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2D Symmetry (1.5 weeks)

2D Symmetry (1.5 weeks). Start with the translation. Add a rotation. lattice point. lattice point. . . lattice point. A translation vector connecting two lattice points! It must be some integer of or we contradicted the basic Assumption of our construction. T : scalar. .

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2D Symmetry (1.5 weeks)

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  1. 2D Symmetry (1.5 weeks)

  2. Start with the translation Add a rotation lattice point lattice point   lattice point A translation vector connecting twolattice points! It must be some integer of or we contradicted the basic Assumption of our construction. T: scalar  p: integer Therefore,  is not arbitrary! The basic constrain has to be met!

  3. B B’ b T T T T   tcos tcos A A’ T T To be consistent with the original translation t: pmust be integer M must be integer M >2 or M<-2: no solution pM cosn (= 2/) b 4 3 2 1 0 -1 -3 -1.5 -- -- -- -2 -1  2 3T -1 -0.5 2/3 3 2T 0 0 /2 4 T 1 0.5 /3 6 0 2 1 0  (1) -T Allowable rotational symmetries are 1, 2, 3, 4 and 6.

  4. Look at the case of p = 2 n = 3; 3-fold  angle = 120o 3-fold lattice. Look at the case of p = 1 n = 4; 4-fold  = 90o 4-fold lattice.

  5. Look at the case of p = 0 n = 6; 6-fold  = 60o Exactly the same as 3-fold lattice. Look at the case of p = 3 n = 2; 2-fold Look at the case of p = -1 n = 1; 1-fold 1 2

  6. Parallelogram Can accommodate 1- and 2-fold rotational symmetries 1-fold 2-fold 3-fold 4-fold 6-fold Hexagonal Net Can accommodate 3- and 6-fold rotational symmetries Square Net Can accommodate 4-fold rotational Symmetry!

  7. Combine mirror line with translation: constrain m m Unless Or 0.5T Primitive cell centered rectangular Rectangular

  8. (1) Parallelogram (2) Hexagonal Net (3) Square Net (4) Centered rectangular Double cell (2 lattice points) (5) Rectangular Primitive cell Lattice + symmetries = plane group (5) (1, 2, 3, 4, 6, m)

  9. The symmetry that the lattice point can accommodate Five kinds of lattice Oblique Rectangular Centered rectangular Square Hexagonal + + + + + 1, 2 m m 4 3,6 Plane group Group theory Here we show that 3D: space group.

  10. Group theory: set of elements (things) for a law of combination is defined and satisfies 3 postulates. (1) the combination of any two elements is also a member of the group; (2) “Identity” (doing nothing) is also a member of the group. “I” aI=Ia=a (a : an element) (3) for element, an inverse exists. a; a-1 a . a-1 = Ia-1. a = I http://en.wikipedia.org/wiki/Group_(mathematics) Example: Group {1, -1}; rank 2 rank (order) of the group = number of elements contained in a set. 1 -1 Another Example: Group {1, -1, i, -i}; rank 4 1 1 -1 -1 -1 1 We will show examples for point groups later!

  11. In a point, there is no translation symmetry! Therefore, consider 2D point group, we only consider rotation and mirror! Put rotation symmetry and mirror together ? m [] [ ] [ ] [ ] [ ] n 1 2 3 4 6

  12. 2+m Example: 2mm: point group L R (4) 1  1: 1 1  2: 2 1  3: A 1  4: 1 (1) m2 R L (2) (3) {1, 1, 2, A} group of rank 4 m1 1 1 2 A Abelian group: a.b=b.a 1 1 1 2 A 1 1 1 A 2 2 2 A 1 1 A A 2 1 1

  13. 6-fold is a subset of 2-fold axis subgroup 3-fold axis

  14. L || if  L    R Chirality not changed: 1 2  Rotation is the right choice! Combination theorem  (3) (4) 2mm (1) (2)

  15. Show it is a group 1 1 2 A 1 1 1 2 A 1 1 1 A 2 2 2 A 1 1 Satisfy 3 postulates? A A 2 1 1 Rank 4 The number of motif in the pattern is exactly the same as the rank (order) of the group!

  16. Notation: Rotation axis Hermann and Mauguin International notation n 1, 2, 3, 4, 6 C1, C2, C3, C4, C6 Cn Schonllies notation C: cyclic group – all elements are “powers” of some basic Operation e.g. http://en.wikipedia.org/wiki/Group_(mathematics)#Cyclic_groups

  17. Mirror plane Hermann and Mauguin International notation m CS Schonllies notation Cnv : Rotational symmetry with mirror plane vertical to the rotation axis. E.g. 2mm – C2v .

  18. (3) L R (1) 2 (2) 1 L m m m The rank of this group is ? m m m m m S:C4v HM: 4mmmmmmmm Only independent symmetry elements.

  19. S:C6v HM: 6mm /6 (3) The rank of this group is 12! 2 (1) (2) 1 (3) R S:C3v HM: 3mm (correct?) The rank of this group is 6! 2 (1) L (2) R 1 2 is not independent of 1. HM (international notation): 3m

  20. So far we have shown 10 point group or specifically 10 2-D crystallographic point group. HM notation , , , , , , , , , ; Schonllies notation , , , , , , , , , . 2-D crystallographic space group 10 2-D crystallographic point group 5 2-D lattices

  21. Oblique Primitive Rectangular Centered rectangular Square Hexagonal 1, 2 m m 4 3,6 Compatible with Compatibility: 2mm, 3m, 4mm, 6mm

  22. 2mm Put mirror planes along the edge of the cell. m m m m Primitive Rectangular Centered rectangular Compatible with m, 2mm Square Compatible with 4, 4mm

  23. 30o Hexagonal Compatible with

  24. Red ones Blue ones Hexagonal Compatible with 6mm

  25. Oblique Primitive Rectangular Centered rectangular Square Hexagonal , , , , ,,, Compatible with

  26. General oblique net. atoms Symbol used to describe the space group Space group: p1 Type of lattice Point group 1 P (for primitive) Upper case P for 3D lower case p for 2D

  27. Primitive oblique net + 2 = p2 (next page first) (1) B (3) (2) plane group: p2

  28. Do this in a more general term B A : (1)  (2); : (2)  (3);  A : (1)  (3). 2 1 /2 /2 /2 (3) x (2) (1) A along the -bisector of at

  29. p4 4 + lattice 4 || || 1 Correct? Combination of A/2 with at

  30. p + 3 = p3 3 120o Combination of A2 /3 with at along the -bisector of T X mass center 30o

  31. (1)(2): A2/3; (2)(3): Translation 60o 60o (2) (3) (1)(3): B2/3; (1) 60o 60o

  32. p + 6 = p6p has to be hexagonal net as well! 2-fold 6 3-fold From 3-fold rotation From 2-fold rotation Combination of A /3 and A- /3 with at

  33. Combination of mirror symmetry with the translation! m + p + c p + m = pm @ (2) L (1) R (3) L   Independent mirror plane

  34. c + m = cm (3) L (2) L not independent mirror plane! (1) R  (1)  (3) Glide plane with glide component  Two-step operation cm m g m g m

  35. p + g = pgpossible? (3) L (2) L (1)  (3)? (1) R g g General form:

  36. c + g = cg possible? g g m g g m rectangular net: pm, pg, cm! cg = cm

  37. p +2mm = p2mm   c +2mm = c2mm  

  38. p4mm p (square) +4mm Red: p4. Blue: pm. Special case of a rectangular. m

  39. p (Hexagonal net) +3m p3 60o 60o two ways  centered rectangular net 60o 60o m || edge m  edge

  40. p3 ||Cell edge p3m1 p31m  Cell edge

  41. Not yet done! Glide plane (or line). 3m 3m 3m 3 p3m1 p31m

  42. p (Hexagonal net) +6mm = p6 + p3m1 + p31m Red Blue Mirror line Glide line p6mm

  43. 2mm compatible with Rectangular! mirror plane? p2mm

  44. This way? Why not? This way? Why not? Leave this unmoved  OK

  45. Center rectangular net (c2mm)? (m ok? ) (g ok? ) p2mg

  46. 2 2mm look of point symmetry: 2mm.

  47. p2gg

  48. OK? Three different ways:

  49. The same result Why not C4gm! p4gm

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