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Crystallography II

Group Theory I-32 point group of the crystal (the first classification)

1.Introduction

Crystal = Lattice structure basis

1)Point symmetry 7 C.S. primitive

1801—1807 lattice

2) Point symmetry Geometric

14B.L. Crystallo-

graphy

Translational symmetry

1818----1839

Crystal = Lattice structure basis

1801—1830

3)Point symmetry 32 point group

of crystals

To describe the physical properties of the

crystals in quantity and quality

.

1Monoclinic 1

1 Lattice2

2

2

basic structure unit

The point group can be regarded as the first classification of the crystal

Same crystal system

Same Bravis lattice

Different basis unit content

Different point symmetry collections

Different crystals

This means we can use the point symmetry collection to classify the crystals. By this way, we get

32 point groups

Other reasons for the point group study

To obtain 32 point system patterns. If adding them to 14B.L., we can obtain some of 230 space groups( the general structure of the crystals)

To symbolize all symmetry operation possessed by crystals.

- To describe the physical properties of crystal.

The essential characteristics of point symmetry collections

- The production of any two operation is also a member of the collection ( closure).
- The collection must include an identity 1 (E)
- For each operation R, there is an inverse R-1, which is included in the collection to meet R.R-1=1
- The multiplication of operations are associative,

(AB)C=A(BC)

In mathematics, this specific collection is called a group

（群）.

The order （阶） of group : the number of the elements

in the group.

Point group (点群) of crystals:

{ 4, 42=2,43, 1…}

{2, 1…}

{1}

In each crystal, there are different point symmetry

collections at different point positions.

The highest point symmetry collection is called the point

group of the crystal.

2. the procedure of 32 point group of the crystal

From the pure math combination theory point of view , all

possibilities of the combination from 1. 2, 3, 4, 6, 1, 2, 3, 4,

6 are

C101+C102+…….+C1010

Then check each possibility

Meet the crystal system condition?

Meet the four basic requirements of the group?

If so, we get one point group of the crystal, or vice versa.

Some disadvantages of combination method :

have enormous possibilities

make some confuses

need a lot of combination laws of the symmetry

operation

In order to overcome these defects, it’s better for us to

discuss the point group for each crystal system

individually.

The procedure is that

adding more symmetries in a certain crystal system, if

these new symmetry elements additions still remain the

crystal condition, and also meet the four basic

requirements of the group . We get one point group which

below to this crystal system.

3. The crystallographic point group for each crystal system

3.1 Triclinic

In this system, we have 1(E) or/ and 1(i) point symmetries.

Any other point symmetry elements addition is not

allowed since this addition will change the crystal

Condition. So we have two point group in Triclinic

They are {1}, {1,1}

Just the same as the symmetry description, we have 4 ways to describe the point group.

Symbology Geometry Mathematics Language

International

Schoenflies

1 C1 general equivalent positions {E} ……….

1 S2

symmetry element distribution {E,i} ……….

Background knowledge: Stereographic projection

极射赤面投影

N pole A 2-D method to describe the

relationship of points, lines,

and planes in 3-D.

general position in N

hemisphere and projection

general position in S

hemisphere and projection

S pole the enantiomorphical position

and it’s projection

1(C1) 1(S2)

,

,

,

,

,

The concept of Holosymmetric point group.

（全对称点群）

the point group in the crystal system with the largest number of symmetry operation.

The holosymmetric point group of Triclinic is 1 (S2).

3.2 Monoclinic

According to the definition of this C.S. , we can discuss the possibilities of the point group in monoclinic .

2 2

Five possibilities

1, 1, 2, 2

1 2 combination {1, 2} a point group in

Monoclinic,Called 2 (C2)

b) 1 2 combination {1,2},……, called m (C1h)

c) 1 2 combination

{1[00]}{2[001]}= -1 0 0 -1 0 0 = 1 0 0

0 -1 0 0 -1 0 0 1 0

0 0 -1 0 0 1 0 0 -1

={2[001]}

A mirror plane to be produced which is perpendicular to

the 2-fold axis.

Add 1, we have a collection {1, 2, 1, m} or {E, C2, i, h }….

,called 2/m (C2h)

d) 1 2 combination

{1[000]}{2[001]} = {2[001]}

All the symmetry elements are included in c) case.

e) 2 2 combination

{2[001]}{2[001]}={1[000]}{2[001]}{2[001]}

= {1[000]} {1[001]}= {1[000]}

Not a new point group

2 (C2), m (C1h), 2/m (C2h)

3.3 Orthorhombic

2, 2, 2 2, 2,

6 possibilities

1, 1; 2, 2

- 1 2, 2, 2 {1, 2[001], 2[010], 2[100]}

A point of group of Orthorhombic, called 222 (D2)

b) 12[100], 2[010}, combination

{1[000]}{2[100]}{2[010]} ={2[001]}

Get a symmetry set{ 1[000]}, 2[100], 2[010], 2[001]}. It ‘s

a point group in Orthorhombic, called mm2 (C2v).

c)1 2, 2, 2 combination

{1[000]]{2[001]}={2[001]} c 2/m

Along the b, c directions, we have the same results

,

,

a b

2/m; 2/m

We get a symmetry set,{ 1[000], 1[000]; 2[001], 2[100],

2[010]; 2[001], 2[010],2[100]}. This collection from a point

group which is below to Orthorhombic, called 2/m2/m2/m

or mmm (D2h)

,

,

d) 1 2, 2 combination

The result was included in the point group of mmm case c)

e) 2 2, 2 combination

The results were included in case b) and c).

f) 2 2, 2, 2 combination

The result was include in the case c).

3.4 Tetragonal

According to the limitation of the crystal condition, we

have

222 (D2) ,mm2 (C2v), 2/m2/m2/m or mmm (D2h)

{ 1[000];2[100], 2[010] 2[001]; 1[000],2[001], 2[010],2[100]}.

222 (D2),

{ 1[000];2[100], 2[010] 2[001]}

mm2 (C2v),

{ 1[000];2[001];2[001], 2[010]}

C.S……….

- B.L. oP,oC(oA,oB),oI,oF
- Point Group (x,y,z)
- B.L. Point Groups Point Space Group Structure of Crystals

222 (D2)

,

mm2 (C2v)

,

,

mmm (D2h)

,

222 (D2)

Totally we have 66 symmorphic space group. They are

CS Symmorphic space group

Triclinic P1,P1

Monoclinic P2,Pm, P2/m; B2, Bm, B2/m

Orthorhombic P222, Pmm2, Pmmm;

C222, Cmm2,Cmmm;Ammm; I222, Imm2,

Immm; F222,Fmm2, Fmmm

Tetragonal P4,P4,P4/m,P422,P4mm,P42m,(P4m2),P4/mmm;

I4,I4,I4/m,I422,I4mm,I42m,( I4m2), I4/mmm;

Cubic P23,Pm3,P432,P43m,Pm3m;

I23, Im3,I432,I43m,Im3m; F23, Fm3,F432,F43m,Fm3m

Trigonal P3,P3,P312,(P321),P3m1,(P31m), P3m1,(P31m)

R3,R3, R32,R3m,R3m

Hexagonal P 6,P6,P6/m,P622, P6mm,P6m2,(P62m),P6/mm

4 4

about 9 possibilities

1, 1, 2, 2, 4

- 1 4 combination

{1[000]}.{4[001]}={4[001]}

The collection is { 1[000], 41[001], 42[001], 43[001]}. ….

formed a point group of this crystal system and called 4

(C4)

b)1 4 combination

{1[000]}.{4[001]}={4[001]}

The collection of { 1, 41, 42=2, 43} was obtained. ……, A

point group was formed, called 4(S4)

c) 1 4 combination

{1[000]}{42{001]}. ={2[001]}

{1[000]}{41{001]}. ={ 41[001]}

A mirror plane and 4-fold symmetry axes were produced.

The symmetry collection was {E,41, 2, 43, i, S41,S43, h}, …

,

,

It formed a point of group in this crystal system, called

4/m (C4h) point group.

d)1 4 combination

The result is the same as the case c)

e)2 4 combination

We have two cases

The first case was include in case a), and the second case

may result in a new point group in this crystal system.

Suppose the 2 along a direction. The 4-fold symmetry will

ask for another 2-fold axes along b direction.

According to {41[001]}{2[100]}={2[110]}, two 2-fold axes

are produced along [110] and [110].

Finally by this combination, we get a symmetry elements

Collection {E; C41,C42=C2, C43;2C2’;2C2’’}……

It forms a point group , called 422 (D4)

f)2 4 combination

We also meet two choices, which are similar to case e).

The first case is include in case b) (1 4 combination).

The second one is that the 2-fold axis is perpendicular to

4 axis.

Since {42[001]}{2[100]}={2[010]} and

{4[001]}{2[100]}={2[110]}

By other combination, no new symmetry operations are

obtained. From the above pattern, we get a symmetry

collection{E; S43, S42=C2, S41;2C2’;2d}……..

We get a point group in this crystal system, called 42m

(D2d)

g)2 4 combination

The result is the same as c)case

,

,

,

,

asked for by 4-fold axis.

Allowed and still not to be discussed.

May form a new point group.

According to the above basic symmetry elements, we can

obtain the following patterns.

,

,

,

,

From the above patterns, we can find other two mirrors

Out to form a symmetry collection, {E;C41, C42=C2,C43;

2v; 2d}, which meets the all needs for the point group.

It is a point group , called 4mm (C4v).

h addition will not change the

crystal condition. The symmetry

collection is different from 4mm

point of group. It is quite

possible to form a new point

group.

,

,

,

,

{4[001]}{2[001]}={43[001]} means 4 axis was

produced along c direction.

{42[001]}{2[001]}={1[001]} means 1 axis was

produced .

Reference the point group of 4mm, the other two mirrors in [110], [1 10] can be founded ( 2d).

The collection may be

{E; i ;C41,C42=C2,C43;S41,S43;h ;2v; 2d? }

Totally there are 16 positions. But in the above collection,

,

,

,

,

So the collection needs other 4 symmetry elements to

form a closure set. Where are they?

{2[100]}{1[000]}={2[100]}

{2[010]}{1[000]}={2[010]}

{2[110]}{1[000]}={2[110]}

{2[110]}{1[000]}={2[110]}

{E; i ;C41,C42=C2,C43;S41,S43;h ;2v; 2d ,2C2’;2C2’’}

It meets the all needs for the point group in this crystal system, called 4/m2/m2/m (D4h).

h)2 4 combination

included in the case c) (4/m )

included in the case f) (42m)

4mm(C4v)

{E; i ;C41,C42=C2,C43;S41,S43;h ;2v; 2d ,2C2’;2C2’’}

{E;C41, C42=C2,C43; 2v;2d}

42m(D2d)

{E;S43, S42=C2,S41;2C2’;2d}

422(D4)

{E;C41, C42=C2,C43; 2C2’,2C2’’}

4/m(C4h)

{E; i ,C41, C42=C2,C43; S41 ,S43,h}

4(S4)

{E;S41,S42=C2,S43}

4(C4)

{E;C41, C42=C2,C43}

Please give a brief summary for tetragonal 7 point groups

by yourselves.

3.5 The point groups in Trigonal

3 3

1, 1, 2, 2, 3 about 9 possibilities

a) 1 3 combination b) 1 3 combination

{1, 31,32} ,….. {1;S65,S64=C32,

3 ( C3 ) point group S63=i,S62=C3,S6}, …..

3 (S6 or C3i) point group

,

,

,

So far, we get all point groups in Trigonal crystal system.

They are 3 (C3), 3(S6 or C3i) , 32(D3), 3m(C3v), 32/m or 3m(D3d).

3.6 Hexgonal

This crystal system is with the main axis, e. g. 6 and or 6, and very similar to Trigonal system. Here we just give a list.

2 6 6/m2/m2/m D6h

Six 2-fold axes, seven mirrors, one E , one i , plus C61,

C62=C3, C63=C2, C64=C32, C65; S35, S31; S61, S65.

3.7 Cubic crystal system

This is a little more difficult to treat than the above

Systems since it is without the principle axis.

a) The first question is that we can obtain a point group just using 1(E) and eight 3-fold axes?

The answer is no, since

{3[111]}{3[111]}=

0 -1 0 0 0 1

0 0 1 1 0 0

-1 0 0 0 1 0

-1 0 0

= 0 1 0 ={2[010]}

0 0 -1

By repeating mutiplcations, no other new symmetry

elements will be produced.

{1, 31[111], 32[111]; 31[111], 32[111] ,31[111], 32[111] ,31[111], 32[111], 2[100], 2[010], 2[001]}….

This is a point group , called 23 (T).

3 , 3 , 3 , 3

23(T)

1 2 4 4

b) Now we try to add 1 at the origin or 2 which is perpendicular to the 2-fold axes,which does not break the crystal condition, and meets……The most important results are the production of 3(S65) along body diagonal directions111.

The collection is

{1, 1; 31[111], 32[111],3[111], 35[111]; 2[100], 2[010], 2[001]

31[111], 32[111],3[111], 35[111]; 2[100], 2[010], 2[001]

31[111], 32[111],3[111], 35[111];

31[111], 32[111],3[111], 35[111];}……

It forms a point group of the cubic system, called 2/m3 or

m3 (Th).

- The mirrors contain both the 3-fold

axes and 2-fold axes in 23 point group.

This addition still remains the

crystal condition, and maybe form a

point group.

,

,

,

,

,

,

The problem is how many mirrors can be added.

6 why?

This combination will produce 4 (S43) along a, b, c axes.

The possible collection is

{1, 31111,32111;2100; 41100,43100; m110}

8 3 6 6

m[110],m[110], m[101], m[101], m[011],m[011]

Totally , we have 24 symmetry elements. It forms a point

group of cubic crystal system, called 43m(Td).

d) 4 –3, 3, 3, 3 combination

The 4-fold axes must be along a, b, c directions. By Matrix production, six 2-fold axes along 110 directions will be obtained.

{1, 31111,32111;2100; 41100,43100; 2110}

8 3 6 6

Totally ,we have 24 symmetry elements. It is a point

group of the cubic crystal system, called 432 (O).

e) 4 3, 3, 3, 3 combination

Since S63=i, there are many groups of the symmetry to

be produced. e.g. , 2100…

We get a point group for cubic system,called 4/m 3 2/m

or m3m (Oh).

1, 31111, 32111= 32111, 33111 =1, 34111= 32111, 35111; 18

41100, 43100;

41100,43100 12

2100 , 2110 9

2100 , 2110 9

4. The application of the point group of crystal.

Neumann’s principle: the macroscopic physical

properties (tensor) of a crystal have at least the

symmetry of the point group of crystal.

<general stress>= <respondingcoefficients><general displacements>

Imported physical varieties

true stress

Outputted physical varieties

true displacements

The polyhedron of physical properties

Of crystal

Since the stress does not cause the rotation of the crystal, we have

ij .a/2=ji.a/2 ij =ji

<ij> is a 2 order symmetric tensor

11 12 13

<ij>= 22 23

33

<T>= ( T1, T2, T3,T4,T5,T6)T

To the strain, we have the similar results

<S>= ( s1, s2, s3,s4,s5,s6)T

According to the elastic law, we have

<Ti>=<cij>< S i >

Physical properties.

elastic coefficients

c11 c12 c13 c14 c15 c16

c21 c22 c23 c24 c25 c26

c31 c32 c33 c34 c35 c36

= c41 c42 c43 c44 c45 c46

c51 c52 c53 c54 c55 c56

c61 c62 c63 c64 c65 c66

We need to obtain 36 numbers to determine elastic

properties of the crystal .

<cij>

For the cubic crystal, we have

c11 c12 c13 0 0 0

c21 c22 c23 0 0 0

c31 c32 c33 0 0 0

<cij> = 0 0 0 c44 0 0

0 0 0 0 c55 0

0 0 0 0 0 c66

Piezoelectricity of the crystals

Physical phomania: When applying a stress on some crystal, some charges will be introduced on the surface of the crystal. This specific characteristics is called Piezoelectricity of the crystals or vice versa.

There are more than 800 crystal with Piezoelectricity.

Only few of them had widely application in practice.

Some examples of crystals with Piezoelectricity

-SiO2 , LiSO4.H2O, LiNbO3, LiTaO3, ZnO2, GaAs,

1940 1950 1960

Pb(Mg 1/3Nb2/3)O3, polymers with piezoelectricity

1970

Physical and mathematic Description Piezoelectricity

ps =d< ij >

polarization intensity, a vector Ps=(p1,p2 p3)T

Piezoelectric coefficient, a tensor

stress, a 2 order tensor

p1

p2 = dijk dijk is a 3 order tensor with

p3 27 components

11

12

13

21

22

23

31

32

33

12

13

21

22

23

31

32

33

d 1.11 d 1.12 d 1.13 d 1.21 d 1.22 d 1.23 d 1.31 d 1.32 d 1.33

d 2.11 d 2.12 d 2.13 d 2..21 d 2.22 d 2.23 d 2.31 d 2.32 d 2.33

d 3.11 d 3.12 d 3.13 d 3.21 d 3.22 d 3.23 d 3.31 d 3.32 d 3.33

p1

p2 =

p3

pi =dijkjk ( j,k=1,2,3)

The physical meaning of dijj

pi=d i.1111 +d i.12 12 +d i.13 13 +d i.21 21+d i.22 22+ d i.23

23+d i.31 31 +d i.32 32 +d i.33 33

d 1.11=p1/11, d 1.22=p1/22, d 1.33=p1/33

d 2.11=p2/11, d 2.22=p2/22, d 2.33=p2/33

d 3.11=p3/11, d 3.22=p3/22, d 3.33=p3/33

p1

d 1.11 d 1.12 d 1.13 d 1.21 d 1.22 d 1.23 d 1.31 d 1.32 d 1.33

p2 = d 2.11 d 2.12 d 2.13 d 2..21 d 2.22 d 2.23 d 2.31 d 2.32 d 2.33

d 3.11 d 3.12 d 3.13 d 3.21 d 3.22 d 3.23 d 3.31 d 3.32 d 3.33

p3

11

12

13

21

22

23

31

32

33

Physical meaning of the other 18 components:

According to jk =kj, we can obtain

pi= (dijk+dikj) jk

j,k =1, 2, 3 j k

Let dijk=dikj, we have dijk= pi/2jk .

p1

d 1.11 d 1.12 d 1.13 d 1.22 d 1.23 d 1.33

p2 = d 2.11 d 2.12 d 2.13 d 2.22 d 2.23 d 2.33

d 3.11 d 3.12 d 3.13 d 3.22 d 3.23 d 3.33

p3

11

12

13

22

23

33

Since 11, 22, 33 are normal stress, we rearrange the above equation as

p1

1112 13

d 1.11 d 1.22 d 1.33 d 1.23 d 1.13 d 1.12

p2 = d 2.11 d 2.22 d 2.33 d 2.23 d 2.13 d 2.12 22 23

d 3.11 d 3.22 d 3.33 d 3.23 d 3.13 d 3.12

33

p3

11

22

33

23

13

12

In order to simplify the foot signs, we let

11, 22, 33, 23, 32, 13, 31, 12, 21

1 2 3 4 5 6

p1

d 1.1 d 1.2 d 1.3 d 1.4 d 1.5 d 1.6

p2 = d 2.1 d 2.2 d 2.3 d 2.4 d 2.5 d 2.6

d 3.1 d 3.2 d 3.3 d 3.4 d 3.5 d 3.6

p3

1

2

3

4

5

6

<din> is more simple than<dijk>, but still not simple

enough. According to Neumann’s principle ,

can be further simplified.

Physical properties can be described by

Xp=(xp1,xp2,xp3)

R

For example, 1 acts on Xp. We get

1 1 1

Xp1 -Xp1 , Xp2 -Xp2 , Xp3 -Xp3

Shortly,

1 -1, 2 -2, 3 -3

Xp’=(xp’1, xp’2, xp’3)

Means the results of R acting on Xp can be indicated by

the foot sign changes.

R

Xijk (+/_) X???

For instance, 1 acts on the dijk

i -1 0 0 -i

j 0 -1 0 =-j dijk =d-i-j-k=-dijk

k 0 0 -1 -k

So dijk =0

This means that if a crystal has 1 symmetry, it has not

piezoelectricity,since< dijk >=0

There are 11 point groups with 1 symmetry. They are

1 (S2, Ci),

2/m(C2h), 2/m 2/m 2/m(D2h),

4/m(C4h) ,4/m 2/m 2/m(D4h),

3(S6,C3i), 3 2/m or 3m (D3d)

6/m(C6h), 6/m 2/m 2/m(D6h)

2/m 3 or m3(Th), 4/m 3 2/m (Oh)432(O)

Totally we have 12 point groups. If a new crystal is not

bellowed the above point group list. There is a chance to

be used as a piezoelectric material.

The limitation to < dijk > resulted from point group .

a) The limitations from 2(C2) point group

According to

Xp1 {2[001]} -Xp1 1 -1

Xp2 -Xp2 shortly 2 -2

Xp3 Xp3 3 3

d123 d-1-23= d123

d133 d-133 =- d133

This means the components in < dijk > with one 3 or

two 3 foot signs or without 3 may be not equal to zero. The other components should be equal to zero.

d 1.11 d 1.22 d 1.33 d 1.23 d 1.13 d 1.12

d 2.11 d 2.22 d 2.33 d 2.23 d 2.13 d 2.12

d 3.11 d 3.22 d 3.33 d 3.23 d 3.13 d 3.12

2(C2)

d 1.11 d 1.22 d 1.33 d 1.23 d 1.13 d 1.12

d 2.11 d 2.22 d 2.33 d 2.23 d 2.13 d 2.12

d 3.11 d 3.22 d 3.33 d 3.23 d 3.13 d 3.12

0 0 0 d 1.23 d 1.13 0 0 0 0 d 1.4 d 15 0

0 0 0 d 2.23 d 2.13 0

d 3.11 d 3.22 d 3.33 0 0 d 3.12d 3.1 d 3.2 d 3.3 0 0 d 3.6

0 0 0 d 2.4 d 2.5 0

b) The limitation of 42m-D2d to < dijk >or < din > .

Since along [001] there is a 2-fold axis,

due to it’s limitation, we have

0 0 0 d 1.23 d 1.13 0

0 0 0 d 2.23 d 2.13 0

d 3.11 d 3.22 d 3.33 0 0 d 3.12

According to

1 {4[001]} -2

2 1 we have

3 -3

d 1.23 d -2.1-3 = d 2.13 , d 1.13 d –2-.2-3=- d 2.23

d 3.11 d -3.-2-2=-d 3.22 , d 3.33 d -3.-3-3 =- d 3.33

d 3.12 d-3.-21= d3.21

Similar to {2[001]}’s limitation, we can discuss the

limitations produced by {2[100]},{2[010]}.

The components with two 1 or without 1 and two 2 or without 2 in foot signs might be equal to zero.

< dijk > for crystal with the point group is

0 0 0 d 1.23 0 0

< dijk >= 0 0 0 0 d 2.13 0

0 0 0 0 0 d 3.12

In a simple form

For which crystal ?

Crystallography III

Group Theory II-230 space groups of the crystal (the general description of the crystals)

1.Introduction

Crystal = Lattice structure basis

1)Point symmetry 7 C.S. primitive

1801—1807 lattice

2) Point symmetry Geometric

14B.L. Crystallo-

graphy

Translational symmetry

1818----1839

Crystal = Lattice structure basis

1801—1830

3)Point symmetry 32 point group

of crystals

To describe the physical properties of the

crystals in quantity and quality

Lay a basis for the structure of the crystal to be

understood.

Crystal = Lattice structure basis

1885—1896

Point symmetry(point group)

4) Translational symmetry(Bravis lattice)

Nonsymmorphic 230 space groups

of crystals

To give a general structure pattern of the crystals

and reflection conditions and other useful information

The space group of crystal is a basic symmetry collection formed by Point symmetry, Translational symmetry and Nonsymmorphic .

2 The symmorphic space group

The symmorphic

point group + Bravis lattice

space group

CS B L Point group Space group

1 P1

Triclinic P

1 P1

2 P2

P m Pm

2/m P2/m

Monoclinic

2 B2

B m Bm

2/m B2/m

Totally we have 66 symmorphic space group. They are

CS Symmorphic space group

Triclinic P1,P1

Monoclinic P2,Pm, P2/m; B2, Bm, B2/m

Orthorhombic P222, Pmm2, Pmmm;

C222, Cmm2,Cmmm;Ammm; I222, Imm2,

Immm; F222,Fmm2, Fmmm

Tetragonal P4,P4,P4/m,P422,P4mm,P42m,(P4m2),P4/mmm;

I4,I4,I4/m,I422,I4mm,I42m,( I4m2), I4/mmm;

Cubic P23,Pm3,P432,P43m,Pm3m;

I23, Im3,I432,I43m,Im3m; F23, Fm3,F432,F43m,Fm3m

Trigonal P3,P3,P312,(P321),P3m1,(P31m), P3m1,(P31m)

R3,R3, R32,R3m,R3m

Hexagonal P 6,P6,P6/m,P622, P6mm,P6m2,(P62m),P6/mm

For these space group we also have four ways to be

described.

Symbology Geometry Mathematics Language

Here we take Pmm2 as an example

Symbol: international Pmm2,

Schoenflies: C12v

Geometric patterns:

general position symmetry position

pattern pattern

,

,

,

,

,

,

,

,

Mathematic description :

{ {2[001]},{2[100]}, {2[010]}, {1[000]}, tn1n2n3}

Some important things we should treat carfully

- Disposition problem

The disposition of the point group symmetry in the Bravis

Lattice. To make sure no extra lattice points to be

produced.

In addition, the 2-fold axis in mm2 could be placed along

a or b axis, but they are equivalent after reselecting the

coordinate axes. Pmm2

However, in some cases, the situation will change.

Example: C +mm2 in Orthorhombic

In Bravis Lattice . A, B, C are the same

c(b, a)

b(a,c)

a(c, b)

We have three ways toput the point group of mm2 into the above Bravis lattice.

(1)2-fold axis along c axis

(2) 2-fold axis along a axis

(3) 2-fold axis along b axis

We get

The mirrors lay on t

he coordinate planes

(1) C lattice 2-fold along c direction

general position symmetry position

pattern pattern

,

,

b

,

,

,

,

,

,

,

,

a

(2) C lattice 2-fold along adirection

general position

pattern

c,a

Different pattern from

case (1). Get a new space

Group, called Amm2

in a,b,c system.

b,c

,

,

,

,

,

a,b

(3) C lattice 2-fold along b direction

general position

pattern

c,a,b

Different pattern from

case (1) but same as (2)

in a,b. c system

No new space

Group to be obtained.

b,c,a

,

,

,

,

,

a,b,c

So for the single face center Orthorhombic lattice, we

have Cmm2, and Amm2 (Bmm2), they are produced by

the difference of the point group patterns position. By the

similar discussion in each crystal system, we get other 6

new symmorphic space groups, (see page 79).

Totally we have 77 symmorphic space groups.

3 The nonsymmorphic space groups.

The nonsymmorphic operation

The existence of The nonsymmorphic

operation

From C mm2 general position pattern, we can find a

new kind symmetry operation. b-glide operation and

a-glide operation.

b

,

,

,

,

,

,

,

,

,

,

a

(1) the nonsymmorphic operation could be produced by

the combination of the point symmetry and translational

Symmetry.

(2) Nonsymmorphic operations also could be resulted

from the periodical regular distortion of the atomic

planes.

4+(0,0,1/4)T

4+(0,0,2/4)T

From the above patterns, we can find the regularities of the distortion.

- A kind of screw type distortion limited by the point symmetry of the crystal.
- A kind of distortion limited by translational symmetry of crystal.

So, the basic characteristics of nonsymmorphic symmetry

is

point symmetry fractional translationvector

1, 2, 3, 4, 6 unit translational vector

1, 2, 3, 4, 6 along the rotation axis

Mathematic description of the nonsymmorphic operation

Seitz Operator of nonsymmorphic

In fact, the operator of the point symmetry and the

translational symmetry can be included in Seitz operator.

For point symmetry , it’s form is {R, 0}.

For translational symmetry , it’s form is {1, tn}.

3. Nonsymmorphic operations

3.1 Screw axes {R, }, R=2,3,4,6 ;=?

Let the axis along c direction, =[0,0,?]T

According to

{R, }n r= {R, } {R, }…. {R, }(R r+ )

n-1

= {R, } {R, }…. {R, }(R2 r+R + )

n-2 0

R2r+2 0

0 0 1

=R n r+ n

(the requirement of the point symmetry)

R n=1

n =mtu ( tu : unit lattice traslational vector ) ,m n

requirement of the translational symmetry

So, {R, }n = {1, n }= {1, mtu}

= m/n

Therefore,

- n=2, m=0, 1

m=0, {2, 0} 2-fold axis ( point symmetry) 2

m=1/2, {2, ½}, 21-screw axis(nonsymmorphic) 21

+1/2

tu

(2) n=3, m=0, 1, 2

m=0, {3, 0} , 3-fold axis ( point symmetry) 3

m=1, {2, 1/3}, 31-screw axis(nonsymmorphic) 31

m=2, {2, 2/3}, 32-screw axis(nonsymmorphic) 32

+2/3 +1/3 +1/3 +2/3

+

+

+

+

+

(3) n=4, m=0, 1, 2,3

m=0, {4, 0} , 4-fold axis ( point symmetry) 4

m=1, {4, 1/4}, 41-screw axis(nonsymmorphic) 41

m=2, {4, 2/4}, 42-screw axis(nonsymmorphic) 42

m=3, {4, 3/4}, 43-screw axis(nonsymmorphic) 43

+2/4 +1/4

+

+

+

+3/4

+

+

+

+1/2

+3/4

+2/4

+1/4

+1/2

+

+

(3) n=6, m=0, 1, 2, 3,4, 5

m=0, {6, 0} , 6-fold axis ( point symmetry) 6

m=1, {6, 1/6}, 61-screw axis(nonsymmorphic) 61

m=2, {6, 2/6}, 62-screw axis(nonsymmorphic) 62

m=3, {6, 3/6}, 63-screw axis(nonsymmorphic) 63

m=4, {6, 4/6}, 64-screw axis(nonsymmorphic) 64

m=5, {6, 5/6}, 65-screw axis(nonsymmorphic) 65

+3/6

+

+4/6

+2/6

+

+

+

+

+5/6

+1/6

+

+

3. 2Glide planes {R, }, R=1, 2, 3, 4, 6 ;=?

Let the symmetry axis be along c direction, we have

0

{ R[001]}= 0

0 0 -1

=(0, 0, )T

{R, }n r= {R, } {R, }…… {R, }(R2 r+R + )

n-2

=-

0=

So this group of combination is not allowed.

R

But , 2 is an exception, since

1 0 0

2= 0 1 0 If 1= ( , 0, 0)T , 2=(0, , 0)T or

0 0 -1 3=(, , 0)T , we have

{2, }2 r =22 r+2

=r +tn=r +tu

={1, tu}

So, we get a new group of nonsymmorphic operation ,

called glide planes.

i= ½ tu

2

3

1

2= (0, 1/2, 0)T

1= (1/2,0,0)T along a axis , called a-glide

along a+b axis , called n-glide

3= (1/2, ½, 0)T

We also have c-glide , and other form’s n-glides

b

,

a

The mathematic description can be written as

{m [100], (0, ½,0)T} r={m[100]}r +

-1 0 0 x 0 -x

= 0 1 0 y + ½ = ½+y

0 0 -1 z 0 z

b

,

a

The mathematic description can be written as

{m [001], (1/2, ½,0)T} r={m[001]}r +

1 0 0 x 1/2 ½+x

= 0 1 0 y + ½ = ½+y

0 0 - 1 z 0 - z

3. 3 Diamond Glide planes

Basing 73 symmorphic space group,

CS Symmorphic space group

Triclinic P1,P1

Monoclinic P2,Pm, P2/m; B2, Bm, B2/m

Orthorhombic P222, Pmm2, Pmmm;

C222, Cmm2,Cmmm;Ammm; I222, Imm2,

Immm; F222,Fmm2, Fmmm

Tetragonal P4,P4,P4/m,P422,P4mm,P42m,(P4m2),P4/mmm;

I4,I4,I4/m,I422,I4mm,I42m,( I4m2), I4/mmm;

Cubic P23,Pm3,P432,P43m,Pm3m;

I23, Im3,I432,I43m,Im3m; F23, Fm3,F432,F43m,Fm3m

Trigonal P3,P3,P312,(P321),P3m1,(P31m), P3m1,(P31m)

R3,R3, R32,R3m,R3m

Hexagonal P 6,P6,P6/m,P622, P6mm,P6m2,(P62m),P6/mm

We can discuss other space groups in each crystal system.

Monoclinic

P2, Pm, P2/m; B2, Bm, B2/m

B21

Ba,Bb

Bc,Bn

…..

P21/m

P2/a

P21/a

.

.

B21/m

B2/a

B2/b

.

.

P21

Pa,Pb,Pc

Pn…….

C34h

5. A page from the International Table

Tetragonal 4/m P4 /n No.85

Origin at 4 at ¼,1/4,0 from 1

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Number of positions, Conditions

Wyckoff notation, coordinates of equavalent positions limiting

And point symmetry possible reflections

General

- g 1 x, y, z; x, y, z ; ½+ x, ½+y, z; ½- x, 1/2-y, z; hkl: no conditions

y, x , z; y, x, z; ½-y, ½+x, z; ½+y, ½-x, z hk0:h+k=2n

00l:No condition

Special :as above , plus

- f 2 0, 0, z; 0, 0, z; ½, ½, z; ½, ½, z hkl: h+k=2n
- e 1 ¼, ¼, ½; ¾, ¾, ½; ¼,3/4, ½; ¾, ¼, ½;

hkl: h,k=2n

- d 1 ¼, ¼, 0; ¾, ¾, 0; ¼,3/4, 0; ¾, ¼,0
- c 4 0, ½, z; ½, 0, z No extra conditions

2 b 4 0, 0.1/2; ½,1/2, ½

hkl: h+k=2n

2 a 4 0, 0, 0; ½, ½, 0

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