3D Symmetry_1
(2 weeks)
Next we would move a step further into 3D symmetry.
Leonhard Euler :
http://en.wikipedia.org/wiki/Leonhard_Euler
Google search: Euler
Spherical trigonometry
Small circle
R<1
For convenience, set R = 1
Great circle
(GC), R=1
Distance: AOB =  (GC)
o

A
B
angle
P
o
A
Well defined
B
B’
Pole 90o to arc AB.
OP  plane defined by OAB
POA = /2; POB = /2; POB’ = /2
Trigonometry: points on a surface of a sphere (directions that
intersect the sphere) are connected using arcs of great circles
Center of the sphere
c
B
A
arc BC = a OB  OC
arc AC = b OA  OC
GC
b
a
arc AB = c. OA  OB
C
angle
Spherical Angles
GC
?
A
B
B’
BAC = B’OC’
o C
C’
B’OC’
A is the pole for plane defined by B’OC’
Polar triangle
A
GC
A
GC
B
B
C
A, pole of arc BC
B, pole of arc AC
C, pole of arc AB
C
ABC and ABC are mutually polar!
Proof: B: pole of arc AC  B is 90o away from point A.
C: pole of arc AB  C is 90o away from point A.
 A:pole of arc BC.
Similarly, B: AC,
C: pole of arc AB.
Proof: BAC = , arc BC = a,   + a = .
B
B
A
A
P
a
Q
C C
B
o
A
C
Q
C
B : pole of arc AC  arc BQ = /2
C : pole of arc AB  arc CP = /2
 arc BQ + arc CP = 
= (arc BP+ arc PQ) + (arc PQ+ arc QC)
= (arc BP+ arc PQ+ arc QC) + arc PQ = a + arc PQ
= POQ = 
Why! See pictures of spherical angle in page 4 (bottom)!
Law of cosines
Plane Trigonometry
c
B
b
A
C
a
length
a 2  b2  c 2  2bc cos A
angle
How about law of cosines in spherical trigonometry?
Triangle is defined as
C is spherical angle at point u.
vOu = a
a, b, c  wOu = b
vOw = c
(3)
w
(4)
b
o
y
O
90o
(1)
(2) a
1
O
(4)
b 1
u z
v
z
(1)
(3)
a
(1)= tana
(3)= tanb
Unit circle
(2)
u
y
u
1/(2)= cosa 1/(4)= cosb
 (2) = seca  (4) = secb
2
2
(From uyz)
tan
a

tan
b  2 tana tanb cosC
yz 
 sec2 a  sec2 b  2 sec a sec b cosc (From oyz)
 2  tan2 a  tan2 b  2 sec a sec b cosc
  tan a tan b cos C  1  sec a sec b cos c
2
 cos c  cos a cos b  sin a sin b cos C
http://en.wikipedia.org/wiki/Spherical_law_of_cosines
Stop here about spherical trigonometry!
We obtain all the relations needed for
further discussion of the 3D point groups!
Combination of two rotation operations in 3 D:
A
C
B
B  A  ?
A : (1)  (2)
B : (2)  (3)
2R
1R
3R
(1) and (3) relation?
c
3-D: translation, reflection,
rotation, and inversion.
B  A  C
must be crystallographic
A
B
A
c
B
c
Locate the position of axis C
A
c
b
C
B
a
A
C
b
a
B
c
Euler construction:
N C’ M’
A symmetry element is the locus of a
point that is left unmoved by an operation.
A
 
A: AMAM’.
B: BNBN’.
C (the point unmoved). OC: the axis
N’ C M
B
(1) A: leave A unmoved.
(2) B: move A to A’.
A
ABC = A’BC = /2
AB = A’B ABC = A’BC
 ACB = A’CB  /2
/2
 /2
N’ C M
A
B  A  C
B
/2
c
b
B
a
C
/2
/2
The law of cosine (spherical trigonometry)
cos c  cos a cos b  sin a sin b cos
cos a  cos b cos c  sin b sin c cos

2

2
A’
A
180o-/2
Polar
B triangle
c
/2 /2
b /2 a
180o-b
180o-a
180o-/2
180o-/2
180o-c
C
Law of cosine to the polar triangle



cos(180  )  cos(180  ) cos(180  )
2
2
2
o
 sin(180 
o
o

2
 cosc 
) sin(180 
o

cos

2
o
) cos(180o  c)


 cos cos
2
2
2


sin
sin
2
2
cosc 

cos


 cos cos
2
2
2
sin  sin 
2
2
cos  cos  cos
2
2
2
cos a 


sin
sin
2
2
cosb 
cos 
 cos cos
2
2
2
sin  sin 
2
2
All the rotation combinations possible in 3D that need to be
B
tested:
1
2
3
4
6
A
1
2
3
4
6
111
112
113
114
116
212
213
214
216
313
314
316
414
416
222
223
224
226
323
324
326
424
426
333
334
336
434
436
444
446
616
626
636
646
666
Axis at
A, B, or C
, , or 
/2
/2
/2
1-fold
2-fold
3-fold
4-fold
6-fold
360o
180o
120o
90o
60o
180o
90o
60o
45o
30o
cos( 2)
sin( 2)
cos( 2)
sin( 2)
cos( 2)
sin( 2)
-1
0
1/2
1/21/2
31/2/2
0
1
31/2/2
1/21/2
1/2
Case: 11n
A
A: 1,  = 360o, cos( /2) = -1; sin( /2) = 0
c
b
B: 1,  = 360o, cos( /2) = -1; sin( /2) = 0
A
B
a
180o
/2
/2
C
C: n,  = 360o /n , cos( /2); sin( /2)
cos
/2
c
B
180o
 cos cos 
b
a
2
2
2
cosc 
180o/n
C
sin  sin 
2
2

cos
1
2
None exist! Except,  = 360 o
cosc 

0
 111
Case: 22n
A: 2,  = 180o, cos( /2) = 0; sin( /2) = 1
A
/2
c
b
B: 2,  = 180o, cos( /2) = 0; sin( /2) = 1
cosc 
cos
 cos cos 
2
2
2
sin  sin 
2
2
cos  0

2
cosc 
 cos
2
1
n2
n3
n4
n6
;
;
;
;
 180o
 120o
 90o
 60o
a
A
90o
b
c
c
B
a
C
/2
/2
C
C: n,  = 360o /n , cos( /2); sin( /2)
B
90o
180o/n

2
; cos / 2  0 ; cosc  0 ; c  90o
; cos / 2  1/ 2 ; cosc  1/ 2 ; c  60o
; cos / 2  1/ 2 ; cosc  1/ 2 ; c  45o
; cos / 2  3/2 ; cosc  3/2 ; c  30o
C
Angle between A and B axis
222
a b
c  90o
A
223
c  60
o
224
c  45
226
c  30o
A
90o
B
60o
B
o
What are a and b?
A
A
45o
B
30o
B
A: 2,  = 180o, cos( /2) = 0; sin( /2) = 1
B: 2,  = 180o, cos( /2) = 0; sin( /2) = 1
C: n,  = 360o /n , cos( /2); sin( /2)



cos
 cos
cos
0  0  cos( / 2)
2
2
2
cos a 

0
1 sin( / 2)
sin  sin 
2
2
 a = 90o.
cos b 

cos


 cos
cos
2
2
2  0  0  cos( / 2)  0
1  sin( / 2)
sin  sin 
2
 b = 90o.
2
C
C
B
A
C
C
45o
30o
60o
90o B
A
A
222
B
223
224
A
226
B
Case: 23n
A: 2,  = 180o, cos( /2) = 0; sin( /2) = 1
B: 3,  =
120o,
C: n,  =
360o
cosc 
cos
cos( /2) = 0.5; sin( /2) =
30.5/2
A
/n , cos( /2); sin( /2)
 cos cos 
cos  0  (0.5)
2
2
2 
2
1 ( 3 / 2)
sin  sin 
2
2
90o
b
c
B
a
C
60o
360o/n
o
o
n

3
;


120
;
cos

/
2

1
/
2
;
cos
c

1
/
3
;
c

54
44'
233
o
o
234 n  4 ;   90 ; cos / 2  1/ 2 ; cosc  2 / 6 ; c  35 16'
236 n  6 ;   60o ; cos / 2  3/2 ; cosc  1 ; c  0o
None exist
The rest of combination does not exist!
Case: 233
A: 2,  = 180o, cos( /2) = 0; sin( /2) = 1
B: 3,  = 120o, cos( /2) = 0.5; sin( /2) = 30.5/2
C: 3,  = 120o, cos( /2) = 0.5; sin( /2) =30.5/2
cos  cos  cos
0  0.5  0.5 1
2
2
2
cos a 


3 / 2 3 / 2 3
sin  sin 
2
2
 a = 70o32.
cosb 

cos


 cos cos
2
2
2  0.5  0  0.5  1


1 3 / 2
3
sin
sin
2
 b = 54o44.
2
[1 1 1]
z
[111]
C
y
x
B
70o32’
000
54o44’
A [100]
54o44’
233
Angle between A and B is
1
[100]  [111]  1  1 3 cosc  cosc 
 c  54o 44'
3
Angle between A and C is
1
[100]  [1 1 1]  1  1 3 cosb  cosb 
 b  54o 44'
3
Angle between B and C is
1
[111]  [1 1 1]  1  3  3 cos b  cos b   b  70 o32 '
3
Case: 234
A: 2,  = 180o, cos( /2) = 0; sin( /2) = 1
B: 3,  = 120o, cos( /2) = 0.5; sin( /2) = 30.5/2
C: 4,  = 90o, cos( /2) = 1/20.5; sin( /2) = 1/20.5
cos  cos  cos
0  0.5  (1 / 2 )
1
2
2
2
cos a 


3 / 2  (1 / 2 )
3
sin  sin 
2
2
 a = 54o44.
cosb 

cos


 cos cos
2
2
2  0.5  0  (1 / 2 )  2

2

1 1 / 2
sin
sin
2
 b = 45o.
2
A [110]
35o16’
z
45o
y
x
B
[111]
000
54o44’
C [100]
234
Angle between A and B is
2
[110]  [111]  2  2  3 cosc  cosc 
 c  35o16'
6
Angle between A and C is
1
[110]  [100]  1  2 1cosb  cosb 
 b  45o
2
Angle between B and C is
1
[111]  [100]  1  3 1cosb  cosb 
 b  54o 44'
3
 Geometry of the permissible nontrivial combination of
rotations:
Combination 2A = 
222
223
224
226
233
234
180o
180o
180o
180o
180o
180o
c
2B = 
90o
60o
45o
30o
54o44
35o16
a
180o 90o
180o 90o
180o 90o
180o 90o
120o 70o32
120o 54o44
2C = 
b
180o 90o
120o 90o
90o
90o
60o
90o
120o 54o44
90o
45o
International symbol
222
322
(1)
422
(2)
(1)
(3)
32(2)
Just like 3m(m)
Only one independent
2 fold rotation axis
622
(2)
(3)
22 operation is basically
on the plane!
n22
Dn
222
D2
32(2)
D3
422 622
D4
D6
dihedral
A
Schonllies notation
different dihedral angle
B3 / 2
C3 / 2
233
23 is enough to specify the symmetry!
A
B3 / 2
23
Schonllies notation: T
Tetrahedral
International symbol
http://en.wikipedia.org/wiki/Tetrahedron
International symbol
35o16’
A
B 2
2
3
45o
C
54o44’
234 or 432
Schonllies notation: O
Octahedron
http://en.wikipedia.org/wiki/Octahedron
11 axial combinations
1 2 3 4 6
222 322 422 622
233 432
11 axial combinations + Extender
Ways to add m:
n
vertical m
n
horizontal m
422
horizontal 
diagonal 
vertical 
for Dn, T, O
Not for Cn
Extender: v, h, d, 1 ! (+ extender  create new rotation axis!)
1
Cnv, Dnv
Tv, Ov
2
3
4
6
222
32
422
622
23
432 4
v
We will explain it later
Cnh, Dnh
Th, Oh
h
Dnd
Td, Od
d
Cni, Dni
Ti, Oi
See reading
crystal4.pdf
1
http://ocw.mit.edu/courses/materialsscience-and-engineering/3-60-symmetrystructure-and-tensor-properties-ofmaterials-fall-2005/readings/crystal4.pdf
1
Cnv, Dnv
Tv, Ov
v
Cnh, Dnh
Th, Oh
h
Dnd
Td, Od
d
Cni, Dni
Ti, Oi
1
2
3
m
2mm
3m
m
2
m
3
6
m
-
-
1
2
m
4
6
222
32
2 2 2
4mm 6mm
6m2
mmm
 mmm
4
m
6
m
2 2 2
mmm
6m2
-
-
-
4 2m
2
3
m
3
4
m
6
m
2 2 2
mmm
2
3
m
4
Let’s look at some cases
Two fold rotation (2) + horizontal mirror (h)
A
R
R
h
C2 h
{1 A  h 1}
L
L
R: right-handedness
L: left-handedness
R
L
R
L
down
2
A   h 
m
2m
up
1 A  h 1
1 1 A  h 1
A A 1 1  h
 h  h 1 1 A
1 1  h A 1
(2)
R
A
xyz
(1) R
h
(3)
L x yz
?: (1)  (3)
 h  A  1
at the point of intersection
Four fold rotation (4) + horizontal mirror (h)
R
A / 2
h
4
A 2   h 
m
R
L
C4 h
1
down
up
L
Four fold rotation (4) + vertical mirror (v)
L
R
L
A 2   v  4mm
R
L
R
C4 v
The mirror that you put in
R
Mirror 45o with respect to the first
L
down
up mirror set
Six fold rotation (6) + horizontal mirror (h)
A / 3
h
6
A 3   h 
m
1
C6 h
Group symmetry elements: 12
R L
R
L
R
L
R
L
R
L
 Down  Up
Six fold rotation (6) + vertical mirror (v)
A 3   v  6mm
C6 v
R
L
Three fold rotation (3) + horizontal mirror (h)
R
(1) 
L
A2 3   h 
 R (3)
L
R
(2) 
L
down
up
New two step operations
A / 3  1  A / 3
A / 3  1  A / 3
Roto-inversion
(1)   (1) 
(1)   (2) 
(1)   (3) 
(1)   (1) 
(1)   (2) 
(1)   (3) 
3
6
m
C3 h
1
A2 / 3
A4 / 3  A2 / 3
h
A / 3  1
A / 3  1
{1 A2 / 3 A2 / 3  h A / 3 A / 3}
Roto-reflection

(2)
R
(3)
L
~
1
~
2
~
3
~
4
~
5
~
6
~
7
~
8
(1) ~
A
R
n~
Roto-inversion
The one used
~
12
~
21
5
~
63
(2)
R

A
(1)
R n
(3)
L
~
36
~
44
7
~
88
Three fold rotation (3) + vertical mirror (v)
A2 / 3   v  3m
C3 v
Three fold rotation (3) + inversion ( 1)
A2 / 3  1  3
(1)
R
(6)
 Down
 Up
C3i
L (5)
L
(2) R
L
(4)
3
3m
m
R (3)
(1)  (1)
(1)  (2)
(1)  (3)
(1)  (4)
(1)  (5)
(1)  (6)
1
A2 / 3
A4 / 3  A2 / 3
1
A2 / 3  1  A2 / 3
A4 / 3  1  A2 / 3
{1 A2 / 3 A2 / 3 A2 / 3 A2 / 3 1}
3
C3v C3h C3i
L
 down  up
1 1  1
1 1
R
21
2?
2
21 
m
You can except 4  1  4
2m
CS
4
S4
(2)
(1) R
L
(4) L
R (3)
Sphenoid (Greek word for axe)
(1) To (1)
(1) To (2)
(1) To (3)
(1) To (4)
1
A / 2
A
A / 2
{1 A / 2 A A / 2}
Equal length
The rest four: equal length.
Not tetrahedron
If n is odd  n  1  n
If n is even  n  1  n
How about 6 ?
L
R
 Down  Up
R
L
3
6
m
R
L
4 is a special one that you have to add to the
11 axial combination
R
L
4
R
L
L
R
Add h 
4
m
R
L
L
R
R
R
R
+ h
R
422
4
m
Add 1 
R R
R
R
R
L
4
Homework!
R L
L L R R
R
L
R
L
L
R
L R
R L
4 2 2
mmm
Look at the ppt file that I send you regarding to
222 + Extender (v, h, d, 1 ) as an example!
all up.
all down.
T
T
up R
down R
23
23
Add a horizontal
mirror plane
R
L
L
R
R
L
Th
L
R
inversion
2
3
m
Create an
inversion center
Symmetry Direction
Crystal
System
Primary
Triclinic
None
Monoclinic
[010]
Orthorhombic
Secondary
Tertiary
[100]
[010]
[001]
Tetragonal
[001]
[100]/[010]
[110]
Hexagonal/
Trigonal
[001]
[100]/[010]
[120]/[1 1 0]
Cubic
[100]/[010]/
[001]
[111]
[110]
Buerger’s book
3D crystallographic point group
2D lattices: chapter 7 (pg. 69-83)
Euler’s construction: pg. 35-43
Some combination theorems: chapter 6
Points group: pg: 59-68