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MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Crystallographic Points, Directions, and Planes.
ISSUES TO ADDRESS...
• How to define points, directions, planes, as well aslinear, planar, and volume densities
– Define basic terms and give examples of each:• Points (atomic positions)• Vectors (defines a particular direction - plane normal)• Miller Indices (defines a particular plane)
• relation to diffraction• 3-index for cubic and 4-index notation for HCP
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
av
bv
cv
Points, Directions, and Planes in Terms of Unit Cell Vectors
All periodic unit cells may be described viathese vectors and angles, if and only if• a, b, and c define axes of a 3D coordinate system.• coordinate system is Right-Handed!
But, we can define points, directions andplanes with a “triplet” of numbers in unitsof a, b, and c unit cell vectors.
For HCP we need a “quad” of numbers, aswe shall see.
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
POINT Coordinates
To define a point within a unit cell….Express the coordinates uvw as fractions of unit cell vectors a, b, and c(so that the axes x, y, and z do not have to be orthogonal).
av
bv
cv
origin
pt. coord.
x (a) y (b) z (c)
0 0 0
1 0 0
1 1 1
1/2 0 1/2
pt.
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Crystallographic Directions
Procedure:1. Any line (or vector direction) is specified by 2 points.
• The first point is, typically, at the origin (000).
2. Determine length of vector projection in each of 3 axes inunits (or fractions) of a, b, and c.• X (a), Y(b), Z(c) 1 1 0
3. Multiply or divide by a common factor to reduce thelengths to the smallest integer values, u v w.
4. Enclose in square brackets: [u v w]: [110] direction.
a b
c
DIRECTIONS will help define PLANES (Miller Indices or plane normal).
�
[1 1 0]5. Designate negative numbers by a bar • Pronounced “bar 1”, “bar 1”, “zero” direction.
6. “Family” of [110] directions is designated as <110>.
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MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Self-Assessment Example 1: What is crystallographic direction?
a b
c Along x: 1 a
Along y: 1 b
Along z: 1 c
[1 1 1]DIRECTION =
Magnitude alongX
Y
Z
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Self-Assessment Example 2:
(a) What is the lattice point given by point P?
(b) What is crystallographic directionfor the origin to P?
The lattice direction [132] from the origin.
Example 3: What lattice direction does the lattice point 264 correspond?
�
[1 12]
�
−112
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Symmetry Equivalent Directions
Note: for some crystal structures, differentdirections can be equivalent.
e.g. For cubic crystals, the directions are allequivalent by symmetry:
[1 0 0], [ 0 0], [0 1 0], [0 0], [0 0 1], [0 0 ]111
Families of crystallographic directions e.g. <1 0 0>
Angled brackets denote a family of crystallographic directions.
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Families and Symmetry: Cubic Symmetry
x
y
z
(100)
Rotate 90o about z-axis
x
y
z
(010)
x
y
z
(001)Rotate 90o about y-axis
Similarly for otherequivalent directions
Symmetry operation cangenerate all the directionswithin in a family.
3
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Designating Lattice Planes
Why are planes in a lattice important?
(A) Determining crystal structure * Diffraction methods measure the distance between parallel lattice planes of atoms.
• This information is used to determine the lattice parameters in a crystal. * Diffraction methods also measure the angles between lattice planes.
(B) Plastic deformation * Plastic deformation in metals occurs by the slip of atoms past each other in the crystal. * This slip tends to occur preferentially along specific crystal-dependent planes.
(C) Transport Properties * In certain materials, atomic structure in some planes causes the transport of electrons and/or heat to be particularly rapid in that plane, and relatively slow not in the plane.
• Example: Graphite: heat conduction is more in sp2-bonded plane.
• Example: YBa2Cu3O7 superconductors: Cu-O planes conduct pairs of electrons(Cooper pairs) responsible for superconductivity, but perpendicular insulating.
+ Some lattice planes contain only Cu and O
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
How Do We Designate Lattice Planes?
Example 1
Planes intersects axes at:• a axis at r= 2• b axis at s= 4/3• c axis at t= 1/2
How do we symbolically designate planes in a lattice?
Possibility #1: Enclose the values of r, s, and t in parentheses (r s t) Advantages:
• r, s, and t uniquely specify the plane in the lattice, relative to the origin.• Parentheses designate planes, as opposed to directions given by [...]
Disadvantage:• What happens if the plane is parallel to --- i.e. does not intersect--- one of the axes?• Then we would say that the plane intersects that axis at ∞ !• This designation is unwieldy and inconvenient.
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
How Do We Designate Lattice Planes?
Planes intersects axes at:• a axis at r= 2• b axis at s= 4/3• c axis at t= 1/2
How do we symbolically designate planes in a lattice?
Possibility #2: THE ACCEPTED ONE
1. Take the reciprocal of r, s, and t. • Here: 1/r = 1/2 , 1/s = 3/4 , and 1/r = 2
2. Find the least common multiple that converts all reciprocals to integers.• With LCM = 4, h = 4/r = 2 , k= 4/s = 3 , and l= 4/r = 8
3. Enclose the new triple (h,k,l) in parentheses: (238)4. This notation is called the Miller Index.
* Note: If a plane does not intercept an axes (I.e., it is at ∞), then you get 0.* Note: All parallel planes at similar staggered distances have the same Miller index.
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Self-Assessment Example
What is the designation of this plane in Miller Index notation?
What is the designation of the top face of the unit cell in Miller Index notation?
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MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Families of Lattice Planes
The Miller indices (hkl) usually refer to the plane thatis nearest to the origin without passing through it.
• You must always shift the origin or move the planeparallel, otherwise a Miller index integer is 1/0!
• Sometimes (hkl) will be used to refer to any other planein the family, or to the family taken together.
• Importantly, the Miller indices (hkl) is the same vectoras the plane normal!
Given any plane in a lattice, there is a infinite set of parallel lattice planes(or family of planes) that are equally spaced from each other.
• One of the planes in any family always passes through the origin.
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
z
x
y
Look down this direction(perpendicular to the plane)
Crystallographic Planes in FCC: (100)
d100 = aDistance between (100) planes
… between (200) planes d200 =a2
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Crystallographic Planes in FCC: (110)
d110 =a 22
Distance between (110) planes
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Crystallographic Planes in FCC: (111)
z
x
y
Look down this direction(perpendicular to the plane)
d111 =a 33
Distance between (111) planes
5
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Note: similar to crystallographic directions, planes that are parallel toeach other, are equivalent
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Comparing Different Crystallographic Planes
-1
1
For (220) Miller Indexed planes you are getting planes at 1/2, 1/2, ∞.The (110) planes are not necessarily (220) planes!
For cubic crystals: Miller Indices provide you easymeasure of distance between planes.
d110 =a
12 +12 + 02=
a2=a 22
Distance between (110) planes
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Directions in HCP Crystals
1. To emphasize that they are equal, a and b is changed to a1 and a2.2. The unit cell is outlined in blue.3. A fourth axis is introduced (a3) to show symmetry.
• Symmetry about c axis makes a3 equivalent to a1 and a2.• Vector addition gives a3 = –( a1 + a2).
4. This 4-coordinate system is used: [a1 a2 –( a1 + a2) c]
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Directions in HCP Crystals: 4-index notation
Example What is 4-index notation for vector D?
• Projecting the vector onto the basal plane, it liesbetween a1 and a2 (vector B is projection).
• Vector B = (a1 + a2), so the direction is [110] incoordinates of [a1 a2 c], where c-intercept is 0.
• In 4-index notation, because a3 = –( a1 + a2), thevector B is since it is 3x farther out.
• In 4-index notation c = [0001], which must beadded to get D (reduced to integers) D = [1123]
Self-Assessment Test: What is vector C?
Easiest to remember: Find the coordinate axes that straddle the vectorof interest, and follow along those axes (but divide the a1, a2, a3 part of vectorby 3 because you are now three times farther out!).
�
1
3[112 0]
Check w/ Eq. 3.7or just use Eq. 3.7
a2
–2a3
B without 1/3
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MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Directions in HCP Crystals: 4-index notation
Example
What is 4-index notation for vector D? • Projection of the vector D in units of [a1 a2 c] givesu’=1, v’=1, and w’=1. Already reduced integers.
• Using Eq. 3.7:
�
[112 3]
�
[ 1
3
1
3
2
31]
Check w/ Eq. 3.7: a dot-product projection in hex coords.
�
u = 13
[2u'−v'] v = 13
[2v'−u'] w = w'
�
u = 13
[2(1)−1] = 13
v = 13
[2(1)−1] = 13
w = w'= 1
• In 4-index notation:
• Reduce to smallest integers:
After some consideration, seems just using Eq. 3.7 most trustworthy.
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Miller Indices for HCP Planes
As soon as you see [1100], you will knowthat it is HCP, and not [110] cubic!
4-index notation is more important for planes in HCP, in orderto distinguish similar planes rotated by 120o.
1. Find the intercepts, r and s, of the plane with any twoof the basal plane axes (a1, a2, or a3), as well as theintercept, t, with the c axes.
2. Get reciprocals 1/r, 1/s, and 1/t.3. Convert reciprocals to smallest integers in same ratios.4. Get h, k, i , l via relation i = - (h+k), where h is
associated with a1, k with a2, i with a3, and l with c.5. Enclose 4-indices in parenthesis: (h k i l) .
Find Miller Indices for HCP:
r s
t
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Miller Indices for HCP Planes
What is the Miller Index of the pink plane?
1. The plane’s intercept a1, a3 and cat r=1, s=1 and t= ∞, respectively.
1. The reciprocals are 1/r = 1, 1/s = 1, and 1/t = 0.
2. They are already smallest integers.
3. We can write (h k i l) = (1 ? 1 0).
4. Using i = - (h+k) relation, k=–2.
5. Miller Index is (1210)
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
Yes, Yes….we can get it without a3!
1. The plane’s intercept a1, a2 and cat r=1, s=–1/2 and t= ∞, respectively.
1. The reciprocals are 1/r = 1, 1/s = –2, and 1/t = 0.
2. They are already smallest integers.
3. We can write (h k i l) =
4. Using i = - (h+k) relation, i=1.
5. Miller Index is
�
(12 10)
�
(12 ?0)
But note that the 4-index notation is unique….Consider all 4 intercepts:• plane intercept a1, a2, a3 and c at 1, –1/2, 1, and ∞, respectively.• Reciprocals are 1, –2, 1, and 0.• So, there is only 1 possible Miller Index is
�
(12 10)
7
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
a1
a2
a3
• Parallel to a1, a2 and a3• So, h = k = i = 0
• Intersects at z = 1
Name this plane…
Basal Plane in HCP
�
plane = (0001)
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
z
a1
a2
a3
(1 1 0 0) plane
+1 in a1
-1 in a2
h = 1, l = 0i = -(1+-1) = 0,k = -1,
Another Plane in HCP
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
(1 1 1) plane of FCCz
x
y
z
a1
a2
a3
(0 0 0 1) plane of HCPSAME THING!*
MSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08
SUMMARY
• Crystal Structure can be defined by space lattice and basis atoms(lattice decorations or motifs).
• Only 14 Bravais Lattices are possible. We focus only on FCC, HCP,and BCC, i.e., the majority in the periodic table.
• We now can identify and determined: atomic positions, atomic planes(Miller Indices), packing along directions (LD) and in planes (PD).
• We now know how to determine structure mathematically. So how to we do it experimentally? DIFFRACTION.
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