Proof: There are only 1, 2, 3, 4, and 6 fold rotation symmetries for crystal with translational symmetry. The space cannot be filled! graphically
Start with the translation Add a rotation lattice point lattice point T: scalar A translation vector connecting two lattice points! It must be some integer of or we contradicted the basic Assumption of our construction. p: integer Therefore, is not arbitrary! The basic constrain has to be met!
T T T tcos b To be consistent with the original translation t: p must be integer M must be integer p M cos n (= 2 / ) b -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 M >2 or M<-2: no solution T TT AA’ B’B 4 3 2 1 0 Allowable rotational symmetries are 1, 2, 3, 4 and 6.
Look at the case of p = 2 = 120 o angle Look at the case of p = 1 n = 3; 3-fold n = 4; 4-fold = 90 o 3-fold lattice. 4-fold lattice.
Look at the case of p = 0 = 60 o n = 6; 6-fold Look at the case of p = 3n = 2; 2-fold Look at the case of p = -1n = 1; 1-fold 12 Exactly the same as 3-fold lattice.
1-fold 2-fold 3-fold 4-fold 6-fold Parallelogram Hexagonal Net Can accommodate 1- and 2-fold rotational symmetries Can accommodate 3- and 6-fold rotational symmetries Square Net Can accommodate 4-fold rotational Symmetry! These are the lattices obtained by combining rotation and translation symmetries? How about combining mirror and translation Symmetries?
Combine mirror line with translation: m Unless 0.5T centered rectangular constrain Or Primitive cell Rectangular m
Symmetry elements in 2D lattice Rectangular = center rectangular?
Two ways to repeat 1-D 2D (1) maintain 1-D symmetry (2) destroy 1-D symmetry 3-2-2. 2-D lattice
Maintain mirror symmetry m
Destroy mirror symmetry
3-2-3. 3-D lattice: 7 systems, 14 Bravais lattices (1)Triclinic system 1-fold rotation (1) b a c
one diad axis (only one axis perpendicular to the drawing plane maintain 2-fold symmetry in a parallelogram lattice) (2) Monoclinic system
(1) Primitive monoclinic lattice (P cell) a b c (2) Base centered monoclinic lattice c b a B-face centered monoclinic lattice
The second layer coincident to the middle of the first layer and maintain 2-fold symmetry Note: other ways to maintain 2-fold symmetry a c b A-face centered monoclinic lattice If relabeling lattice coordination A-face centered monoclinic = B-face centered
(2) Body centered monoclinic lattice Body centered monoclinic = Base centered monoclinic
So monoclinic has two types 1. Primitive monoclinic 2. Base centered monoclinic
(3) Orthorhombic system a b c 3 -diad axes
(1) Derived from rectangular lattice to maintain 2 fold symmetry The second layer superposes directly on the first layer (a) Primitive orthorhombic lattice a b c
(b) B- face centered orthorhombic = A -face centered orthorhombic c a b (c) Body-centered orthorhombic (I- cell)
rectangular body-centered orthorhombic based centered orthorhombic
(2) Derived from centered rectangular lattice (a) C-face centered Orthorhombic a b c C- face centered orthorhombic = B- face centered orthorhombic
(b) Face-centered Orthorhombic (F-cell) Up & Down Left & Right Front & Back
Orthorhombic has 4 types 1. Primitive orthorhombic 2. Base centered orthorhombic 3. Body centered orthorhombic 4. Face centered orthorhombic
(4) Tetragonal system a b c One tetrad axis starting from square lattice
(1) maintain 4-fold symmetry (a) Primitive tetragonal lattice First layer Second layer
(b) Body-centered tetragonal lattice First layer Second layer Tetragonal has 2 types 1. Primitive tetragonal 2. Body centered tetragonal
(5) Hexagonal system a = b c; = = 90 o ; = 120 o b a c One hexad axis starting from hexagonal lattice (2D) a = b; = 120 o
(1) maintain 6-fold symmetry Primitive hexagonal lattice b a c (2) maintain 3-fold symmetry a b c 1/3 2/3 1/3 2/3 a = b = c; = = 90 o
(6) Cubic system 4 triad axes ( triad axis = cube diagonal ) Cubic is a special form of Rhombohedral lattice Cubic system has 4 triad axes mutually inclined along cube diagonal b a c a = b = c; = = = 90 o
(a) Primitive cubic = 90 o b a c a = b = c; = = = 90 o (b) Face centered cubic = 60 o a = b = c; = = = 60 o
(c) body centered cubic = 109 o a = b = c; = = = 109 o
cubic (isometric) Primitive (P) Body centered (I) Face centered (F) Base center (C) Special case of orthorhombic with a = b = c a = b c Tetragonal (P) Tetragonal (I)? Cubic has 3 types 1. Primitive cubic (simple cubic) 2. Body centered cubic (BCC) 3. Face centered cubic (FCC)
http://www.theory. nipne.ro/~dragos/S olid/Bravais_table. jpg = P= I = T P