a set of three noncoplanar vectors

a chosen origin O.

in S, then for any vector r in S there exists a unique set of real numbers x,

y

and

I

such

that

xa + yb + IC is called

o=

a

=

r

linear

{a,b,c} is called a basis.

thoroughly written

later.

concisely

that x,

Triples:

An

IC.

combination

of

expression

like

{a,b ,c} and the set

We shall study the nature of bases more

Using set notation,

=

{xa

+ yb + zc

these statements

about S can be

1

x,y,z

E

R}

.

is read ItS is the set of all vectors xa + yb + zc such

y and z are elements of R,

Since each vector

combination

yb +

as

S This expression

xa +

r

in

S

the real numbers."

can be expressed uniquely as a linear

r = xa + yb + zc , r can

respect to the basis 0

=

be unambiguous ly determined

with

{a,b,c} by the three scalars x, y and z.

When

we write these scalars in a vertical

11

column enclosed in brackets

as

The set of all such triples is denoted by R3.

they form a triple.

That

is,

([;]1

',Y,'

r in S can be assigned

Hence each vector

respect to the basis D.

, R) .

to a unique

Since this triple

triple

representative

pendent on which basis D is chosen, we denote it by [rID'

[;]

where D

and on Ly if

xa + yb +

r

of r is deThat is

IC

,

{a,b,c}.

The decision taking

if

in R3 with

as to which basis

into account

is to be used

the natural geometry

if a crystal like a-quartz

is usually

of the crystal.

made

by

For example,

is to be studied, then it would be convenient

for each atom in the structure to have a triple that defines its position in the crystal relative to three noncoplanar vectors like a, band radiate

from a common

origin, 0, and which

rections along which the structure terval.

When

lie along

C which

well-defined

di-

is repeated in a relatively short in-

.referring to molecular

customary to speak of the coordinates

and

crystal

structures,

it

is

of a point rather than its triple.

In this book, we shall use the words

"triples"

and "coordinates"

of

a

point interchangeably.

Space lattice:

If

D == {a,b,c}

is a set of three noncoplanar

vectors in S, then the subset of vectors

LD == {ua

defines an important

+

vb

+

wc

(points)

I

u,v,w

E

Z}

set of points LD called a space lattice.

The vectors

comprising LD are special in the sense that each is of the form

12

nonzero

•

•

•

•

Figure 1.11 (top): {a,b,c} nitely

in all

passing

of the space lattice

The pattern

of points

LO

generated

in this figure is assumed

by the

vectors

to continue

0 =

indefi-

directions.

Figure 1.12 (bottom): by

The points

of a-quartz.

lines

A line

parallel

representation

to a, band

of the space lattice

c , respectively,

13

in Figure

1.11 constructed

through each point of the figure.

ua + vb + wc where u, lattice 1.11.

associated

v and ware

with

A drawing

structure

c were chosen

respectively,

Figure

1.12.

placed

in the lattice.

each

In Figure

parallelepipeds

point

the same.

of the

This

three

to the vectors

Note how the contents

is exactly

in Figure

The lattice can be seen

in the space

1.13, a drawing

space

in this case to be

lines that are parallel

through

of the

is displayed

repeat units in the structure.

more easily by drawing c,

the a-quartz

The vectors a, band

of the shortest

integers.

a, band

lattice as shown in

a-quartz

structure

is

of each of the resulting

is true

regardless

of

the

placement of the origin of the lattice in the a-quartz structure, provided the orientation

of the lattice is not changed.

Also the atomic pattern

at each lattice point in the structure will be exactly the same regardless of the choice of the origin. crystal

can

be

parallelepipeds

by

This illustrates

description

contents of a relatively

framework for describing

of an a-quartz

anyone

it indefinitely

of

over

how space lattices

to a theoretically

infinite

these

and

over

enable us to

crystal

by

the

Such parallelepipeds

are

they contain at least one complete unit of the

repeating pattern of the crystal.

a-quartz.

structure with

small parallelepiped.

called unit cells because

permits

starting

and by simply repeating

again to fill space. give a precise

Hence, the whole

constructed

Hence space lattices provide a natural

the periodic patterns of atoms in a crystal like

Besides the overall geometry of the crystal, the space lattice

a simple description

of other properties

of the crystal.

For

example, any line that passes through the origin and another lattice point defines

a direction

repeated.

Moreover,

(zone) in the crystal along which the distance

countered along this zone measured

between

the structure is

the first lattice

point

en-

from 0 defines the repeat unit of the

"-

structure along this direction in the crystal. directions

along which

also parallel

the structure

Not only do zones parallel

of a crystal is repeated, but they

the lines along which the

faces

of

such

a

crystal

may

intersect.

(El.5)

Example

of a-quartz:

- Finding

triples (coordinates)

Consider the vectors r, s, t, u,

of vectors V

and

W

in the lattice emanating

from

the origin of the space lattice LO associated with the a-quartz structure (Figure 1.14) and find [r)O'

Solution:

An

examination

u = i a + 2b -+ Oc and w = -

[U)O

and [W)O'

of Figure 1a

1.14 shows that r = 2a + ib + 2c,

Ib - ic are vectors

thus 14

contained

in

L 0'

Figure 1.13 (top): sentation. into

an

which the

Note

A drawing

'that

indefinite

contains

each

number at

least

of

the

crystal

parallelepiped of

one

disjoint complete

structure

of

the

identically and

lattice

of

a-quartz

partitions

constituted

representative

and

its

the

a-quartz

unit

unit

of

the

cell

lattice

volumes

repeating

represtructure each

pattern

of of

structure.

Figure vector

1.14 (bottom): p

P ;; Pia coefficients

+

is

The

contained

+

P2b of

Plc. each

in Thus,

are

space

LD the

lattice, if

there

vectors

LO' exists r,s,

...

integers.

15

of a-quartz three ,W are

generated

integers each

PI,

contained

0 = (a,b,cj.

by P2

and in

p ,

such

LD because

A that the

o

(Pl.2)

Problem:

VELD

(Figure 1.14) and (2) locate the vectors r1,

(1)

[t10

[SlO'

Find

and [V10 for the vectors

r2, r],

r,

E

S,

t,

LO that

satisfy the following equalities:

(c)

(b)

Vector Spaces:

[r]lO

Hl·

[r210

In our earlier discussion

of geometric three-dimensional

we defined the sum r1 + r2 of any two vectors

space, S, Figure

1.9) and the scalar product xr

1. 10) .

In each case the resulting

algebraic

of any x vector

E

r1,

r2

rES

Rand

is uniquely

E

S

determined.

system that is formed using these two operations

(see

(see Figure The

is an example

of what is called a vector space.

In general, scalars,

a vector

space consists

the two operations

and a list of properties

of a set of vectors,

a set of

of vector addition and scalar multiplication

that the two operations

must satisfy.

For our

vector spaces, the set of scalars will always be taken to be the set of real numbers and thus will be real vector spaces. be defined in various ways. definition ometric

properties low.

(Dl.6)

R

In earlier discussions,

of these operations.

interpretation,

we gave a geometric

Of course, if vectors do not have a ge-

other definitions

must be devised.

The required

of a vector space are listed in a formal definition

Notice that each of these properties

Definition of a vector

denote

The two operations can

space:

ement in V, and the operation

given be-

by S.

Let V denote a nonempty set and let

the set of real numbers.

dition denoted by + that combines

is satisfied

Consider

an operation

any two elements in

of scalar multiplication

16

V

of vector ad-

to yield

an el-

that combines an

in R with

element

with r is denoted xr). and x, Y

R

E

rEV

Rand

E

Then

V

then the scalar product

of

is a vector space if for all r, s, t

E

X V

the following rules hold:

=

(1)

r +

(2)

r + (s + t)

(3)

There

(4)

Corresponding

V

in V to yield an element in V denoted by

an element

(i.e., if x

juxtaposition

S

S

+ r (Commutative Law);

=

exists

(r + s) + t (Associative Law); a vector

U

E

V

such

that

v +

V

U

for all

V (The existence of an identity element u);

E

to each vector V

such that V + W =

where

U

istence of an inverse

w

E

V,

there exists a vector W

u is the identity element (The ex-

for each V

E

V);

(5)

x(r + s)

=

xr + xs (Left Distributive Law);

(6)

(x + Y) r

=

xr + yr (Right Distributive Law);

(7)

(xy)r = x(yr) (Associative Law);

(8)

lr

=

r (Unit Element).

Besides the vector space S of all geometric vectors,

there are many

For example, the set R3 defined in (1.1)

other important vector spaces. with operations

and

for all x,

Xl,

in R3

equality

if and only if

(Pl.3)

Problem:

qualifies

Yl,

ll'

X2,

Y2,

12

E

Xl] X

[

Yl II

=

[XX1] XYl

XIl

R is a real vector space.

Note that

is defined to be

Xl

Y2,

Show that R

3

and

II

satisfies

as a vector space.

17

the

properties

of

Dl.6

and

Vector

Space

Bases:

We observed

vectors

in S, then each vector V

that

b

if a,

and care

S can be expressed

E

noncoplanar

as a linear com-

bination xa + yb + zc where x,y,z

v When each vector

R .

E

(1. 2

in a vector space can be expressed

tion of a set of vectors,

0 = {a,b,c},

as a linear combina-

we say that 0 spans the vector

space.

If

V

distance

=

determined are

xa + yb

that

V

by z.

uniquely

+ ZC, where a, b, and Care

lies from the plane Hence

defined

Z is uniquely

determined.

By uniquely

Zl

=

Z2.

If 0 is a set of vectors

determined

linear

combination

Consequently,

of 0,

then 0 is

independent.

said

to

as a basis. assumed

However,

be

unless otherwise

Yl = Y2

=

and

=

independent.

{a,b,c}

both spans

in a vector space that is called a basis

vectors in S qualifies

stated, the bases we use will be

(see Appendix

c, a, ~ and ~, two choices of {a,b,c}

linearly

independent

Thus, any three noncoplanar

to be right-handed

if v

x2,

in only one way as a

vectors 0

Any set of vectors

both spans the vector space and is linearly for the vector space.

we mean that

in a vector space such that each vector

a set of three noncoplanar

S and is linearly

x and y

Xl =

then

of 0 can be written

is a linear combination

then the

Similarly,

determined.

xla + ylb + zlc and V = X2a + Y2b + Z2C,

that

noncoplanar,

by a and b is completely

4).

For example,

are possible

given a, b,

- one left-handed

and

one right-handed.

In this book a, b, c, a, ~, ~ will refer unambiguously

to the right-handed

one.

i, j and k are mutually referred

in R?

perpendicular

to as a cartesian basis.

be difficult following

A commonly u~ed basis for S is C = {i,j,k} where

to visualize

example

unit length vectors.

the set of vectors

illustrates

Usually

C

is

In vector spaces other than S, it may

how to determine

that forms a basis. whether

The

a set of vectors

is a basis or not.

(El.7) Example for R l:

- A demonstration

that a set of vectors qualifies as a basis

Show that the set of vectors

18

forms a basis for R'.

Solution: V

E

R',

To show that 0 is a basis, we must show that given any vector V can be expressed in D.

of the vectors for x,

solution

in exactly

one way as a linear combination

Hence, we shall show that there exists

a unique

y and z in the equation

V

This equation

is equivalent

to the system of equations

x

y

x + y

= v,

z This system can be written

in matrix

-1 1

form (see Appendix

2) as

o o

(1.5;

o Using any of the methods the unique

solution

x

in Appendix

=

2, it is found that the system has y

vl/2 + V2/2,

=

-v1/2

+ v2/2,

Z = V3'

Since

(1.3) has a unique solution for each choice of the vector V, 0 is a basis o

for R'.

(Pl.4)

Problem:

described

Let 0 denote the basis given in E1. 7.

in the solution

to El.7, write

Hl as a linear combination

of the vectors

19

in D.

Using the method

(Pl.5)

Problem:

Show that

is a basis of R:",

As demonstrated for Rl,

is a basis

n

some

V

E

R

Ax = V has no solution

l,

> 3, then A has more columns than rows implying

=

Hence, unless n

unique.

Furthermore,

{V1,V2,

=

... ,V } n V has a

[V11 ' 0 If n < 3, then A has fewer columns than rows implying

[V210"",[vnlO' for

0 =

whether

for all V where the columns of A are the triples

unique solution

that

in E1.7, to determine

one must show that the matrix equation Ax

if

n

= 3,

(see Appendix that solutions

3 there is no possibility

Ax = V has

a unique

If

are not

that 0 is a basis.

solution

det(A) # 0 (that is, when A-I exists (see Appendix

2).

2)).

if and only if In summary, we

have the following theorem:

(Tl.8)

Theorem:

{V1'V2'Vl}

All bases for R

is a basis for R?

3x3 matrix with columns

if

[VdO'

l

have 3 vectors.

Furthermore,

0

=

and only if det(A) # 0 where A is the

[V210

and [Vll ' O

The first statement in Theorem TI.8 can be generalized to any vector space that has a finite set of vectors as a basis. vector space with a basis consisting of n vectors,

V must have exactly n vectors. of

V is n.

(Pl.6)

Hence Sand

V

is a

then every basis for

In this case we say that the dimension

R' both have dimension 3.

Problem: Determine which of the following sets of vectors qualify

as a basis for

(a) 0

(c)

That is, if

n>,

[:] , [=;] , [=;]

(b) 0

0

(d) 0

20

The one-to-one correspondence between 5 and Rl:

Let 0 =

note a basis for 5 and let W denote any vector in 5.

[::] where

the

unique

vectors of 0 is w

representation

=

w1a +

W2b

in 5, there is a unique triple

Rl

is the triple

to a unique

of

Hence corresponding

[W)O

Rl.

in

Conversely,

in 5.

spondence

between 5 and

or written

another way:

Consequently,

Rl

correspondence

it V

geometric

Rl

As such, each has a vector .one important property

"preserves" £

these

vector

of the space

x[U)o'

W +-+

To formalize

if

u

£

5

and

[w) 0 correspondence

is that

and justify

For

R,

if

u,

add the triples and then

vector

£

example,

Hence, to add two vectors

triples,

x

vector defined

of using the clumsy parallelogram

Similarly,

can be

and scalar product

+ [V)O.

that sum into its corresponding

the necessity vectors.

Rl

are both three-dimensional

operations.

in 5 we first find their corresponding

reason why

space 5.

addition

5, then u + v corresponds to [U)O

convert

corre-

-[;:1

is the fundamental

We have already seen that 5 and

on it.

in

given by

used to model three-dimensional

spaces.

to each vector each vector

we have the one-to-one

w,a + w,b + w,e

This one-to-one

of the

of 0 and hence gives rise

of some linear combination

vector

de-

,

as a linear combination

W

+ wlc.

{a,b,c}

Then

in 5.

This eliminates

rule for adding geometric then

xu

corresponds

to

these remarks we have the following

theorem.

(Tl.9) Theorem:

Let 0 =

lowing two statements

{a,b,c}

denote

are true:

21

a basis of

5.

Then the fol-

(1)

[u +

(2)

[xU]D~

Proof:

Let

u2,

ui,

=

V]D

V

t

U

+

t

Rand

V

U t

t

S;

S.

S.

Since D is a bas is of S, there exist real numbers

V2

and

Vi,

via + v2b + V3C.

for all u,

[V]D

for all x

x[U]D

U,

U3,

+

[U]D

such

V3

that

U

=

uia + u2b + u3c

and

=

V

Hence

(uia + u2b + u3c) + (via + v2b + V3C)

V

(ui + vi)a + (u2 + v2)b + (u3 + V3)C

.

(1. 6

In R3 we have

and

But according Hence

[u

+

to equation

V]D

(Pl.8)

Problem

Estimate

[r + U]D

last

triple

is

also

[u + V]D'

[V]O'

o

Prove part (2)

of Tl.9.

- Calculating vectors

we found the triple 1. 14.

+

= [U]D

(P1.7) Problem:

(1.6), this

corresponding

in the lattice of ex-quartz:

to several

r + U using the parallelogram

from the figure.

and compare your answers.

vectors

In E1.5 in Figure

rule and then determine

[r + U]D

Now calculate

labeled

using Theorem

Use the theorem to calculate

T1.9

each of the 'fol-

lowing: (a)

Theorem

[-r]O

(b)

[6r + 2U]D

(c)

[3r -

SU]D

T1.9 shows that, as vector spaces, Sand

The mathematical

statement

and that the mapping

of this fact is that Sand

of S to R3 that takes W to [W]D

If we choose an origin 0 and a basis served earlier the vectors a frame of reference vectors,

we

observe

Oa + lb + Oc and c

=

on S. that

0, a, band

D

=

{a,b,c}

R3 are identical. R3 are isomorphic is an isomorphism. in S, then as ob-

c are fundamental

to establishing

To find the triples that correspond 0

=

Oa + ob + Oc,

Oa + ob + 1c.

Hence

22

a

=

to these

la + ob + Oc,

b

=

[:],

the basis of RJ

In particular,

m'

[biD

that corresponds

[eiD

to 0 is

(1.7)

Consequently, tween a, band

no matter how peculiar the geometric

C may be, the corresponding

This simplicity

is a great help

in conveniently

S with respect to the established

as the set IJ

R

are very simple.

describing

frame of reference.

be-

features

For example,

in the

in the space lattice defined by 0 is represented

set of all vectors RJ

triples in

relationship

J

of all triples with integer entries.

However,

in

caution

must be exercised when lengths of vectors and angles between vectors are considered

since,

for example,

in spite of the great similarity

three vectors in (1.7), their corresponding greatly

in length.

vectors a, band

of the

C may vary

We shall discuss how to overcome this problem later

in the chapter. Vectors in S that lie in the direction simple form xa for some x written

o =

in the

{a,b,c}

R.

E

form yb and

of a can be written

Similarly, vectors along band respectively,

IC,

where y,

I

forms a basis, any vector V E S can be written V = xa + yb +

combination

Hence,

IC.

geometrically

E

in the

c can be

R.

Since

as a linear

V can be

pictured

as the sum of three vectors each in the direction of a basis vector Figure

1.lS(a)).

can decompose

As the correspondence

a vector in

[xalD

R

J

[:],

Coordinate

Axes:

[~l'

[yblD

of a-quartz,

pictured

Let 0

a, band

=

Sand

in a similar manner.

and since these are scalar multiples have the correspondence

between

R

J

(see we

Since

[,eID

of the basis vectors

in Figure

suggests,

in (1.7), we

1.lS(b).

{a,b,c} denote a basis for S.

As in the case

c will usually be chosen so that each lies in an 23

(a)

Figure

1.15.

geometric

A graphical

S

space

axes X,

dinate

are displayed there

exists

Y and Z,

three

[V]O

R'.

[b]O

the

basis

placed

numbers x,

y

for each vector

along X,

and V

t

S,

there

of

this

and [V]O

other

than as a model

figure

is

= to

a set

and Z.

that exists

for Figure

+

v

axes

of these vectors

For each point can

in (a)

along coor-

be

a triple

written

v as

representative

t

S,

=

v

of

of triples)

there

significance

1.15(a).

should

The vectors

is no intrinsic

vector or the angles between vectors

z[c]O

show the correspondence

No geometrical

t R'.

+ y[b]O

x[a]O

an angle can be defined between any two nonzero vectors just

c are directed

representatives Y

such

I

and coordinate

such that

V t S

is

vectors a, band

In (b), the triple

and [C]O

[V]o The purpose

of

In (a), basis vectors

respectively.

real

Likewise,

e R'

representation

(b)

with [a]O'

xa + yb + Ie.

v,

and

in R J

•

24

that

exists

be attached

between vectors

to Figure

l.IS(b)

in S have length and direction in S.

On the other

meaning to the notions

hand,

and

as R

J

of the length of a

important

direction

relative

Since these directions

to the structure

will be important

points

lying

defined

to be the set of points

a.

That

shown

in

including and

is

I r2

{r2b

X-axis

along them coordinate

E

The

r

{[':]I

=

through

is

in

R3

0

and

{r1a I rl E defined

I r3

R}.

E

r1a + r2b + r3c where r2

=

the triples for the points on the X-axis

Similarly,

Y-axis

is defined to be {r3c

of the vectors

x

the X -axis is

For example,

is the set of vectors

1.15 (a) .

and the Z-axis

R}

consists

to us, we call the set of all

axes.

lying on a line passing

is, the X-axis Figure

of the crystal under study.

=

to

R}

be

As the

r3

=

0,

are (see Figure 1.15(b)):

r , e R}

the triples for the points on the Y- and Z-axes

in

R3

are

Y

respectively.

By convention,

dinate axes are denoted

CL

the interaxial

angles between these coor-

fl = «Z:X)

= «Y:Z),

and.

= «X:Y).

LENGTHS AND ANGLES

Inner Product:

space S, the length of a given vector or

In geometric

the angle between two vectors can actually be measured. in

R3

has no intrinsic

angle between

real world by R ometric

in

a method

R3.

Since the objective

R3

in

R3.

is to model the

must be devised so that the length of a ge-

vector V and the angle between

from the coordinates tors in

length nor is there a natural way to define the

two vectors 3,

However, a vector

such vectors

Consequently,

must be based on information

can be

calculated

the lengths assigned to vec-

about the geometry

of the basis

D. The necessary information is best described by the inner product (this product

is also referred

to as the dot product).

the inner product of two vectors v • W where 8 is the angle between

=

V

and

W

In geometric

(1.8

vwcos 8 ,

v and W such that 00 25

space,

is defined to be

~

8 ~ 1800 and where

oL ~_j_

Figure 1.16: The projected length of a vector on a unit vector W is given by vcos a where e = -...-.

~~ o '" " '~C~] " .... ~ rJ'JQ)j:Q~

B §.nQ)~"::';' 0

~

-

O

.J:l

f()

.cOO .µ U of"'(

"

II-I'rl CIl

o

..

,......,

1::'-" ts

• .-1

~.~.~ ~.~e 0 A. ~

t::

o

~ :>

'.-i..--i

.µ

CCI

c

t::

'+-I

0

'... VI OO.µ

t:: ~

a::I'.-j

CO ~

Q)

'rl

e: ~

~

'.-1

CCI

~

~ " ....~

.c

•• E-< N

~

M N § 1:

7

.. "~~"

ro

~t

f()t .J:l

•

L.

--8 o

u

of.

::::,o,?Il

.J:l

~ u .. :: :::].9)ftl

N

,

0

" 1.

=

where N

n. I

=

#(C),

I

I

1/

The basic strategy for establishing

this theorem will be to find

two distinct ways of counting the nonidentity will be two expressions rotations of C. expressions

classes of pole points,

and v . = # (Coo) .

# (c. (C))

Proof:

is the number of equivalence

t

each equaling

This will establish

will be precisely

rotations of C.

twice the number

The result

of

nonidentity

the result since these two

those appearing

in the equation.

equal

We begin

each pole point poo in P(C) one at a time and counting the 1/ number of nonidentity rotations of Cleaving poo fixed. The sum of these 1/ numbers taken over all the pole points in P(C) will equal twice the number

by taking

of nonidentity

rotations

in C because

each of these rotations

exactly two pole points fixed and hence is counted twice. nonidentity

leaving poo fixed is if(Coo)

rotations

II

1/

The number of

1 = v . - 1.

-

leaves

Thus, for

I

the pole points in ei(C) we have

#(nonidentity

rotations

leaving Pil fixed)

v. -

#(nonidentity

rotations

leaving Pi2

V. I

1

I

fixed)

1

-

equations

nj

#(nonidentity

rotations

leaving p.

v. -

fixed)

In.

1

I

I

Summing up these numbers

we find that the contribution

ni(vi - 1) for each 1 SiS

t.

e/C)

from

Adding the contribution

is

from each of the

t t equivalence

classes,

we find that the sum is

L

n.(v. -

1=1

N -

1 is the number of nonidentity

the number of nonidentity tablished

I

1).

Since

I

rotations in C, it follows that twice

rotations

in C is 2(N

-

1) and so we have es-

that t 2 (N -

From

etc;

E5.2

(322))

If ( [ 10012)

=

=

3

and

P5.1,

and

n3

2 and v 3

=

=

1)

we liCe

0 if ( [ 11 12)

L5.8 in this case by observing

2(6

-

1)

n.t», -

L i=l

see 3

I

that

(322)) =

=

2.

o

1)

I

n1

=

3.

Since N

ff(e1(322))

Also

=

V1

11(322)

= =

that

2(3 - 1) + 3(2 - 1) + 3(2 - 1)

163

=

#(3)

2,

=

3,

n2 V2

6, we can verify

(P5.4)

Problem:

(including

(T5.9)

T

Theorem:

denote

coset

Let

7.

Since We

=

information

p

and q

have

developed

the equation

denote

p

that map

pole

q.

to

points

Then

q22

about for q22.

in L5.8

C-equivalent

of C

of cosets and related to q,

p is C-equivalent shall

q.

you

verify

T

and let is a left

.

P

A discussion

q.

the

to P5.2)

the set of all elements

of C

Proof:

t(p)

Using

the solution

show

Since

g(p)

=

there

T = gC ' p

that

topics

can be found

exists

9

Let t

C

E

T.

E

such

in Appendix that

=

g(p)

By definition

T,

of

q, we have

9

-1

-1

9

t(p)

(q)

p -1

9

Hence

t

C.

E

P

h = gk where

Then

k

Therefore

t

C

Hence

E

p

.

gC

E

h(p)

.

P

=

Conversely,

h

suppose

gC

E

.

P

gk(p) g(p) q

Hence

T.

h E

T

Consequently

One consequence

of T5.9

o

is that

if C

such that fI(C)

> 1 and p is a pole point

correspondence

between

C-equivalent be the C

p

in C.

the

same

cosets

to p.

number

Recall

of C

established

(see the of elements

partition p that

In the

=

3,

V2

case

=

VJ

=

if p

C

of 322,

of

of C, C

ei(C),

E

in Ci(C),

points

#(C)

V1

cosets

Hence,

of pole

number

the

proof

n.

p

C

p

then there

and

the

, that

cosets

pole

which number

that

each

=

is defined

to

of cosets

of

coset

of

p

has

Since

C

the

I

v.

each having

niV i for each 1

N

are

elements

we have

I

we observe

2 and that

that

is v . elements.

I

N

is a one-to-one

n, i

point group

points

the

then equals

of TA7.l3)

as does

into

is a finite proper

6

=

s

that since

nivi

164

i

s

n1

t

=

in all three

(5.1)

2,

n2

=

cases.

nJ

3 and

(P5.5)

N for each 1 :5 i :5 3 in the case

Show that n » .

Problem:

I

I

of

q22. (T5.10)

Theorem:

Let t denote the number of equivalence classes of pole

points of a finite proper point group C where N = if (C)

> 1.

Then t = 2

or 3 and (1)

if

2,

(2)

if

3,

C C

is a monaxial

group and

is a polyaxial

group such that

and 2/ClJ_v_1 + 1/v2

flCC)

Proof:

+ l/vJ

-

1)

By L5.8, we have

2 (N -

and by (5.1), we have N

=

t E i=l

1)

n.(v. I

-

the left side of (5.2) by N

Dividing

ni~'

and the right side of (5.2) by nivi,

(5.2)

1)

I

we obtain t

2 -

2/N

E

=

(1 -

I

Since vi is the order of the stabilizer

E

of a pole point, vi ~ 2.

t

t (1 -

(1/v.))

i=l

~

E

!)

(1 -

i=l

I

(5.3)

l/v.)

i=l

=

t E

(!)

Hence

t/2

i=l

Also, since N ~ 2, 2 > 2 - 2/N Hence, from (5.3) 2 > t/2

Hence 4 > t.

Therefore,

t can only equal 1, 2 or 3.

We can thus conclude

that there are no rotation groups having more than 3 equivalence of pole points.

We now examine each of these three cases for the value

of t. Case where t

=

1:

In this case, Equation t

2 -

classes

2/N

E i=l

(1 - l/v.) I

165

(5.3) becomes

or 1 - 2/N

The

left member of this equation

is always nonnegative

but the right member is always negative tradiction.

Therefore,

t cannot

because

V1

~

N ~

because

2, which

2,

is a con-

equal 1, from which we conclude

that

P(C) must contain more than one equivalence class of pole points. Case where

t

=

2:

In this case Equation

(5.3) becomes

2

2/N

2 -

L i=l

=

(l - l/v.)

By a little algebraic manipultion

From Equation

+ (l - 1/v2)

(l - 1/v1)

I

we find that

(5.1) we have that N/v

ni,

i

and so the above expression

simplifies to

Because n1

and n2

are positive

integers,

is the only possible solution. we have two equivalence together

classes

we conclude

that n1

pole points

Therefore,

=

consisting of one pole point each.

monaxial

groups given

=

1 2,

Al-

one and only

those groups with two equivalence

must be the proper

=

n2

for a rotation group with t

C has a total of two pole points, which defines

one rotation axis. of

Hence,

classes

in Table 4.2.

The number of elements in each of these possible monaxial groups is equal to the order of the rotation axis, #(C)

Case where t = 3: 2 -

=

V1

In this case Equation

=

V2'

(5.3) expands to

2/N

Rewriting this result we see that

1 +

2/N

Since N ~ 2, it follows that 1

(5.4)

+ 2/N

> 1 and so

Table

Symbol for G =

5.1:

III(C)

Possible

finite

II(C1(C))

= N

proper

III(C'(C))

= N/Vl

"1\)2\»)

polyaxial

point

groups.

Group

I II(C,(C))

= N/vz

= N/v,

Name

I 2

222

4

2

2

322

6

2

3

3

422

8

2

4

4

522*

10

2

5

5

622

12

2

6

6

n22*"

2n

2

n

n

Dihedral

332

12

4

4

6

Tetrahedral

432

24

6

8

12

Octahedral

532*

60

12

20

30

Icosahedral

~.. non -crys

t

a l Iog r aph Lc

*.,,: non-crystallographic

when

The group 332 is usually designated

n > 6.

designated

by 23 and the

group 532 is usually

by 235.

(5.5)

Solving for

N

in (5.4) we obtain

Using part V1 ~ V2 ~ VJ,

(2) of T5. 10 where,

we

o

2/(1/V1 + 1/v2 + l/vJ - 1)

N

fI(C)

shall

polyaxial point groups.

construct

for

all

convenience,

of

the

we

possible

assume finite

that proper

Note that if VJ > 2, then each of the fractions

l/vi would be less than or equal to 1/3 for each i and so (5.5) would not be satisfied. or V2

equal

=

to

2 or 3.

(5.5).

Hence VJ

t

=

2.

If V2 > 3, then 1/V1 + 1/v2

and so (5.5) would

Suppose

V2

=

VJ

These groups, denoted

=

2.

fI(n22)

=

3 and VJ

=

is less than Therefore,

Then any value of V1 > 1 would satisfy

n22 when n

=

V1, are called the dihedral

of which there are an infinite number.

groups

If V2

again not be satisfied.

Using (T5.10) we see that

= 2n

2, then if V1 > 5, (5.5) would not be satisfied.

the only groups of this type are 332, 432 and 532. finite proper point groups are recorded Table 5.1.

Hence

All of these possible Note that we have

~

Figure 5.2:

I

dicular

The

2-fold

ordinate

system

[010]2 along

orientation

axes

in

of

222.

v..ith

2

Each

2-fold

b.

the

These

lying

three

axes

along axis

mutually

define c,

is

pe rpen-

a natural

[100]2

along

coa

and

by a diad

represented

symbol.

Figure 5.3:

Multiplication

table for Mp(222).

Mp(222)

Mp(1)

Mp(2)

M ([100]2)

Mp(l)

Mp(l)

Mp(2)

M

Mp(2)

Mp(2)

Mp(1)

M ([010]2)

M ([100]2)

M ([100]2) P M ([010]2) P

p

M ([010]2) p

M ([010]2)

p

p

p

(! 100]2)

M ([010]2) p

M ([100)2)

p

M ([010]2) p

M ([100]2) p

yet to show that each of these possibilities

p

Mp(1)

Mp(2)

Mp(2)

Mp(1)

actually

occurs

as a point

group.

CONSTRUCTION

In Appendix dicular

6 we proved,

to each 2-fold

group must intersect

n22 is actually

OF THE DIHEDRAL

for n22, that the n-fold

symmetry

axis and adjacent

at an angle of 180jn

a group, we shall define

the elements of n22 as described then form the multiplication

Hence the metrical

2-fold

axis is perpenaxes

in

this

(TA6.1).

To confirm that each

a basis

of S for each, write

in TA6.1 with respect

to this basis and

table to check closure.

The construction of 222: Since ular, we define P = {a,b,c} to tually perpendicular

GROUPS

the 2-fold axes are mutually be a basis where a, band

perpendic-

c are also mu-

such that each lies along an axis (see Figure 5.2). matrix G for

P

is

168

With respect to this choice of basis, the half-turns

[010]2, 2 along a, b, c, respectively.

are denoted

[100]2,

By Table 5.1, 222 can only have

4 elements and so we conjecture that

is a group.

The matrix representation

of each of these can be found em-

ploying the approach used for 322 in Chapter 3.

These matrices

are described

in Table

group, we form the mUltiplication Since no new entries resulted

Since the mapping

from Mp(222)

the operation,

that 222

To confirm

5.2.

is

a

table of Mpl222) shown in Figure 5.3.

in the formation of the table,

closed under matrix multiplication

preserves

Hence

Mp(222) is

and, since it is finite, it is a group. to 222 that maps Mp(a)

the multiplication

tion can be obtained by deleting the Mp(

to a for each a

table of 222 under

composi-

) for each element (Figure 4.2).

Hence, 222 is a group.

The construction p

=

of 322:

As in Chapter

of the two-fold axes and b = 3(a) G of

3 (see Figure 3.10), we choose

{a,b,c} where c coincides with the 3-fold axis, a coincides with one

P

(Figure 5.4).

Hence the metrical matrix

is

G

911

o Since 60

=

180/3, there are two-fold

the one coinciding with a.

axes at 600 intervals

Hence, a two-fold

axis lies along b.

that this is the same basis that was used for the monaxial 6,

6

and 6/m.

With respect to P,

169

starting with Recall

groups 3, 3,

Table 5.2:

The nonzero entries of the matrix representations

II ([i11]3) p

ll1 = l'2 = -1; 1"

IIp(2)

II ([100]2

P

11([ill]

= 1

p

II ([iil]3-1

p

1]1:;;:

P

IIp ([i01]2

-1 3)

1;

=

(a,b.c).

=

112

III

=-1

1J2

=

l21

=

III

(1)

= 1; 121

=

1)2 =-1

1;

=-1

IIp ([iil]3 ) :

)

H ([010] p 2)

II ([011]2

=

tlJ

for

Mp(a)

rotation isometries a groups 1, 2, 4, 222, 422, 23 and 432 for P

tll

:;

1;

t12

)

IIP ([li1]3 ) :

=-1

)

II ([lil]3-1

)

I1p(4)

P

)

121

=

1"

=

=

1; lIZ

-1

II ([Oil] p 2)

In 322,

Figures

3.14

and 3.15,

respectively,

the

multiplication

are displayed.

From these

tables tables,

for M (322) and O we observed that

322 is a group.

The construction c coincides 4(a).

of q22:

with the 4-fold

The metrical

one coinciding

The elements

axis,

matrix G for

Since 45 = 180/4, there the

As in Chapter 4, we choose P a

P

are two-fold

with a.

is

along

{a,b,c}

a 2-fold

axis

where

and b

=

is

axes at 450 intervals

Hence, there

of q22 with respect

=

to P,

5.5) 170

is

a two-fold

starting axis

given in P3.14 are

with

along

(see

b.

Figure

Table

5.3: for

Non-zero

the

entries

rotation

of the

isometries

matrix

representations

ex of point

p = (a,b,c)

P

and

groups

*

3,

Mp(ex) and

6, 322.

and

Mp*(ex)

622 for

= (a * ,b * ,c * ).

-I

2.)]

"-,J :;;

Np(3)

-1

ill

:;;

Np(3-1) 1; 2.)]

:;; 1;

t12

2.11 ;; i,,:;;

Np(6)

:;; -1

2.)]

ill;:

2.21

1;

-1

:;;-1

:;::;.2.22

= 2.21 :;.-1

t11

i,,:;;

:;;

:;;

1;

:;;-1

2.12

1; 2.21=-1

~lp"(l)

~lp*(2)

:

ill:;;

t22

:;; ill:;;

til:;;

t22

:;;

N ,.,([100]2) P

:

N ,,([210]2) P ~I ,,([110]2) P

:

N ,,([120]2) p

-1;

I

1

ill

::;;:. 1

til:;;

1;

2.21

til:;::

i12

:;;

1; 2.22:;;

1; t]]

;;;;

2.21

=

2.12

:;;

t21

:;;

2.22

= 1;

2.22

til:;;

:;;

;:

2.]]

:;;-1

2.22:;;

~I ,([110]2) p

2.21

~lp"(3)

-1

2.)]

N ,.,([010]2) P

~lp"'(3-1 )

~lp"'(6)

:;;-1

Figure The

5.4:

3- fold

three

at

of

..!&

I

a diad

a

the

:;;

2.12

so

axis

symbol.

171

angle

axes that

is

:;;

::;;

i,,:;;

of

the

and

The

axis,

b is

represented

i12

:;;

-1

-1

axes to

axes

basis

to

a

in

01012

[

and

of

plane

C is lie

along

dealong

another

I0012

3 parallels

a triad

the

to

lie

322.

in plane

vector

defined

+ band by

:;;-1

rotation

a is

Thus, a

12.22

-1

1; 2.21:;;

defined

b = 3(a). parallels

1;

2-fold

600•

of

;:;:

:;;

perpendicular

adjacent

3-fold

2.21

ill

is

2.12

til:;;

:;; 1;

:;;

group

vith

the

[110/2

a

J-fold

an

along 2-fold

to

parallels The

of

at lie

the

1200

::;; i12

2.11

axes

intersecting to

2.)3

:;::

:;;-1

-"-12 :;; 2.11

:;; i,,:;;-1

1;

The orientation axis

2-fold

fined

£21

til:;;

:;;t12

:;; i'l:;;

t12

~I "'(6-1) P

:;;-1

tll

one

t21

1;

c, { parallels

each

2-fold

b . by

Figure

5.5

The orientation

The 4-fold four

axis

2-fold

axes

intersecting along at

a

5.6: The

2-fold

axes

at 30°. a is as c,

The orientation 6-fold

in 322 IOO I2,

parallels

to

C is

lie

of

the rotation

perpendicular 2-folds

defined

along

to

one of

to the

lie

along

the

Ie l s -a + b.

I20

+ b,

l2

{

The

parallels

6-fold

by diad

is

+ 2b

and

the

= 4(a). as

in

-a + b. 2-fold

axes,

is

defined

is

defined

a to

lie

The 4- fo ld by a diad

along

parallels

l

{110 2

222,

=

one

axis

is

to

{I0012 a

symbol.

..........

a

•

table for Hp(q22) and (q22) were found in P3.15.

An

table of q22 shows that q22 is closed

and hence is a group.

with a.

Thus,

Since

axes at intervals of 300 starting with the there

is a two-fold

axis along b.

The

elements of 622 (see Figure 5.6) are

622

(P5.G)

Problem:

Find all of the matrices

in ~lp(622). Confirm your re-

sults with those given in Table 5.3.

(P5.7)

Problem:

that Hp(622)

Prepare a multiplication

table for ~lp(622) and observe

is closed under multiplication,

172

demonstrating

+

represented

-e If _q~I~"; A

and

lie

another

paral-

a hexad

of

plane

I

180/6, there are two-fold coinciding

c,

parallels

The construction of 622: We choose the basis used for 322 for 622. 30

the

.-:-.

and {210l2

(110l2 by

of the multiplication

under composition

in

vector

'122.

in

a plane

axes

defined Thus,'

to

six

6 parallels

in 322 and

of

6-fold

and b

represented

each

basis

axes

rotation

2-fold

and b is

symbols.

The multiplication examination

axis

a

The

b

the

perpendicular

intersecting

Hence, as

that

is

adjacent

disposed

para lIe Is

plane

plane

2-folds

with

2-folds

so

are

of

group

axes of group a

in

at 120· to a so that b = 3(0). 110 { l2 and {Ol012 are disposed 2a

2-folds

is

adjacent

The vector

defined

{

the

axis

with

a

I2

(11

the

by a tetrad

I

Figure

to

{~'012 0

and

Ar

622.

90·

of

the

45°.

at

one

and

of

that 622 is

b

a group.

(P5.8)

Problem:

For 522 no basis exists that gives all of the pole points

and matrices with integer entries.

Therefore,

a cartesian basis is used

k is along the five-fold axis and i is along one of the two-fold

where axes.

Find the matrices

showing that Me(522)

in M (522)

and form the multiplication

e

is closed and hence 522 is a group.

Note that 522 is not a crystallographic a lattice into self-coincidence. that the representation pings consist

entirely

crystallographic

group and hence does not map

This is why no basis can be found such

of the pole points and the matrices of the mapof irttegers.

point groups.

all of this type.

table

This fact

In particular,

is true

for

all

non-

n22 groups with n > 6 are

However, using a cartesian basis as in P5.8, each of

these can be shown to be groups.

CONSTRUCTION

In the construction

OF THE

pole points are constrained

the pole point

of the constituent

computations.

GROUPS

rotations.

Since the

to be on a unit ball, the triple representing

can be somewhat

[i3/3,i3/3,i3/3]t.

[l11], we shall use

AXIAL

of these groups we shall need to map pole points

using the matrix representation

a given pole point

CUBIC

complicated.

For example,

As this pole point

[lll]t to represent

consider

lies on the zone

the pole point to simplify

the

We shall see that each pole can be easily represented

in

this manner.

(E5.11) of

222,

Example: form

the

Using the zone symbols equivalence

classes

to represent

of

the pole points

pole points with respect

to

222-equivalence.

Solution: points.

Since 222 has three 2-fold Recall that by the way the basis

rotation

P

=

axes,

it has

six

pole

{a,b,c} for 222 was defined,

the triples for these pole points on the unit ball Bare

{a/a,-a/a,b/b,-b/b,c/c,-c/c} The representatives

formed from the zone symbol associated with these pole

173

points

The

are

equivalence

I

{g(P)

9

E

class

222}.

of

Starting

a

given

[alp'

with

pole the

p

point

equivalence

is

the

[a]

class

set

a

of

is

[a]

I

{Mp(g)[a]p

9 E 222}

Similarly,

and

This

is consistent

serve

that

classes

the

with

pole

form

the

Construction satisfied

occur

Using

of

are

your

partitioned

symbols

pole

third-turns

whose

(recall

shall

discover

Once these

composition

composition

together

and hence

which would

violate

in 332 whose

rotations

by noting

with

we

part

third-turn the

would

have

«3:3) «3:2)

would

-1

group

Hence, there exist In Appendix

(1/3)" 70.530, and -1 r: = cos (>'3/3)" 54.740

174

about

be a closed

2, then

= cos

332 not

must been

be es-

that must and

show

exist

two

then the set of all third

a polyaxial

is a half-turn.

=

does

is a third-turn

identity

have

angles

construct

of to

5.1.

conditions

conditions

if there

points

respect

to Figure

what

and then

that

(2) of TS.10.

c6mposition

that when 33

we ob-

equivalence

with

the interaxial

is a half-turn,

that the inverse

end of the axis)

3

the pole

points

by referring

we shall use them to determine

We begin

5.1 where

into

to represent

of

answer

We

the generating

in Table

c

classes

332(=23):

given

each.

if 332 is to be a group.

between

and PA6.5)

of 222

the zone

Confirm

it is a group.

turns

points

information

equivalence

322-equivalence.

tablished,

the

t~o pole points

(P5.9) Problem: 322,

with

of the

the other set

under

form

333

two third-turns

6, we showed

(PA6.2

Figure

5.7:

332=23

given

The placement «3:3)

that

cos -1(13/3):::::54.

74°.

are mutually

a

is

of

perpendicular

oriented the

C is

at

3-fold

cos -1(13/3)

with

choice

of

3-fo1d a -

A convenient basis

P

=

{a,b,c}

along

b

a

of

the

e

is the

=

= J

lie

along

to

the

designated

this

to

each

angle

of

an

angle

of

other.

For

designated

/11113

and

this

2.

that

the along

(11113.

of these

axes is the basis whose metrical matrix G is

where we orient the two-fold axis along the zone [001], and the three-fold axes along the zones [lll] and [111]. shown

in the

next problem.

The fact that this can be dOne is

This placement

of axes is illustrated

in

Figure 5.7.

(P5.10)

Problem:

Show that

« [111]: [001])

«[111]:

[001])

and that « [111]: [111])

To find the remaining search

cos

-1

(1/3) ~ 70.530•

axes belonging

for the remaining

to the rotations

pole points.

point, the stabilizer of that pole point is determined and we add these to our list of elements of 332. that there

of 332, we shall

Each time we obtain

a new pole

according to T5.5

By Table 5.1, we know

is a total of 14 pole points, 8 of which are associated

with

3-fold axes and 6 with 2-fold axes. To begin with we have the following associated

with 3-fold axes in Figure 5.7:

representation

of the half-turn

list representing {[111],[11lj}.

about [001] is

175

=

axis,

an

is

the orientation

=

group

2-fold

respect

and

c

for

«3:2)

90° and a = b the

at

3-folds

along

for

with

oriented

respect

+ b + c is

for describing

to

2-fold

+ C is designated

rotations and

chosen

cos -le13/3) b

of

with

basis.

=

with a

and

one

generating

vectors

defined

angle axes

cos-1C-I3/3)=125.3°

a

the

The basis

By convention,

c.

of

= cos-I(l/3)=70.53°

pole points The matrix

Hence

also represent

pole points.

with 3-fold axes.

By T5.5, these pole points

point associated with an n-fold axis, then point associated

with the n-fold

8 pole points.

While [111] and takes

place,

[11113.

[111], ll1lj,

[iii]

we

In general,when

Applying Mp([111]3)

also represents

a pole

Hence we have found all of the

l11l], [IiI],

the

monaxial

possible,we

are

[11i], [111], [Ill]}

group

along

this

direction

as

shall choose the zone symbol with a

Hence, we have found the following 3-fold axes:

[11113

=

{1,[111]3,[111]3-1}

[11113

=

{1,[111]3,[111]3-1}

[11113

=

{1,[111]3,[111]3-1}

[11113

= {1,[111]3,[111]3-1}

To find the 6 pole points

[111]3

[uvw]

a pole

both designate the same axis about which a 3-fold

denote

positive third component.

matrix for

axis.

Their zone representatives

{[Ill],

are associated

It is helpful to note that if [uvw] represents

associated

with the half-turns,

and apply it to the pole point represented

to this new pole point we obtain

176

[010].

we write the by [001].

Including

[ill]

)io] - ~-

0-

-G-'

)\

[010)

[IIi]

[Iii]

I

Figure

[ooi]

the negatives

of these,

with the two-fold

we have

5.8:

A diagram

the following

of the rotation

axes

6 pole points

for 23.

associated

axes:

{[100],

[010], [001], [IOO] , [010], [DOl]}

Hence we have the following

two fold axes in 332,

{1,[100]2} {1, [010]2} 2 As predicted

in Table

turns, 8 third-turns

23

=

332

{1,2}

5.1, we have found a total

or negative

third-turns

12 rotations,

and an identity.

3 halfHence

= {1,2,[100]2,[010]2,[111]3,[111]3-1

[111]3,[111]3-1,[111]3,[li1]3-1,[111]3,[111]3-1} Figure 5.8 shows the placement

of the rotation axes of 23 in terms of the

basis p.

177

(P5.11)

Find ~!p(23).

Problem:

Check your results

with those given in

Table 5.2.

We can show that Mp(23) table and observing

is a group,

closure.

This is a tedious but straightforward

=

and when completed will show that 23

of q32:

Construction

by forming its multiplication

The strategy

for showing that there

is a group of

the form q32 will be similar to that followed in the case of 23. TA6.4

it

«4:3)

~ 103.84

can

be 0 •

shown

that

However,

turn angle of 156.094 I

composition

43

at -103.840,

in q32.

sition 43 in q32 must yield either a 4 or a 2.

the fixed quarter-turn. tinct rotations. compositions 43

=

2.

From Appendix

cos

cannot be a quarter-turn 0

one quarter-turn

of the form 43 using

some of the 43

Let 4 and 3 be such that the composition

/3 cos -l (3/3) 12 cos -1 (2/2)

«3:2)

163.158

rotation has a

6 we see that

«4:2)

turns whose

then

law these must all be dis-

Since there are only 6 quarter-turns,

must be half-turns.

happens,

Hence, each compo-

Fixing

we obtain 8 rotations

By the cancellation

«4:3)

There

3

the resulting

which is impossible

4, since q32 has 8 third-turns,

=

Using

423 is formed about the

when the composition

same pair of axes intersecting 0

if

task

332 is a group.

axes intersect

at

-1

16 (6/3)

~ 54.740

,

= 450

along the 2-fold 0

a 45

which does not exist in q32.

and

~ 35.260

compose

axis since two quarter-

to

yield

a rotation

of

Using the same basis as in 23, we

place the 4-fold axis along [001], the 3-fold along [111] and the 2-fold along [101].

(P5.12)

(P5.13)

Problem:

Problem:

Show that -1 rz: (v2/2)

«[001]:

[101])

cos

«[101]:

[111])

cos-1(i6/3)

= 450 and ~ 35.260

Find the 8 pole points belonging

to the 3-fold axes.

(The answer is the same as found for 23 except they all occur in the same q32-equivalence

class).

178

~ ...J, [i~T] \

f

I

l\

[101] [001]

(P5.14) Problem:

~

"

........

[iii]

[all]

[IIi]

Figure

5.9:

A diagram

of

Show that the zone representations

the

rotation

axes

of

432.

of the pole points

associated with the 2-fold axes are {[101], [011], [iOl],

[Oil],

[110], [i10],

[ioi],

[Oli],

[101], [011], [lio],

(P5.15) Problem:

Show that the zone representations

[lio]}

of the pole points

associated with the 4-fold axes are {[001], [100],1010],

(P5.16) Problem:

Enumerate

(P5.17) Problem:

Enumerate

100i],

I icoj

the rotations

the rotations

,

loio]}

in each of the

3-fold

axes

in each of the 4-fold axes q,

[l001q and [0101q in q32. (P5.18) Problem:

Enumerate

the rotations

(P5.19) Problem:

Enumerate

the rotations

the distinct

rotations

in q32 by collecting

found in P5.16, P5.17, and P5.18.

of 23, one need only form a multiplication closure to conclude that q32 is a group. axes of q32 defined

in each of the 2-fold

together

As in the case

table for ~ID(q32) and observe Figure

in terms of its basis vectors.

179

axes,

5.9 shows the rotation

Table

5.4:

The 32 crystallographic

as derived

from the proper

point

groups

crystallographic

The 21

improper

and their point

orders

groups.

crystallographic

point groups I

The 11 proper

Halving

crystallographic

Groups

point

I

containing

groups

G

I

II(G)

I

H

G U

none 2

I

i

Not containing

J

(centrosymmetrical)

2

ct

II(C

U ~

-2--1

1

I

(C \ H)i)

4

m none

3

j

6

Q

4

2

Qlm

8

II(H

U (G \ H)i)

none

21m

3

none

(H U

Ii

4

6

6

3

61m

12

6

6

222

4

2

mmm

8

mm2

4

322

6

3

321m

12

3mm

6

Q22

8

Q

Qlmmm

16

Qmm

8

622

121

61mmm

24

222

6 322

332 = 23

121 24

Q32

7i2m

8

6mm

12

62m

12

none

21 ~3

24

:one

23

Qlm321m

48

Q3m

The noncrystallographic

group 532

=

24

235 will be discussed

at the end

of the chapter.

CONSTRUCTION

OF THE

IMPROPER

CRYSTALLOGRAPHIC

POINT

GROUPS

The construction

of these groups

the Improper Point Group Generating 5.4, each of the proper (1) of T4. 28 yields

will be accomplished

Theorem T4.28. point

groups

one centrosymmetric

point

group C U Ci,

(2) of T4.28 requires

crystallographic

point group.

are listed. listed

Part in

The application

a list of the halving groups of each proper Since any subgroup

point group is again a crystallographic can be found in column one.

applying

In column one of Table

crystallographic

column 4, from each proper point group C from column 1. of part

by

of a crystallographic

point group, these halving groups

For example,

322 has 6 elements.

Examining

column two, we see that the group 3 has 3 elements and hence is a candidate to be a halving

group.

Since 3 is a subset of 322, it is a halving group

180

and hence is listed opposite 322 in column 3. the candidates 6.

Since 23 has 12 elements,

for the halving groups of 23 are 322 and 6, both of order

However, neither of these are subsets of 23 and hence 23 has no halving

groups.

A similar analysis of the remaining

the results shown in column 3. halving

groups from column one gives

Applying part (2) of T4.28 to each

C

with

groups, we obtain the results shown in column 6.

The name given to each of the improper crystallographic is derived

from the names of the resulting

along the axes of the generators

monaxial

of the possible

point groups,

groups

that occur

proper crystallographic

groups.

(E5.12)

Example:

of 322 are 3,

The generators elements

Hence

C

Consider

= 322.

[100]

2,

Then

[110]

2.

C U

ct

322i

= 322 U

.

In 322 U 3221, the symmetry

lie along the three zones: -

-1

-

({1,3,3

[100]

({1 i. [100]2, [l00]m) which is the monaxial group 21m)

[ 110]

({1,i,[110]2,[110]m}

the full symbol

,i,3,3

--1

[001]

which

is the monaxial

group 21m)

for 322 U 322i is 3(2Im) (21m).

Haugu Ln symbol

for this group

the elements

3

of

} which is the monaxial group 3)

is 3m.

The

Hermann-

This symbol is reasonable

and any of the mirrors generate the remaining

since

elements

of the group.

As indicated

in Column

3,

H

3 is a halving

group

Hence, we construct

H U (C \ H)i =

3 U (322 \ 3)i

{1,3,3-1, [100]m, [110]m, [010]m}

181

of

C

322.

Along

[001],

associated

we have with

3,

perpendicular

mirror

Hence,

Note

the

that

[110] full

3 and

lye have symbol

planes

3mm.

is

m generates

either

Find

C = 422.

Consider \ 222)i

222 U (422

Solution:

222

q22 U 422i and show that

Find

its

full

Hermann-Mauguin

(qlm) (21m) (21m).

is

(E5.13) Example:

We begin

The halving

and derive

by generating

the

its

full

elements

of

groups

of

Hermann-Mauguin

the

q and

Care

symbol.

set

U (422 \ 222)i

As a 4-rotoinversion

{1,[100]2},

parallels

4 U (422)

{1,4,2,4-1},

axis, [100]

Hermann-Mauguin

(P5.21) Problem: for

and

0

(P5.20) Problem:

the

3m.

is

[100j

group.

symbol

222.

m.

operations

The Hermann -Naugu i.n symbol the

to

symbol

parallels

and a mirror of

Construct

this

plane

group

is

[001],

a 2-fold

is perpendicular

axis, to

[110],

i2m.

and determine

the

full

C = 622,

show that

o

Hermann -Naugu i n symbol

\ 4)i.

(P5.22) Problem:

Given

that

622

U 622i = 61mmm

{1,6,3,2,3-1 ,6-1 ,[100]2,[210]2,[110]2,[120]2,[010]2,[110]2, . -6, 3 " m -3-1, -6-1, [100]m, [210] m, [110]m, [120]m , [010]m, [110]}m

I,

(P5.23) Problem:

Given

that

C

=

432,

show that

q32 U q32i = (4Im)3(2im)

182

i ,4,m,'4-1,

[100]4, [100]m, [100]'4-1,

Using the procedures one can construct According

followed

[010]'4, [010]m, [010]'4-1,

in the preceeding

examples

to T4.28, these are all of the possible improper point groups.

Hence, in summary, we have found 11 proper crystallographic 11

centrosymmetric

centrosymmetric

crystallographic

molecule

Problem: displayed

by

point groups, and

groups

10

non-

point groups for a total of 32

point groups.

Determine the point symmetry C of the tricyclosiloxane in Figure

1.2.

and show that the coordinates (interchanged)

point

improper crystallographic

possible crystallographic

(P5.24)

and problems,

all of the improper point groups shown in Table 5.4.

Find the elements of th.e set ~le(C)

of the atoms in Table

the elements

molecule that are C-equivalent

of Ne(C). to 01,

1.2 are

Ascertain

Hl, and Silo

permuted

the atoms in the

Determine

which of

the atoms are on special positions.

(P5.25)

Problem:

are presented C.

Stereoscopic

pair drawings of C-equivalent

ellipsoids

in Figure 5.10 for each of the 19 polyaxial point groups

Study each of these drawings and confirm the point symmetry of each.

THE CRYSTAL

SYSTEMS

It is customary to organize the 32 crystallographic classes according to geometrical

considerations.

point groups into

The geometry

involved

in each point group C is used to determine a natural basis with respect to which the matrix representation

of the elements of C are simply written

(so that all of the entries of the matrices are either 0, 1 or -1).

toto, we use only 6 different bases.

In

In Table 5.5 the metrical matrix

for each of these bases is listed together with a list of all of the point groups using the basis.

This gives rise to the 6 crystal

183

systems whose

names are given in column 3. the monoclinic

The metrical matrix given in Table 5.5 for

system is that for the so called first setting where c is

chosen along the axis of order 2.

In the second setting, which is more

commonly used, b is chosen along the axis of order

2, resulting

metrical matrix Figure

5.10:

ellipsoids

Stereoscopic

pair drawings

for each of the 19 polyaxial

222

-~/

I~

32

184

of C-equivalent pOint groups

C.

in the

422

622 ----

\~l?\~~ /~

!1~

23 185

432

~! ~v J~ J~~ mmm

~v ~\V~ /l~ /]~ 3m 186

4/mmm

6/mmm

2/m 3

187

-

4/m 32/m

mm2

~/

~I \

\' 3m

188

4mm

42m

6mm 189

62m

43m

190

Table

5.5:

METRICAL

Metrical

Matrices

for the

MATRICES

912 912

crystallographic

POINT

9"

92'

913

groups.

CRYSTAL

CROUPS

9"j

922

point

SYSTEM

I,T

Triclinic

2,m,2Im

Monoclinic

911 [

9'2

o

222,mm2,mmm

o

Orthorhombic

q ,ii,qlm qn ,qmm,q2m

gl1

Tetragonal

u Lmmm

-

3,3,32,3m,3m

911 [

6 ,6,6/m,622

-9~ 1/2

Hexagonal

,6mm

62m,6/mmm

0

23 ,m3 ,q32

o

911

43m,m3m

o

0

911

o

Cubic

glJ] o

o Schoenflies Symbols: groups used here,

Schoenflies Table

there

symbolism, 5.6.

e1,e2"",en groups

1,3,

J

Besides the Hermann-Mauguin

between the Schoenflies in

gJ

.

The

is

another

important

symbolism for the point symbolism,

which is widely used by chemists. symbols and the Hermann-Mauguin proper

in the Schoenflies

cyclic

groups

symbolism

the

symbols are given

1,2, ... ,n

whereas

called

The equivalence

are

the improper

denoted cyclic

'ii, {; are denoted c., e3i, es' Sq' e3h, respectively. i signifies that the group contains the inversion isometry

m,

The subscript

and h signifies

that it contains

to a rotation axis. for reflection

a horizontal

mirror plane perpendicular

The subscript s stands for the German word Spiegelung

and signifies that

monaxial groups 21m, 31m, ... ,nlm

es =

m.

The improper

are symbolized

as

centrosymmetric

e2h, e3h, ... ,enh.

Table 5.6:

5choenflies

(5)

and Hermann-Mauguin

point group

HM

I

5

HM

The

noted

21m

C2h

3

C,

3

C3i

q

C.

qlm

Cqh

Ii

s.

61m

C6h

6

C3h

mmm

D2h

mm2

C2v

32

0,

3m

D3d

3m

C3v

q22

D.

I

a tmmm

0,

23

T

q32

a

I

m3m

235

I

I

m35

61mmm

I

ii2m

D2d

umm

Cqv

D6h

62m

D3h

Th

6mm

C6v

°h

ii3m

Dqh

m3

with

vertical

Td

Ih

noncentrosymmetric

,nmm

monaxial

mirror

planes

2mm(=mm2)

groups are

whereas

Th'

the improper

are denoted 02h'

the presence

of diagonal

The tetrahedral

Td and the octahedral

denoted 0 and 0h'

respectively.

m3S are symbolized

as I and I , h

icosahedral V

J

point groups:

= 2.

crystals because

Regular

dihedral

03d'

02d'

mmm,

groups

0qh'

reflection

,

e ' 2v are de-

symbolized

The proper dihedral groups 222, 322(=32),oo.,n22

° ,°

= 3 and

I

s

O2

adj acent 2-fold axes.

V2

C

C,

script d denotes

The

m

6

622

T,

I

222

,On 2 3"" u Immm , 6/mmm

denoted

5

I

C2

e3v,oo"env' 42m,

HM

I

C.

2

improper

3mm, ...

I

I

5

t

C1

(HM)

symbols.

3m,

06h'

The sub-

planes

bisecting

cubic groups 23, m3,

q3m

cubic groups 432

m3m

and

are are

Finally, the icosahedral groups 235 and respectively.

These point groups occur when icoshedral

arrangements

are

not

they require a five-fold axis of symmetry,

Vl

=

found

5, as

and b~cause

a set of regular icosahedra cannot be packed together so as to fill space. However,

the combination

skutterudite,

CO.AS12,

of an

icosahedral

unit

as the

AS12

unit

in

with interstitial Co atoms can fill space to form

192

21m3

a cubic crystal with dominates

point symmetry.

Also, an icosahedral

in the structure types (allotropes) of crystalline

B12 icosahedra are packed together inefficiently voids.

Al though the B 12 groups

On the other hand, evidence

geometry and cation-anion

to explain the "super" dense packing ture types

(Moore, 1976).

alloy

AI.Mn

in the glaserite,

The icosahedron

icosahedral

an

electron- diffraction

the presence

and so the periodicity

of a five-fold

struc-

in the world of

with icosahedral

symmetry has added new importance

Of course,

KJNa(S04)2

also occurs

The recent discovery by Schectmann

yields

for

mixed layer packings has been used

viruses where virus particles pack together metry (PA6.4).

boron where

leaving regularly spaced

in boron can be viewed as being close-

packed, only 37% of space is occupied. icosahedral

unit

point sym-

et al. (1984) that the

record that conforms

with

to this type of symmetry.

axis of symmetry violates

of the atoms in the alloy cannot

T4.17

be represented

by a lattice.

Until now solids have been classified as either crystalline

or amorphous.

However, the work on AI.~ln suggests a new state of matter

called quasi-crystals.

In such substances

the atoms are believed

to be

arranged in rows as in a crystal but the spacing within and between these rows is believed to exhibit a more complicated clustering intervals.

However,

maintained

as

scientists

as the directions

in a crystal,

orientational

matter.

rapport.

The result is a

of atoms that is repeated over and over again but at irregular

rather than translational

and mathematicians Some

argue

that

of the bonds

quasi-crystals

atoms

it is clearly

important

this

in such matter

pyritohedron,

Finally,

the

fact

that we devote

that

new

skutterudite

some

it may have passed through a quasi-crystalline crystallization,

material form

of

Whatever

time

to

the

of substances

crystallizes

which closely resembles the dual of an icosahedron,

its point symmetry is a subgroup of the icosahedral

exhibit

are quasiperiodic ordering.

icosahedral groups so that we may gain a better appreciation like AI.Mn.

to

Physicists,

studying

whereas others argue that they exhibit incommensurate the outcome,

in each cluster are

believed

symmetry.

are actively

the

are

as

a

and that

group, suggests that

state prior to its final

leaving a remnant AS12 iscosahedral

unit in the struc-

ture.

(P5.26) Problem:

Use TA6.2 to show that the composition

turns cannot be a half-turn.

193

of two fifth-

Figure

k

5.11:

bounded

by

vectors

C:;

{i,j,k}.

fifth-turn,

a

turn

faces

bisector

(P5.27)

Problem:

fifth-turns

Use TA6.4

axes is approximately

(P5.28)

Problem:

fifth-turns

Use

= 3,

TA6. 4 to

is a fifth turn, 55

perpendicular

each

edge.

to

through

that

that

of

if

of

of 235

basis are

The rotation

each each

pentagonal corner

a half-turn

the

set

rotations

and an half-turn.

passes and

of an icosahedron)

a cartesian

passes

composition

then the angle between

face,

where

a

axis

three

through

of

that of the

two

the fifth-turn

=

show

that

if the composition

of two

5, then the angle between the fifth-turn

116.5650•

Note that since 63.435° given

of

with

63.435°.

axes is approximately

results

faces

The generating

meet

to show

is a third turn, 55

is

third-turn

these

dodecahedron(dual

a third-turn

of a fifth of

A regular

12 pentagonal

in P5.27

and 116.565°

and P5.28

are supplementary

are compatible.

Hence

angles,

the

the 532 groups

must be such that the angle between two of the five-fold axes is 63.435°. Let

e

denote a cartesian

coordinate

system.

We place two five-fold axes

in the j, k plane such that the angle between them is bisected by k (Figure 5.11).

Hence the angle between

the unit vectors

k and each of these

in the directions

(0, sin(31.718),

is 31.718°.

Hence

of these axes are

cos(31.718))t

=

(0, .52574,

.85065)t

and (0, sin(-31.718),

cos(-31.718))t

As will become apparent 532, certain

numbers

of these repeating

t

the "golden mean". five-fold

as we generate

occur

numbers,

=

= (0,

.85065)t

the pole point representatives

again and again.

for

In order to take advantage

we define t to be

(1 + 15)/2

=

1.618034

=

Using t, representatives

axes becomes

~.52574,

ctn(31.718)

of the pole points of these

(O,l,t)t and (O,-l,t)t which are at a distance

194

of

~

~ 1.902113

our calculations

(P5.29)

units

from the origin.

,2 = ,

is that

Problem:

Confirm

One

fact that

will

aid us

in

+ 1.

that

1 , -

.309017

-.809017

.809017 [ -.500000

by using the unit vector

1

(0, .52574,

.500000]

.500000

.309017

.309017

.809017

720 in (A3.l).

and p

.85065)t

Also

show that

,

t

-

1

-1

-1

1

1

1 -

1

[ 1 -

.309017

1

1

-.809017

.809017

.500000

-.500000] -.309017

[ .500000

by using

(P5.30)

the unit vector

Problem:

(0, -.52574,

-.309017

.85065)t

and p

Show that

1 1-

-1 1 -

1

1

-.309017

-.809017

.809017

-.500000

.500000] -.309017

[ .500000

For example

the

(1,2) entry

.309017

is found

.809017

as follows:

195

.809017

720 in (A3.1).

=

1.12

i[(t - l)(-t) + (-t)(l) + (1 - t)] t[-t2

t + 1]

-

t[-(t + 1) - t + 1]

=

t[-2t]

-t/2

Analyze this matrix and show that

(P5.31)

Problem:

[

t

r -

-1

1

1

[-.500000 '"

(P5.32)

Problem:

r

-t

r ~ 1

]

=

1 t

-

.309017

Me([t,

1,

t

+ 1]2)

1

.809017]

.309017

-.809017

.500000

.809017

.500000

.309017

Show that

o Setting A

=

Me(

[Oh]5)

and B

=

M ([ttt]3) e

and recalling

that

we

have already [1,0,

found the pole point representatives [Olt]t for the 5-fo1ds, t t t r + 1] and [rt t ] for the 3-folds and [t,l,t + 1] for the

2-folds, we can find the remaining pole point representatives

in the first

octant as follows: (1)

5-fold pole points: t

{[Olt] , B[Olt] (2)

t

,B

-1

t

[Olt]}

3-fold pole points:

196

{[Olt]

t

t t ,[t01] ,[l,t ,0] }

t

t

B[t , 1,t

t ,B[l,O,t + 1] } t t + 1,1] ,[t + 1,l,0] }

,[O,t

t

+ 1] ,A[t,l,t t

-1

+ 1] ,B(A

=

{[t,l,t

[2t,0,0]2,

Byapplying

point representatives resulting

t

show that M (532),

e

no proper polyaxial

=

-1

-1

t

-1

+ 1] ), B (A

,[2t,0,0]

t

t

+ 1] ,

[t,l,t

t

[t , 1 ,t + 1] )}

+ 1,t]t,

,[0,2t,0]

t

}

[0,2t,0]2 and [0,0,2t]2 to each of the pole

in the first octant and then including the negatives

representatives,

points are obtained.

group n22 with n

[r , 1,t

t

+ 1] ,A

+ 1]t,[0,0,2t]t,[l,t

[r + 1,t,1]

the

t

,A[ttt]

2-fold pole points: {[t,l,t

of

t

+ 1] ,[ttt]

{[O,l,t (3)

t

+ l] ,[ttt]

{[O,l,t

representatives

of all of the pole

One can now find all 60 matrix representatives and hence 532, is a group. subgroup

and

Note that since we have

of 532 of order 30 (note that the dihedral

15 is not a subset of 532).

Hence the only improper

group created from 532 is

(532) U (532)i = 53(21m) The ITFC (Hahn, 1983) denotes 532 by 235 and 5321m by m35.

197

CHAPTER

THE

For some minutes directions

BRAVAIS

6

LATTICE

Alice stood without

over the country

TYPES

speaking,

- and a most curious country

were a number of tiny brooks running straight and the ground

between

looking out in all

was divided

There

across it from side to side,

up into squares

green hedges that reached from brook to brook. out just like a large chess-board,"

it was.

by a number

"1 declare,

Alice said at last.

" ...

of

it's marked all over the

world - if this is the world at all." -- Lewis Carroll INTRODUCTION

In Chapters 4 and 5, all of the possible point groups were found that contain isometries

with turn angles of multiples

To confirm that all of these are crystallographic

of either 60° or 90°. point groups, a lattice

must be found for each that is left invariant under the group.

In this

chapter, we not only show that such a lattice exists for each of these groups,

but also we find a description

for each group. fundamental

(1842), the first to study lattices,

derivation that there are 15 distinct lattice types. Auguste Bravais

(1849) undertook

that there are only 14 types. lattice types in his honor. has published

an English

our derivation

a more rigorous

are

concluded

in his

Several years later,

derivation

and

showed

These lattices were named the 14 Bravais The American

translation

1850) that makes interesting

Zacharias en

5

In toto we shall find

of these lattices.,

lattice types.

Frankenheim

Diffraction

all such lattices

We shall see that the bases chosen in Chapter

to the discovery

14 different

that includes

of

reading.

Crystallographic Bravais'

Association

derivation

(Bravais,

However, the strategy followed in

of the lattice types is more closely akin to that used by

(1945)

in his

beautiful

book

entitled

Theory

of X-ray

in Crystals. LATTICES

We recall

from Chapter

1 that any basis 0

lattice LO where

199

{a,b,c}

generates

a

+ vb + we I

LO = {ua

(T6.1)

Theorem:

of S,

Then Oland

in

01

is

tegral

an integral

[CdO

'

2

combination

01,

of

and

2

the

same lattice

[a210

LOI

is,

[b 10 '

'

1 -

2

O2,

a,

can

be

vectors

O2,

in

of

integral

Similarly

01,

of

basis

an

of

vectors

as

vector

O2

in

from 01

matrix

O2

in

bases

each vector

and only

all

is if

an

in-

[a

d02'

Z3.

in

of

integral

01

the

vectors

in

combinations

be written in

and vice

O2,

to

if

combination

and c2 can

each

vector

denote if

a 1.

integral

b2

a2,

and only

are

1

be written

Now suppose

combination

change

as

c , can

b , and

Likewise,

binations

written

{a2,b2,c2}

if

= L02

[C210

1

Cons ider Hence

=

O2 and each

of

That

Z}

1:

and O2

{a1,b1,c1}

O2 generate

combination

[bdO

=

01

Let

u, v ,W

as

can

integral

be written

versa.

of

Let

comas

T denote

an the

Then

T

Note is

that, an

[VIOl

V

1:

by assumption,

integral

L02'

so

the ,

Z3 by definition

1:

L01

Therefore

constructed and

matrix)

T

from -1

[a2101,

entries

of T are

Let

denote

L

of

V

'

is

a subset

[b21

01

a

Hence

01

and

=

02

'

in

T[V1

01

Similarly,

02

[C21

integers

vector

[V1

L

of

all

,

it

01

is

an

(that

is,

L01'

Then

Z3

1:

T

since integral

and

so

T-1 is matrix

[V1

02

L02 . (D6.2)

o

Definition:

A

an integral

matrix.

unimodular

over

If

T is

integral

matrix If,

the

T all in

or,

entries

det (T) =

addition

integers

and det(T)

of whose

for

= 1, then

our T is

are

±1,

purposes, said

integers T

is

simply

is said

called to

to be a proper

unimodular

matrix.

(P6.1)

Problem:

Show that

the

product

200

of two proper

be

unimodular.

unimodular

matrices

is

again

a proper

unimodular

(T6.3)

Theorem:

Let

only

the

of basis

if

change

Furthermore, and only

if

Proof: the

In the

T

-1

By TA2.24,

,

the

and N is

if

Since det(T)

if

(E6.4)

also

C.

an

det (TN) and det (T) agree

of S.

the

cross

in sign,

signs

01

=

{b,

O2

=

{(2/3)a

+ (1/3)b

+ (1/3)c,

-(1/3)a

-

+

of

change

) are

integers.

and

so to

±1

T

is

O2 is Suppose

a right-handed

of basis of

O2

from

T from 01

matrix

from

det (H) and det (TN)

01

Since

section

generate

change the

(TA2.25).

change

we have

a basis

-1

if

matrix

from O2 to

< O.

product

denote

(l/3)a

the

L02'

basis

matrix

det(T)

Example - When two bases

{a,b,c}

if

only

and

and the

matrix

matrix

(N),

if

of

if

=

L01

det (T) =

integral

Then nl is

L02

and det (T

of basis

det (TN) = det (T)det

the

matrix.

same handedness

integral

change

of basis

> 0 and disagree (see

is

Therefore, change

change

basis

same handedness

only

is

the

coordinate

C.

to

T-

the

=

Hence if

Hence det (T)

) = l/det(T).

1

then

= L02

agree

-1

O2

to

O2 and the .

if and

a unimodular

the

LOI

that

integral.

to

integral.

Now suppose

cartesian

01

both

01

from

of

= L02

matrix.

we showed

01 is

T from 01

det(T

unimodular,

01

matrix

to O2 is

01 and O2 are

then

Then L01

of S.

T from 01

matrix

of T6.1,

basis

are

O2 be bases

unimodular

from O2 to

unimodular.

'L

= L02

a proper

matrix

01,

01 and

proof

of

matrix

of basis to

T is

change

basis

LOI

if

matrix.

of

O2

and

Chapter

established

are

of

1) if

the

the same lattice:

and

theorem.

Let

0

0

Consider

+ (2/3)b

+

(2/3)c,

(2/3)a

+ (l/3)b

+

(l/3)c}

and

Show

that

01

and

O2

(2/3)b

generate

-(1/3)a

+ (1/3)b

+ (1/3)c,

(1/3)c}

the

same

lattice

and

are

of

the

same

handedness.

Solution: we can circuit

0

Since

find

the

01

and O2 are

change

of

basis

both

expressed

matrix

diagram.

201

from

in

01

to

terms

O2

of

using

0,

the

basis

the

following

-T

From

the

diagram,

we see

that

T

From

the

definition

of

O2,

we see that

01

we

have

2/3]

1/3

1/3

2/3

2/3

[: From

P2.19,

when

0

was denoted

T2 =

01

by

.

1/3

O2

and

by

1 0 1]

[o

-1

1

1

-1

1

Hence

T

Since

det(T)

matrix.

=

Hence

1 and T is an

01

and O2

integral

generate

matrix,

the same

T is a proper

lattice

unimodular

and are of the same

handedness.

We

shall

three-dimensions. a one-dimensional

o

now

consider

generators

Let a denote lattice

I

u

let a and b denote non-collinear

a two-dimensional

lattice

lattices

vector.

of

Then

0

one-,

=

two-

or

{a} generates

L ' O LO = {ua

Next,

of

a nonzero

L ' O

202

1:

l}

vectors,

then

0

{a,b}

generates

Note

that

the

dimensional

term

LO = {ua

+ vb I

"lattice"

will

u,v

l}

1:

be used

lattice unless otherwise

by us to denote a three-

indicated.

In any lattice

(of di-

mension

1, 2 or 3) there exists shortest nonzero lattice vectors (Newman,

1972) .

That is, there exists a lattice vector V

whose length is less

than or equal to the length of every other nonzero vector in the lattice. The next theorem uses the existence of these vectors to find a basis for the lattice.

(TG.5) Theorem: lattice LO'

Let a denote the shortest vector

Then {a} generates

LO'

collinear vectors in a two-dimensional LO'

Let a, band

Let a and b be the shortest nonlattice LO'

c be the shortest non-coplanar

lattice LO'

mensional

in a one-dimensional

Then {a,b} generates vectors in a three di-

then {a,b,c} generates LO'

The proof of this theorem would require a geometric digression

that

we do not have space for in this book. We will now consider

the situation where one lattice is contained

in another.

(DG.G) Definition:

Let L denote

we mean a subset of

L

By a sub lattice L'

a lattice.

of L,

that is a lattice in its own right.

We will show that every lattice left invariant under a point group

C

has a sublattice of the form Lp where P is a basis of the type defined

for C in Chapters containing

4 and 5.

group.

is contained

of LO since lattices

in L O.

of LOILp.

Note that if

As groups, L p

lattices

is a

Hence LOILp

a

normal is a

between Lp and LO by considering V is a vector

then [Vip must contain fractional coordinates.

type of fractional coordinates

for

Suppose LOis

are abelian groups.

We will study the relationship

the elements

Lp'

searching

L p that are also left invariant under C.

lattice such that L p subgroup

Hence we will be

in LO and not in

We can restrict the

used if we note that LOILp

of SILp

and use the equivalence

relation

group.

We call this equivalence

relation Lp-equivalence

is a subgroup

associated with this

factor

and explicitly

state the relation in the following definition.

(DG.7) Definition:

Let Lp denote a lattice and let V and

203

W

denote vectors

in S.

We define

the relation

~ on S by

V ~ W W -

If V ~ W, we say that

equivalence

relations.

Let V

integer

Then

By the way the u.'s

were defined,

for each i. ordinates

Lp

Hence each vector

with

respect

to

P

groups and their

S and suppose [Vip

1:

where P = {a ,b ,c}. I

to W.

of factor

u S vi' i

such that

U 1:

Lp

1:

V is Lp-equivalent

See Appendix 7 for a discussion

ui be the largest

V

and V -

Consequently

U.

1t

[W1,W2,W3

in S is equivalent are greater

Let

Consider

W =

=

[Wlp

=

related

[v1,v2,V31t.

W

~

v.

where 0 S w. < 1 I

to a vector

whose co-

than or equal to zero and less

than 1.

(D6.B) Definition: unit

cell

Let LO denote a lattice

{V

1:

S

I

[VlO

Consequently, Lo-equivalent

of L. unit

Theorem:

< 1, 0

L ' O

to a lattice

S I

< 1}

each vector

in

S has

an

in U O.

Show that 01

S Y

< 1, 0

t

1:

if

[VlO

[-13.7,

12.3,

61, then V ~ W where

U ' O

Let L denote a lattice

Each element of the factor

and let

group LILO

L 0 denote a sublattice

has a representative

in the

U0 of LO'

cell

Proof: LOin

0.3,

x

relative

vector

Problem:

[WID = [0.3,

(T6.9)

Then the

1,2,3}

{xa + yb + Ie lOS

(P6.2)

where 0 = {a,b,c}.

U 0 of LO is

The elements L.

equivalence

of the factor

By EA7.8 we see that classes

of the vectors

group LjL

the right in

L

204

are the right cosets of O cosets of LOin L are the

with respect

to Lo-equivalence

(when applying

EA7. 8 recall

notion

is

a"

T6.9,

each

each

"b -

vector

right

coset

T6.9

enables

LO by listing

(EG.l0)

in

additive

in

L

has

us

to

the

with

from

of two right

+ 0

t

[t,t,il .

Theorem:

invariant

under

Proof:

with

L

A lattice

is

is

invariant

invariant

C.

under

(C \ H)i

and H.

is

left

Proof:

of

C

then

each

of

Let C if

terms

of

a

sublattice

components.

LO'

lattice:

=

0

The body-

{a,b,c},

is

the

0

of

these

right

cosets

under

a point

are

invariant T4.28)

if

and only

if

group

L

is

of left

C.

L is

invariant

operation

in

under

L

is

of

=

and

(C \ H)i, Hence

L is

invariant

under

a

in-

C \ H.

that

L

C U Ci.

Since

invariant

under

L is

C,

C is left

H U (C \ H)i.

group. left

i,

under

\ H)ili

a subset

a point if

[(C

Since

L is

that

Now suppose

invariant

C denote

Suppose

Ci and hence

under

Ci and H is

and only

by C U Ci.

invariant

under

o

L

Then a lattice

invariant

under

each

is

left

of the

C.

Let {gl ,g2'"

invariant

left

Since

Hence L is

under

generators

to

(see

invariant

invariant

a subset

(TG.12) Theorem:

is

in

fractional

+ (ta + tb + tc)

C = H U (C \ H).

Then L is is

under

0

a lattice

respect

a lattice

L is

under

(C \ H)i

invariant

Hence

C.

C U Ci,

it

U ' O

in

cosets:

and L 0

by H U (C \ H)i.

variant

preceding

Hence

L denote

Let of

under

some vector

L

a lattice

form C U Ci or H U (C \ H)i

subset

to

discussion

in U O.

describe

representatives

and

(T6.11) the

By the

Lo-equivalent

few vectors

L0

L

notation).

L constructed

consisting

[0001

-1 "ba " in multiplicative

statement

a representative

relatively

lattice

lattice

't

is

the

Example - The two cosets of a body-centered

centered

Hence

that

it

is

under

. ,gn}

denote

invariant each

of

the

under

generators

each

of {gl,9 , 2 Let

{91,g2, ...,9n}. 20S

If L is

of C. ... g

,gn}' 1:

C.

Lnvar Lant

Now suppose

Then

9 is a

finite

h1h2· .. h, where hi

1:

product {gI,g2,

(composition)

... ,gn}'

of

,gn}'

{gl""

say

9

Then

geL)

hI

(L)

0

L

Given a point isometry a and a lattice LO we need a strategy for determining whether

LO is left invariant by a(O)

is the lattice generated

a(ua + vb + we)

=

under a or not.

{a(a),a(b),a(c)}

Note that

a(L ) O

since

ua(a) + va(b) + wa(c)

Hence LO = a(LO) if and only if 0 and a(O) generate the same lattice. Using T6.1, to show that LO is left invariant under a, we need only show that each vector a(O)

in 0 can be expressed

as an

integral

combination

of

and vice versa.

(T6.13)

Theorem:

a basis of S.

Let C denote a point group and let 0

=

{a,b,c} denote

Then LO is invariant under C if and only if MO(a)

is an

integral matrix for each generator a of C.

Proof:

Recall that

[[.(',ID which is the change of basis matrix from a(O) CA3.8)

that det(NO(a))

and only if NO(a)

(E6.14)

=

to O.

Since we know

±I, T6.3 yields the result that LO

=

(see

La(O)

if

is integral.

o

Example - A lattice left invariant

a and ~ are generators

under a point group C:

for some point group C,

206

0 = {a,b,c}

Suppose

denotes

a

basis for Sand

=

-a, and ~(c)

Solution:

a(a)

=

=

b, a(b)

and MO(~)

-b,

e, ~(a)

Show that LO is invariant under

-c.

To show that the lattice LOis

that NO(a)

=

-a + c, aCe)

~(b)

C.

invariant under C, we must show

are both unimodular.

Since

0 -1 0]

[o -1

are both integral LO is mapped

(PG.3)

and det(MO(a))

=

into self coincidence

Problem:

detUI0(~))

=

0

0

0-1

+1, we may conclude that

by each point isometry in C.

Use T6.13 to show that

Lp

(for the appropriate

o

choice

of P) is left invariant by each of the point groups whose generators are represented

by the matrices

We have established

in Tables 5.2 and 5.3.

some results

about properties

insure that a given lattice is left invariant under C. emphasis

and ask what properties

C

of

that would

We now shift the

L

must hold in a lattice

in order for

it to be left invariant under a given point group C.

(TG.15)

Theorem:

whose rotation

Suppose that a is either a half-turn

or a third-turn

L

denote a lattice

axis ~ passes through the origin.

that is left invariant

under a.

along ~ and a two-dimensional

Let

Then there is a nonzero

lattice plane perpendicular

lattice vector to ~ passing

through the origin.

The proof of T6.15 is given in Appendix

A DERIVATION

OF THE

14 BRAVAIS

4.

LATTICE

TYPES

In this section, we shall consider the proper crystallographic

point

C, we shall determine the structure of a lattice Lp left invariant under C such that any lattice L left invariant under C will have a sublattice with the same structure as Lp' For convenience we will denote Lp by P. The fact that both the lattice and its basis are denoted by P will not cause confusion since the context will groups.

For each such group

207

always make the meaning of P clear.

L a centered

lattice with respect

We will now determine

If L contains P and L ~ P,

we call

to P.

the possible

lattice types

for each of the

point groups.

Lattices invariant so no conditions tion the basis

under 1:

All lattices are left invariant under 1 and

beyond those of being a lattice are needed.

P

=

{a,b,c}

for the lattice

P

By conven-

is chosen to be such that

a is the shortest nonzero vector in the lattice, b is the shortest lattice vector

not

collinear

with

a and c is the shortest

coplanar with a and b chosen so that the resulting right-handed.

=

under 2:

invariant

system

is

no further

Hence there is only one lattice type for 1 which

is denoted P and is called the primitive

P

P,

Since all lattices satisfy the condition for

lattices need be sought.

Lattices

lattice vector not

coordinate

lattice type

(see Figure 6.1).

As in Chapter 5, for 2 we take the basis

{a,b,e} such that c is along the 2-fold axis and a and b are per-

pendicular

to c so that

P

is right-handed.

Then the lattice

P

is left

invariant under 2 because Np(2) consists only of integral matrices

(see

T6.13 and Table 5.2). Let L denote exists a nonzero

any lattice

left invariant

lattice vector

under 2.

By T6 .15, there

along the 2-fold axis

lattice plane passing through the origin perpendicular

and

Since no geometric

plane, no preference note

the

{a,b,c}

shortest

have been placed

to the 2-fold

on this lattice

will be given to one basis over another. nonzero

is a right-handed

of the same type as

constraints

P

vector

along

system.

described

the

is a

to the 2-fold axis.

Let {a,b} denote a basis for the lattice plane perpendicular axis.

there

two-fold

axis

Let c deso

that

The lattice generated by {a,b,e} is in the previous

paragraph.

denote both {a,b,c} and the lattice generated by it by P.

Hence

If P ~ L,

we then

P is a proper subset of L and so there are vectors in L whose coordinates with respect to

P

are fractional.

we shall use P-equivalence

In order to facilitate the discussion,

to discuss the vectors in

each vector in L has a representative

whose coordinates

respect to P are such that 0 S fi < 1. [f 1,t «.t

31

t

where 0 S fi < 1.

Then

208

Suppose f

1:

L.

Hence, by T6.9

fl'

f2 and f3 with

L such that [flp

=

-1 0 0] [ff21] = [-f-f21] [ oo t, r,

lp

[2(f)

0

0

1

+ f, then

2(f)

If e is defined to be e

-1

eEL

and

[elp

Since c is the shortest vector

2f3

a mUltiple of c and so define e to be e

=

1:

axis in L, e must be

along the two-fold

Z.

Hence,

eEL

f - 2(f), then

f3 =

i.

0 or

and [elp

= [2f1,2f2,0It.

{a,b} is a basis for the lattice plane perpendicular

2f 2

1:

Z.

Therefore,

combinations nations dictions.

that

f1

= 0 or

t

and

f2

= 0 or

of these fractional coordinates violate

the

choice

By the discussion

of

i.

Similarly,

if we Since

to c, we have

2fl'

We tabulate the various

in Table 6.1.

Those combi-

the basis P are noted as contra-

following D6.6, LIP

is a group.

Hence, the

only combination of the four possible fractional vectors that are suitable are those that form groups modulo P. cyclic group modulo

PIP

AlP

P

Each of these vectors

I [:]

liP

[:]

No other groups can be formed from the four possible This

can

be

seen

by

observing

that combining

fractional vectors yields an impossible vector. [~ 2,0,2.l]t

generates

a

as shown below:

+

[.l1]t_[.l.l]t 0,2,2 -

2,2,0

209

[:] [:]

fractional vectors.

any two of the nonzero For example,

(modulo P)

the choice of {a,b}.

which, as noted in Table 6.1, violates

Table 6.1: Possible fl

Combinations

Coordinates f2

Impossible

I

f3

of fractional

fl

coordinates

Coordinates f2

f3

for 2.

Contradictions

0

0

0

i

0

0

0

i

i

0

i

0

choice of b

i

0

i

0

0

i

choice of c

i

i

i

.1

.1

0

choice of {a,b}

2

2

choice of a

Show that [i,o,t]t and [i,t,ilt cannot both be in L by forming their sum modulo P. Show that [O,i,i]t and [i,t,ilt cannot both

(P6.4)

Problem:

be in L. Consider the lattice A = (P + 0) U (P + [O,i,ilt). basis

c1

=

from

P

= {a,b,c}

to

Pl

= {a ,b ,c 1

1}

1

where a ,

-c, then the change of basis matrix from P to Pl

T

1

0

o o

0

=

If we

change

b, b ,

= a

basis

for the

and

is

-1

Hence PI is another

which is a proper unimodular

matrix.

lattice P and, furthermore,

PI satisfies

the criteria to qualify as a P

forms a basis

for the lattice plane perpen-

basis.

That is, al

dicular

to the 2-fold axis and cl

the axis.

and bl

is the shortest

lattice vector along

Since

(modulo P l)

the A lattice of P is the same set of points Similarly , it can be shown

as the B

lattice of PI'

(see P6. 5) that the I lattice also occurs as

a B lattice with respect to a different

210

allowable basis.

Hence if we

take the A lattice with respect to every allowable basis, we will obtain all of B and I lattice types as well.

(P6.5)

Show that the A

Problem:

is the I lattice with respect to the P2 a + b, that

c2

P2

-

satisfies

the conditions

plane perpendicular

{a2,b2,c2}

[O,t,t]t with respect to

that {a2,b2}

centered

a2

=

a,

b2

=

is a basis for the lattice

P

becomes

is a right-handed

lattices.

basis

[t,t,t]t with respect to

Hence, we need only use one centering possible

where

to the 2-fold axis and that c2 is the shortest vector Then show that P2

a long the axis.

=

basis

Show that P 2 is a bas is for the lattice P and

(Hint:

c.

to the P

lattice with respect

and

that

P2).

for each basis to obtain all

When c is along the rotation axis, as in our

case, the ITFC (Hahn, 1983) gives as the first choice

[O,t,t]t.

Thus we

obtain the lattice type

A

= (P +

0) U (P + tb + tc)

{ua + v(tb + tc) + we I u,v,w

Hence a basis for A is 0 = {a, tb + tc, c}. coset P + f is called

a colattice

that if f is not an element of P,

o

is not an element CpCfIJ2J3)'

geometric vector f. of

resemblance

to

P

In the case of A,

CP (0,t, t)

lie

and is denoted then CpCfIJ2J3) However,

since

If

1:

[f]p

=

the

[fIJ2J3]t,

by CpCfIJ2J3)'

Note

is not a lattice since

CpCfIJ2J3)

it is an image of

A = P U Cp(O,i,t).

in planes perpendicular

Z}

bears

P

a strong

displaced

by the

Note that the points

to the 2-fold axis,

located

halfway between the planes containing the lattice points of the sub lattice

P.

To show that A is invariant under 2, we need to show that N (2) is

an integral matrix.

O

Since

211

A

is invariant under 2.

variant by 2. by

P.

Thus, there are two types of lattices left in-

The first type is called a primitive

P

Every lattice plane of

lattice point on the axis. called an A-centered the 2-fold

lattice.

axis alternate,

perpendicular

The second

to the 2-fold axis has a

lattice

by A,

is

of A perpendicular

to

type denoted

The lattice planes one having

lattice and is denoted

a point on the axis (i.e., in P)

and the next not on the axis (i.e., in Cp(O,t,i)).

(P6.G)

Problem: (1)

Given a basis 0

=

{a,b,c}

with b taken along the 2-fold axis

and a and c perpendicular

LO

show that the lattice

to

b

forming a right-handed

generated by

0

system,

is left invariant under

[010J2. (2)

Let L denote any lattice type left invariant that the coordinates

of the fractional

spect to 0 left invariant by [010]2 [O,i,i]t,

(3)

by [010J2.

vectors

are

Show

in L with re-

[O,O,olt,

[t,i,O]t,

[t,t,i]t.

Using the four fractional generators,

construct

vectors

obtained

multiplication

in the part (2) as

tables

for

the

LILO

groups for each of the following choices of L:

LOILO = CILO

{[O,O,O]

t

}

= {[O,O,O]t,[t,i,O]t}

{[O,O,O]t, [O,i,i]t} {[O,O,O]t, [i,t,t]t}

LILO

(4)

Show that no

(5)

Show that the A-centered

groups contain

to 0 can be realized other allowable

[t,i,O]t and [O,t,t]t.

and I-centered lattices with respect

as C-centered

bases for L O.

lattices with

respect

to

When b is chosen to lie along

the 2-fold axis, the ITFC (Hahn, 1983) gives as the first choice [t,i,O]t.

This results in the lattice type

C = (L 0 + 0) U (L 0 + (ia + tb)) {U(ia + ib) + vb + we

with the basis DC

{ta + tb, b, c}.

212

I

u,v,w

1:

Z}

Show that C is left in-

variant by [010J2.

With the completion of this problem, we may

cone lude that there are two types of lattices left invariant 010 12: P and C. The lattice type denoted C is called a

by [

C-centered one-face

lattice type (see Figure 6.1).

centered

lattices

This setting for the

in the monoclinic

systems

is

the

choice used in the ITFC (Hahn, 1983).

Lattices

invariant

under 3:

5, we choose the basis P

As in Chapter

=

{a,b,c} for 3 such that c is in the positive direction of the 3-fold axis, a and b are perpendicular to the axis such that «a:b) a right-handed

is left invariant 3.

Since Np(3) consists

system.

under 3.

Let

L

=

1200 and

Then b = 3 (a)

length.

lattice plane and its length is also the shortest.

vector

along

the

{a,b,c} is a right-handed

direction

basis.

as the lattice P discussed

of

P

Hence

in Lunder

f in L of the form [f]p

=

is in the

Hence {a,b} is a basis

shortest

the

3-fold

nonzero axis.

lattice Then

P =

is a lattice of the same form

in the previous

of 2, we cons ider the vectors vectors

Let c be the

(T6.5).

positive

to the

Let a be a lattice vector in the lattice

to c of shortest

for the lattice plane

Lp

lattice

axis and a lattice plane perpendicular

axis passing through the origin. plane perpendicular

forms

denote a lattice left invariant under

By T6.15 all lattices left invariant under 3 have a nonzero

vector along the three-fold

P

of integral matrices,

paragraph.

P-equivalence.

[fIJ2J3]t

As in the

case

Hence we seek

where 0 S fi < 1.

Since L

is invariant under 3, f + 3(f) + 3-1(f) is in the lattice, and, (see Table 5.3)

Since c is the shortest an integer.

Hence f3

=

vector

in

L

along the 3-fold axis, 3f3

0, 1/3 or 2/3.

Also f - 3(f) is in Land

fl+f2] [f-3(f)]p

[

213

2f2

-

fl

o

must be

Since {a,b}

is a basis for the two-dimensional

the

axis, fl + f2 and 2f2 - fl are integers.

3-fold

lattice perpendicular

to

Hence there exist

integers u and v such that

=

fl + f2 -f I + 2f2 Using the technique

of Appendix

U

= V

2, we solve this system by row reducing

the augmented matrix

[.:

1 2

I I

~J

obtaining

[~ Hence fl

=

=

1

=

u - tl3 and f2

tegers and 0 S

1/3, f2

I u - (u + V)/3] (u + v)/3 I

0

0
R be defined

(EA 1. 7) Example:

by a(z)

2z.

= a(z2)'

By the definition

Show that

a is

one-to-one.

Solution: a, 2z1

Let Zl,Z2

=

Note that exist that

Z £ Z such that

a mapping a:A->B

a(zd

of positive

a(z)

is onto is to

A such that

£

(EA 1.8) Example:

=

a(a)

=

S.

of o

numbers.

there

The technique

choose an arbitrary

b

Prove that

by a(r)

=

By the definition

er

=

equation

real

numbers, r

we have r

=

=

of a, this

£n(p).

£n(p) exists, a(r)

B and find

a is onto.

Then we must find an element r

this

£

er where R+ is the

Let p e R+.

a(r)

does not

for showing

b.

Let a: R->R+ be defined

real

Solution:

p.

Z such that

the mapping a of EAl.7 is not onto since

an integer

an element a

set

£

Hence, Zl = Z2, and so a is one-to-one.

2z2.

would mean that

Since £n(p) is defined

£

R such that

= p.

Solving

for all positive

and

=

a(£n(p)) e£n(p) ) p.

Hence a is onto.

(PA1.l)

Problem:

(PA 1.2) Problem: and onto.

Find a

0

Show that

Showthat -1

the a defined

in EA1.8 is one-to-one.

a:R->R defined bya(r)

.

307

3r + 5 is one-to-one

APPENDIX

MATRIX

(DA2.1)

A matrix

Definition:

between brackets

2

METHODS

is a rectangular

enclosed

and arranged in rows and columns as displayed below:

First

mth

column

column {-

{-

I all a12 ... aIm

First row'"

a

a21

22

...

a2m

a

nth row

In this matrix, we have n rows and m columns. of the ith

array of numbers

row and ith column is referred

nm

The number at the intersection

to as the (i ,i) -entry.

Note that

it by a... A matrix with n rows and m columns is called an 'I (read n by m matrix).

we have denoted

n x m matrix (DA2.2)

Definition:

Two n x m matrices

b.. for all 1 :5 i :5 nand 'I This definition

1 :5

i

A and B are said to be equal

if

a 1/..

In this example, by the definition

of

:5 m.

enables us to solve matrix equations such as

-1

cp -sp

for p where cp equality, represents C

=

cp

=

=

sp

cp

o

o

0

o

cosp and sp

0, sp

=

1.

=

sinp.

Hence, p

a 900 counterclockwise

=

90°.

In fact, the matrix on the right

rotation with respect to a cartesian basis

{i,i,k} about k (see Appendix 3).

309

OPERATIONS Two n

x

m matrices A and B can be added according to the following rule

°11 °12, .. 01m °21 °22"

A + B

=

.02m

I

.b2m

b~1 bn2··

.bnm

.

+ °nl

°n2"

vs=Si2H3,

---->

and vg=Si2H4. V6 = V7 = Vs = Vg = L(S'

-----J>

----io

31) Let vI=oSil,

VI V2 V4 Vs -

V3 = V3 = V2 = V2 =

. ) _ -I ( Ii 0IS12 - cos

V6 • V7

) _

II V6 1111V7 II -

i

([vslhGc[vslc)i

L( 3 .)2 4 -1 ( H S1 H = cos

• Vs ) II vsVs 1111 Vs II

34) The metrical

matrix

cos

-1 ( [v61hGc[v71c) (1.646)(1.646)

= 1.466.4 0

= 111.271 .

for a-quartz

to Si2 are 01,03,04,

____..

---+

is given on the top of page 33. The four

and 06.

-----+

_ ° - 136.673 ,

= 1.466.4

~

~

~

Let vI=oSi2,

VS=006, v6=Si203, v7=Si204, vs=Si206 and v9=Si20I. bond distances and angles are shown. below

-----+

Calculations

[0.5301] [-0.1160] 0.5301 = -0.2620 0.3333 -0.2145

0.2681] 0.4141 0.5479

-

[0.5301] [-0.2620] 0.5301 = -0.1160 0.3333 0.2146

[vsln = [vsln - [vl]n =

[v61n = [v31n -

=

R(0ISi2)

=

0.5859] [0.5301] [0.0558] 0.8540 - 0.5301 = 0.3239 [ 0.2145 0.3333 -0.1188 0.8540] 0.5859 0.4521

-

[

[0.5301] [0.3239] 0.5301 = 0.0558 0.3333 0.1188

= ([v91bGn[v91n)~

-0.1160]t -0.2620 -0.2145

([

lvilo

[

---+

V4=004,

made to determine

-

[v71n = [v41n - [viln =

___.

V2=OOI, V3=003,

0.4141] [v91n = [v21n - [vlln = [ 0.2681 0.1188

II v911 =

V7=0IS12,

1.0212i - 1.2914j, -1.6289i + 0.2387j -0.4137i + 0.7164j + 1.2098k -0.4137i + 0.7164j - 1.2098k

R(Si2H4) =11vsll=

oxygen atoms nearest

----:-+

----+

V6=0ISil,

Then

,R(Si2H3) =11Vs 11= ([vslhGc[vslc)

Pl.ll (page

----t

----t

v2=oSi2, V3=OOl, v4=oH3, vs=oH4'

[

24.1474 -12.0737 0

-12.0737 24.1474 0

0] [-0.1160]) 0 -0.2620 29.2573 -0.2145

i

= 1.611.4

II V7 II = II vs II = II v611 =

R(04Si2)

= ([v71bGn[v71n)

i=

R(06Si2)

= ([vslbGn[vsln)

~ = 1.608.4

R(03Si2)

= ([v61bGn[v61n)i

V9 • V7 = [v91bGn[v71n V9 • V6 = [v91bGn[v61n V7. V6 = [v71bGn[v61n L(03Si204) = cos "! L(03Si201)

= cos "!

L(04Si201)

= cos ?

= 1.608.4 V9 • Vs = [v91bGn[vsln V7. Vs = [v71bGn[vsln vs • V6 = [vslbGn[v61n

= -0.870 = -0.903 = -0.829

CI:.W;,II) = CI:.i~IV"'II) = CI:'I~IV"'II) =

1.611.4

109.591

0

L(03Si206)

= cos "!

110.4080

L(04Si206)

= cos "!

108.674°

L(06Si201)

= cos "!

= -0.830 = -0.903 = -0.844

CI:.i~IV".II) = CI:,I~IV".II)= ("::I~IV",") =

0

108.687

0

110.411

109.064°

the

:>l.12 (page

---t

34) The metrical matrix is given on page 33 of the text.

'3=;;sG, V4=e>:;SI; and vs=04Si3.

Let Vl=004,

---t

v2=oSi2,

Calculations made to determine the bond angle follow

[v41n = [V2- vlln = [0,2620

-0.21461t

0.1160

[vsln = [V3- vl]n = [-0.2681

0.11881t

0.0558

IIV4 II= R( Si204) = ([v41bGn[v41n)'; = 1.611.4 IIvs II= R( Si304) = ([vs)bGn[vs)n)';

= 1.608.4

V4• Vs = [v41bGn[vsln = -2.087.4 0

L(Si304Si2) = cos "! CI v:41~I:Ss II) = 143.664 'l.13 (page latrix

•

34) Given the cell dimensions of anorthite, it is possible to calculate the metrical a. a G = [ b. a c. a

C]

a. b b. b c. b

a. b. c c. c

=

[66.7979 -2.3128 -50.5870

et VI=;;Q, v2=;;si, v3=;;M, v4=oSi and vs=DAi.

s follows

[v41n = [V2-

v.l»

-2.3128 165.6112 -9.8898

-50.5870] -9.8898 200.6472

Calculations to determine the bond angle are

= [0.1622

-0.0383

[vsln = [V3- viln = [-0.1567

-0.02341t 0.04831t

0.0188

IIv411 = R( SiO ) = ([v41bGn[v41n);

= 1.601.4

IIvs II= R( AIO ) = ([vslbGn[vsln);

= 1.728.4

V4· vs = [v41bGn[vsln = -2.697.4 L

'0) -1 (s: AI = cos

(

V4· vs ) ° IIV4 1111 Vs II = 167.148 .

l.14 (page 37) Given the cell dimensions of protoamphibole, the following relationship is tablished a=9.335i, b=17.880j, c=5.287k. A perpendicular to the plane defined by 04, 05 ---->

---->

id 06 is given by the cross product, V4 X Vs, where V4=040S and V6=060S' Then [v41n = -0.0285 -0.1280 -0.25971t and sln = [-0.0030 -0.0088 -0.50351t. Since the transformation matrix from the D basis to the .rtesian basis C is A = [[ale

[ble

[cle 1 =

9.335 0

[

o

0 17.880 0

0] 0 5.287

is possible to rewrite the vectors r4 and rs in terms of the cartesian basis, evaluate the cross oduct and then make the necessary calculations to determine the angle of misfit as shown below [v41e = A[v41n =

9.335 0

0 17.880 0

9.335 0

0 0] [-0.0030] [-0.028] 17.880 0 -0.0088 = -0.157 0 5.287 -0.5035 . -2.662 -0.0281 -0.157 = 5.876i - 0.670j - 0.022k -2.662

[

[vsle = A[vsln = [ i V4 X rs = V6 =

j Ik

o

o

-0.266 -2.287 -1.373

0] 0 5.287

IIV6 II= ([v61hGe[v41e) ~ = 5.914.4 V6• a = [v61hGe[ale = 5.876.4 1

L(V6:a) = cos-

(II

v:61~1:

II)

= 6.506°.

[-0.0285] -0.1280 -0.2597

=

[-0.266] -2.289 -1.373

0

Pl.15 (page 40) The cell dimensions of a-quartz are given on page 32. Since (3 = 90 and c lies on k, a is perpendicular to k. Then a lies in the ij-plane. Then the i, j components of a are given 0 by the projection of a onto i and of a onto j. Since b lies on j and 'Y = 1200, L( a:j)= 120 • Since a lies 120 degrees from j which lies 90 from i on the ij-plane, it follows that L( a:i)= 120-90=30 As a result, a=acos(30)i+acos(120)j=av'3/2i-a/2j. As discussed on page 39, the volume of the unit cell is 0

0

v=a.bxc=

•

ob o

0 0 c

I=T v'3a b _ c -

v'3 a2c = 113.lX 2

CHAPTER 2 P2.1 (page 49) By construction D* is such that s = ha* +kb* +/c* is a vector perpendicular to the (hkl) stack of planes and p es = 1 for any vector p whose terminus lies on the plane hz+ ky+lz = 1. Now b" = Oa" + Lb" + Oc" so b* is perpendicular to the (010) stack of planes. Since the terminus of b lies on the plane Oz + ly + Oz = 1, it follows that b • b" = 1. The (010) planes are parallel to the ac-plane and therefore perpendicular to a x c. Since b* is perpendicular to the (010) stack of planes, b" is parallel to (a X c). Then b" = 7'a X c for some r E IR. Now b" • b = 1 = ra x c e b. Then (axc.b)-l = r. So b" = (axc.b)-laxc. But (ax c e b) = -(bx c e a] and ax c = -cxa. So b' = (c X a)(b x c e a)-i. P2.2 (page 49) The (001) planes are parallel to the ab-plane and therefore perpendicular to a X b. Since c" is perpendicular to the (001) stack of planes, c" is parallel to (a X b). Then c" = ra X b for some 7' E IR. Now c" • c = 1 = 7'a X b. c. Then (a X b • c)-l = r. So c" = (a X b. c)-la X b. P2.3 (page 50) In the case that h f= 0, k f= 0 and I f= 0 the plane crosses the z-axis at alh, the y-axis at blk and the z-axis at cll. Then alh, blk and cll all have termini which lie in the plane so that (alh). s = (blk). s = (cll). s = 1. And so 1 = s e blk 1 = s e ef l

= (tla' + t2b' + t3C'). = (tla' + t2b' + t3C').

which implies that t2

=k

and t3

blk = (1/k)(tla'. b + t2b'. b + t3C'. b) = (1/k)t2 cll = (111)(tia' • c + t2b' • c + t3c' • c) = (111)t3

= I.

P2.4 (page 51) In the case that h = 0, k f= 0 and I f= 0, the z-axis is parallel to the plane and the plane crosses the y-axis at blk and the z-axis at cll. Then a is parallel to the plane and blk and e] I both have termini which lie in the plane so that a • s = 0 and (b I k) • s = (cl I) • s = 1. And so 0= s e a

=

(tla'

+ t2b' + t3C') • a = (tla' • a + t2b' • a + t3C' • a) = tl + t2b' + t3C'). blk = (1/k)(tla'. b + t2b*. b + t3C'. b) = (1/k)t2 + t2b' + t3C'). ell = (111)(tla' • c + t2b' • c + t3C*• c) = (111)t3

1 = s e blk = (tla' 1 = s e e]! = (tla' which implies that tl

= 0, t2

= k and t3

= I.

P2.5 (page 51) In the case that h f= 0, k = 0 and 1 f= 0 the y-axis is parallel to the plane, th, plane crosses the e-axis at al h and the z-axis at e] I. Then b is parallel to the plane and e] h ant e]] both have termini which lie in the plane so that b. s = 0 and (al h) • s = (cl I) • s = 1. And s, 1 = s e alh = (tla* + t2b' + t3C'). alh = (1Ih)(tla'. a + t2b'. a + t3C'. a) = (1Ih)tl 0= s e b = (tla' + t2b' + t3C'). b = (tla'. b + t2b'. b + t3C'. b) = t2 1 = s e cll = (tla' + t2b' + t3C') • cll = (1/1)(ha' • c + t2b' • c + t3C' • c) = (1/1)t3 which implies that tl = h, t2

= 0 and

P2.6 (page 52) Given that r = ria

= I. + r2b + 7'3Cit t3

+ r2b + r3c) • b" = ria. + r2b + r3c). c" = ria.

r. b' = (rIa r. c" = (ria

follows that b" + r2b • b' + r3c. b" = 7'2 c" + r2b. c" + 7'3C. C' = 1'3.

P2.7 (page 52) Proof: Since D' is a basis for IR3 there exist scalars 1'1, 1'2, and 1'3 E IR such that r = ria' + r2b' + r3c'. Then r. a = (ria' r. b = (7'la' r e c = (ria' So (r. a)a'

+ (r.

b)b'

+ r2b' + r3c') + r2b' + r3c') + r2b' + r3c').

+ (r.

c)c* = ria'

• a = rIa' • a + r2b' • a + 7'3C'• a = 1'1 • b = 7'la' • b + r2b' • b + r3c' • b = 1'2 c = ria'. c + r2b'. c + r3c'. c = 1'3.

+ r2b' + 7'3C' = r.

P2.8 (page 63) By T2.13, det(G*) = (a')2 (V')2 = a'b' cos('Y') cos({3') 1 a'c' = (a*b*c*)2(1-

II'.

Then

a'b' cos('Y') (b')2 b'c' cos(a')

a'c' cos({3') 1 b'c' cos(a') (c')2 cos2 a' - cos2 {3' - cos2 '"(' + 2 cos a' cos{3' cos-y")

so that '* '" V'" = abc

'" (1

- cos2 a '" - cos 2 {3'" - cos 2 '"('"

*)1 + 2 cos a '" cos {3'" COS'"(

P2.9 (page 65) The metrical matrix and volume (det(G)i)

k

X

i = (det G)-i

I~

G[i)n

G[k)n

2.

are given on page 65. Then by T2.14

I = 0.0895a -

0.097b - 0.003c.

P2.10 (page 68) Given the cell dimensions with respect to the Dl basis

c, =

a~ al hI cos '"(1 aiel COS{3I] [ 127.374 bl CIc~s al = 0.000 bl al cos '"(1 h~ [ cl -65.097 Clal cos {3l Clbl cos a 1

0.000 208.456 0.000

-65.097] 0.000 99.142

.

The transformation matrix, T, that maps vectors written in terms of the Dl basis into vectors written in terms of the D2, basis can be written as -5/2 5/2

o

-1] -1

T

-1 _

-

[.25 -.2

o

2

.25 .2 0

.25] 0 .5

By T2.12, G2

=

T-tGl T-l

=

[

16.299 -0.377 -0.176

-0.377 16.299 -0.176

-0.176] -0.176 16.472

.

Evaluating the entries of this matrix the cell dimensions are determined as a2 C2= 4.059.4, '"(2= 91.3270 and {32= a2 = 90.6160•

4.037.4,

P2.11 (page 69) The vectors

are perpendicular to the stacks of planes (111), (111), (111), and (111), respectively, which parallel the closed-packed monolayers of oxygen atoms of the sapphirine structure. The transformation matrix which transforms vectors written in terms of the Dl basis into vectors written in terms of the D2 basis is given in P2.10. By (2.22) Tt is the transformation matrix that transforms vectors written in terms of the Di basis into vectors written in terms of the D; basis. Then the vectors can be written as shown below. Tt[slln; Tt[s31n;

= =

[slln; [s31n;

= =

[4 0 olt [-4 0 41t

Tt[s21n; = [s21n; Tt[s41n; = [s41n;

= =

[0 [0

5 21t -5 21t

Since these vectors are perpendicular to the (100), (052), (101), and (052) stacks of planes in the sapphirine face-centered subcell, it follows that the monolayers of oxygen atoms listed for the sapphirine structure parallel the monolayers listed for the face-centered sub cell. P2.12 (page 70) Given the cell dimensions of the Dl basis it follows that

Gl =

[

a2 abcos'"( ba cos '"( b2 ca cos {3 cb cos a

accos{3] be cos a c2

=

[95,648 0.000 14.205

0.000 316.840 0.000

14.205] 0.000. 27.668

Given the vectors of the D2 basis written in terms of the Dl basis, the transformation matrix, T-l, that maps vectors written in terms of the D2 into vectors written in terms of the Dl, is formed as follows

By theorem, G2

=

T-tGl T-l

=

[

94.905 0.00 -13.462

0.00 316.840 0.00

-13.462] 0.00. 27.668

Analyzing othe matrix the following cell dimensions are obtained: a2 = 9.742.4, b2 = 17.800.4, C2= 5.260A, 'Y2= 90 a2 = 90 and {32= 105.231 Since T-t transforms vectors written in terms of the D~ basis into vectors written in terms of the Di basis, the vector [sln; = [-2 0 11t which is perpendicular to the Dl stack of planes (201) can be written as 0

0

0

,

•

[sln;

=

[1 0 -1]~ [-2]~ [-3]

= ~ ~

T-t[sln;

=

~

which is perpendicular to the D2 stack of planes (301). Then the (201) stack of planes in Dl is oarallel to the (301) stack of planes in D2. Since T-I transforms vectors written in terms of the D2 basis into vectors written in terms of the Dl basis, the vector [sJn, = [0.29 0.08 O.OI]t can oe written in terms of DI as

[sln, = T-l[sln,

=

[

01 -1

0 1 0

0] [0.29] 0 0.08 1 0.01

= [0.29] 0.08. -0.28

:>2.13 (page 71) The metrical matrix G2 is given on page 68. 1) V2 = (det(G2))~ = (4372.5513); = 66.125.43. 2) If V2 = 16V2 rg, it follows that ro = (66.12526/16V2)* = 1.430. 3) Given the assumptions, the metrical matrix, G2, for an ideally closed-packed sapphirine cell is 16,362 0.000 0.000

G2 =

[

0.000 16.362 0.000

0.000] 0.000. 16.362

By (2.22) Gl

= TtG2T =

[

130.896 0.000 -65.448

0.000 204.525 0.000

-65.448] 0.000. 98.172

4) Then analyzing the entries of this matrix the cell dimensions are as follows: al = 11.441.4, bl = 14.301.4, Cl = 9.908.4, 'Yl = 90 al ::.::90 and {3l = 125.264 agreeing quite well with the calculated cell dimensions. 0

0

,

'2.14 (page 74) 1) The transformation matrix, B, from the reciprocal basis, D*, to the Cartesian basis, C, is given by bll B = [[a*le [b*le [c*le 1 b [ 21

o

2) The circuit diagram below shows that G* = BtB. [rln G*T [rlD

[rle B

-+

1I [rle

3) Given that G* =BtB it follows that:

G* =

so that, equating matrix entries, using some algebra and T2.8 (page 55) the cell dimensions are determined to be b22 = b*, b2l = a: cos 'Y*,b23 = c" cos o", bll = a" sin 'Y*, b13 = _cO cos {3sin a" and b33 = c" sin o" sin {3. f>2.15 (page 79) Since the (100) and (110) planes are perpendicular to the vectors Vi = [1001b· snd V2 = [110Jb. respectively, then it follows that the angle between these two vectors is the same ss the angle between the two stacks of planes. Given the cell dimensions of amblygonite, we obtain

G* =

[

.0444 -.0123 -.0004 -l(

L(Vl : V2) = cos

II

- .0123 .0264 .0134

- .0004] .0134 .0469

Vl.V2) Vi 1111 V2 II

= cos

-1 ([vllb.G*[v21n.) II Vi 1111 V2 II

4 = 4 .936

0

P2.16 (page 86) The transformation matrix A is given on page 86. It has the property A -l[rle = [rln. Then it follows that A[rln = [rle. Then A[r31n = [14.196

2.900

A[r41n = [10.967

2.901

A[rsln = [15.830

5.616

that

t 1.8741 = [r31e t 1.8741 = [r41e t 1.8721 = [rsle

where r3, r4' and rs denote vectors that radiate from the origin to the positions of M3, M4 and M5 respectively. P2.17 (page 86) (1) It is known that A is the matrix that transforms vectors written in terms of the direct basis into vectors written in terms of the cartesian basis, C. By previous work it is also known that AtA=G. Then 2 0 bacos 'Y ca cos {3] [ a b2 cbcosa = 0 c2 ac cos {3 0 be cos a

2 a AtA = G = [ abcos'Y accos{3

AtA =

[all a~l

0 a22

0

v

a13] [all a23 0 a33

a13

a2l

1- [

o

o

an a23

a~,+al,a

a2lall

a33

ca c~s {3]

+ a23 13

c2 all a21 + a13a23 aL

a33al3

+ a~2 + a~3 a33a23

a"a~ ] a23~33 a33 2

'

Equating entries we have a~l + a~3 = a2, alla2l + a13a23 = 0, a~l + a~2 + a~3 = b , a33a13 = ac cos {3, a33a23 = 0 and a~3 = c2. Then with some algebraic manipulation, it follows that a33 = c, a23 = 0, a13 = acos{3, all = asin/3, a2l = 0 and an = b. So that, as expected,

A=

a sin{3 0 0 b a cos {3 0

[

0] 0 . c

(2) Substitute the appropriate values into the variable A-matrix to get:

A=

8.980 .000 [ -2.840

0.000 8.562 0.000

0.000] 0.000 5.219

Let [rlln, [r21n, [r31n, [r41n, and [rsln denote the vectors that radiate from the origin to the position of Si, 01, O2, 03, and 04, respectively. The following results are obtained: A[rlln = [1.880 3.481 3.4361\ A[r21n = [3.511 3.628 3.4431\ A[r31n = [1.250 2.029 3.2951t, A[r4Jn = [1.317 4.221 4. 7721t, and A[rsln = [1.317 4.341 2.1631t. P2.18 (page 89) (1) Let [rdn, [r2ln, [r31n, [r41n, and [rsln denote the vectors that radiate from the origin to the position of Si(3), 0(A3), 0(B3), 0(C2), and 0(C3), respectively. To transform the atomic coordinates (Table 2.2) for pyroxferroite from the direct basis to the cartesian basis, premultiply by the matrix A to obtain: 2.666] [rl]c = A[rlln = [ 12.125 , -5.822

[r21e = A[r21n =

3.665] [r31e = A[r31n = [ 11.123 , -6.536

[r41e = A[r41n =

[rsle = A[rsln =

[

[

2.666] 12.124 , -4.207

[

1.151] 11.681 -6.336

and

2.892] 13.626 -6.422

(2) In the cartesian coordinate system, with metrical matrix G=I3, the computed distances are obtained as follows: .

II 0(A3)Si(3) II 0(B3)Si(3) II 0(C2)Si(3) II 0(C3)Si(3)

II =11[rile II =11[rile II =11[rl]c II =11[rde

- [r2]c 11=11 [.000

-1.6151t 11=1.615.4

.001

- [r31e 11=11 [-.999

.7141t 11=1.585.4

1.002

.5131t 11=1.660.4

- [r41e 11=II [1.515

.444

- [rsle 11=11 [-.226

-1.501

.6001t 11=1.632.4

agreeing somewhat with the results shown below, published by Burnham, 1971:

II 0(A3)Si(3) 11=1.616.4, II 0(B3)Si(3) II 0(C3)Si(3) 11=1.630A.

11=1.585.4,

II 0(C2)Si(3)

11=1.661.4,

and

:>2.19 (page 89) 1) Let A be the transformation

matrix from D2 to DI. That is,

A = [[a21n,

[b21D1

1 [c2ln,1 = [ -1

o

0 1 -1

1]

1 . 1

Since A -1 transforms vectors written in terms of the basis Dl into vectors written in terms of the basis D2, then it follows that A -1 = [[adn,

[blln,

[clln,l

=

.667 .333 .333

[

-'.333 .333 .333

-.333] - .667 .333

Given the definition of a triple in a specified basis, the desired result is obtained. 2) Given the cell dimensions we obtain the metrical matrix for the basis D2: G2 =

[

254.434401 -127.217200 .000000

-127.217200 254.434401 .000000

.000000] .000000 52.417600

Calculating the lengths with respect to the D2 basis of [alln" [bi]o, and [clln" respectively, the following results are obtained: al = bl = ci = 9.520.4. Calculating the inner products with respect to the D2 basis of [alln2 and [blln" [blln, and [cdn, and [clln, and [adn, respectively the following results are obtained: 'Yl = al = {3l = 9.5200 as was expected,

CHAPTER 3 P3.1 (page 98) r = l(r)

u = mer)

s = 6(r)

v = a-l(r)

t = 3(r)

w = T\r)

P3.2 (page 101) The set of all isometries that describes the symmetry of the plane figure is given by the set {I, 6, 3, 2, 3-\ 6-1} =6. Since the turn angle is 600 and since 360/60=6, then the rotation axis, c, perpendicular to the object, is a 6-fold axis. P3.3 (page 105) Proof: Let 'Ybe a rotoinversion that leaves the origin fixed. By definition a rotoinversion is the composition of some rotation, a and the inversion, i. By E3.8, the inversion is a linear mapping and by E3.7, the composition of two linear mappings is also a linear mapping. Then if it can be shown that a is a linear mapping, it follows that 'Yis a linear mapping. By the work done on pages 102-103, any rotation whose axis of rotation passes thru the origin is a linear mapping. Then proof that the axis of rotation of a passes thru the origin is proof that a is a linear mapping. Suppose that the axis of rotation of a does not pass thru the origin. Since 'Y= ai = ia either 'Y(O) = ia(O) = i(a(O)) = i(x) = -x or 'Y(O)= ai(O) = a(i(O)) = a(O) = x for some xf=O. Either case ends with a contradiction to the definition of'Y which states that 'Yleaves the origin fixed. Then a must have an axis of rotation that passes thru the origin. Therefore, a is a linear mapping. It follows immediately that 'Yis a linear mapping. P3.4 (page 108) Mn(a)

= [[a(a)ln

[a(b)ln

[a(c)lnl

= [-~ 1

-~ 0

-~] 0

P3.5 (page 108)

Mn(6)=[[6(a)ln

6(r)

=

6(b)ln

6(c)lnl=[a+b

[o1 -1 0] [1'1] = [rl 1

0 0

0 1

1'2

7'1

1'3

1'3

-a

cl=

[1o -1 0] 1

0 0

0 1

1'2]

P3.6 (page 110)

Mn({3a) = Mn({3)Mn(a)

f= [~

1 0 -1

[ -1 ~

=

-i]

-1 0 0

-~][-~

0 -1 o0] = [0 1 1 1 -1

0 1 0

-u

= Mn(a{3)

P3.7 (page 110)

Mn(.Ba) = Mn({3)Mn(a)

Mn(a{3) = Mn(a)Mn({3)

= [-1 -~

=

U

0 1 0 -1 0 0

J]U 0] o [-1-1 1 0

-1 0 0 0 1 0

0] 0=0 1

[-1 0

~] = [ .: -1 0

1 1 0 -1 0 0

J] J]

P3.8 (page 110) Using the guidelines of (RA3,2): (1.) det(Mn(a{3))=-1 which implies that the operation is a rotoinversion (2-3.) The operation has a turn angle of 1800 since tr(Mn(a{3)) = 1 and p = cos-l Cr(Mn~a:)) + 1) = cos-l«1 + 1)/( -2)) = cos-l( -1) = 180 0

•

The symbol for a{3 is m.

"3.9 (page 110) Using the guidelines of (RA3.2): (1.) det(Mn({3a))=-1 which implies that the rperation is a rotoinversion (2-3.) The operation has a turn angle of 180 since tr(Mn({3a)) = 1 md 0

l'he symbol for {3a is m. ~3.10 (page 110) Using the guidelines of (RA3.2): (1.) det(Mn({3a))=1 which implies that the iperation is a rotation (2-3.) The operation has a turn angle of 180 since tr(Mn({3a)) = -1 and 0

l'he symbol for {3a is 2. ~3.11 (page 120) Suppose w = Ba = 'Ya where w, 'Y,a, {3E 322, {3and 'Yare elements that appear n the guide column of the multiplication table for 322 (see Figure 3.15) and a is an element that .ppears in the guide row, Then both {3a and 'Ya appear in the same column (under a) of the table. lut each element of 322 appears exactly once in any given column. Then both {3a and 'Ya appear n the same row as well as the same column. Since both 'Yand {3are guide column elements that .re in the same row then 'Y= {3. Then right cancellation holds for elements in 322. ~3.12 (page 120) 3 1 3 3-1 [010J2 1 [010]2

I

I I

I

1

3

3-1

[100J2

1 3 3-1

3 3-1 1

3-1

1 [100]2

1

[010]2

[110}2

1 [010]2

[010]2

1 [110]2

1 3

1

I

I I

1

[100]2

1 [100]2

[100]2

1

[110]2

1

[110]2

1 [110]2

I

T,

1

1

~3.14 (page 121) The result of each operation in the group 422 is shown in figure 3.16. Examination f each figure shows that

Mn(l)

=

U

Mn(4)

=

[!

Mn(2)

=

[ -1 ~

l

Mn(4- )

= [ -~

0 1 0 -1 0 0

~] ~] ~] ~]

0 -1 0 1 0 0

Mn([100]2) Mn([010]2)

=

U

0 -1

0 0 [ -1 = ~ 1 0

J] J]

Mn([110]2) = [01 01 . 00]

o

_

Mn([110]2)

'3.15 (page 121) Since 422 is isomorphic to Mn(422), lot be presented due to limitations of available space.

=

[

0 -~

0

-1 -1 0 0

J]

a multiplication table for Mn(422)

will

422 1 4

[110]2

[ilo] 2

[100]2

[010]2

[110]2

[il0] 2

1

[110]2

[110] 2

[010]2

[100]2

4

[010]2

[100]2

[110] 2

[110]2

[110] 2

[110]2

[100]2

[010]2

4

2

4-1

[100]2

1

4

2 4-1

4-1

4

2 4-1

[010]2

1

2 4-1

2 4-1

1

[100]2

[100]2

1 [110] 2

4 [010]2

2 [110]2

1

2

4-1

[010]2

[010]2

[110]2

[100]2

[110] 2

2

4

4 4-1

4 4-1

1 4-1

1

2

4

2

1

[110]2

[110]2

[100]2

[110] 2

[010]2

[110] 2

[110] 2

[010]2

[110]2

[100]2

The rotations of the 4-fold axis are 4 = {I, 2,4,4-1}, those of the 2-folds are {100J2 = {[lOO]2, I}, {l10J2 = {[110]2, I}, [110J2 = {[110]2, I} and {010J2 = {[010]2, I} and that of the identity is 1 = {I}. The multiplication table for each group is shown below. P3.16

(page

4 I 1 4 2 4-1

1

4

1

4

4 2 2 4-1 4-1 1

{010J21

122)

{ 100 J21

2 4-1 4-1 1 1 4 4 2

1 [ 100]2

I

1

[ 100 ]2

1 [100]2

[ 100 ]2 1

[ 010 ]2

{110J2

I

[ 010 ]2

[ 110 ]2 1 1 [ 110 ]2 [ 110 ]2 1

(page

122)

1

1 1 [010]2[010]2 P3.17

4-1

2

Mn(l) Mn(6) Mn(3) Mn(2-l)

Mn(,1"l) Mn( 6-1)

Me(l) Me(6) Me(3) Me(2-1) Me(3-1) Me(6-1)

1

Mn(l) Mn(6) Mn(3) Mn(2) Mn(3-1 ) Mn(6-1)

Me(l) Me(6) Me(3) Mc(2)

Me(3-1 ) Me(6"l)

Mn(6) Mn(3) Mn(2) Mn(3-1 ) Mn(6-1) Mn(l)

1

[110 J 2

[ 110 ]2

Mn(3) Mn(2)

1 [110]

Mn(2)

Mn(3-1) Mn(6-l)

Mn(3-l) Mn(6-l)

1

~

1

1 2 [110]

Mn(3-l) Mn(6-1)

2

[110]

2

[110]

2

1

Mn(6"l) Mn(l) Mn(6) Mn(3) Mn(2) Mn(3-1 )

Mn(l) Mn(6) Mn(3)

Mn(l) Mn(6) Mn(3) Mn(2)

)

Mc(21

Me{3-l

1

Me{6"l

Me(6) Me(3) Me(2) Me(3-1)

Me(3) Me(2) Me(3-1)

Me(2) Me(3-1)

Me(3-1)

Me(6-l)

Me(6"l)

Me(6"l)

Me(l) Me(6)

Me(l) Me(6) Me(3) Me(2) Me(3-1)

Me(l)

Mn(l) Mn(6)

Me(6"l)

MeWl)

Me(l) Me(6) Me(3)

Me(l) Me(6) Me(3) Me(2)

Since there are no new elements in the multiplication table shown below, it follows that 6 is closed under composition. Since Ig=g for every gE 6, it follows that 1 is the identity element for 6. Finally, since 1 appears in each column and exactly once in each row of the table, it follows that for every element gE6 there exists a unique element g-1 E6 such that gg-l=l. That is, each element has a unique inverse. Then by definition of a group, 6 is a group under the binary operation of composition.

6

I

2

3-1

6-1

3

2

2

3-1 6-1

6-1

3-1 6-1

1

6

3

1 6 3

6 3

2

2

3-1 6-1

3-1 6-1

3-1 6-1

1 6 3

2

1

3-1 6-1 1 6

1 6 3

1 6 3

2

'3.18 (page 122) Due to limited available space, a multiplication table for i isomorphic to Mn(7i)

7i

I

1

6

3

m

3-1

61

1

6 3

3

m

6

m

3-1

6 -1

3

3

m

3-1

6 -1

1

1

m

3-1

6 -1

3-1

6 -1

6 -1

6 -1

1

1

6

7i is presented

61

1

m

2 3-1

3-1

6 3-1

1 6 3

1

6

6

3

6

3

m

3

m

3-1

since it

CHAPTER 4 P4.1 (page 124) All operations qualify as binary operations except for (7), which is not close, under the given operation, P4.2 (page 127) All systems qualify as binary operations except for: (3) Every element does no have an inverse and (4) The operation is not defined on the given set. P4.3 (page 128) Let P denote the set of all point isometries that leave an origin fixed and mal B into self-coincidence. Let (3, a E P. (1) Then (3a(B) = (3(a(B)) = (3(B) = B and (3a(o) = (3(a(o)) = (3(0) = 0 so that P is close. under the operation of composition. (2) Since every point isometry is a mapping then by TA1.4 P obeys the law of associativity. (3) There exists an identity in P, the point isometry 1. (4) Since a is one-to-one and onto, a-l exists and a-l(B) = a-l(a(B» = I(B) = Band a-leo) = a-l(a(o)) = 1(0) = o. Then a-l E P so that for every gEP there exists a unique invers g-l EP such that gg-l=l. Then P is a group under the operation of composition by definition. P4.4 (page 132) There are eight elements in the group 422. Therefore #(422)=8. P4.5 (page 135) (1) Since det(Mn({3)=det(Mn(a))=-1 then det(Mn(a{3))=det(Mn(a))det(Mn({3))=1. By TA3. a{3 is a rotation. (2) Since det(Mn({3))=-1 and det(Mn(a))=1 then det(Mn(a{3»=det(Mn(a))det(Mn({3))=-1. B: TA3.8 a{3 is a rotoinversion, an improper operation. P4.6 (page 137) Let G be a group defined under the operation *. Let gE G. Let gi,gi E< g:: where i, k E Z. Then gigi = gi+i E< g > by D4.24 since i + j E Z, It follows that < g > is closet under the operation of composition. Now let gk E< g >where k E Z. Then gi(gigk) = gi(gi+k) = gi+(i+k) = g(i+i)+k

= (gi+i)gi = (gigi)gk

which shows that < g > obeys the associative property. Since s" E< g > and for every gi E< g >, g-' E< g > it follows immediately from the discussion in D4.22 that properties (3) and (4) of D4.2 hold true. Then it follows that < g > is a group itself and therefore is a subgroup of G. P4.7 (page 140) It is known that 2={1, 2} so that 2i={i, m} and 2U2i={1, 2, i, m}. Since #(2)=2, then #(H)=I, where H is a halving group of 2. Then H =1 since it is the only subgroup of 2 with only one element. Finally 1U(2\1)i={I, m}. P4.8 (page 141) Using T4.28 we have that (1) 1={1} so that 1i={li}={i}. Then 1U1i={I}U{i}={I, i}. (2) There is no halving group for 1, since there is no group of order ~. P4.9 (page 141) It is known that 6={1, 6, 3, 2, 6-1, 3-l}. Since ~(#(6))=6/2=3 and 3 is th only subgroup of 6 with order three, it follows that 3={1, 3, 3-l} is the halving group of 6, Usin T4.28 we have that 6U6i= 6U {i, 6, 3, m, 31, 6-l} = {I, 6, 3,2,6-1, 3-1, i,6,3, m, 31, 'ifl}. 3U(6\3)i= 3U({6, 2,6-l})i= {I, 3, 3-1,6, m, 'ifl}. P4.10 (page 144) Mp(1)={Mp(1 )} Mp( l)UMp(1)Mp( i)={Mp(I), Mp(i)} There is no halving group for Mp(l). Mp(2)={Mp(I), Mp(2)} Mp(2)UMp(2)Mp(i)={Mp(I), Mp(2), Mp(i), Mp(m)} The halving group is Mp(1). Mp(1)U(Mp(2)\Mp(1))i ={Mp(I), Mp(m)} Mp(4)={Mp(I),

Mp(4),Mp(2),

Mp(4)UMp(4)Mp(i)={Mp(I),

Mp(4-l)} Mp(4), Mp(2), Mp(4-l), Mp(i), Mp(4), Mp(m), Mp(4

"»

The halving group is Mp(2).

={Mp(I),

Mp(2)U(Mp(4)\Mp(2)i

Mp(2),Mp(4),

Mp(4

Mp(3-l),

Mp(6)={

"» 1

Mp(I), Mp(6), Mp(3), Mp(2), Mp(6- )} Mp(6)UMp(6)Mp(i) ={Mp(I), Mp(6), Mp(3), Mp(2), Mp(3-l), Mp(6-1), Mp(i), Mp(m), Mp(3"-l), Mp(ij-l)} The halving group is Mp(3). 1 Mp( 3)U(Mp( 6)\Mp( 3)i ={Mp(I), Mp(3), Mp(3-1), Mp(m), Mp(6), Mp(6- )} P4.11 (page

o

= {Me(I)[rle,

0 1 0] 0 0 1

= { [~:] ,

Hi :

Me (4 -l)[rle}

Me(4)[rle, Me(2)[rle,

[7'I] [ -10 7'3 0 1'2,

1 0 0

7'12,] [ -10

0] 0 [ -1 1'3

0

-10 0

0] 0 1

[1'1] [ 01 1'2, 1'3

0

-10 0

0] 0 -1

[7'1]} 1'2

1'3

[=~~],[=~:], [~n }

1.875] orb e([r]c) = -0.237, { [ 4, 1.089

[-0.237] -1.875, -1.089

[-1.875] 0.237, 1.089

[

0.237]} 1.875 -1.089

-0.237] -1.875, -1.089

[-1.875] 0.237, 1.089

[

0.237] 1.875, -1.089

[

1.875]} -0.237 1.089

-1.875] 0.237, 1.089

[

0.237] 1.875, -1.089

[

1.875] [-0.237]} -0.237 , -1.875 1.089 -1.089

1.875] orb e([r]c) = 0.237, { [ 4, -1.089

[

0.237] [-1.875] -1.875 , -0.237, 1.089 -1.089

H2 :

orb e([rle) 4,

=

H3 :

orb:4e([rle) ,

=

H4 :

_

Mp(3),

147) Let [r]c denote the triple of the coordinates of a hydrogen atom. Then

orb4,e([rle) _ - { [10

_

Mp(6),

{ [

{ [

[-0.237]} 1.875 1.089

P4.12 (page 147) Let [rle denote the triple of the coordinates of a silicon atom. Then [rle = [0 0 olt and orb ,e([r]c) = {[O 0 On which implies that the silicon is on a special

4

position and its orbit consists of a single point. P4.13 (page 150) Let [r)p. denote the triple representative of the Miller indices of a plane. Then

P4.14 (page

150)

Let q=[;;siln, r=[;;o;ln,

s=[~ln,

t=[~ln,

and u=[;;o;Jn, Then

Mn(4/m)={ Mn(I), Mn(4), Mn(2), Mn(4-1), Mn(i), Mn(4), Mn(m), Mn(41)} =

{

la,

0-10] 1 00, o 01

[

[-1 0 -10 00] , [010] -100 0 01 001

orb4/m,D(q) = { q, [

- .3085] .0118, .1921

{

0] 0] , [0-1 OJ} . 0 , [10 01 0 1 0 0 00-1 0 0-1 [.3085] -.0118, -.1921

[

.0118] .3085, -.1921

[- .3085] } .0118 -.1921

[,3024] } .0400 .0000

-.1754] .0488, .2684

[-.0488] [.1754] [.1754] -.1754 , -.0488 ,-s, -.0488, .2684 .2684 -.2684

[

.0488] .1754, -.2684

[-.1754]} .0488 -.2684

-.4023]

[-.1324] [.4023] -.4023 , -.1324 ,-t, .1934 .1934

[

.1324] .4023, -.1934

[-.4023]} .1324 -.1934

= { s, [

orb4/m,D(t) =

00-1

[- .0118] [ .3085] -.3085 , -.0118 ,-q, .1921 .1921

-.3024] orb 4/m D(r) = { r, -.0400 ,-r, [ .0000 s orb 4/m,D(s)

,-13, [01 -10

t, [ .1324, .1934

- .3676] [ - .0977] [ .3676] orb4/m,D(u) = { u, -.0977 , -.3676 , -.0977 ,-u, [ .3062 .3062 .3062

II q 11=3.647 =114(q) II lis 11=2.892 =114(s) II II u 11=4.751 =114(u) II q. s = 9.550 = 4(q). 4(s) q. U = 16.632 = 4(q). 4(u) r. t = 13.389 = 4(r). 4(t) s e t = 12.142 = 4(s). 4(t) t. u = 19.269 = 4(t). 4(u) L(q: s) = 25.1170 = L(4(q: 4(s) L(q: u) = 16.2540 = L(4(q : 4(u) L(r: t) = 31.4410 = L(4(r: 4(t) L(s : t) = 28.9160 = L(4(s: 4(t) L(t : u) = 32.2550 = L(4(t : 4(u)

.[ .4023] -.1324, -.1934

[.3676 ]" [- .0977] -,0977, .3676, - .3062 - .3062

II r 11=3.272 =114(r) II lit 11=4.796 =114(t) II q. r = 10.680 = 4(q). 4(r) q. t = 16.808 = 4(q). 4(t) r e s = 5.879 = 4(r). 4(s) r. u = 13.241 = 4(r) • 4(u) s e u = 12.062 = 4(s). 4(u) L(q: r) = 26.481 = L(4(q: L(q: t) = 16.058 = L(4(q: L(r: s) = 51.597 = L(4(r: L(r: u) = 31.5950 = L(4(r: L(s: u) = 28.613 = L(4(s: 0

0

0

0

4(r) 4(t) 4(s) 4(u) 4(u)

[ - .3676] } -.0977 - .3062

CHAPTER 5

P5.1

(page

159) C2(322) = {P21,P22,P23}

[P211= {g(P2l)

Ig

since

E 322}

= {1(P2l),3(P2i),3-1(P21),[100] 2(P21),[OI0] 2(P2l),l1l0] 2(P2 I)'} = {P2l,P22,P23} [p22] = {g(P22) I g E 322} = {1(P22), 3(P22), 3-1 (P22),[100] 2(P22),[OI0] 2(P22),[1l0] 2(P22)} = {P2l, P22, P23} [P23J = {g(P23) I g E 322} = {1(P23), 3(P23), 3-1 (P23), [100]2(P23), [010]2(P23),[llO] 2(P23)} = {P2l, P22, P23}

and C3(322) = {P3l,P32,P33}

since

[P311 = {g(P3l) I g E 322} = {1(P3l),3(P3l),3-1(P3l),l100]

2(P3l),l°1O] 2(P3l),[llO]2(P31 ),} = {P3l,P32,P33} E 322} = {1(P32), 3(P32), 3-1 (P32),[100] 2(P32),[OI0] 2(P32),l1l0] 2(P32)} = {P3l, P32, P33}

[P321= {g(P32)

Ig

[P331 = {g(P33)

Ig

E 322}

= {1(P33), 3(P33), 3-1(P33),[100] 2(P33),[010] 2(P33),[1l0]2(P33)}

= {P3l, P32, P33}

P5.2 (page 159) A diagram of the rotation axes of 422 is given in Figure 5.5. Then Cl(422) = [Plll = {Pll,P12}, C2(422) = [P211 = {P21,P22,P23,P24} and C3(422) = [p3d = {P3l,P32,P33,P34}' P5.3

(page

161)

{Mn(3)Mn(h)Mn(3-1)

Ih

E (322b}

= {Mn(3)13Mn(3-1),Mn(3)[100]2Mn(3"1)}

O]}

= { [01 -1 -1 0] 0 [10 01 0] 0 [-1 -1 10 0] 0 , [01 -1 -1 0] 0 [10 -1 -1 00] [-1 -1 01 0 o 0 1 0 01 0 0 1 0 0 1 0 0 -1 0 0 1

OJ}

= {[100] 0 1 0 , [-10 -1 1 0 o01 0 0 -1

P5.4 (page 164) n2 = #(C2(422)) #(G3j) = 2. Then which verifies L5.8

= {Mn(I),Mn([OI0]2)}

= Mn«322)22)

From previous work it is known that N=#(422)=8, t = 3, nl = #(CI(422)) = 2, = 4, n3 = #(C3(422)) = 4, Vi = #(Glj) = 4, V2 = #(G2j) = 2 and V3 = 2(N-l)=2(8-1)=14 and L::=l ni(vi - 1) = nl(vl-l)+n2(v2-1)+n3(v3-1) = 14 for 422.

P5.5 (page 165) Values for N, ni and Vi (for each 1 ~ i ~3) are known from previous work. Substitution of the appropriate values gives nl Vi = 2 X 4 = 8 = N, n2v2 = 4 X 2 = 8 = N and n3v3=4x2=8=N. P 5. 7 (page 172) Since 622 is isomorphic to Mp( 622), a multiplication table for 622 will be presented due to limitations of available space.

622

1

6

3

2

3-1

6-1

1

1

6

3 2 3-1 6-1

2 3-1 6-1

3-1 6-1

6-1

1

6

1

6

3 3 2 3-1 2 3-1 6-1 6-1 1 [100]2 (il0] 2 [110]2 [210]2 [010]2 [120]2 [110]2 [010]2 [210]2 [100]2 [120]2 [110]2 6

6

3 2 3-1 6-1 100]2 110]2 010]2 [10] 2 210]2 120]2

1

6 3 [010]2 [120]2 [100]2 (il0] 2 [110]2 [210]2 [120]2 [110]2 [110]2 [010]2 [210]2 [100]2

1

3 2 3 3-1 2 [110]2 [210]2 [010]2 [120]2 [100]2 [110]2 [210]2 [100]2 [120]2 [110]2 [110]2 [010]2 6

[100]2 [110]2 [010]2 rr 10] 2 [100]2 [210]2 [110]2 [120]2 [010]2 (il0] 2 1 3 3-1 6-1

[110]2 [120]2 [010]2 (il0] 2 [100]2 [210]2 3-1

6

1 3 2 6-1

2

6

[210]2 [120]2 [120]2 [010]2 (ilo] 2 [100]2

(il0] 2 [100]2 [210]2 [110]2 [120]2

[100]2 [210]2 [110]2 [120]2 [010]2

[210]2 [110]2 [120]2 [010]2 (il0] 2 [100]2

3 3-1

6

6-1

2

2 6-1

6

6-1

[010]2 (il0] 2

1

1 3 3-1

6

2 6-1

[210]2 [110]2

2 3-1 1 3

6

3 3-1 1

) 5.8 (page 173) Define ten vectors rl to rlO each of which has its head at a common origin, 0, .nd its tail at a point of intersection with the unit circle centered at o. Let consecutive vectors be eparated by an angle of 18°. Let rl denote the basis vector i. Since the basis is cartesian, G = la. [ote that by construction for 1 ::; i ::; 10 II r, 11=1 and ri = aia + {3ib for some ai,{3i E IR. Then

rl.ri=a.ri=llallllrillcosL(a:ri)=cosL(a:r;)a.ri=[1

r6 Hi = b Hi =11b 1111 r,

'0

II cosL(b:

ri)b Hi = [0

0

Ol[~:]

=ai

1

01 [~:]

= (3i

that ai = cos L(a : ri) and (3i = cos L(b : ri). Then the following coordinates are obtained

al = cos L(rl : rl) = cos(OO)= 1 a2 = a3 = a4 = as = a6 = a7 = as = a9 = alO =

cos L(rl cos L(rl cos L(rl cos L(rl cos L(rl cos L(rl cos L(rl cos L(rl cos L(rl

: r2) = cos(18°) = .95 : r3) = cos(36°) = .81 : r4) = cos(54°) = .59 : rs) = cos(72°) = .31 : r6) = cos(900) = 0 : r7) = cos(108°) = -.31 : rs) = cos(126°) = -.59 : r9) = cos(144°) = -.81 : riO) = cos(162°) = -.95

50 5={ 1,5,5-1,5/2,5/2-1} :-.81 .59 012,[-.81 -.59

(3l = cos L(r6 : ri) = cos(900) = 0 (32= (33= (34= (3s = (36= (37= (3s = (39= (3lO=

cos L(r6 cos L(r6 cos L(r6 cos L(r6 cos L(r6 cos L(r6 cos L(r6 cos L(r6 cos L(r6

: r2) = cos(72°) = .31 : r3) = cos(54°) = .59 : r4) = cos(36°) = .81 : rs) = cos(18°) = .95 : r6) = cosfO"] = 1 : r7) = cos(18°) = .9.5 : rs) = cos(36°) = .81 : r9) = cos(54°) = .59 : rlO) = cos(72°) = .31

and the set of half-turns in 522 are [1 0 012, [.31 .95 012, 012and[.31 -.95 012. The matrix for each of these operations

is as follows

M;

=

Mn(l)

= 13

M2

=

Mn(5)

=

M3

M4

Ms

[ .31 .~5

=

~] ~] ~] ~]

-.59 -.81 0 .59 [ -.81 Mn(5/2-1)= -059 -.81 0 -.95 [ .31 MnWl) = .~5 .31 0

= Mn(5/2) = =

-.95 .31 0

[ -.81 .~9

M6

=

M7

=

Ms

=

M9

=

MlO =

Mn([1

0

ob)

0 -1 0

= [~

~J

.59 .81 0 -.95 [ .31 Mn([ -.81 .59 Ob) = -095 -.31 0 .95 [ .31 Mn([ -.81 -.59 Ob) = .~5 -.31 0 -.59 [ -.81 Mn([ .31 -.95 Ob) = -059 .81 0 .95

Mn([·31

Ob)

=

[ -.85 .~9

~J ~J ~J ~1]

Given that 522 is isomorphic to Mp( 522), a multiplication table for 522 (using the labels given above) will be presented due to limitations of available space. 522

Ml M2 M3 M4 Ms

Me M7 Ms M9 M10

Ml Ml M2 M3 M4 Ms M6 M7 Ms M9 MIO

M2 M2 M3 M4 Ms Ml Ms M9 M10 M6 M7

M3 M3 M4 Ms Ml M2 MlO M6 M7 Ms M9

M4 M4 Ms Ml M2 M3 M7 Ms M9 MlO M6

Ms Ms Ml M2 M3

~

Mg MlO

14 M7 Ms

M6 M6 M9 M7 M10 Ms MI M3 Ms M2 M4

M7 Ms M7 Ms MIO M6 Ms M9 M6 M7 M9 MIO M4 M2 Ml M4 M3 Ml Ms M3 M2 Ms

M9 M9 M7 MlO Ms M6 Ms M2 M4 Ml M3

MlO MlO Ms

14 M9 M7 M3 Ms M2

~ Ml

P5.9 (page 174) Since 322 has one 3-fold axis and three 2-fold axes it has eight pole points. By the way the basis P = {a, b, c} for 322 is defined the triples for these pole points on the unit ball Bare { a/a, bib, -~ - ~, ~ + ~, -e] a, -bib, cl c, -c /c }. The representatives formed from the zone symbol associated with these pole points are

{[a/alp, [b/blp, [-~

-

~L,[~+~L

,[-a/alp,

[-b/blp,

[c/clp, [-c/clp}

= {[100], [0101,[TIo], [110],[1001,[010], [001],[001]}.

The equivalence class of a given pole point p is the set {g(p) I g E 322}. Then the equivalence classes of 322 are as follows

[al = {Mp(g)[alp) I g E 322} = {[alp, [blp, [-a - blp} [a + b] = {Mp(g)[a + blp) I g E 322} = {[-alp, [-blp, [a + b]»] [cl = {Mp(g)[clp) I g E 322} = {[clp, [-clp}

which is consistent with the information given in Table 5.1.

P5.10 (page

175) We are given that G = gllIa and G* =

11[111111= ([1111g;1113

11[111111= ([1

-1

UJ);

= j3fu

g;/Ia.

Then

11[001111=([0011g;/Ia

[~J);

=..;gu

[-u)~

11g;11I3

= j3fu

[1111. [0011= gll =11[111111 11[001111cos(L([1111 : [001])) = j3fu..;gu

cos(L([1111 : [001]))

= V3gll cos(L([1111 : [001])) [1111. [1111= gll =11[111111 II [111111cos(L([1111 : [111])) = j3fu

j3fu

cos(L([1111 : [111]))

= 3gn cos(L([1111 : [111]))

which implies that L([1111 : [001]) = cos-l(ta)

= 54.740 and L([1111 : [111]) = cos-l(~)

g;/Ia.

f>5.12 (page 178) We are given that G = gllI3. Then G* = md II [001)11= ..;gu. Furthermore,

11[101111= ([1011gll13

= 70.530•

II [111111= y'3fu

From P5.10,

[~J)~ y'2fu =

[001] • [101) = gll =11[001111 II [101)11cos(L([0011 : [101])) = =

[1011. [1111= =

v'fu y'2fu cos(L([0011 : [101])) hgll cos(L([0011 : [101])) 2g11 =11[101] II 11[111111cos(L([1011 : [111])) = y'2fu j3fu cos(L([1011 : [111])) V6g11 cos(L([1011 : [111]))

which implies that L([0011 : [101]) = cos-1(1)

= 35.260•

= 450 and L([111] : [111]) = COS-l(~)

:>5.13 (page 178) Since 432 has four 3-fold axes it has eight associated pole points. The pole ioints associated with these third-turns are all 432-equivalent. Now the pole point represented by 1111is associated with a third-turn. Since Mp(I)[1

1

1]t = [1

Mp(4)[1

1

11t = [-1

Mp(2)[

1

1

Mp(4-

1

[1

)

11t = [1

Mp([OlO]4)

[1

1

1]t = [1

Mp([OlO]2)

[1

1

11t = [-1

-1

1]t

Mp([100]2)

[1

1

1]t = [1

-1

1]t

Mp([OlO]2)

[1

1

11t = [-1

1]t

1

11t = [-1 1

11t

1

-11t

1

-11t

1

-11t

-1

-11t

-1

hen each of the pole points represented above are associated with a third-turn. :>5.14 (page 179) Since 432 has six 2-fold axes it has twelve associated pole points, The pole Joints associated with these half-turns are all 432-equivalent. Now the pole point represented by 1011is associated with a half-turn. Since Mp(I)[l

11t=[1

0

Mp(4) Mp(2)

[1

0

1]t=[O

[1

0

1]t = [-1

1

Mp(4- )[1

0

11t = [0

0

Mp([lOO]4)[1

1]t

Mp([lOO]4-l) [1

lilt 0 -1

Mp([OlO]4) [1

0

1]t=[1

Mp([0IO]2)[1

0

1]t = [-1

Mp([ill]3)[1

11t i ]'

0

-11t 0

-11t

0 0

Mp«(ill]3- )[1

Mp([ll1]3-1)[1

1

r = [1

1]t = [0 0

1

Mp([111]3) [1

11t=[1

0

0

11t = [0

hen each of the pole points represented above are associated with a half-turn.

olt

1

olt

-1

1]t=[-1 1]t = [-1

0

-1

_IJt 1

-1 1

OJt 0 Jt _IJt

P5.15 (page 179) Since 432 has three 4-fold axes it has six associated pole points. The pole points associated with these quarter-turns are all 432-equivalent. Now the pole point represented by [0011 is associated with a quarter-turn. Since

Mp([OI0]4)[0

0

11t=[1

0

olt

Mp([OI0]2)[0 Mp([OI0]4-1)

0

11t = [0

0

-11t

[0

0

11t = [-1

0

olt

Mp(l) [0 0 1]t = [0 0 11t Mp([100]4)[0 0 11t=[0 -1 olt 1 Mp([lOO]4- ) [0 0 11t = [0 1 olt

then each of the pole points represented above are associated with a quarter-turn. P5.17 (page 179) The rotations are as follows: [100J4 = {1,ft°0] 4,ft°0] 4-1 ,ft00]2}, l = {I, 4, 4-1,2} and [01OJ4 = {I ,[010]4,[010]4-1 '[010]2}. P5.18 (page 179) The rotations are as follows: [101J2 = {1,[101]2}, [011J2 = {1,[011]2}, 701J2 = {1,[101]2}, [011J2 = {1,[011]2} [110J2 = {1,[110]2} and [111J2 = {1,[110]2} f'5.19 (page 179) The rotations are as follows: 432 = {I, [100]4, [100]4-1, [100]2, [010]4, [010]4-1, 010]2" 4 4-1 " 2 [101]2, [011]2, [101]2, [011]2, [110]2, [110]2, [111]3, [ill]3 , [111]3, [111]3, [111]3, 111]3, [111]3, [111]3}. :>5.20 (page 182) To determine the full Hermann-Mauguin symbol first discover the three non.quivalent axes by finding the equivalence classes of the pole points of 422 U 422i under 422.quivalence where 422 U '22i = {I 4 4-1 2 [100]2 [110]2 [010]2 [110]2} U {I 4 4-1 2 [100]2 [110]2 [010]2 [ll0]2}i "" , " """, = {I 4 4-1 2 [100]2 [110]2 [010]2 [110]2 i '4 -4-1 m [100]m [110]m [010]m [110]m} " " , , , ," " , , ,

.,

[J]} [!],[-~],[-n} {U] ,[-i] ,[-iJ '[=i]} .

{MP(g)

[~]

Ig

E 422U

422i}

= { [~] ,

{MP(g)

[~]

Ig

E 422U

422i}

= { [~] ,

{MP(g)

[iJ

Ig

E 422U

422i}

=

'hen three non-equivalent axes are along the zones [001], [1001 and [1101. The subgroups associated rith each are 41m, 21m and 21m, respectively. Hence the HM symbol is 41m21m21m. '5.21 (page 182) To determine the full Hermann-Mauguin symbol first discover the three nonquivalent axes by finding the equivalence classes of the pole points under 4 U {42214}i-equivalence rhere 4 U {42214}i = {1,4,4-l, 2} U {[100]m,[110]m,[010] m,[110]m} , {1,4,4-1, 2,[100]m,[110] m,[OIO] m,[110] m}

hen three non-equivalent axes are along the zones [001], [1001 and [1101. The subgroups associated ith each are 4, 21m and 21m, respectively. Hence the HM symbol is 4mm.

P5.22 (page 182) To determine the full Hermann-Mauguin symbol first discover the three nonequivalent axes by finding the equivalence classes of the pole points under 622 U 622i-equivalence where 622 U 622i = {I, 6, 3, 2, 3-1,6-1,[100]2,[210] 2,[110]2,[120]2,[010]2/110] 2} U{I, 6, 3, 2, 3-1,6-1,[100]2,[210] 2,[110]2,[120]2,[010]2,[11O]2}i = {I, 6, 3, 2, 3-1,6-1,[100]2,[210]2,[110]2,[120] 2,[010]2,[110]2, i, 6, 3, m,

31,'61 pOOlm,[210] m,[l1O] m,[120] m,[010] m,[110] m}

Then three non-equivalent axes are along the zones [001], [1001and [2101. The subgroups associated with each are 6/m, 2/m and 2/m, respectively. Hence the HM symbol is 6/m2/m2/m. P5.23 (page 182) The elements of 432 were determined in P5.19. Therefore it is easily verified that 432 U 432i is as shown on pages 182-183. To determine the full Hermann-Mauguin note that [001], [1111and [1001 are three non-equivalent axes since equivalent pole points, at the very least, must be associated with point groups of equal order. Then the subgroups associated with these three zones are 4/m, 3 and 2/m, respectively. Hence the HM symbol is 4/m712/m. P5.24 (page 183) The point symmetry is 322. Since P = {ap,bp,cp} and C = {ac,bc,cc} then The change of basis matrix T from P to C is derived as shown below bp • be ap • be

T-l =

= cos(30 = :I,} = 0

)

=

cos(120

=

0 )

[aplc = [aclc = [1001t [cplc = [cc]c = [0011t [bplc = [~ o]t

t (3 [01011a[a{30J= -0.5 = [1001I3[a{30j1= a

¥

1 _1 0] [ o 40

0

3

0

For each a E 322 Me = TMp(a)T-l.

Mc(3) Mc(3-l)

Then

= 13 =

[ -.500 .86~

[ -.500 = -.86~

MC([100]2) -.866 -.500 0 -.866 -.500 0

---t ---t

~] ~]

= [~

0 -1 0

MC([010]2)

=

[ -.500 -.86~

MC([110]2)

=

[ -.500 .86~

J] -.866 .500 0 .866 .500 0

-->

=

orb322,c(oSi2)

= orb322,c(oSi3) =

oSil,oSi2,oSi3

orb322,C(001)

---t

=

orb322,C(002)

=

orb322,c(003)

001,002,003

orb322,C(~)

=

orb322,c(~)

=

orb322,C(~)

---t {---t ---t

= orb322,C(oH6) =

j] j]

---t {--> -->} ---t = {---t ---t ---t}

orb322,C(oSil)

---">

[bc]c

~ o 0] :Ji 1

0 1

Mc(l)

f=

= orb322,c(~) --->

---t ---t ---t}

oHl,oH2,oH3,oH4,oHs,oH6

= orb322,c(~)

=

t

[010J

It follows that the atoms on special positions are Sil, Si2, Si3, 01, O2 and 03. P5.26 (page 193) Suppose the composition of two fifth-turns is a half-turn.

By TA6.2

which does not exist. Therefore the composition cannot result in a half-turn. P5.27 (page 193) Let al and a2 be fifth-turns P3 = 120. By TA6.4 2 L( al ,,a2 ) -- cos -1 [pd )

and a3 a half-turn.

cos(p2/2) ± COS(P3/2)] sin(pd2) sin(p2/2)

Then PI = P2 = 720 and

_- cos -1 [.6545 ± .5] __ 63435 . .3455

where the negative sign is chosen since choice of the positive sign proves to be an impossible choice. Then Pi = P2

P5.28 (page 193) Let al, a2 and a3 be fifth-turns.

=

L(al : (2)

.6545 ± .8090] .3455

cos"? [

=

= P3 =

72

0 ,

By TA6.4

cos-l( -0.4472) ~ 116.565

where the negative sign is chosen since choice of the positive sign proves to be an impossible choice. P5.29 (page 195) Let lij denote the entry in the ith row and the lh column of Mc([Ol'T]5)where P = 720 and L = [0 .52574 .850651t. Then cp = .309, sp = .951 and by (A3.1) 1~(1 - cp) + cp = ,309 h/2(1 - cp) -/3Sp = -.85065(.951) = -.809 11/3(1 - cp) + 12sp = .52574(,951) = .5 1211(1- cp) + 13sp = .85065(.951) = ,809 122 = 1~(1 - cp) + cp = .525742(.691) + .309 = ,5 123 = /213(1 - cp) - liSP = .52574( .85065)( .691) = .309 131= 1311(1- cp) -/2sp = -.52574(.951) = -.5 132 = 13/2(1 - cp) + hsp = .85065(.52574)(.691) = .309 133 = 1~(1 - cp) + cp = .850652(.691) + .309 = .809 111 = 112= /13 = 121=

The work is similar for Mc([O -1 r]5) P5.30 (page 195) Upon multiplying the matrices it is seen that 1-

MC([Olr15)Mc([0 -1 r15) = ~ [ 2

T

1

T

-T

-1 T -

1

as expected. The methods described in RA3.2 are used to analyze this new matrix. (1) The matrix represents a rotation since it is the product of two matrices that represent two rotations. -3) Let M denote the new matrix. Since the tr(M) = 0 it follows that p = cos-l Cr(~)-l) = 1200 which implies that it is a third-turn. ' (4) By 4b 11 = (/32 -/23)/2sp = (T-l)/v'3, 12 = (113-13l)/2sp = 0 and 13 = (/21-112)/2sp = T /v'3. Note that TJ3(~, 0, = (1,0, T + 1) which implies that the two vectors are collinear. It follows that the matrix represents a third turn-rotation about the vector (1 ,O,T+ 1) which is denoted by MC([l 0 r+113).

7a)

P5.31 (page 196) Upon multiplying the matrices it is seen that the product is as shown on page 196. The methods described in RA3.2 are used to analyze this new matrix. (1) The matrix represents a rotation since it is the product of two matrices that represent two rotations.

(2-3) Let B denote the new matrix.

Since tr(B) = -1, it follows that p = cos ?

which implies that it is a half-t;:.:u=r=n.:_. __ (4) By 4a

11

= /~( =l- + 1) = = /±Kr"21 + 1) =

= (/32 - 123)/2sp

.5,12

= (/13 - 13d/2sp

=

Cr(~)-l)

/±~(7' +

0

= 180

1) = .309 and

13 = (/21 - 112)/2sp .809. Note that (7",1,7" + 1)/ II (7",1,7" + 1) 11= (.5, .309, .809) which implies that the two vectors are collinear. Then it follows that the matrix represents a half-turn rotation about the vector (7",1,7" + 1) which is denoted by MC([ -r 1 r +1]2). (page 196) Upon multiplying the matrices it is seen that the product is as shown on page 196. The methods described in RA3,2 are used to analyze this new matrix. (1) The matrix represents a rotation since it is the product of three matrices each representing a rotation. 0 (2-3) Let B denote the new matrix. Since tr(B) = 0, it follows that p = cos"? = 120 P5.32

Cr(~)-l)

which implies that it is a third-turn. (4) By 4b 11 = (/32 - 123)/2sp = 12 =

7s' = (~'7a'7a)

(113 -

13i)/2sp

= ~

and 13

=

(121 - 112)/2sp

= ~.

Note that ;7a(7",7",7") which implies that the two vectors are collinear. Then it follows that the matrix represents a half-turn rotation about the vector (7",7",7") which is denoted by Mc([ -r-r-r ]2).

CHAPTER 6 P6.1 (page 200) Proof: Let M, and M2 be proper unimodular matrices. By D6.2, det(Ml) = det(M2) = 1 and the entries of MI and M2 are integral. Then each entry of M1M2 and M2MI is the sum of the product of integers which is again an integer. That is, MIM2 and M2Ml are both integral matrices, Furthermore, det(M1M2) = det(Ml) det(M2) = 1 and det(M2MI) = det(M2) det(Ml) = 1. Then by definition MIM2 and M2Ml are both proper unimodular matrices which proves the product of two proper unimodular matrices is again a proper unimodular matrix. P6.2 (page 204) By definition of Ln-equivalence, v ~ w since [wln - [vln = [0.3

0.3

01- [-13.7

12.3

61 = [14

-12

-61 E Ln.

P6.3 (page 207) For each of the point groups choose basis P as described in Chapter 5. Then as shown in Tables 5.2 and 5.3, Mp(a) is an integral matrix for all rotation isometries a. (Obviously, this implies rotoinversions are also integral matrices). Then by T6.13 for such a P Lp is left invariant by each of the point groups. P6.4 (page 210) As shown in Table 6.1 [o~or and [~oor are impossible fractional coordinates 'or 2 and therefore cannot be in L/ P. Suppose [~O~lt, [~Hlt E L/ P. Then since 1

1

2"

(modulo P)

2"

md L/ P is a group it follows that [O~olt E L/ P, a contradiction. .hen [0 which implies that [~

~ 0

H+[~ ~

~lt=[~

0

olt

Similarly, if[0Hlt,

[~Hlt E L/ P

(moduloP)

O)t E L/ P which is also a contradiction.

:>6.5 (page 211) The transformation matrix from P to P2 is

vhich is a proper unimodular matrix. Therefore by T6.3, P2 is another basis for the lattice P since ~p, = Lp. Since P2 has the same handedness as P, P2 is a right-handed basis. Furthermore, P2 [ualifies as another P-basis since {a2' b2} forms a basis for the lattice perpendicular to the 2-fold ixis and C2is the shortest lattice vector along the axis. Finally, the A lattice of P is the same as he I lattice of P2 since

'6.6 (page 212) (1) By T6.13 Ln is invariant under [010]2 since [010]2generates [010]2 and

o 1

o

J]

an integral matrix. (2) By T6.15 there exists a nonzero lattice vector along the 2-fold axis and here is a lattice plane passing thru the origin perpendicular to the 2-fold. Let b denote the shortest onzero vector along the 2-fold axis and let {a, c} denote a basis for the lattice plane perpendicular ) the 2-fold so that {a, b, c} is a right-handed system. Then the lattice generated by {a, b, c} . the same as D described in (1). If Ln f= L, then Ln is a proper subset of L and so there are ectors in L whose coordinates in Ln are fractional. By T6.9, each vector in L has a representative i

Ii ~

vhose coordinates II, 12, and fa with respect to D are such that 0 ~ f]n = [/11213]1 where 0 ~ t, ~ 1. Then

o [[OI0]2(f)] n = Mn([010]2)[f]n

= [-~

1. Suppose f ELand

0] [11] [- h· h]

o

1

o

12 fa

-1

=

-fa

f e is defined to be [010]2(f) + f then eEL and [eln = Mn([010]2)[f]n + [f]n = [0 212 0). Since ) is the shortest vector along the 2-fold axis in L, then e is a multiple of b and so 212 E Z. Then f2 = ! or O. Similarly if e is defined to be f - [010] 2 (f) then eEL and [eln = [211 0 2fa 1 Since :a,c} is a basis for the lattice plane perpendicular to the 2-fold axis if follows that 2h, 2/3 E L, hen h = ! or 0 and fa = ! or O. Then the combinations of fractional coordinates that arise are rs follows Possible

Contradiction

Impossible

h

h

fa

0 1

0 1

0 0 1

'2

I I '2

0 1

'2

h

h

fa

'2

0 1

0 0 1

0 0 1

I '2

'2

0 0

'2

choice of a choice of b choice of c choice of {a, c}

I '2

3) Ln/Ln

I

[OOolt

I

G/Ln

[HOJt

[OOolt

-

[OOolt

A/Ln

[Ooolt [OHlt

I I

I

[OOOlt

[OOOlt [Holt

I I

[Ooolt

[OHlt

J/Ln

[OOOlt [OHlt

[OHlt [Ooolt

[Ooolt [HW

4) As shown in the table above [!O!lt are impossible HOJt, [OHlt E L/ Ln, Then [!O!lt E L/ Ln since [Holt

+ [O!H

I

[000]1 [Holt

[Holt [OOOlt

[Ooolt

[H!lt

[Ooolt [HW

[H!lt [OOO)t

fractional

= [!OH

coordinates

for [010j2.

Suppose

modulo D,

~contradiction. It follows that both vectors cannot be in L/ Ln. 5) Consider Pi = {a - c, b, -a} and P2 = {a - c, b, c}. The change of basis matrix from D to Pi md the change of basis matrix from D to P2 are

o 1

o

-~]

o 1

o

-1

espectively. Both change of basis matrices are unimodular to qualify as a P basis. Since

~]

and both the bases satisfy the criterion

iecessary

o 1

o

-1] _~![0]

=

[!] ~

modulo Pi

[1 ~

o 1

o

1]

mm =

modulo P,

2

hen the A lattice of D is the same set of points as the C lattice of Pl. Similarly the J lattice of ') is the same set of points as the C lattice of P2• To show that C is left invariant by [010j2 recall hat Mnc([010]2) = T-1Mn([010]2)T where T is the change of basis matrix from Dc to D so that

o 1

o

J]

which is unimodular. Therefore by T6.13, since [010J2 generates [010J2 and Mnc([OI0]2) modular matrix, it follows that Lnc is invariant under [010J2.

is a uni-

P6.7 (page 215) PIP

[OOOlt

I

[OOO)t

I

[OOOlt

P

[OOolt

[EW

[~Hlt

[OOOlt

[OOOlt

[EW

aHlt

[~Hlt

[EW

aHlt

[OOolt

[~Hlt

[~Hlt

[OOO]t [EW

R(obv)/

[OOolt

[~Elt

[~Hlt

[OOolt

[~Elt

[~Hlt

aHlt

[~Hlt

[~Hlt

[OOolt

[~Hlt

[~Hlt

[OOolt

[~H]1

P6.8 (page 215) Let DR = {anR' bnR, cnR} as is defined in the problem. By definition of R( obv) ~iven in P6.7, it is obvious that every vector in DR is in R(obv). Also every vector in R(obv) .s an integral combination of DR. In particular, [000]1 = O[alnR + O[blnR + O[clnR' [HW = )[alnR + O[blnR + l[clnR and [~Hlt = l[alnR + O[bJnR + O[clnR' P6.9 (page 215) Let A be the matrix whose columns are the vectors given as a possible basis Dr this problem. Then det(A) = 0 which implies that the possible basis has linearly dependent rectors. Since basis vectors must be linearly independent, it follows that the given set of vectors is iot a basis for R( obv). :>6.10 (page 217) The definition of I is given on page 217. It is obvious, given this definition .hat each vector in D t is a member of the set of all vectors that make up the lattice 1. Also note hat [OOolt= [OOOlb,and that = [0011b,. Then every vector in I is a linear combination of he vectors in D[. It follows that D[ is a basis for I.

[HW

"6.11 (page 217) By T6.3 4(Dl) natrix from 4(Dl) to Dl is

and Dl generate the same lattice since the change of basis

T =

[0 -1 -1] 1 001

0

0

rhich is unimodular. '6.12 (page 220) By T6.3 6(P) and P generate the same lattice since the change of basis matrix rom 6(P) to P is -1

o o

~]

rhich is unimodular. '6.13 (page 221) A basis for C is Dc = Ha + ~b,b,c} since every vector in Dc is in C and very vector in C is a linear combination of the vectors in Dc. To show C is invariant under 222 ; is only necessary to show that C is invariant under 2 and [100]2 the generators of 222. Now (Dc) = {-ta - ~b, -b, c} and [100J2(Dc) = Ha - ~b, -b, -c} so that -1 Mc(2) = [ ~

o -1

o

o -1

o

~] .

-1

By T6.13, since both matrices are integral, it follows that C is invariant under 222.

= {aDF, bDF' = [BOr so that

P6.14 (page 221) To see that DF [bDFJDF

= [~O~)t

and [CDFJDF

[OOOJt= 0 X [OB)t = 1 X

[aDFJDF

[~O~f = 0

[aDFJDF

[HO)t

X

[aDFJDF

= 0 X [aDFJDF

CDF} is a basis for F note that [aDFJDF every vector of DF is in F. And

+0 X +0 X +1X +0 x

+0 X +0X +0 X +1X

[bDFJDF [bDFJDF [bDFJDF [bDFJDF

=

[oBf

[CDFJDF [CDFJDF [CDFJDF [CDFJDF

so that every vector in F is an integral combination of vectors in DF.

Then DF is a basis for F.

P6.15 (page 221) To show F is left invariant under 222 it is necessary to show F is invarianl under 2 and [100J2, generators of 222. Now 2(DF) = {-~b + ~c, -~a + ~c, -~a - ~b} and [100J2(DF) = {-~b - ~c, ~a - ~c, ~a - ~b} so that 1

-1

o

o

-1

1

-1]

1 .

o

By T6.13, since both matrices are integral, it follows that F is invariant under 222. P6.16 (page 222) A basis for Dc was found in P6.13. Using that same basis, let PI = {b, c, ~a + = Ha + ~b,c, -b}. Then the change of basis matrix from Dc to PI and the change of basis matrix from Dc to P2 are such that

~b}and P2

0

1 0 1 0

o [

and

:] m ~[!]

o o

[i

-1

modulo P,.

Since both change of basis matrices are unimodular matrices, by T6.3, it follows that the lattices generated by Dc, PI and P2 are equivalent. As seen above both the A and B lattices can b¤ expressed as a C lattice. P6.17 (page 222) To show R( obv) is invariant under [100J2 it is only necessary to show that it is invariant under [100J2, a generator of the group. Now [10oJ2(R(obv» = aa - ib - ~c,-~a - ibic, ia + ~b- ic} so that

~]

-1

o o

.

-1

By T6.13, since the matrix is integral, it follows that R(obv) is invariant under [100J2. It was already shown that R(obv) is invariant under 3. Therefore, R(obv) is invariant under 322 by T6.13. P6.18 (page 222) Since 4 and [100J2 are generators of 422, then only those lattices invariant under 4 and [100J2 can possibly be invariant under 422. As discussed on page 217 there are only two lattices invariant under 4, P and I. Then the only lattices that are possibly invariant under 422 are P and I. Orient C along the positive direction on the 4-fold axis, a along a 2-fold axis and b such that b = 4(a), then P is left invariant under 422 since as shown in Chapter 5, Mp(422) consists of integral matrices. In P6.11, it was shown that I is left invariant under 4. To see that I is left invariant under [100J2 note that [100J2(D/) = {a, -b, ~a - ~b- ~c} so that

o -1

o

~] .

-1

By T6.13, since the matrix is integral, it follows that I is invariant under [100J 2. It follows imrru diately that P and I are invariant under 422. P6.19 (page 222) Since 6 is a subgroup of 622, then only those lattices left invariant under can possibly be left invariant under 622. But only P is left invariant under 622. As in Chapter 1 choose the same basis P for 622 as was chosen for 6. Then as shown in Table 5.3 all matrices ar integral and so P is invariant under 622. P6.20 (page 223) As stated on page 222, the only lattices possibly invariant under 23 are tho! which are invariant under both 222, a subgroup of 23, and [111J3. The lattices invariant under 22 are the P, F, I and C lattices. Now P is defined on page 222 and is the SaIne P as was defined i Chapter 5. Then since Mp(23) are integral, it follows that P is invariant under 23. To see that th F and I lattices are invariant under 23 it is necessary to show that they are invariant under [ll1J< Now [lllJ3(DF) = {!a + ~c, ~a + ~b, ~b + ~c} and [lllJ3(D/) = {b, -a - b + 2c, c} so that

o o

-1 -1

1

2

By T6.13, since both matrices are integral, it follows that F and I are invariant under 23. P6.21 (page 223) To see that the C lattice based on P is not left invariant under 23 note tha for Dc as defined in P6.13

ml

Mc(,

3) ~

[1

o o

1

Then by T6.13, C is not invariant under 23. P6.22 (page 223) Since 23 and 4 are subgroups of 432 the lattices based on P that are invariar under 432 must be invariant under both 23 and 4. Now the lattices based on P invariant unde 23 are P, F and I. Since P is as defined in Chapter 5, as seen in Table 5.2, Mp(432) are integra matrices which shows that P is left invariant under 432. To see that the F and I lattices ar invariant under 23 it is necessary to show that they are invariant under 4. Now 4(DF) = {-~a~c, ~b + ~c, -~a + ~b} and 4(D/) = {b, -a, -~a + ~b + ~c} so that 1

-1

o o

o

o

-1]o . 1

By T6.13, since both matrices are integral, it follows that F and I are invariant under 432. P6.23 (page 227) Suppose there exists a unimodular change of basis matrix T that transform the matrices of Mp{42m) into those of Mp(4m2). Then for some Q E {Mp(a) where a E 4m2] TMp([100J2)T-I = Q. By DA3.3, Mp([100J2) ~ Q and so by TA3.5 tr(Mp([100J2» = tr(Q). Tha is, the rotation angle of the transformed matrix is unchanged. Furthermore, det(TMp([100J2)T-l) = det(Q). That is, a rotation is mapped to a rotation and a rotoinversion i mapped to a rotoinversion. Then Q is a half-turn. Suppose Q = Mp([110J2). Then

or so that t31 = 0, tll = t21, tl3 = -t23 and tl2 = -t22' Since T is unimodular, all of its entrie are integral and so tll(t22t33 - t23t32) is an integer. Furthermore, det(T) = ±l. But det(T) = 2tll(t22t33 - t23t32) which implies that tll(t22t33 - t23t32) = ±~, a contradiction. Then no such 1 can possibly exist for this choice of Q. Suppose then that Q = Mp([110J2). Then

But then for G = 4T it follows that GMp([100]2)G-1 earlier. Then no such T exists.

=

Mp([110]2) which is impossible as see

P6.24 (page 228) Suppose 3m1 and 31m generate the same lattice. Then there exists a un modular matrix T that transforms the matrices of Mp(3m1) into those of Mp(31m). Then fc some Q E Mp(31m), TMp([100]m)T-1 = Q. By DA3.3, Mp([lOO]m) ~ Q and so by TA3. tr(Mp([lOO]m» = tr(Q). That is, the rotation angle of the transformed matrix is unchangec Furthermore, det(TMp([100]m)T-1) = det(Q). That is, a rotation is mapped to a rotation and rotoinversion is mapped to a rotoinversion. Then Q is a mirror. Case 1: Suppose Q = Mp([210]m Then TMp([lOO]m) = Mp([210]m)T

so that t31

=

0, tl3

=

=

0, tll

-2tI2, and t21

=

-t12.

Then

and no conclusion can be made given the limited number of constraints on T. Since T must als map the third-turns of Mp(3m1) into those of Mp(31m) it follows that TMp(3)T-1 = Mp(3) ( TMp(3)T-I = Mp(3-1). Case 1a: Suppose TMp(3)T-I = TMp(3). Then -t23]

-t23

or

TMp(3) = Mp(3)T

t33

=

so that t22

-tI2, t32

T =

[

=

-2t12 -tl2

0]

t12 -tl2

o

=

0 and t23

0

O. So that which gives

0 t33

det(T) = t33(2t~2

+ t~2)

= t33(3t~2)'

But since T is unimodular, det(T) = ±1 which implies that t33(t~2) = ±~, a contraction sine every entry of T is integral. Then T must be such that TMp(3)T-1 = Mp(3-1). Case 1b: Suppo: TMp(3)T-1 = Mp(3-1). Then

0] [

or so that t22

=

2tl2, t32 -2t12

T =

[

-tl2

o

=

0 and t23

t12 2tl2 0

0]

0 t33

=

t23 t33

=

t12 2tl2 0

-t12 + t22 -t12 - tn t32

t23] 0 t33

O. So that which gives

det(T) = t33( -4t~2

+ t~2)

= t33( -3t~2)'

But since T is unimodular, det(T) = ±1 which implies that t33(t~2) = ±i, a contraction sinevery entry of T is an integer. Then T must map Mp([lOO]m) to one of the other mirrors. Case Suppose Q = Mp([lIO]m). Then TMp([IOO]m)T-1 = Mp([lIO]m) which implies Mp(3)TMp([IOO]m)T-IMp(3-1)

= Mp(3)Mp«(110]m)Mp(3-1)

= Mp([210]m).

But then for G = Mp(3)T it follows that GMp([IOO]m )G-1 = Mp([210]m) which is impossible: seen in Case 1. Then it must be that T maps Mp([lOO]m) to some other mirror. Case 3: Suppo Q = Mp([120]m). Then TMp([100]m)T-1 = Mp([120]m) which implies Mp(3-1)TMp([lOO]m)T-IMp(3)

= Mp(3-I)Mp([120]m)Mp(3)

= Mp«(llO]m).

But then for G = Mp(3-I)T it follows that GMp([IOOJm )G-1 as seen in Case 2. Then no such T exists.

P6.25 (page

= Mp«(lIOJm)

which is impossible

228) The basis for the lattice R(obv) is given on page 215. Note that

0 1] =! -; =1

!

noi

MD.(,

2) ~

[

3

3

so that by T6.13 R(obv) is not left invariant under 312 since [210J2is a generator of 312. P6.26 (page 228) The generators for 4m2 and 42m are given on page 227. From previous work it was discovered that 422 leaves the lattices P and I invariant. Since 4,(100]2 E 422 it follows that P and I are invariant under 4 and [100]2. Then since 4" = -1(4) and [IOO]m= _1([100J2) it follows that 4" and [lOO]malso leave P and I invariant. But these are precisely the generators of 4m2 and 42m. So P and I are invariant under 4m2 and 42m.

CHAPTER

7

P7.1 (page 231) Let p and q define Tl, and let q and r define T2. Let T3 = T2Tl' For an arbitrary xES, T3(X) = T2(Tl(X», The distance and directions of pq and pr equal those of XTl(X) and Tl(X)T3(X), respectively. Hence the angle Lpqr equals that of LXT1(X)T3(X),We can conclude that the triangles pqr and XTl(X)T3(X)are congruent. This congruency implies that the lengths of pr and XT3(X) are equal. Since the directions of pq and qr equal those of XTl(X) and Tl(X)T3(X), respectively, the directions of XT3(X)and pr must be the same. Thus T3 is the translation from p to r. P7.2 (page 231) Let 1 be the identity mapping. Then for any translation T E I', TO 1 = T = lOT. i.e., for any pES, T(l(p» = T(p), Then l(p) = T-I(T(l(p») = T-I(T(p» = p. Hence 1 maps P to p. P7.3 (page 231) Let p and q define T-l( q) = p =l(p). Thus T-lT =1.

T

and let q and p define T-l. Then (T-lT)(p) = T-l(T(p»

P7.4 (page 232) Suppose Tl, T2 E I' with translational vectors tl and t2 respectively. TlT2(X) = x + (tl + t2) = x + (t2 + t1) = T2Tl(X) for all xES. Thus TIT2= T2Tl· P7.5 (page 232) Let G =< g > be a cyclic group. Then for gn,gm E G, n,m E Z, gn+m = gm+n = gm * s" (by D4.22(1». Thus G is abelian.

=

Then

s" * gm =

P7.6 (page 232)

Mp([110J2)Mp(3)

= [~

Mp(3)Mp([110J2) = [~

:/=

[110]2 3

1 0 0 -1 -1 0

-1 -1 -1 -1 0 0 ~] = [~ 0 1 0] o = [-1-1 1 0 0 0 0 -1

J] [~ ~][~

~] = Mp([100J2) -1 ~] = Mp([010J2) -1

3 [110]2. Thus 322 is nonabelian.

P7.7 (page 232) l(x) = x = x-l-O for all xES.

Thus the translational vector of 1 is O.

r 6 P7.8 (page 233) (TrT6)(x) = Tr(T6(X» = Tr(X + st) = x + st + 1't = x + (1' + s)t = T + (X) where T E I' and 1',s E R and t is the translational vector for T. Hence r" T6 = Tr+6. P7.9 (page 234) Let Tl,T2 E I' with translational Then hT2nX)

P7.10 hold:

vectors tl and t2 respectively and let rE R

= x } 1'(tl + t2) = x + 1'tl + 1't2 = (x +1't2) + 1'h = Tr(X + 1't2) = Tr(T;(X» = TrT;(X).

(page 234) r is a vector space if for all r, s, t E rand

z , y E R the following propertie

(1) rs = sri (2) rest)

= (rs)t;

(3) there exists a translation u E r such that vu = v for all v E I'; (4) for each v E I' there exists a translation w such that vw = u here u is the identity; (5) (rs)Z = rZsz; (6) rZ+Y = rZ + rY; (7) rZY = (rY)Z; A?C

(8) rl = r. P7.12 (page 242)

P7.13 (page 242) Let {Mit} be an invertible matrix. Then the augmented matrix

m3l

m12 m22 m32

0

0

[mn m2l

m13 tl m23 t2 m33 t3 1 0

I 1 I 0 I 0 I 0

1

0 0

0 0

0

a12 an a32

a13 a23 a33

0

0

0

1

!]

can be reduced to 0

1

[!

0 0

0 0

1

0 0 0

0

1

I all I a2l I a3l I 0

sS2 ] S3 1

by using elementary row operations. So {Mlt}-l = [A]s] implies that {Mlt}{Als} = {MAIMs + t} = {I310}. Manipulating the equations MA = I3 and Ms + t =0 we find that A = M-1 and s = -M-lt. P7.14 (page 246) [rJD,(o,j {I.I-p}

1

[rJD,(o,)

{TIO} ------>

[rJD,(o,)

1

{I.I-p}

{TIO} ------>

[rJD,(o,)

f-

{T-'lo}

The above circuit diagram yields

P7.15(page

246)

(1) See figure 7.4. (2) Refer to (1). (3) See Table 7.1. P7.16 (page 250) By D7.21, D is a basis for r and T = {Tl'T2T3'lu,v,w,E Z}. Let 0 E S. Then 01'~(0) = {Tl'T2T3'(0)lu,v,w E Z} and LD(o) = {UTl(O)+ VT2(0) + WT3(0)lu,v,w E Z} = {Ti'T2T3'(0)lu,v,w E Z} = o1'bT(o). P7.17 (page 250) Let T be the set of translational isometries in G and let 0 E S. Since G is a group, 1E G. Thus the zero translational isometry, is in T. Suppose al = TIl and a2 = T21 are in T. Then ala2(r) = (TI1T21)(r) = (Tl1T2)(1(r» = (TJ1)(1(r) + T2(0» = (Tl)(l(l(r» + 1(T2(0)) = l(l(r» + 1h(0» + TI(O) = r + T2(0) + Tl(O) for vector r in S. Thus ala2 E T. Since all = -TIl is a translation, all E T. Thus T is a subgroup of G. P7.18 (page 250) The proof of T7.14 can be converted to the proof of P7.18 by replacing every occurrence of I with G and every occurrence of r with T, the translation group of G. With these modifications, the proof of T7.14 is exactly that of P7.18. 436

P7.19

(page

251) To prove that Mn(G) is a group, show the following:

(i) The identity element is in Mn(G), (ii) Mn(G)

is closed,

(iii) Inverse elements exist, and (iv) Associativity holds. PROOF:

(i) IE G since G is a group, therefore Ia E Mn( G). (ii) Suppose Mn(a), Mn(.8) E Mn(G). If a = {AltI} and d = {Blt2}, then a.8 = {ABIAt2 Now Mn(a)Mn(.8) = AB E M(G) since a.8 E G. (iii) Suppose Mn(a) E Mn(G). If a = {Alt}, then a-I Mn(a-l) E Mn(G) since a-I E G. (iv) Mn(a)[Mn(.8)Mn(l)l

= Mn[a(.8-y)l = Mn[(a.8hl

= {A-11_ A-It}.

Now Mn(a)-l

= [Mn(a)Mn(.8)lMn(l)

+ tl:

= A-I,

for Mn(l),Mn(a

Mn(.8) E Mn(G). Thus Mn(G) is a group. P7.20 (page 251) Note that Ao(G) is represented follows that Ao(G) is a group.

by Mn(G).

By P7.20, Mn(G)

is a group;

P7.21 (page 254) Let G denote a crystallographic space group, let hl, ... ,hn E Ao(G) be sue that hlh2 .,. hIt = hk+l ... hn. Let D denote a basis for I' and let 0 E S. Let M, = Mn(hi) an suppose {MlltI}, ... ,{Mnltn} E Rn(o)( G). Then MlM2 ... M" = Mk+l ... Mn· Now {Mlltl}{M2It2} ... {M"lt,,} = {Ml ... M"lsl} E Rn(o)(G) for some Sl E R3. Similarl {Mk+1ltk+l}{Mk+2Itk+2} ... {Mnltn} = {Mk+l'" Mnls2} E Rn(o)(G) for some S2 E R3. Li S = Sl - S2. Then by T7.26, {I3Is} E Rn(o)(T(G». Hence {Ials}{Mk+l ... Mnls2}

P7.22

(page

=

{Mk+l ... Mnls2

+ Sl -

s2}

=

{MI",

M"lsI}'

255) Let a,.8 E Ao(G).

O(a)O(.8) = Rn(T(G»){M",lt",}Rn(T(G»){M{3lt{3} = Rn(T(G»){M",lt",}{M{3lt{3} = Rn(T(G»){MaM{3IM",t{3 + tal = Rn(T(G»){Ma{3lta{3} where t",{3= Mat{3 + t", and Mal' = M",M{3

= O(a.8) Thus 0 is a homomorphism. Suppose a = .8 E Ao(G). Then O(a) = Rn(T(G»){Mltl} and 0(.8) = Rn(T(G»){Mlt: where M = Mn(a) = Mn(.8) and tl, t2 E R3. By T7.26 {!altl - t2} E Rn(T(G», so 0(.8) Rn(T(G»){I3Itl - t2}{Mlt2} = Rn(T(G»){MltI} = O(a). Hence 0 is well defined. Suppose Rn(T(G»){Mlt} E Rn(o)(G)/Rn(T(G)). Then M = Mn(.8) for some .8 E Ao«( and since 0 is well defined, 0(.8) = Rn(T(G»){Mlt}. Thus 0 is onto. Ke1'O = {l}. By tl Fundamental Theorem 0 is an isomorphism and Ao(G) ;0::: Rn(o)(G)/Rn(T(G». Note that Rn(o)(( and Rn(T( G» are the group of matrix representations of elements in G and T(G) respectively. V conclude that Ao(G);o:::G/T(G). P7.23 (page 260) [ll1J32 consists of a third-turn along [1111 and a translation of ~a + ~b + ~c ~(a + b + c) since a + b + c is a shortest vector in the [111 1 direction. Hence the Seitz notation f [111]32 is {Ial[~, ~, ~n{Mp([111J3)I[OOolt} = {Mp([11IJ3)1[~,~,

~n·

417

P7.24(page 260)

{MI[i,i,H}

i(

The translation component is ~a + ~b nonzero vector in the direction [111J.

+ ~c)

=

ia + ib + ~c since ~a + ~b + ~c is the shortest

7.25(page 260)

o o

1

0

1 0 o 0 P7.26(page 260)

P7.27(page 260)

P7.28(page 260)

P7.29 (page 261) Consider the space group Gin E7.36with the lattice type P in the first setting. Then M = Mp(m) = origin to p= [0,0,

J; J!

[1o 0 0]

[2tl]

l

1 0 and Nt = 2t2 and so tl = 0 or ~, t2 = 0 or Shifting the 0 -1 0 transforms t3 to O. The following is a table of the resulting glide planes.

0

tl

t2

0

0

m

reflection

"2

I

0

a

a-glide

0

1 "2

b

b-glide

n

n-glide

1

"2

Symbol of Glide Operation

1

"2

P7.30 (page 262) Choose a basis for the A-lattice

(e.g.

[1 01 0]1

Type

A = {a, b, ~b

+ ~c}).

In this case

and Nt = [2tI, 2t2 + t3, OJ. SOif we change the origin to p= [0,0, J;J! then 0 -1 t3 = 0, tl = 0 or ~ and t2 = 0 or~. Hence the glide-translations remain unchanged. Using a similar process it can be shown that the translations are unchanged if a B~lattice is used. MA(m) =

0

o

P7.31 (page 264) Suppose A = {I3IsI}{Mlr}i where {I3IsI} E Rp(TL)'

AA-l

= {I3Is1}{Mlr}i{I31- Mo(a)-i(s

= {I3Ist}{I3IMi[-Mo(a)-i(s = {I3Ist}{I31 - Mo(a)-i+i(s = {I3lsl - Mo(a)(s

+ sl)}{Mlry(a)-i

+ sl)]}{Mlr}i{Mlry(a)-i + sl)}{Mlry(a)-i+i

+ st)}{Mlry(a) 438

Then

by (7.24)

= {hlsl - Ia(S + sl)}{hls} by (7.23) =

{I3lsl - (s

+ Sl) + s}

= {laIO}. Thus A-I = {lal- Mo(a)-i(s + sl)}{Mlr}o(a)-i. Since {I31- Mo(a)-i(s + Sl)} E Rp(TL) and {Mlry(a)-i

P?32 (page 264) Suppose {I3It} M,-lr+ ... +Mr+r}. So

= {I3IsI}{Mlr}i

E {{Mlr}ili = 1, ... ,o(a)},

where 1 :::;i :::; o(a).

{Mlr}i

A-I E j,

= {MilMir

.

{I3IsI}{Mlr}i = {laMilMir + ... + r + sI} = {I3It} Since laMi = la, Mi = Ia. Thus i = o(a). It is given that {IalsI} E Rp(TL) and since r consistent with ao(a) =l, {Mlr}o(a) = {lals} E Rp(TL)' Since Rp(TL) is closed, {laISl}{hls} {I3It} E Rp(TL)' L is left invariant by H, the point group of G, therefore the components of translation in G are all integers. Thus Rp(T(G» 92 E 4.

0 is

one-to-one ant

INDEX A ~-cristoba1ite (see cristobalite) ~-quartz (see quartz) A-centered lattice 209-212, 262 a-glide 261 abelian group 232, 243 aenigmatite 332 A16Mn 193 amb1ygonite 79 amphibole 37, 70 angle of misfit 37 angles 25-37 angles between zones and face poles 76-83 anorthite, 34, 45-46, 76-79 antipodal points 371 aragonite 36-37 associative law 17, 25, 119, 125, 305 augumented matrix 312-314, 316-317, 323, 331, 333, 350, 362 axial ratio 79-83

8 B-1attice type 209, 262 basis 18-20, 141-149, 201-202, 250, 314, 316, 330, 340, 357 binary operation 119, 123-124, 136, 391, 395 binary relation 379 body-centered cubic lattice 260 body-centered lattice (see I-centered lattice) bond length and angle calculation 28-34 boron 193 Bragg equation 48 Bravais lattice types 199, 207-228, 252

c

C-centered 209, 289, 385 C-centered lattice 212-213, 221, 254 c-glide 261 cancellation law 398 cartesian basis 18, 61, 72-75, 329, 339, 344, 352 cell dimensions 320-326 centered-lattice 257 centrosymmetric group 140 chalcanthite 79-83 change of basis 57-62, 66-71, 258, 292, 352 change of basis matrix 57, 61, 66-67, 235-236, 245-246, 246-249, 257, 335, 329, 341, 353, 385

change of origin 244, 246-249, 2! 258, 265-266, 273, 284, 291, 301 circuit diagram 61, 73 clockwise rotations 311 closest-packed 66-72, 332 closure 125 coesite 39-40, 246-248, 294-296, 3 cofactor method 337, 353 co1attice 211 commutative law 17, 232, 305 compositions 95-96, 104, 108-10 127, 206, 232, 237, 278, 305 conformable for addition 310 conformable for multiplication 311 conjugate 162 conjugation 162 consistent with a rotation 25 258-259, 263, 278-281, 291, 296-29 300 coordinate axes 23-25 coordinates 12 cordierite 385 cosets 164, 204-205, 255, 267-26 270, 283-284, 289, 385-398 cristoba1ite 7, 289 cross product 34-35, 62-65, 80, a 357-358 crystal 3 crystal face form 149 crystal structure 3, 83-85, 87-8' 379 crystal structure drawing 83-8~ 87-89 crystal systems 183, 224 crystalline solid 3 crystallographic group 129 crystallographic isometry 129 crystallographic point group 12~ 132, 135, 252 crystallographic restrictions 129-1~ crystallographic space group 250-25 crystallographic space group operatioll 258-262 crystallographic translation grou 249-250 cubic axial groups 173-179, 371, 37: CuSO~.5H20 79-83 cyclic group 136-137, 276, 397-398

D d-spacing 46-47, 320, 368-369 determinant 35-38, 206, 326-328, 343 349, 355 diad axis 115 diffraction data for protoamphibo1 321

diffraction data for quartz 325 diffraction experiment 335 dihedral groups 167-173, 227, 371 dimension of a vector space 20 diopside 70 dioptase 149-150 direct basis 51 direct lattice 51, 156 direction cosines 343, 372-373 distance preserving 91, 96 distributive laws 17, 26, 35 dodecahedron 194 dot product (see inner product) drawing crystal structures 83-85, 87-89

generators 203, 205-206, 2 250-251, 255-256, 265-266, 273-2 277, 292, 298-299 geometric three-dimensional space 11, 21 229 glaserite 193 glide operation 261 glide plane 261 glide translation 261 golden mean 194 goniometric data 79-83 group of isometries 382, 395 groups 120, 125-128 guide column 117 guide row 117

E

H

eigenvectors 348 elementary row operations 312, 331 end-centered lattices 208-212 equal mappings 303 equal matrices 309 equation of a plane 2, 42-47, 81, 362-363 equivalence class 146, 158-160, 165, 174, 382-383, 385, 392 equivalence relation 145-146, 162, 203-204, 256-257, 265, 371, 379-394 equivalent planes 144, 147-150 equivalent points 144, 203-204 euclidean algorithm 365-366 Euler's theorem 376-377 exsolution 70

H4Si04 1, 28, 146-147 H6Si303 2, 29, 183 half-turn 94-95, 107-119, 142 halving group 139, 389 handedness of the basis 18, 35, 99, 201, 256, 259, 355-357 hemimorphite 335-336 Hermann-Manguin symbols 191-192 herpes virus 373

F F-1attice type 220-223 face poles 69, 147-149 face-centered cubic structure 341 face-centered lattice 221 factor group 204, 389, 390-391 feldspar 246-248 finite property of a crystallographic point group 132 form 149

G G-equiva1ent 144-155, 185-191, 379, 398 GL(3,R)

158-164,

126

general bases 345 general cartesian rotation matrix 339, 352 general cartesian rotoinversion matrix 339, 344 general equivalent positions 271-272, 274-275, 284, 294 general linear group 127 general position 146, 271

457

I-centered lattice 205, 212, 217, icosahedral group 389 icosahedral point groups 192-197 icosahedral rotational symmetry 3; ideal crystal 3, 9 identity 17, 94, 99, 110-119, 1 142, 231, 306, 354, 386 identity matrix 110-119, 317, 3 380 image 92, 303, 306 improper point group generating theo 138-139 improper point groups 139-1· 180-184 improper operation 136 index 388 indices 335, 337 25, 80 inner product 337 integer matrix 200-202, 206, 2 integral matrix 256 interaxia1 angles 6, 25, 49, 56 interplanar spacings 47, 320 intersection angles 371-378 invariant 100 inverse 20, 114, 125, 306, 330-332 inversion 96-97, 99, 104, 138-142 inversion matrix 339 invertible 384

invertible matrix 331, 337, 346-347, 381 isometry 91, 96-98, 128-129, 229-230, 237, 345-346 isomorphism 21-22, 161-162, 251, 255, 264, 272, 276, 395-398 i~ostructura1 379

J jadeite

27, 86-87

N

K kyanite

mirror plane 98, 261, 340 modular arithmetic 380 monaxia1 point groups 142, 151-155: 165 monaxial rotation groups 134, 140 monosilicic acid (see H4Si04) motions 9 multiplication table 117-120, 209, 212, 221, 390 mUltiplicative identity 330

66-69, 70-71, 330

L L-equiva1ent 251-253 Lagrange's Theorem 389 lattice 12-16, 128-134, 205, 358-359 lattice plane 2, 42-47, 207, 359, 361-370 lattice points 361, 364 lattice vectors 14-16~ 363 least-squares estimates 323, 326 least-squares method 320-326 left cancellation law 120,238 left coset 387 left handed basis 99 lengths 25-34 linear combination 11, 314, 316-317, 333 linear component 238, 250-251 linear equations 312, 321, 332, 334 linear mappings 101-104, 108-109, 238 linearly independent 18

M nagnitude of a vector 9-11 napping 95-96, 303, 395, 397 natrix 309 natrix addition 310 natrix equality 354 natrix equations 309 natrix groups 225-228 natrix methods 309-338 natrix multiplication 310 natrix representations 105-110, 141-144, 105-107, 170-171, 225-228, 256 latrix representations of point isometries 339-356 leasurement error 321 letric tensor (see metrical matrix) iet r Lca I matrix 26-33, 27, 42, 47, 54, 63, 68, 73, 191, 224, 369, 385 [iller indices 44-46, 48, 50, 59, 63, 69, 82, 362-370 tinimum variance 322 tinors 326

458

n-glide 261 nxm matrix 309 nth-turn screw 259 nth-turn 93 narsarsukite 100-101, 150 negative nth-turn 93-95 negative quarter-turn 93, 108 negative third-turn 93 non-abelian group 232 nonbonded radius of oxygen 71 noncrysta11ographic rotation groups 371 normal equations 323-324, 326 normal subgroup 242-243, 250, 252, 270, 388, 391 normality 252

o

obverse setting 89, 215-216 one dimensional lattice 203 one-generator point groups 262-276 one-to-one and onto mappings 96, 305-306, 395 one-to-one mappings 305 onto mappings 305 orbit 146, 149, 249 order 137, 306, 389 orientation of rotation axis 354-356 orientation symbols 95 oriented point groups 225-228, 252, 255, 258 origin 9-11, 96, 244-252

p pairwise coprime 365 parallelogram rule 10, 21, 22 permutation 306 plane 42-43 point groups 120, 128, 151-155, 180, 252 1 138, 142, 180, 208 1" 141-142, 180 2 138, 140-142, 180, 208-213, 253-254 m 141, 261 3 138, 140, 142, 180, 213-216

3

140, 142, 180 101-102, 137-139, 142, 180, 217, 223 4 139-140, 142, 146-147, 180, 93-95, 122, 127, 138, 142, 180, 6 220, 264 6 122, 141-142, 147-149 mmm 180 180 mm2 140-141, 180 21m 180 21m3 21m35 197, 180 124,134-135,167,168, 222 173-174, 180, 220-222 23 167-169, 174-178, 180, 222-223 ?..35 167, 180, 192-197 3m1 227-228 31m 227-228 321m 180-181 3mm(=3m) 181-182 3m1 227 31m 227 312 227-228 321 227 110-120, 125, 132, 134, 322(=32) 167, 169-171, 222, 158-165, 225-227 iim2 (see ii2m) ii2m 180, 182, 227-228 ii3m 180 41m 139, 150 u lmmm 180, 182 41m321m 180, 182-183 umm 180 422 121-122, 124, 132, 134, 167, 170-172, 180, 222, 227 432 167, 178-180, 222-223 522 167, 173 532 (see 235) 6m2 227 62m 180, 227 61m 141-142, 180 61mmm 180, 182 6mm 180, 227 622 167, 172, 180, 222, 227 point isometry 91, 258, 340, 342-344, 349, 371 point of general position 146 point of special position 146 point symmetry 128 pole points 158-168, 173-179, 371 po1yaxia1 crystallographic groups 371 po1yaxia1 point groups 157-168, 184-190 positions 271 primitive hexagonal lattice type 208, 216,-253, 255, 260 primitive lattice type 208

4

459

principal representative of C with reo spect to P 270 principal representatives 271-272, 283-284, 295 projection 26 proper crystallographic operatioI 135, 140 proper monaxia1 group 138 proper point groups 224 proper po1yaxia1 point groups 157-191 proper unimodular matrix 200, 210, 266 protoamphibo1e 37, 320-324 pyritohedron 193 pyroxene 70, 86-87 pyroxferroite 64-65, 83-86

Q Q-va1ues 320 quadratic formula 377 quarter-turn 94-95, 143 quarter-turn inversion 93, 109-110 quarter-turn screw operations 259 quartz 3, 6, 14, 22, 31, 40, 45-46, 110-120, 286-289, 314, 325 quasi-crystals 193

R 321 random variable 362 rational numbers 16-25 real vector space 47-65, 69, 73-83, reciprocal basis 86, 171, 325 reciprocal lattice 51-56 54-56 reciprocal metrical matrix 317-318, reduced row echelon matrices 350 reflection 98-99 reflection isometry 97-99, 340 reflexive property 45, 145, 379 relations 255, 258, 263, 279, 281, 296-298, 300, 379 relatively prime 366 reverse setting 90, 215-216 rhombohedral lattice type 89-90, 201-202, 222, 215-216, 228 right cancellation law 120 right coset 386 right-handed basis 99 rotation 92-95 rotation axis 92, 207, 341, 349-350, 355, 358-359, 371-372 rotation isometry 94-95, 303, 339, 342-344, 346-347, 349, 352, 354 rotoinversion 94, 96-99, 105 row operation 317, 362

s

tricyclosiloxane

sanidine 246-248, 294 sapphirine 68-69, 71-72 scalar 9, 16 scalar multiplication 10, 11, 23, 102-104 Schoenf1ies symbols 191-192 screw and glide operations 258 screw translation 259 Seitz notation 240-241, 260, 294 self-coincidence 100, 129 similar 346, 384 similar matrices 380 singular 332 sixth-turn 94-95, 110, 143 skutterudite 192-193 312, 316, 324 solution set 12 (see lattice) space lattice spans 18, 316 288 special positions spinel 341, 351 stabilizer 160 stereogram 75 stereographic projection 72-83 subgroup 120, 134-135, 250, 386 sub1attice 203, 278, 299 super dense packings 193 symmetric matrix 27, 312 symmetric property 145, 379 symmetry 100-101, 128 symmetry element 99, 259 system of linear equations 313

triple scalar product 38-39 triples 12-14, 21, 23-24 turn angle 92-95, 311, 343, 346, 354: 371, 379, 383 two-dimensional lattice 203 two-fold axis 115 two-generator point groups 276-278: 293, 296

2,

29,

38

(see

H6SiJOJ)

T tetrad axis 100 third-turn 93-94, 110-119, 143, 207, 384 267 third-turn screw operation 203 three-dimensinal lattice three-fold axis 114-115 three-generator point groups 296-297 three types of solution sets 320 tilt angle 36 trace 133, 343-344, 346-351 transformation matrix (see change of basis matrix) transitive property 145, 379 translation group 231, 242-243, 249-252, 254-255, 258 translation vector 3, 6, 8, 231-236 translational component 238, 259, 285 translational isometry 268 translations 229-237 transpose of a matrix 311 tremolite 70 triad 114

460

u

unimodular over the integers 200, 221 unit cell 14, 204, 288, 330 unit cell volume 39, 40, 63, 329

v vector addition 10, 16 vector space 16-17 viruses 193 volume (see triple scalar product)

Z

zone symbols 14, 63-64, 76-78, 173, 179 zussmanite 101

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