Mathematical Crystallography

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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 Titles NEW to the REVISED EDITION SOLUTIONS TO PROBLEMS INDEX 456 Page 1 Titles 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 Titles A MATHEMATICAL DESCRIPTION OF THE GEOMETRIES Page 9 Page 10 Titles o -./2v o !V Vector addition and scalar multiplication: Page 11 Titles S = {xa + yb + zc 1 x,y,z E R} . Page 12 Titles ( [;]1 ',Y,' , R) . Space lattice: LD == {ua + vb + wc I u,v,w E Z} Page 13 Titles • • • • 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 Titles (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 Titles [:], [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 Titles 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 Titles 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 Titles 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 Titles u = a m, Page 5 Titles [s]~ = [hk£]G-1 Page 6 Titles y z hx + ky + tz Page 7 Titles (E2.3) Example - Calculation of the d-spacing for a plane in a crystal: Solution: RECIPROCAL BASIS VECTORS Page 8 Titles 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 Titles ['10* = [:] hx + ky + t z = 1 s s • • a s so Tables Table 1 Page 11 Titles (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 Titles r • a ~ (P2.6) Problem: Show that r • b (T2.5) Theorem: r • a r • b'" r • c Tables Table 1 Page 13 Titles a • b b • b c • b a • c b • c c • c Proof: 0* be ~ ~ Tables Table 1 Page 14 Titles (0 ) b"· (a")" = c"· (a")" = 0 a. {a,b,c} . Page 15 Titles (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 Titles * * * ,~ CHANGE OF BASIS * G[rlD = [rlD* Page 17 Titles By T1.9, we have [r]D = rdazjD + r,[bdD + rdcdD. But then Tables Table 1 Page 18 Titles [ [adO, [bdO, I [']0, Page 19 Titles D~ = {a~,b~,c~}. By T2.5, [ r] ,~ [ r] ,~ * ... " Tables Table 1 Table 2 Page 20 Titles 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 Titles 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 Titles * 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 Titles 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 Titles -1 -1 r = ua + vb + we [100] [110] [010] Page 6 Titles Proof: Tables Table 1 Page 7 Titles 1 i2(r) = 2(r) Page 8 Titles 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 Titles (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 Titles Solution: Since i(r + s) = -(r + s) = -r - s = i(r) + i(s) and i(xr) = r~ Page 16 Titles , , , , I , I , I , I I I I Page 17 Titles [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 Titles 2(r) 2(xa + yb + ze) = -xa - yb + (x + y + z)e . (P3.5) Problem: Page 19 Titles [1 1 1] Page 20 Titles THE CONSTRUCTION OF A SET OF MATRICES DEFINING THE RO­ Page 21 Titles (a) (b) Page 22 Titles 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 Titles . [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 Titles 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 Titles CHAPTER FOUR THE MONAXIAL CRYSTALLOGRAPHIC POINT GROUPS INTRODUCTION ALGEBRAIC CONCEPTS Page 2 Titles 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 Titles (E4.6) Example - (Z,-) is not a group: The set of integers Z under Page 5 Titles -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 Titles (P4.3) Problem: point symmetry group of 8. Page 7 Titles (D4.12) Defi nition: A crystallographic group is a group of isometries CRYSTALLOGRAPHIC RESTRICTIONS (T4.14) Theorem: Page 8 Titles 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 < . ---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