Crystallography
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C6H2
PART IIA and PART IIB MATERIALS SCIENCE AND METALLURGY Course C6: Crystallography
C6H2
Basic Crystallography The Weiss Zone Law This law expresses the mathematical condition for a vector [uvw] to lie in a plane (hkl). This condition can be determined through elementary vector considerations: c C Normal to (hkl) (hkl) c l O
b k
B
a h
b
A a
Consider the plane (hkl) in the above diagram. The vectors defining the unit cell of the crystal are a, b and c. A general vector, r, lying in (hkl) can be expressed as a linear combination of any two vectors lying in this plane, such as AB and AC, i.e.,
r = λAB + µAC for suitable λ and µ. Hence, expressing AB and AC in terms of a, b and c, it follows that µ λ +µ r = λ b –a +µ c –a = – a+λ b + c k h l h h k l If we re-express this as r = u a + v b + w c, i.e., a general vector [uvw] lying in (hkl), it follows that u =–
λ +µ µ , v = λ, w = h k l
and so hu + kv + lw = 0 which is the condition for a vector [uvw] to lie in the plane (hkl): the Weiss Zone Law. It is evident from this derivation that it is valid for any crystal system.
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Reciprocal lattice vectors These are particularly useful in the description of electron diffraction patterns. The reciprocal lattice vectors a*, b* and c* are related to a, b and c through the equations a* =
b× c , a.b× c
b* =
c ×a , a.b× c
c*=
a× b a.b × c
Normals to planes can be expressed conveniently in terms of these reciprocal lattice vectors. To see this, consider the plane (hkl) shown on the previous page. A vector normal to this plane, n, must be parallel to the cross product AB × AC. Hence,
n || b – a × c – a , i.e., n || b× c – b × a – a× c k h l h kl hk hl and so after some straightforward mathematical manipulation, making use of the identities a×b = – b×a and c×a = – a×c, this shows that n is parallel to the vector ha* + kb* + lc*, i.e., n = ξ (ha* + kb* + lc*) for some ξ. If we wish the magnitude of n to be related to the interplanar spacing of the (hkl) planes, dhkl, it is convenient to consider the diagram below: O a h
A
ψ
(hkl)
^n N
Here, N is a point on the plane defined by A, B and C where the normal to the plane (hkl) passing through the origin, O, meets the plane. Hence, |ON| = dhkl. Also, 2 OA.ON = OA ON cos ψ = ON 2 = dhkl
since the angle ONA is a right angle. If we write the vector ON in the form ON = n = χ ha* + k b* + lc* for some χ, it follows that the dot product OA.ON = χ since OA = a/h. Hence, 2 χ = dhkl
Now, ON.ON = n.n = χ 2 ha* + k b* + lc* 2
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but we also know 2 ON 2 = dhkl
and of course |ON|2 = ON.ON. Therefore,
dhkl 2 = dhkl 4 ha* + k b* + lc* 2 and so we obtain the important result that
ha* + k b* + lc* =
1 dhkl
i.e., in words, the magnitude of the reciprocal lattice vector ha* + kb* + lc* is inversely proportional to the spacing of the hkl planes. It is also evident from this derivation that the above equation is valid for any crystal system.
It also follows that if the normal to the (hkl) set of planes is simply taken to be the vector n = ha* + kb* + lc* then a vector r =[uvw] lying in the plane must make an angle of 90° with this vector., i.e., r.n = 0. Writing out this dot product explicitly, we obtain the result
hu + kv + lw = 0 i.e., the Weiss Zone Law.
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Hexagonal Indices: the four-index notation The conventional three-index notation does not show symmetry explicitly, e.g., [100], [010] and [110] in hexagonal close packed metals are related by symmetry, but a simple permutation of the indices does not show this. The four-index notation for hexagonal (and rhombohedral) crystals enables planes and vectors related by symmetry to be more easily recognised than in the three-index notation. The ‘trick’ is to define a redundant additional axis lying in the x-y plane, so that the x-y plane is spanned by three vectors, a1, a2 and a3, as on the diagram below. These three axes make an angle of 120° with respect to one another. For a three-index plane (hkl), the intercepts on the three axes are, respectively, a/h, a/k and a/i, where a is the magnitude of the unit cell side in this plane. The intercept on the c-axis, c/l remains unaffected by the introduction of this redundant axis. Thus, on the diagram below, h > 0, k > 0, but i < 0, because the intercept on the a3 axis is along the – a3 vector. a3
a k
O 60˚ 60˚
a h θ
A
a i G
(120˚ – θ)
(60˚ – θ)
B
a2
Trace of (hkl) in the x-y plane
a1
The three-index plane (hkl) is then written in the four-index notation as (hkil). Simple geometry can be used to determine the relationship between h, k and i. To avoid confusion with negative signs, it is convenient in the above figure to let the (positive definite) magnitude of OG be a/p for a positive p.
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Then by the sine rule in triangles OAB and OGA, we have: sin θ = sin 60Þ – θ a/k a/h
sin θ = sin 120Þ –θ a/p a/h
and
respectively. Now sin (120° – θ) = sin (180° – (120° – θ)) = sin (60° + θ), and so these two equations can be rearranged in the forms
k sin θ = sin 60Þ –θ h and p sin θ = sin 60Þ + θ h
Using the identity sin (60° + θ) – sin (60° – θ) = 2 cos 60° sin θ = sin θ, it follows from subtracting the first equation from the second that
p – k = h, i.e.,
p = h + k.
Therefore the intercepts on the three axes are a/h, a/k and –a/(h + k). Hence, in the four index notation
h+k+i=0 when re-indexing three index planes (hkl). Thus, the three three-index planes (100), (010) and (110) related to one another by symmetry in hexagonal crystals re-index in the four-index notation as (1010), (0110) and (1100). Thus, simple permutation of h, k and i can be used to show planes related to one another by symmetry in hexagonal crystals in the four-index notation.
For vectors, the situation is less straightforward. Since three vectors overdetermine a plane, we need to introduce a constraint when defining four-vector equivalents [UVJW] of vectors [uvw] defined conventionally in the three-index notation. The constraint is: U+V+J=0
so that for example, the relationships tabulated overleaf between [uvw] and [UVJW] hold.
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Three-index notation
Four-index notation
[100]
1 3
2110
[010]
1 3
1210
1 3
[11 2 0]
[110] 1 3
[uv0]
1 3
[uvw]
2u – v, 2v – u, – u + v , 0 2u – v, 2v – u, – u + v , 3w
Permutations of U, V and J produce symmetrically equivalent directions. The dot product between a normal n = ha* + kb* + lc* and a vector r = ua + vb + wc using conventional a, b and c crystal axes is r.n = hu + kv + lw
in the three-index notation. In the four-index notation this becomes r.n = hU + kV + iJ + lW
since from the above table, hU + kV + iJ + lW = 1 h 2u – v + 1 k 2v – u + 1 h + k u +v +lW = hu + kv + lw 3
3
3
Hence the condition for a vector [UVJW] to lie in a plane (hkil) is simply hU + kV + iJ + lW = 0
which is the four-vector Weiss Zone Law. Another useful feature of the four index notation is that the vector [UVJ0] is normal to the (UVJ0) set of planes. The proof is straightforward. Referring to the conventional three index hexagonal axes a and b, i.e., a1 and a2 on the diagram on C6H2 page 4, a* is parallel to the vector 2a + b and b* is parallel to the vector 2b + a (remember a*.b = b*.a = 0 from the definitions of a* and b* on C6H2, page 2). a* and b* are equal in magnitude. Hence the normal to the (UVJ0) set of planes is parallel to the vector r = U(2a + b) + V(2b + a) = Ua + Vb – (U + V)(–(a + b)),
i.e., the vector [UVJ0] expressed with respect to a and b.
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The Symmetry Elements of a Cube I.
Rotation axes of symmetry
Three tetrads normal to the cube faces. Four triads along the body diagonals. Six diads through the midpoints of opposite edges.
II.
Planes of symmetry
Three mirror planes parallel to (100), (010) and (001), shown in the diagram opposite. In addition there are six {110} mirror planes shown in the diagrams below: (a) (011) and (011); (b) (101) and (101); (c) (110) and (110).
III.
Summary
Symmetry elements
Tetrads
Triads
Diads
Mirror Planes
Orientation
{100} and {110}
Note that for any crystal system: are all the directions related by the point group symmetry to [uvw].
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{hkl} are all the sets of planes, or faces, related by the point group symmetry to (hkl).
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We now need a way of conveniently depicting these symmetry elements. A start can be made by imagining the cube to be placed at the centre of a large sphere. The rotation axes of symmetry of the cube can be extended as radii of the sphere. At the point of intersection with the sphere the symmetry elements can be marked. Similarly, planes of symmetry extended to intersect the sphere can be indicated by great circle lines on the surface. An example of each is shown here:
The sphere with all the symmetry elements marked on its surface will then look like:
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PART IIA and PART IIB MATERIALS SCIENCE AND METALLURGY Course C6: Crystallography
C6H3
Stereographic Projections
Representation of directions and plane orientations In studying crystallographic and symmetry relationships we are concerned with directions and plane orientations, not with positions. As seen in C6H2 for symmetry elements, we use the intersections with a sphere of directions and planes passing through the centre of the sphere. This is shown here for an arbitrary plane (hkl) and its normal P. The plane intersects the sphere in a great circle, which represents the shortest distance between points such as Q and R on the surface of the sphere. The angle between the directions represented by Q and R is most easily measured along the great circle. Figure 1
Mapping the surface of a sphere We are now faced with the problem of mapping onto a 2D sheet the points and lines on the surface of a sphere. This is a difficult problem, although it is at least familiar from the mapping of the world. The lines of longitude (or meridians) on a globe are great circles. The equator is also a great circle, but other lines of latitude (or parallels) are small circles (intersections with the sphere of planes not passing through the origin). Possible ways of viewing the world in three different projections are shown in Figures 2, 3 and 4.
Figure 2 The world as viewed from a distance (the orthographic projection)
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The Mercator Projection
Figure 3 This is probably the most familiar projection of the world, having the advantage of giving a continuous map, with the lines of longitude and latitude always orthogonal to each other, as on the surface of the globe. However, great circles cannot be plotted directly.
The Stereographic Projection The stereographic projection shows one hemisphere, here chosen to correspond to the side of the globe shown in Figure 2. The angular scale is more convenient than on the ‘orthographic’ projection. Note that the lines of longitude and latitude are always orthogonal. As will be seen next, the stereographic projection is particularly useful for plotting angular relationships because great circles are always arcs of circles and are easily constructed.
Figure 4
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Construction of the stereographic projection A direction of interest intersects the surface of the sphere at point P (see Figure 5 below). Such a point is called a pole. A plane of projection passing through the origin of the sphere is then chosen. The normal to the projection plane intersects the sphere at the point of projection. A straight line from the pole P to the point of projection passes through the plane of projection at some point p – the projected pole of P. The array of projected poles p on the plane of projection corresponding to an array of points such as P forms the stereographic projection (or stereogram) of the poles on the surface of the sphere, and therefore the stereographic projection of the set of directions represented by the radii generating those poles. Lines on the surface of the sphere (e.g., great circles or small circles) can be projected point-bypoint to give lines in the projection.
Figure 5
Figure 6
Instead of using an equatorial plane as a plane of projection, a plane parallel to it can be used, as in the example above in Figure 6 in which the plane is a tangent plane. Here, the stereographic projection will be identical to that shown in Figure 5, but will be linearly twice the size. The shadow projector used in the first examples class to demonstrate the stereographic projection uses the principle of obtaining the stereographic projection as a shadow on a plate, with a transparent sphere and a light source at the point of projection. Properties of the stereographic projection The most important properties of the stereographic projection are: 1.
Angular truth is conserved, i.e., the angle between lines on the surface of the sphere is equal to the angle between the projections of those lines.
2.
Circles on the surface of the sphere project as circles on the plane of projection (‘circles project as circles’). [This is true for both great circles and small circles.]
These two properties in particular make the stereographic projection appropriate for representing angular relationships in three dimensions.
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The stereographic projection: the projection of a pole – further considerations
The diagrams above show a pole P on the surface of the sphere and its projected pole p in the corresponding stereographic projection when (from left to right), P is in the upper hemisphere, in the lower hemisphere and in the plane of projection. A pole in the lower hemisphere projects outside the primitive circle if L is the point of projection. It projects inside the primitive circle at p' if U is used as the projection point. In this case, the projected pole is represented by an open circle, rather than by a dot.
The first two diagrams above show the vertical section of the projection sphere through U, L and P for a pole P in the upper and lower hemispheres respectively. The projected pole p lies at the intersection of LP and QR. The projected pole p' lies at the intersection of UP and QR. The third diagram is a stereogram showing the projected poles p and p' which represent two directions OP related to one another by reflection in the plane of projection.
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sketch of sphere
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stereogram
A direction OP can be plotted on a stereogram if the two angles θ and ψ are known, where θ = UOP and ψ is the angle that the plane QUPRL makes with a reference plane TUML. In the stereographic projection Op = OL tan (θ/2) and the angle MOR = ψ. The stereographic projection: great circles
projection sphere
stereogram Figure 7
In general, a great circle projects within the primitive circle as an arc of a circle, as in the example above (Figure 7). Great circles pass through the centre of the sphere of projection. Hence, any two non-diametrically opposite points A and B on the sphere are sufficient to define the plane of the great circle. A great circle containing A and B must also contain the diametrically opposite points AO and BO. It follows that the projection of a great circle intersects the primitive circle in two diametrically opposite points, C and CO. Given two projected poles such as a and b on the stereogram, it is possible to find the great circle on which they lie using the Wulff net (or stereographic net) – see page 8.
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Figure 8 (projection sphere)
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Figure 9 (stereogram)
The normal OP to the plane of a great circle meets the sphere at a point P called the pole of the great circle (Figure 8). It makes an angle of 90° with every direction in the great circle. If the great circle meets the primitive at the opposite ends of the diameter CCO (Figures 8 and 9), the pole must lie in the vertical plane perpendicular to the line CCO; its projected pole must lie on the diameter EF of the primitive. P must be perpendicular to the pole B which corresponds to the line of intersection of the great circle and the vertical plane. The projected pole of P is therefore at p, lying on the line EF at an angular distance of 90° from b, the projected pole of B. The distance bp can be measured conveniently using the Wulff net. Conversely, the great circle normal to a direction whose projection is the pole p can be drawn by first plotting the pole b which lies on the diameter of the primitive through p where the distance bp = 90°, measured with the Wulff net. The projection of the great circle passes through b and through opposite ends C and CO of the diameter of the primitive which is normal to the diameter through p.
Finally, a great circle projects as a straight line if it passes through the point of projection L, as in the example on the right here. Such a great circle represents a plane which is normal to the plane of projection.
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The stereographic projection: small circles
QTR and DTV are examples of great circles and AB is an example of a small circle. The small circle is the intersection of the cone AOB with the surface of the projection sphere whose centre is at O. The stereographic projection of a small circle AB is the circle ab on the equatorial plane, as on the diagram below. The axis of the cone AOB is OP. The small circle represents the locus of points which lie at a given angle to the direction OP. Because the angular scale on the stereographic projection is non-linear (see page 8), the pole p does not lie in the middle of the small circle ab unless the centre of the small circle is centred at O.
section of projection sphere
stereographic projection
Finally, on the right, a special case: a smallcircle projects as a straight line if it passes through the point of projection L. (If that part of the small circle which lies in the lower hemisphere is projected using U as the projection point, its projection will be an arc of a circle).
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The Wulff (or stereographic) net On a globe, lines of longitude and latitude are used to provide angular scales for defining positions. Figure 4 on page 2 shows a hemisphere of the globe in the stereographic projection. This projection was made from a pole on the equator. The resulting arrangement of the lines of longitude and latitude provides a useful net for angular measurement on the stereographic projection. This is the Wulff net (or stereographic net), shown in more detail below.
In this standard form of the net, 2° angular intervals are used. In using the net to measure angles on the stereographic projection, the diameters of the primitive circle of the projection and of the net must match and the centres of the net and of the projection must remain coincident. The net may be rotated about the centre. When this is done, any two points (poles) on the projection can be made to lie on the same great circle, which is therefore found. In general, the angle between the two directions represented by the projected poles can then be measured along the great circle. Occasionally, small circles are used for measuring angular intervals on the stereographic projection, but great care is needed.
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Uses of the Wulff net The following methods are applicable irrespective of whether the projection is drawn on tracing paper pinned to the centre of the net or the projection is drawn on opaque paper and a transparent net is used on top of it. 1.
To plot a pole a given angle around the primitive circle
Measure around the circumference of the net.
2.
To plot a pole a given angle from the centre
Measure along one of the diameters of the net.
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3.
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To plot a small circle around a pole within the primitive circle
Note that along the diameter of the projection passing through the projected pole of the centre of the circle, opposite ends of the diameter of the projected pole must be at equal angular distances from the projected centre. In general, the geometrical centre of a projected circle is NOT coincident with the projected centre and is displaced towards the primitive circle.
4.
To locate a pole at specified angles from two other poles
Construct two small circles of appropriate radii around the two poles. Usually, there will be two solutions, occasionally just one or none.
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5.
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To locate the great circle linking two poles and to measure the angle between them
For this, rotate the net about its centre until the two poles lie on the same great circle. Measure the angle along the great circle.
6.
To locate the pole of a great circle
The pole must be 90° away from every point on the great circle including the points where it crosses the primitive circle and the point which is itself 90° from those points on the primitive.
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7.
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To measure the angle between great circles
The best method is to locate the poles of the two great circles and then to measure the angle between those poles (use procedures 5 and 6 above).
Note:
on a 5" (127 mm) diameter stereogram it is possible to achieve an accuracy of better than ± 2° in procedures involving several of the above methods in sequence provided reasonable care is taken.
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The Stereographic Projection applied to the cubic system The stereographic projection shows {100}, {110} and {111} in the holosymmetric (= most symmetric) cubic class, m3m.
The poles of (100), (010) and (001) are coincident with the poles of the reference axes, x, y and z respectively. The poles of the form {110} can be plotted using (100) : (110) = (010) : (110) = 45°, etc. (111) lies at the intersection of the zones [(100), (011)], [(010), (101)] and [(001), (110)]. Its pole is found by drawing the great circles which represent these zones. It is useful to note that these particular great circles can easily be drawn with compasses centred on the primitive.
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Stereographic Projection showing the symmetry elements of a cube
The stereographic projection is particularly useful for displaying symmetry elements. Here it is illustrated for the symmetry elements of a cube. This projection can be compared with the depictions of the symmetry in C6H2, pages 7-8 of either a drawing of the cube or of the sphere showing symmetry elements. This projection is much more convenient to use than such drawings.
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The Stereographic Projection applied to non-cubic systems
The stereographic projections on pages 15-17 are examples taken from (i) the hexagonal system, (ii) the orthorhombic system and (iii) the monoclinic system. In each case the construction of the stereogram is logical, but the precise details of where general poles are located on the stereogram is specific to the particular crystal structure.
Standard (0001) projection for zinc showing normals to planes plotted as poles Zinc has the h.c.p. crystal structure with c/a = 1.86. Thus, for example, the angle (0001) : (1011) = c 2c tan–1 3a = 65.0° and the angle (0001) : (0112) = tan–1 3a = 47.0°. (hki0) poles on the primitive can be plotted straightforwardly by noting that they plot in the same positions as the vectors [hki0], as shown in C6H2. The positions of other poles can be confirmed from straightforward calculations.
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Standard (001) projection for an orthorhombic crystal showing normals to planes plotted as poles The x-, y- and z- axes plot as for cubic crystals. The angle θ between (hk0) and (100) is tan–1 ba. kh , enabling the pole hk0 to be plotted. Likewise, we can locate poles such as 0kl by calculating the angle between (0kl) and (001). A pole hkl then lies at the intersection of the great circle [(100), (0kl)] and {(001), (hk0)].
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Standard (001) projection for a monoclinic crystal showing normals to planes plotted as poles Fortunately, this is non-examinable, but again it demonstrates the logic inherent in stereographic projections. Here, the convention is to plot the crystal z-axis at the centre of the stereogram. The 010 pole and the crystal y-axis are coincident on the primitive since the crystal y-axis makes an angle of 90° with the x- and z-axes of the crystal. The 100 pole, i.e., the normal to the (100) planes, can also be plotted on the primitive as shown. The crystal x-axis makes an angle of β – 90° with the normal to the (100) planes and is a southern hemisphere pole, as it makes an angle of β with the z-axis. Likewise, the 001 pole is β – 90° away from the z-axis towards 100 and is a northern hemisphere pole.
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PART IIA and PART IIB MATERIALS SCIENCE AND METALLURGY Course C6: Crystallography
C6H4
Geometry of Single Crystal Slip
Figure 1
Suppose that in a specimen undergoing a tensile test, a slip plane lies with its normal at an angle φ to the tensile axis and a slip direction within the slip plane is inclined at an angle λ to the tensile axis. A three dimensional sketch of the specimen is shown in Figure 1. Figure 2 is a stereographic projection which shows the direction of applied force, F, the slip plane, together with its normal, N, and the slip direction, S. The area of the slip plane is A/cos φ, where A is the cross-sectional area normal to the tensile axis. The force acting in the slip direction is F cos λ, and so the resolved shear stress τ acting on the slip plane in the slip direction is τ = F cos φ cos λ A The factor (cos φ cos λ) is called the Schmid factor.
Figure 2 (stereogram)
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Figure 3 If no constraints are placed on the crystal, the two ends of the crystal would move relative to each other during slip. In a tensile test the two ends of the crystal are held rigidly above and below one another. In those parts of the crystal that are free to move, there is a rotation about the direction normal to both the tensile axis and the slip direction. The change in orientation of the crystal during deformation is illustrated in Figure 3 schematically above. The situation in an actual experiment is illustrated in Figure 4 below.
Note that the crystal has deformed, the slip direction S has moved towards the tensile axis. This continues throughout the deformation and can be illustrated on the stereogram (Figure 2) by the movement of the pole S towards the pole F, following the great circle linking the two poles
Figure 4
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Diehl’s Rule Once the slip systems characteristic of a particular crystal structure have been established, the general approach to establishing which slip system will operate in a tensile test on a single crystal is to compute the Schmid factor cos φ cos λ for each distinct slip system, and for each orientation of the tensile axis as deformation proceeds. As the applied stress is increased, the system with the largest Schmid factor, i.e. that experiencing the greatest resolved shear stress, will be that on which slip first occurs as the critical resolved shear stress (C.R.S.S.) is exceeded. Diehl’s rule is a simple stereographic method for finding the slip system with the highest Schmid factor, without calculation, and can be used for (i)
c.c.p. crystals slipping on {111} < 110>
(ii)
b.c.c. crystals slipping on {110} < 111>
(i)
Use of Diehl’s rule for {111} < 110 > slip (e.g., c.c.p. metals)
1.
Sketch a cubic stereogram displaying standard triangles, i.e., showing all poles of the forms {100}, {110} and {111} for the holosymmetric cubic class and the great circles connecting them to form 48 right-angled spherical triangles, as on the figure overleaf. [Each of the small spherical triangles in the figure is bounded by mirror planes and its contents can, in turn, be reflected into all the other 47 standard triangles covering the surface of the projection sphere, i.e., any standard triangle comprises the smallest angular region necessary for considering any property variation with angle in the holosymmetric cube.]
2.
Locate the standard triangle containing the pole of the tensile axis (T.A.), e.g., in the 001 – 101 – 111 spherical triangle overleaf.
3.
The identity of the slip plane is found by taking the {111}-type pole in this standard triangle and forming its reflection in the side of the triangle opposite to it, e.g. (111) is the reflection of (111) in the great circle 001 – 101 – 100.
4.
Similarly, the identity of the slip direction is found by forming the mirror image of the -type pole of the triangle in the side opposite to it. For example, [011] is the reflection of [101] in the great circle 001 – 111 – 110.
NOTE: (a)
Care is needed when considering triangles bordering the primitive, since one of the poles calculated using this procedure may be in the lower hemisphere of the stereographic projection, e.g. for T.A2 the slip system is (111) [011].
(b)
If the T.A. lies at the boundary (junction plane) of two (or more) standard triangles, then the Schmid factors are equal on the corresponding two (or more) systems.
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With the initial position of the tensile axis 1 (T.A1 ) in this case, ( 111) [011] is the slip system with the highest Schmid factor.
(i)
Use of Diehl’s rule for {110 } slip (e.g., b.c.c. metals)
For b.c.c. metals slipping on {110} the slip plane / slip direction indices are interchanged. Thus, for T.A1 the slip system would be (011)[111], etc. This affects the direction of rotation of the T.A. on the stereogram once slip has begun.
The T.A. always rotates towards the slip direction. Thus, for c.c.p., T.A1 would move towards [011]. For b.c.c., T.A1 would move towards [111].
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Proof of Diehl’s rule and OILS rule
The proof is straightforward. Consider a tensile axis [uvw] where |u| = |v| = |w|. For simplicity, we can choose [uvw] to lie within, or on a side, of the standard stereographic triangle defined by 001, 011 and 111 poles. The most general case is where [uvw] lies within the standard stereographic triangle, so that 0 < u < v < w, i.e., [uvw] is of the form [LIH] referred to the use of OILS rule given in the Part II Data Book. For c.c.p. we have twelve possible slip systems (A - L), and for each we can specify the Schmid factor cos φ cos λ, as in the Table below:
Slip system
Slip plane and slip direction
|cos φ cos λ| ( × 6 u 2 + v 2 + w 2 )
A
(111) [110]
(u + v + w) (v – u)
B
(111) [101]
(u + v + w) (w – u)
C
(111) [011]
(u + v + w) (w – v)
D
(111) [110]
|w – u – v | (v – u)
E
(111) [101]
|w – u – v | (u + w)
F
(111) [011]
|w – u – v | (v + w)
G
(111) [110]
(u + w – v) (u + v)
H
(111) [101]
(u + w – v) (w – u)
I
(111) [011]
(u + w – v) (v + w)
J
(111) [110]
(v + w – u) (u + v)
K
(111) [101]
(v + w – u) (u + w)
L
(111) [011]
(v + w – u) (w – v)
We now have to identify the row in the far right hand column of this table for which cos φ cos λ is a maximum. A comparison of the Schmid factors for A, B and C (which have a common factor (u + v + w)) shows that the one for B is the largest of the three for 0 < u < v < w. Likewise, for 0 < u < v < w, (i) the Schmid factor for F is greater than D or E, (ii) the Schmid factor for I is greater than G or H, and (iii) the Schmid factor for K is greater than J or L. Thus, we have only have to determine which of B, F, I and K has the largest Schmid factor. Comparing I and F, which have a common factor (v + w), it is apparent that I has the greater Schmid factor if u > 0.
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Looking more closely at the Schmid factors for B and K, we have: B: | cos φ cos λ| ( × 6 u 2 + v 2 + w 2 ) = w2 – u2 + v(w – u) K: | cos φ cos λ| ( × 6 u 2 + v 2 + w 2 ) = w2 – u2 + v(w + u) and so the Schmid factor for K is greater than that for B for u > 0. Finally, comparing the Schmid factors for K and I, we have: K: | cos φ cos λ| ( × 6 u 2 + v 2 + w 2 ) = w2 + vu + vw – u2 I: | cos φ cos λ| ( × 6 u 2 + v 2 + w 2 ) = w2 + vu + uw – v2 and since vw + v2 > uw + u2 for 0 < u < v < w, (i.e., [LIH],) it follows that K is the slip system with the greatest Schmid factor, i.e., the slip system (111) [101]. This is of course the slip system found using either Diehl’s rule or OILS rule for a pole within the standard stereographic triangle in c.c.p. crystals. Thus, Diehl’s rule and OILS rule can both be proved using this methodology. A simple extension of this methodology enables special cases, e.g. where u = v > 0, to be considered when two or more slip systems have the same Schmid factor.
C6H5
PART IIA and PART IIB MATERIALS SCIENCE AND METALLURGY Course C6: Crystallography
C6H5
Point Group Symmetry A point group is a consistent set of symmetry elements relating vectors from a single point. The point group of a crystal describes the symmetry of the crystal on a macroscopic (mm scale). It describes the shape of an ideal crystal and relates values of physical properties, such as thermal expansion, in different directions. Since directions from a single point are related by a group of symmetry elements, we can draw a stereogram showing the operation of the symmetry elements of a point group on a single pole. For example, the operation of a tetrad gives:
The possible symmetry elements in a point group are:
and
rotation axes, mirror planes, centre of symmetry inversion axes.
Crystallographic rotation axes of symmetry (1, 2, 3, 4 and 6), mirror planes, m, centres of symmetry and inversion axes 1, 2, 3, 4 and 6 were all described in Part IA Materials and Mineral Sciences. Inversion axes (IA revision)
The operation of an inversion axis is equivalent to a rotation through 2π/n (for a n-fold inversion axis) followed by inversion through the origin (the point through which all the symmetry elements of the point group pass).
1 ≡ centre of symmetry
2 ≡ mirror plane
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Stereograms showing the operation of 3, 4 and 6 axes on a general direction together with perspective drawings of a crystal displaying the corresponding symmetry and showing all the faces of a single form {hkl}. 3
4
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Rotation axes consistent with translational periodicity (IA revision)
In the above diagram, let A, B and C be lattice points, separated by a distance t. Suppose that there is an n-fold rotation axis normal to the paper through A, B and C. The rotation angle θ of this rotation axis is 2π/n by definition. The axis through A takes B to B', which by definition must also be a lattice point. The axis through C takes B to B", which is also by definition a lattice point. Since B'B" is parallel to ABC, it follows that for consistency B'B" must be an integral multiple of the distance AB = BC = t, i.e., 2t – 2t cos θ = mt, where m is an integer Hence it follows that cos θ = 1 –m/2 Now –1 = cos θ = 1, and so possible values of m are:
m
0
1
2
3
4
cos θ
1.0
0.5
0
– 0.5
– 1.0
θ
2π/1
2π/6
2π/4
2π/3
2π/2
n
1
6
4
3
2
name
monad
hexad
tetrad
triad
diad
All other rotation axes with n = 5 or n = 6 are inconsistent with lattice translation, and so cannot occur as symmetry elements describing translational symmetry in crystalline materials. However, they can be found in diffraction patterns from Penrose tilings (see C6H10) and in individual molecules, such as ‘buckyballs’, C60, although not of course in the symmetry elements describing translational symmetry in the crystalline form of buckminsterfullerene.
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The thirty two crystallographic point groups
There are thirty two different possible combinations of symmetry elements in three dimensions compatible with translational symmetry. These are known as the crystallographic point groups. Only certain combinations of symmetry elements are allowed in crystals. Some point groups are symmetry elements by themselves: 1, 2, 3, 4, 6, 1, 2, 3, 4 and 6. The remainder fall into two groups philosophically: 1.
A single rotation axis or inversion axis may be combined with a perpendicular diad, a perpendicular mirror plane or parallel mirror plane.
2.
Four triads may be combined with diads, tetrads or mirror planes.
In general, if two symmetry elements operate on a direction, a third symmetry element is generated. For instance, if one symmetry element relates direction A to direction B, and another relates A to C, then B and C must be equivalent through a suitable symmetry operation. For example, in the orthorhombic system, if there is a diad parallel to [100] relating a general (hkl) to (hkl) and one parallel to [010] relating (hkl) to (hkl) and (hkl) to (hkl), then there must also be one parallel to [001] relating (hkl)) to (hkl) and (hkl ) to (hkl), as on the schematics below, defining the point group 222.
Diad paral to [100]l
Dial parallet [010]
The point group 222 Representations of point groups
Stereograms showing either the symmetry elements and/or the operation of the symmetry elements on a single direction are one of the simplest ways of representing point groups. These are shown overleaf. The nomenclature is consistent and conforms to the rules: (i)
m is used in preference to 2 ;
(ii)
a mirror plane normal to a symmetry axis is indicated by X/m where X is 2, 3, 4 or 6;
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Up to three symbols or combination of symbols can be used to describe a point group, e.g. 3m, 23, 432 and 6/mmm are all point groups. The rules governing the symbols for the various crystal systems are as follows:
Position in symbol
First
Second
Third
Cubic
Tetragonal, hexagonal, trigonal
Orthorhombic
[100]
[010]
[001]
Monoclinic
[010]
–
Thus, 32 is a trigonal point group, whereas 23 is cubic to conform to these rules. Crystal class
All crystals that display the same macroscopic symmetry are said to belong to a crystal class. The class is given the same symbol as the corresponding point group. Forms
The form {hkl} is the set of faces related to (hkl) by the symmetry elements of the point group. A face belongs to a general form if its orientation is not related in any special way to any of the symmetry elements of the point group, e.g., if hkl = (123). A face belongs to a special form if its orientation is related in a special way to any of the symmetry elements of the point group so that the number of faces in the form will be less than in the general form. Thus, in point group 222 the form {123} is a general form and the form {100} is a special form.
Symmetry elements in 222
{123}
{100}
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The triclinic, monoclinic, orthorhombic and tetragonal crystallographic point groups
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The trigonal, hexagonal and cubic crystallographic point groups
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PART IIA and PART IIB MATERIALS SCIENCE AND METALLURGY Course C6: Crystallography
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Space Group Symmetry In crystal symmetry on the unit cell scale, that is on the Å or nm scale, the symmetry elements of the crystal’s point group may be combined with the translational symmetry of the lattice. Rotation axes may become screw axes and mirror planes may become glide planes. With a screw axis, given the symbol np, each rotation through 2π/n is coupled with a translation of p/n of the lattice spacing along the rotation axis in a right-hand screw sense. After n operations an atom is back in its original position in a unit cell p lattice points along the axis. For example, if a crystal has a rotation tetrad axis in its point group, its crystal structure may show a rotation tetrad, a 41, a 42, or a 43 axis, as shown in the diagrams below.
In cuprite, Cu2O, which is cubic, point group m3m (≡ m3m), there are 42 axes at 1 , 0, z and 0, 1 , z . 2 1 1 1 1 1 3 2 The screw axis at A relates the Cu atom at 4 , 4 , 4 to those at (i) 4 , – 4 , 4 ,
(ii) 34 , – 14 ,
1 4
and (iii) 34 , 14 , 34 . y 1 4
3 4
1 4
1 2 3 4
3 4 1 2
1 4
3 4
1 4
Oxygen Copper 1 4
3 4
1 4
1 2 3 4
x
3 4 1 2
1 4
3 4
1 4
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The different types of screw axes are drawn below.
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With a glide plane the reflection operation is coupled with a translation vector equal to one half a lattice translation parallel to the glide plane. After two operations on an atom the atom is in its original position relative to a lattice point.
r 2
glide plane
A
A A
The possible three-dimensional glide-reflection elements are summarised below. Note that in addition to axial glide, diagonal glide (denoted by n) and diamond glide (denoted by d) can occur.
Translation component Glide plane element
Direction
Axial glide
|| to a axis
1 /2
a
a
Axial glide
|| to b axis
1 /2
b
b
Axial glide
|| to c axis
1 /2
c
c
a + 1/2 b, 1/2 b + 1/2 c,
n
Diagonal glide
|| to face diagonal
Magnitude
1 /2
1 /2
Diamond glide
|| to face diagonal for a face centred cell
Diamond glide
|| to face diagonal for a body centred cell
1 /4
Symbol
c + 1 /2 a
a + 1/4 b, 1/4 b + 1/4 c, 1 /4 c + 1 /4 a
d
1 /4
d
a + 1 /4 b + 1 /4 c
In cuprite (page 1), there are mirror planes parallel to {110} and diagonal glide planes parallel to {100}. The diagonal glide planes parallel to (100) intersect the x-axis at x = 1/4 and 3/4. Reflection across the glide plane is followed by the translation 1/2 b + 1/2 c. Likewise, the diagonal glide planes parallel to (010) intersect the y-axis at y = 1/4 and 3/4. Reflection across the glide plane is followed by the translation 1/2 a + 1/2 c. Thus, the copper atom at 14 , 14 , to the oxygen atom at 12 , 12 ,
1 2
1 4
is related to that at 34 , 14 , 34 . and the oxygen atom at 0, 0, 0 is related
by the (010) glide plane at y = 1/4.
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The full space group designation of cuprite is P 42/n 3 2/m. Here, P denotes the Bravais lattice of cuprite, and the following three symbol combinations denote symmetries parallel to , and respectively. This is space group no. 224, more commonly referred to as Pn3m (≡ Pn3m), a more full description of which is shown below.
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Overall there are 230 space groups in 3D (17 in 2D). Space groups are tabulated in the International Tables for X-ray Crystallography. Two further examples are given here from the tables, one for a space group describing the symmetry of the orthorhombic crystal, ZrTiO4, the structure of which is discussed in Worked Example #1 on C6H11, and one for the space group describing the symmetry of sphalerite.
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Relationship between the point group and the space group of a structure
The point group is determined by the space group. Point groups deal with the symmetry of directions from a point (e.g., face normals) and these are not affected by the lattice repeats or by the translational component of glide planes or screw axes. A screw axis in the space group acts as a rotation axis in a point group and a glide plane in the space group acts as a mirror plane in the point group. Thus, [Space group] – [translation] = [point group] So, for example, (i)
Cuprite has the space group P 42/n 3 2/m (Pn3m) and the point group m3m.
(ii)
ZrTiO4 has the space group P 21/b 2/c 21/n (Pbcn) and the point group mmm.
(iii)
Sphalerite, ZnS, has the space group F43m and the point group 43m.
(iv)
The h.c.p. crystal structure has the space group P63/mmc and the point group 6/mmm.
C6H7
PART IIA and PART IIB MATERIALS SCIENCE AND METALLURGY Course C6: Crystallography
C6H7
Texture If the grains in a polycrystal have a non-random distribution in the orientation of their crystallographic axes, then we have preferred orientation, or texture. Measurement of texture This is most commonly based on X-ray diffraction, the most usual method being the Schulz reflection method:
The diffractometer has source and detector angles set to receive a particular Bragg reflection (i.e., the angle 2θ is defined). The sample is then tilted to permit measurement of the reflection intensity at various sample orientations. The sample is tilted by an angle φ and rotated by an angle ψ about the rotation axis. The angular range of φ is limited (φ cannot get too close to 90°), but ψ is usually varied in the full range 0° – 360°.
An alternative method uses the scanning electron microscope. Rather than being scanned, the electron beam can be rocked about a fixed point on the surface (within one grain) to obtain electron channelling patterns. These arise from Bragg reflection of electrons in the sample and show the crystal symmetry directly.
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Channelling pattern for copper. This map is over a wide orientation range and is made by combining several photographs. It is plotted in a standard stereographic triangle, and can be used to identify orientations in smaller photographs taken over a limited angular range.
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A typical individual channelling pattern — this time from silicon. A 〈111〉 direction is close to the centre.
Representation of Texture Pole figures show the orientation distribution of a crystallographic axis on a stereogram plotted with respect to sample axes.
In this example, the distribution of 〈220〉 poles is shown for a thin film of silver deposited on a rock salt (NaCl) substrate. The relationship between the crystallographic axes of the film and substrate shows that the film is epitaxial, with [100]Ag // [100]NaCl, [010]Ag // [010]NaCl, etc.
Inverse pole figures show the orientation distribution of a sample axis on a stereogram plotted with respect to crystallographic orientations.
In this example, the sample is a rolled sheet with its significant axes being the rolling direction (RD), the transverse direction (TD) and the normal direction (ND). The distributions of each of these directions in a cold-rolled steel is shown on a standard stereographic triangle (referred to the crystallographic axes).
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Crystal orientation distribution functions
For samples which are nearly single crystals (like the thin-film example) pole figures can be unambiguous, but for orientation distributions, the information on the true orientation distribution (requiring the specification of three independent angular variables) can only be obtained by combining information from more than one pole figure (each individual figure specifying only two angular variables). Computer processing then gives 3D orientation distribution plots, usually shown as 2D sections. The Euler angles ψ, θ and φ are used to specify fully the orientation of crystallographic axes with respect to sample axes for a rolled sheet, as shown on the diagram here.
Various crystal orientations, (hkl)[UVW], plotted in terms of the Euler angles ψ, θ and φ.
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Example of a crystal orientation distribution function (CODF) determined for cold-rolled steel.
To judge from the intensity contours in the above diagram, there is a strong texture at ψ ~ 40°, θ ~ 65° and φ = 26.6°. Reference to the texture diagrams on page 3 shows that this indicates a (211)[011] texture. Here, since ND = (211), φ = [210] : [100] = 26.6°; φ = [001] : [211] = 65.9°; ψ = [215] : [011] = 39.2° ([215] is the position on the great circle whose pole is ND for which ψ = 0°: it must be 90° away from [211] and lie on the same great circle as [001], [211] and [210].
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Origins of Texture Solidification
Inverse pole figures showing the orientations of columnar grains in lead (of two compositions) after directional solidification. Zone-refined lead shows 〈111〉 directions near the heat flow (growth) direction; the alloy sample shows 〈100〉 directions near the heat flow direction.
Mechanical deformation
Inverse pole figure from an extruded aluminium rod. There are two main components in this texture: 〈001〉 // rod axis and 〈111〉 // rod axis.
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Rolling textures:
(a)
(b)
(a) 〈111〉 and (b) 〈200〉 pole figures for cold-rolled copper (c.c.p.). These pole figures indicate that the rolling texture in copper is composed mainly of the (123)[412] and (146)[211] type orientations, rather than the so-called silver or brass type rolling texture (110)[112].
(In the diagram above and the one below CD = cross direction ≡ transverse direction TD.)
(a) 〈100〉 and (b) 〈110〉 pole figures for cold-rolled molybdenum (b.c.c.). Here, there is a strong (001)[110] texture.
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Annealing
The above is a 〈200〉 pole figure for cold-rolled copper (with texture as in earlier example), then annealed for 5 min at 200 °C. Here there is a strong (001)[100] texture with further local intensity maxima caused by twinning on {111} planes – annealing twins. The filled squares indicate (122)[212] texture consistent with such twinning.
Annealing textures in b.c.c. crystals are more complicated, e.g. (111)[ 211], (001), 15° from [110], (112), 15° from [110] being examples of textures quoted in the literature. In the diagrams above, open circles are (100)[011] and crosses are (111)[112].
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Effects of Texture
Mechanical properties
The effect of anisotropy is most easily seen in the phenomenon of ‘earing’ in deep drawing. This is the formation of a wavy edge on top of a drawn cup which then requires extensive trimming to produce a uniform top. Depending on the degree of preferred orientation of the sheet, two, four or six ears will usually be formed. Deepest cups that can be drawn from various metals.
Magnetic properties
e.g., Grain-oriented silicon steel (GOSS) in which the texture is manipulated to minimise core loss in transformers. The diagram below shows the variation of magnetic power loss with angle around the sheet for GOSS.
C6H8
PART IIA and PART IIB MATERIALS SCIENCE AND METALLURGY Course C6: Crystallography
C6H8
Crystallography of Interfaces
Example of a coincidence site lattice: This shows two c.c.p. crystals rotated by 38° about a common [111] direction normal to the paper with a common grain boundary. Atoms in white represent coincidence sites. This is a Σ = 7 coincidence site lattice. In this simple picture of the boundary the atoms are taken to be hard spheres and assumed to remain in coincidence site relation.
Definition of Σ Let two lattices interpenetrate to fill all space, and assume that there is a common lattice point. In general, there may be then other lattice points which coincide, and the set of such points forms the coincidence site lattice. Σ is the reciprocal of the density of coincidence lattice sites relative to ordinary lattice sites. In the example above, the coincident sites in the planes of the paper are white. One in seven of the lattice sites of a particular grain are white (remember that for c.c.p. crystals each atom position can be taken to coincide with a lattice point), and so Σ = 7. This will also be true for each (111) layer superimposed on this layer, and so Σ = 7 in three dimensions as well. Further theorems show that Σ must be an odd number in cubic materials. As a generalisation, low Σ (e.g, Σ = 21 and Σ = 3n for integer n (i.e., multiple twinning)) are important crystallographically in c.c.p. metals.
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Structure of high angle grain boundaries in c.c.p. metals Computer simulations of high angle grain boundaries in metals tend to be restricted to symmetrical tilt grain boundaries with low index rotation axes, e.g. or . However, such simulations have been very useful in developing our ideas of how grains fit together.
One feature that these calculations show is the tendency for certain polyhedral shapes to occur at grain boundaries, such as those shown below. Five of these were found by Bernal in hard-sphere models of liquid structures.
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Relaxed structures of (a) (115) and (b) (113) symmetric tilt [110] grain boundaries in Al seen in projection along the [110] tilt axis. Triangles and crosses distinguish atoms in successive (220) planes. The boundaries run from left to right in the middle of each picture. Two ‘A’ structural units are identified in (a), each comprising an irregular pentagon, labelled ‘p’ and a tetrahedron, labelled ‘t’. Two ‘B’ structural units, each comprising a capped trigonal prism are marked in (b).
In (a) the rotation about [110] is θ = 31.59° and Σ = 27. In (b) θ = 50.48° and Σ = 11. Symmetrical tilt grain boundaries with values of θ given by 31.59° < θ < 50.48° will have structures built up of ‘A’ and ‘B’ units, e.g. when θ = 38.94° and Σ = 9 and the boundary plane is (114), the repeat unit is AB, i.e., one ‘A’ unit and one ‘B’ unit.
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In more general boundaries, such as the (16, 16, 61) boundary, in which the repeat sequence of units comprise 13 A and 19 B units, the strip method of quasicrystallography can be used, as shown here. Further details of this strip method can be found in Worked Example #7 on C6H11.
Martensitic Transformations
The Bain correspondence, useful for the description of the martensitic transformation in steels, is one in which a b.c.t. unit cell is identified in a c.c.p. crystal structure. Suitably strained, this b.c.t. unit cell can be reshaped to form the b.c.t unit cell of martensite.
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In the crystallography of martensitic transformations, the overall strain is an invariant plane strain: one plane, the habit plane, is neither rotated nor distorted as a consequence of the transformation form the high temperature parent phase to the low temperature martensite phase, as in steels and shape memory alloys, such as those based on Cu-Zn-Al and Ni-Ti. Thus, in a suitable reference frame, the strain matrix formally describing the martensitic transformation is of the form λ1 0 S = 0 λ 2 0 0
0 0 1
for λ1 > 1 and λ2 < 1. Note that the definition of principal strains ε1, ε2, and ε3 used in the Tensor Properties course are then (1 – λ1), (1 – λ2) and 0.
For almost all martensitic transformations, the requirement of an overall invariant plane strain transformation (which is driven by considerations of minimum strain energy) means that in addition to the lattice variant shear which reshapes the crystal structure (e.g. forming the b.c.t martensite in steels from the c.c.p. austenite), lattice invariant shears (slip, faulting or twinning) are required in the martensite to enable it to fit as best it can in the volume originally occupied by the high temperature parent phase.
Thus, in the schematic above, the square can be considered to be a volume of parent phase. In (a) a lattice variant shear is applied to the whole of the square, creating martensite and transforming the square into a parallelogram by a simple shear, S. In (b) the martensite is faulted parallel to the base of the square, enabling it to reduce the strain on the surrounding material, and in (c) twinning has occurred, also enabling S to be reduced.
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Deformation twinning crystallography
Twinning, a common lattice invariant shear mode in martensites, can be described using deformation twinning crystallography. The geometry of simple shear is relevant to this. In the diagram below the twin plane is denoted K1. All points on the upper side of are displaced in the shear direction η1 by an amount u1 proportional to their distance above K1, so that u1 = g x2 where g is the strength of the simple shear. The plane containing η1 and the normal to K1 is termed the plane of shear, S. The vector η2 in S is rotated but unchanged in length after the shear process has been applied for a shear of amount g = 2 tan α. K2 , the plane normal to S containing η2, is termed the second undistorted plane.
In compound twins K1, η1, K2 and η2 all have rational indices. For example in a highly symmetric structure such as the c.c.p. crystal structure, an example of {111} deformation twinning is: K1 = (111), η1 = [112], K2 = (111) and η2 = [112] with g = 0.707.
In reflection twins (Type I twins), K1 and η2 have rational indices, but η1 and K2 have irrational indices. In rotation twins (Type II twins), η1 and K2 have rational indices, but K1 and η2 have irrational indices. Example: lattice invariant shears in martensite
Equiatomic nickel-titanium martensite, which has a monoclinic crystal structure, can exhibit both Type I twins and Type II twins. Thus, for example, one Type I twinning mode in Ni-Ti martensite has K1 = (111), while the Type II twinning mode, which is actually the most commonly observed twinning mode, has η1 = [011] and K1 = (0.72054, 1, 1).
C6H9
PART IIA and PART IIB MATERIALS SCIENCE AND METALLURGY Course C6: Crystallography
C6H9
Crystallography of Diffraction You have already met this in Part IA Materials and Mineral Sciences. For X-rays, electrons and neutrons, the structure factor for hkl reflections, Fhkl, is given by Fhkl =
∑ fn exp {2πi(hun + kvn + lw n)} n atoms
where fn is the scattering factor of atom n. Hence, it follows from this definition that for a facecentred lattice, systematic absences occur when h, k and l are mixed odd and even, and for a bodycentred lattice systematic absences arise when (h + k + l) is odd. In addition, the motif can also make reflections have zero structure factors. For example, in h.c.p. metals, F001 = 0 because the atoms at height z = 1/2 interfere destructively with those at z = 0. Silicon, used as a standard for X-ray diffraction, has zero structure factors when h + k + l = 4n + 2, so that the N (= h2 + k2 + l2) values of the first few lines to be seen are 3, 8, 11, 16, 19 and 24. RbBr and KCl, both of which have the NaCl structure, have absent reflections when h, k and l are all odd for all practical purposes because the scattering factors of the anions and cations in these two examples are almost identical. Note that the c.c.p. crystal structure has a reciprocal space b.c.c. lattice, and visa-versa. Ewald sphere By definition the radius of the Ewald sphere is 1/λ. An initial undiffracted X-ray can be represented by a point on the sphere. Diffraction occurs when the X-ray is scattered through an angle of 2θ relative to the forward direction of the X-ray by the hkl set of crystal planes. Since this is an elastic event, the diffracted X-ray and initial X-ray have the same magnitude of their wave number (=1/λ), and so the diffracted X-ray can also be represented by a point on the surface of the sphere.
The angle between these two points subtended at the centre of the circle is 2θ, and the chord between these two points is the magnitude of the reciprocal lattice vector ghkl. This is 1/dhkl, as in the diagram below:
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1/ 2dhkl , i.e., we derive the Bragg equation λ = 2dhkl sin θ . 1/ λ Electron diffraction patterns typically have θ ˜ 1°, since for 100 kV electrons λ = 0.037 Å and interplanar spacings are typically in the range 1-2 Å. Hence, it follows that sin θ =
In addition, for electrons, spots close to the Ewald sphere have finite intensity because of the thin ~ 1000 Å crystals through which the electrons beam is transmitted to form the diffraction pattern. Conventional ‘spot’ electron diffraction patterns are formed by a parallel beam of electrons as the incident beam. These can have intense double diffraction: intensity can be scattered from one beam to another because of the small angles required for Bragg diffraction. This is a consequence of the dynamical nature of electron diffraction, more of which in C20, Part III. A consequence of double diffraction is that reflections in electron diffraction patterns which have zero structure factors because of the motif, can have finite, strong intensities when electron beams are suitably oriented. Thus, for example, in electron diffraction it is common for 002 spots to be strong in the silicon [110 ] zone through double diffraction from the 111 and 111 spots., since vectorially 111 + 111 = 002, and the 111 and 111 reflections have non-zero structure factors. The diagram below demonstrates this.
111
002
000
111
Conversely, in the silicon [100] zone, 002 reflections will be systematically absent, as there is no double diffraction route capable of generating intensity at the 002 position in reciprocal space.
Convergent beam electron diffraction (CBED) patterns
These are what the terminology implies: in these patterns the electron beam is defined by a convergence angle, α, and so the electron beam is no longer a parallel beam. This gives rise to discs rather than spots in the Zero Order Laue Zone (ZOLZ), i.e. the zone for which hu + kv + lw = 0 for the hkl spots in this zone when the electron beam direction is [uvw], as in conventional spot patterns.
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Within the discs, as in the example below from silicon, information from higher angles is present as Higher Order Laue Zone (HOLZ) lines. These lines contain information about the lattice parameter, the accurate electron beam direction and any strain present in the crystal. The lines arise from inelastically scattered Bragg electrons which are subsequently Bragg diffracted by HOLZ planes. CBED patterns are particularly useful in symmetry determination, strain determination and composition analysis (see for example, Williams and Carter, Transmission Electron Microscopy). Symmetry in electron diffraction patterns
Convergent beam electron diffraction (CBED) can be used to demonstrate that axes in silicon have three fold symmetry, rather than the six fold symmetry which could be inferred from an examination of conventional selected area ‘spot’ electron diffraction patterns (i.e. ones obtained without using convergent beam). The Higher Order Laue Zone (HOLZ) lines exhibit the three fold symmetry clearly: note the equilateral triangle in the centre of the 000 disc formed by some of them.
Expanded Zero Order Laue Zone (ZOLZ) region of a CBED pattern of silicon taken with the electron beam parallel to [111].
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Point group symmetries of CBED patterns can be determined as in the four examples below. More advanced techniques enable space groups of crystal structures to be determined – see for example discussion in D.B. Williams and C.B. Carter, Transmission Electron Microscopy (Plenum 1996).
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For relatively small ( < 500 mm) camera lengths convergent beam patterns show higher order Laue zone (HOLZ) rings, as in the example below.
Experimental CBED pattern showing the first order Laue zone clearly.
The terminology to label the centre of the pattern and the rings is as follows: ZOLZ – Zero order Laue zone: hu + kv + lw = 0 for the hkl spots in this zone when the electron beam direction is [uvw]. FOLZ – First order Laue zone: hu + kv + lw = 1 for the hkl spots in this ring. SOLZ – Second order Laue zone: hu + kv + lw = 2 for the hkl spots in this ring. A straightforward calculation in reciprocal space shows that the radius of the nth HOLZ ring, rn*, is given by the formula rn * = k 2 – k – nt* 2 1/2 - 2knt* 1/2 where k = 1/λ and t* is the reciprocal of the magnitude of the lattice vector [uvw]. In making this calculation, use is made of the result that for the nth ring, hu + kv + lw = n = |p*| |uvw|, where p* is the projection of the vector ha* + kb* + lc* lying in the HOLZ onto the direction [uvw]. Hence, |p*| = n / |uvw| ≡ nt*.
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PART IIA and PART IIB MATERIALS SCIENCE AND METALLURGY Course C6: Crystallography
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Non-Crystalline Materials Materials which do not possess translational periodicity in three dimensions may nevertheless not be amorphous like inorganic glasses. Two interesting classes that we shall consider here are liquid crystals and quasicrystalline materials. There are other very interesting cases such as incommensurate crystals, but these are outside the scope of this course. Liquid Crystals These are classified as either ‘nematic’, ‘cholesteric’ or ‘smectic’.
A typical liquid crystalline molecule, 4-methoxybenzylidene–4'–butylaniline (MBBA). The transition temperature from the crystalline phase to the nematic phase, TK-N, of this compound is 20 °C, while it transforms from nematic to isotropic at 74 °C, TN-I.
Schematic diagrams showing the arrangement of rod-like molecules in crystalline, liquid crystalline and isotropic phases. In the liquid crystalline phase, the vector about which the molecules are preferentially oriented, n, is known as the ‘director’. Twisted nematic devices are used extensively in liquid crystal devices (LCDs) for watches, pocket calculators, car dashboards, mobile phones, etc.
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Two representations of a cholesteric (chiral nematic) texture. On a local scale the molecular arrangement is similar to nematic, but here the molecules twist gradually about an axis normal to their length. The twist is the result of the presence of a chiral centre in the molecule.
A molecule which forms a cholesteric phase. A chiral centre is marked as a *. The molecule can have left and right handed versions as a result of the chiral centre.
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Comparison of nematic and smectic textures. In addition to long range orientational order, the smectic phase has one dimensional long range positional (crystalline) order. The molecules are segregated into layers, which in this example are normal to the director. The layer spacing is of the same order as the molecular length. As the summary on page 5 indicates, there are many different types of smectic phase.
A molecule, 4-cyanobenzylidene-4'-n-octyloxyanaline (CBOOA), which forms a smectic phase below 83 °C and a nematic phase between 83 °C and 109 °C. The layers are stabilised by the longer aliphatic tail and the tendency for aliphatic and aromatic parts to segregate.
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Nematic Liquid Crystals When imaged between crossed polars, thin films about 10 µm thick of nematic liquid crystals exhibit schlieren textures or structures à noyaux, as in the example below. The black brushes originating from the points are due to line singularities perpendicular to the thin layer, and are regions where the director is either parallel or perpendicular to the plane of polarisation of the incident radiation. Note that some points have four brushes and others have two.
The defects where the brushes meet are called disclinations. The strength, s, of a disclination is one quarter of the number of brushes. Possible molecular orientations in the neighbourhood of a disclination are shown below.
Examples of disclinations in nematic liquid crystals
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Smectic Liquid Crystals There is a wide variety of structures taken by smectic liquid crystals as the structural classification in the table below demonstrates.
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Schematic representations of the molecular arrangement in (a) smectic A and (b) smectic C.
(c)
(d)
(a) - (d): Various disclinations in smectic A.
(a) - (d): Various types of edge dislocations composed of disclination pairs in smectic A.
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(a) and (b): Examples of ‘pincements’ composed of disclination pairs in smectic A.
Flexibility of smectic A layers: only such deformations which preserve the interlayer spacing take place readily.
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Quasicrystalline Materials These exhibit symmetries in reciprocal space which are inconsistent with translational periodicity, as in the example below taken from an Al-Mn melt-spun alloy rapidly cooled at 106 K–1.
The point group symmetry in reciprocal space of this alloy is m35, as in the stereographic projection below.
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Patterns can be made which have the same symmetry in reciprocal space and non-periodic repeats of reciprocal lattice vectors. One example is given below. The diffraction pattern of this has 10mm point group symmetry.
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PART IIA and PART IIB MATERIALS SCIENCE AND METALLURGY Course C6: Crystallography
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Worked Examples in Crystallography The aim of these worked examples is to help you with the course in general as well as with the question sheets and examples classes. Question 1 At high temperatures, the low-loss dielectric zirconium titanate, ZrTiO4, has a random arrangement of zirconium and titanium ions on cation sites. Neutron powder diffraction data show that the structure is orthorhombic, with a = 4.804 Å, b = 5.483 Å and c = 5.031 Å. Ion positions are as follows: O:
x, y, z; – x, – y, – z;
1/
– x, 1/2 – y, 1/2 + z; 1/ + x, 1/ + y, 1/ – z; 2 2 2 2
1/
2+ 1/ – 2
x, 1/2 – y, – z; x, 1/2 + y, z;
– x, y, 1/2 – z; x, – y, 1/2 + z.
with x = 0.27, y = 0.10 and z = 0.07. Zr, Ti :
0, v, 1/4;
0, – v, 3/4;
1/
1 2 , /2
+ v, 1/4;
1/
1 2 , /2
– v, 3/4.
with v = 0.265. (a)
Draw an accurate plan of 2 × 2 unit cells of this structure projected on (100). Specify the Bravais lattice of this crystal structure.
(b)
Determine the nearest neighbour oxygen anions of the oxygen anion at x, y, z. Hence or otherwise show that the oxygen anions form a distorted hexagonal close packed array on (100). Relative to this array of oxygen anions, where are the cations located?
(c)
On cooling ZrTiO4 below 1200 °C, a phase transition occurs which is associated with partial ordering of the cations, but otherwise the crystal structure remains essentially unchanged. Stating your reasons, specify whether you think this phase transition is likely to be (i) reconstructive or (ii) displacive.
Solution (a)
A plan of the structure projected on (100) is shown overleaf. The Bravais lattice is P.
(b)
There are twelve nearest neighbour oxygen anions of the oxygen anion at 0.27, 0.10, 0.07. Six of these anions are at heights x ~ 0.25, three at a height ~ 0.75 and three at a height ~ – 0.25. A ‘unit cell’ of the distorted hexagonal close packed array on (100) is shown shaded in the diagram overleaf of the crystal structure. Cations are located in half the octahedral interstices of this close packed hexagonal array of oxygen ions, as in the example shown in the ‘unit cell’.
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Plan of the structure of ZrTiO4 projected on (100). [001] 0.27
0.73
y
0.23
0.23
0.77
0.27
0.73
Hex
0.77
x
[010]
Hex
Oxygen ion at x ~ 0.75 Oxygen ion at x ~ 0.25 Metal ion at x = 0 Metal ion at x = 0.5
(c)
The phase transition is reconstructive. In order to induce partial ordering of the cations, cations will have to change places, i.e. break bonds and form new ones. This contrasts with what happens in a displacive phase transition where bonds remain unbroken, but changes in local environments occur, e.g. in the normal → ferroelectric phase transition in BaTiO3.
Question 2 ZnO has the wurtzite structure, with u not exactly 1/8 and dependent on temperature. Experimental work on ZnO shows that at 300 K, the lattice parameters are a = 3.24992 Å and c = 5.20658 Å with u = 0.1181.
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(a)
Draw an accurate plan of 2 × 2 unit cells of ZnO projected on (001), labelling the positions of the Zn atoms and the O atoms. Mark on the (210) and (120) sets of planes on this diagram.
(b)
Write down the indices of all the planes related by symmetry to the (100) planes in ZnO.
(c)
What the distance in Å between the oxygen ion at 1, 0, 0 and the zinc ion at 1 3 , /3 , u ?
2/
(d)
What is the interplanar spacing in Å of (i) the (110) planes and (ii) the (110) planes?
(e)
What is the volume in Å3 of the unit cell of ZnO?
(f)
Does ZnO exhibit piezoelectricity?
Solution
(a) y 1 2
1 2
+u
+u
1 2
+u (210) planes
x 1 2
1 2
u 1 2
u 1 2
+u
1 2
+u
Zinc
u 1 2
+u
+u
Oxygen
1 2
u 1 2
1 2
+u
–
(120) planes
1 2
+u
(b)
(100), (100), (010), (0 10), (110), (110).
(c)
1.97453 Å (note that the dimensions of the unit cell have been given to five decimal places, so the Zn-O distance can also be specified to five decimal places).
(d)
(i) 2.81451 Å; (ii) 1.62496 Å.
(e)
47.62429 Å3.
(f)
Yes – the structure is not centrosymmetric, so it will exhibit piezoelectricity. (Piezoelectricity can arise in all non-centrosymmetric crystals unless they have a point group of 432 - see J.F. Nye, Physical Properties of Crystals, O.U.P., 1985, pages 123-124. Zinc oxide has a noncentrosymmetric hexagonal crystal structure.)
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Question 3
At room temperature, the lattice parameter of silicon is 0.54306 nm and the lattice parameter of NiSi2 is 0.5406 nm. NiSi2 has the CaF2 structure. When a layer of NiSi2 is deposited on the (111) plane of single crystal silicon, it can take one of two orientations with respect to the silicon. In orientation A all planes and directions are coincident in the two crystals. In orientation B the two crystals have the (111) plane and [111] direction in common, but the [110] direction in the silicon crystal corresponds to a [114] direction in the NiSi2 layer. Draw an accurate stereogram centred on (001) including the [111], [110] and [114] directions of silicon. By constructing a suitable small circle show that if the stereogram is rotated by 180° about [111] the new position of the [110] pole coincides with the original position of the [114] pole. Assuming that the NiSi2 layer nucleates by the formation of a close-packed monolayer of Ni atoms on the Si (111) plane, suggest reasons for the occurrence of the two growth orientations A and B. Solution
A suitable stereogram is shown below. [110] : [111] = cos-1 {2/ 6 } = [111] : [114] = 35.26°. Furthermore, [110], [111] and [114] are coplanar, since by inspection they all lie in the (110) plane.
001 010 [114] [111]
[110] 100 Accurate stereogram centred on (001) showing selected poles and a small circle of angular radius 35.26˚ drawn about [111]
From the calculations and the construction of a small circle of angular radius 35.26° about [111], it is evident that a rotation of 180° about the vector [111] brings the vector [110] into the position where [114] used to be, and visa-versa.
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The stacking sequence of the (111) planes in silicon can be described in the form ...ABCABCABC..., since silicon has the cubic F Bravais lattice. However, A, B and C in this case do not represent close packed planes, but rather pairs of nonclose packed planes, because of the motif of Si – an atom at 0,0,0 and an atom at 1/4, 1/4, 1/4 (see Data Book). The stacking sequence of (111) planes of NiSi2 can likewise be described in the form ..ABCABCABC.., since it too has a cubic F Bravais lattice, but again because of the motif, A, B and C do not represent close packed planes, but rather three sets of planes (2 planes each with Si atoms in between planes containing only Ni atoms). Suppose the Si (111) planes terminate in a C layer. Then we can place either an A or a B layer of the NiSi2 stacking sequence on top of this Si C layer (if we assume that a NiSi2 C would be unfavourable as it would be directly on top of C). The four possible sequences are therefore: (i)
..ABCABCABCABCABCABCABCABCABC..
(ii)
..ABCABCABCABCBCABCABCABCABCA..
(iii)
..ABCABCABCABCACBACBACBACBACB..
(iv)
..ABCABCABCABCBACBACBACBACBAC..
Of these four sequences, (i) and (ii) reproduce the same sequence in the NiSi2 as in the Si, whereas the NiSi2 sequences in (iii) and (iv) are in effect mirror images of the ..ABC.. stacking sequence in the Si. Thus, two distinct growth orientations arise depending on the exact stacking sequence chosen by the NiSi2 layer. These are twinned with respect to one another by a twinning operation of 180° about the (111) plane normal. Directions in one twin orientation are thus reflected in the (111) plane to their positions in the other orientation, a mirror operation which is equivalent to a 180° rotation about the [111] direction followed by an inversion (2 = m).
Question 4
GaAs has the sphalerite structure given in the Data Book. Draw sketch stereograms centred on 001 of the environments of (a) Ga atoms around As atoms, and (b) As atoms around Ga atoms in sphalerite. Specify the nature of the tetrad axis parallel to [001] in GaAs and (ii) the orientation of the mirror planes parallel to [001]. What is the point group of GaAs? If it were possible to grow single crystals of GaAs what would be the shapes of the {100}, {110} and {111} forms?
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Solution
The environments of (a) Ga atoms around As atoms, and (b) As atoms around Ga atoms in sphalerite are both tetrahedral, with the tetrahedra for (a) and (b) inverted with respect to one another. Suitable sketch stereograms centred on (001), with the x–axis down the page and the y–axis to the right and with respect to sphalerite are drawn below:
As atoms around Ga
Ga atoms around As
The environments of As atoms around Ga atoms are identical for all gallium atoms (since all the lattice points are able to coincide with the gallium atoms). For the same reason, the environments of Ga atoms around As atoms are necessarily identical for all arsenic atoms. The tetrad axis parallel to [001] is an inverse tetrad. (110) and (110) are the mirror planes parallel to [001]. The point group of GaAs is 43m. Shapes of the various forms: {100}: cube; {110}: rhombic dodecahedron (a regular figure with twelve rhombus faces); {111}: since there are 2 symmetrically distinct sets of planes of the form {111}, the shape is a tetrahedron. There are two possible tetrahedra for the {111} form inverted with respect to one another.
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Question 5
Rutile is tetragonal with a = 0.459 nm and c = 0.296 nm. Further details about the crystal structure of rutile are given in the Data Book. Draw a plan of this structure on (001). Hence or otherwise confirm the Bravais lattice. Determine (i) the nature of the screw tetrad along [001] in the structure and (ii) the nature of the screw axes along and (iii) the glide planes parallel to {100}. Confirm that {110} planes are mirror planes and that there are diad axes parallel to directions. What would be the most complete description for the space group of rutile? Solution
[010] 1 2
1 2
1 4 1 2
1 2
1 2
1 4
1 2
1 2
1 2
1 2
1 2 1 2
mirror planes
1 2
[100] [110] 1 4
1 4
KEY Ti
4+
O
2-
3A
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Note that it not necessary to know the details of the lattice parameters to answer this question – the fact that the structure is tetragonal is sufficient. A suitable plan to scale of 2 × 2 unit cells is shown on the previous page, with some (but not all!) symmetry elements present. 1/2,
(i)
There are 42 screw axes along [001] intersecting the x-y plane at positions.
(ii)
There are 21 screw axes along , such as the one along [100] intersecting the y-z plane at 0, 1/4, 1/4.
0, 0 and equivalent
(iii) The glide planes parallel to {100} are diagonal glide planes (denoted by n) intersecting the [100] direction at x = 1/4 and 3/4. The most complete description for the space group of rutile would be P 42 /m 21 /n 2/m. The conventional description is P42 /mnm , space group no. 136. Question 6
Cadmium has the h.c.p. crystal structure with a = 2.98 Å, c = 5.62 Å. It slips on {0001} . Figure 1 is a stereographic projection with the z axis at the centre mark and the +y axis as shown. Using a Wulff net, mark on the directions , , and which lie in the northern hemisphere. A single crystal of cadmium grown in the form of a long cylinder of length 50 mm with a diameter of 3 mm is stressed parallel to its length. The axis T of the cylinder is marked on the stereographic projection. (i)
Using a Wulff net, measure the angle φ between T and [0001] and the acute angles λ1, λ2 and λ3 between T and the possible slip directions.
(ii)
Suppose that the crystal cylinder is subjected to an increasing tensile stress parallel to the axis T. From a consideration of the Schmid factor for each of the three possible slip systems, decide which will be the first one to operate. Taking the critically resolved shear stress for cadmium as 0.15 MPa, calculate the tensile stress and the actual load when slip begins. (g = 9.81 ms–2)
(iii) Draw on your stereographic projection the way in which the tensile axis changes direction as the crystal slips, i.e. draw the path of its pole. (iv) From the relevant equations given in the Data Book, calculate the new values of φ and λ when the crystal has extended to 75 mm. Plot on the stereogram the new position of the axis of the cylinder.
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z
+y
T
Figure 1. Stereographic projection for a cadmium single crystal
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Solution
Directions on the primitive are straightforward to plot. The stereogram below shows the directions , , and directions which plot as poles in the northern hemisphere.
The angle [1123] : [0001] is given by θ = tan–1 (a/c) = 27.9°. This enables the directions to be plotted. (i)
From the stereogram, φ = 52°, λ1 = 54°, λ2 = 41° and λ3 = 80°, where λ1 = T : [2110], λ2 = T : [1120] and λ3 = T : [1210].
(ii)
The resolved shear stress τ = σ cos φ cos λ. At the onset of slip, τ = τcrit = constant, and so slip occurs first on the system with highest Schmid factor, i.e. the highest value of cos φ cos λ. Since φ is the same for all three slip systems, it follows that slip occurs first on the system for which cos λ is greatest, i.e. for the slip system [1120](0001) for which λ2 = T : [1120] = 41°. τcrit = 0.15 MPa, so therefore from the formula τ = σ cos φ cos λ, it follows that when φ = 52° and λ = 41°, σ = 0.32 MPa.
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Now σ = F/A where F is the applied force and A the area over which the force acts. Thus the mass m required to cause slip is given by 0.32 × 106 × π 1.5 × 10–3 m = σA = g 9.81
2
kg = 0.230 kg = 230 gm
(iii) The path of the tensile axis as a function of increasing crystal slip is drawn on the stereogram as the thick black line between T and [1120]. (iv) The relevant equations are cos φ0 l1 = l0 cos φ1
=
sin λ0 sin λ1
where the ‘0’ subscript denote the original values of φ and λ and the length l, and where the ‘1’ subscript represents their values after slipping has occurred. Here, l1/l0 = 1.5 and so it follows that φ1= 66° and λ1= 26°. The new position of the axis of the cylinder after the crystal has been extended to 75 mm is denoted by TT on the stereogram.
Question 7
In higher dimensional projectional geometry used in quasicrystallography, two orthogonal low dimensional spaces V (of n dimensions) and W (of m dimensions) are defined in a suitable higher dimensional space H of m + n dimensions in which the unit vectors spanning the space are orthonormal (i.e. of unit length and orthogonal to one another). Tilings can then be generated in V space by projecting the hypercube of H centred at the origin onto W to form the corresponding projected shape R in m-dimensional space and then selecting only those points on the integer lattice in H which project into R to form a vertex of the tiling in V space by its projection onto V. The principle of this can be demonstrated graphically in 2D. V and W are then perpendicular lines. The 2D hypercube centred at the origin is a square with vertices at (1/2, 1/2), (– 1/2, 1/2), (1/2, – 1/2) and (– 1/2, – 1/2). Suitable templates then have the 2D integer lattice marked on with a suitable origin and a square centred at the origin. Using templates with a 10 mm square mesh on which dots are placed to represent the 2D integer lattice, examine the periodicity of the pattern generated in V if: (i) (ii)
W is the rational direction [230] (so that V is the direction [320]); W is the irrational direction [1τ0], where τ = 5 2+ 1 .
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Solution
(i)
(ii)
If W is the rational direction [230] (so that V is the direction [320]), then the repeat sequence along V is LLSLS, where ‘L’ is a long vector and ‘S’ a short vector (see page 13). This is a periodic sequence. If W is the irrational direction [1τ0], where τ = 5 2+ 1 , then the sequence along V is not periodic (see page 14).
Thus, if W is rational, the sequence generated along V is periodic, whereas if W is irrational, the sequence generated is not, although there are ‘patches’ arbitrarily large in size which can be identified in the sequence, e.g. in the example on page 14, SLSLL occurs, but not periodically. In fact for this particular choice of W, there cannot be SS or LLL in the sequence along V. The sequence along V in (ii) is very interesting – in reciprocal space there are peaks of intensity from this sequence, but the peaks are spaced aperiodically. A small computer program can be written easily to demonstrate this.
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Question 7, (i): periodic sequence along V.
S
W
y
L
S
L
x
L
S
L
S
L
L
S
L
S
L
L
S
L
S
L
L
S
L
S
L
L
S
L
S
L
L
S
L
V
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Question 7, (ii): non-periodic sequence along V.
S
W
y
L
S
L
x
L
S
L
S
L
L
S
L
L
S
L
S
L
L
S
L
L
S
L
S
L
L
S
L
S
L
L
S
L
L
V
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PART IIA and PART IIB MATERIALS SCIENCE AND METALLURGY Course C6: Crystallography: Data Book Entry
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BRACKET CONVENTIONS
hkl
diffracted wave: reciprocal lattice point
(hkl)
crystal face, lattice planes
{hkl}
crystal faces or lattice planes related by symmetry to (hkl)
[UVW]
lattice direction, zone axis
〈UVW〉
lattice directions or zone axes related by symmetry to [UVW]
a n [UVW]
a cubic lattice vector (components n U, etc.) STRUCTURE TYPES
STRUCTURE TYPE
FORMULA
SYSTEM LATTICE
LATTICE MOTIF
h.c.p.
-
hexagonal
P
0,0,0 ;
2/
1 1 3 , /3 , /2
diamond
C
cubic
F
C: 0,0,0 ;
1/
caesium chloride
CsCl
cubic
P
Cl: 0,0,0 ;
Cs:
sodium chloride
NaCl
cubic
F
Cl: 0,0,0 ;
Na: 0,0,1/2
sphalerite (zinc blende)
ZnS
cubic
F
S: 0,0,0 ;
Zn:
wurtzite
ZnS
hexagonal
P
S: 0,0,0 ; 2/3 ,1/3 ,1/2 Zn: 0,0, 1/2+u ; 2/3 ,1/3, u
1 1 4 , /4 , /4 1/
1/
1 1 2 , /2 , /2
1 1 4 , /4 , /4
(u ≈ 1/8 ) nickel arsenide
NiAs
hexagonal
P
As: 0,0,0 ; Ni: 1/3 ,2/3 ,1/4 ;
2/
1 1 3 , /3 , /2
1/
2 3 3 , /3 , /4
fluorite
CaF2
cubic
F
Ca: 0,0,0 ;
F: ± (1/4,1/4,1/4)
rutile
TiO2
tetragonal
P
Ti: 0,0,0 ;
1/
1 1 2 , /2 , /2
O: ± (u,u,0); ± (1/2+u,1/2 -u,1/2) (u ≈ 0.30 ) perovskite
CaTiO3
cubic
P
Ti: 0,0,0 ;
Ca:
O:
0,1/2,0 ;
1/
2,0,0
;
1/
1 1 2 , /2 , /2
0,0,1/2
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Crystal structures of some metals at room temperature c.c.p. metals
h.c.p. metals
Ag
a/Å 4.09
Al
b.c.c. metals
Cd
a/Å 2.98
c/Å 5.62
Cr
a/Å 2.88
4.05
Co
2.51
4.07
Fe(α)
2.87
Au Cu
4.08 3.61
Mg Zn
3.21 2.66
5.21 4.95
K Na
5.33 4.29
Ni
3.52 CRYSTAL SYSTEMS, LATTICES AND SYMMETRY
CRYSTAL SYSTEM
TRICLINIC
MONOCLINIC
ORTHORHOMBIC
TETRAGONAL
DEFINING SYMMETRY
monad
1 diad
3 diads
1 tetrad
a≠b≠c α = γ = 90o β ≥ 90o
a≠b≠c
CELL
a≠b≠c α≠β≠γ
α = β = γ = 90û
a=b≠c α = β = γ = 90û
CONVENTIONAL
P
P, C
P, C, I, F
P, I
(rotation or inversion) CONVENTIONAL UNIT
LATTICE TYPES
CRYSTAL SYSTEM
TRIGONAL
HEXAGONAL
CUBIC
DEFINING SYMMETRY
1 triad
1 hexad
4 triads
a=b≠c α = β = 90û γ = 120û
a=b≠c α = β = 90û γ = 120û
a=b=c α = β = γ = 90û
(rotation or inversion) CONVENTIONAL UNIT CELL
CONVENTIONAL LATTICE TYPES
P R † (trigonal only)
P, I, F
The symbol ≠ implies that equality is not required by symmetry †
Crystals of either system may have the hexagonal P lattice. The R lattice is only possible in the trigonal system, and for it a (primitive) rhombohedral unit cell may be used (a = b = c, α = β = γ < 120û), though more commonly a (triple) hexagonal cell (a = b ≠ c, α = β = 90û, γ = 120û) is chosen.
C6H12
-3-
C6H12
Geometry of single crystal slip Resolved shear stress
τ = σ cos φ cos λ
Elongation
l1 cos φ0 = l0 cos φ1
Bowing stress
τ = L
=
sin λ0 sin λ1
Gb
Slip systems c.c.p. metals
{111}
b.c.c. metals
{110} (other systems occur at high temperatures)
diamond C, Si, Ge
{111} (at high temperatures only)
NaCl*
{110}
CsCl*
{110}
h.c.p. metals
{001} (other systems can occur)
* and other ionic crystals with these structures OILS rule
This rule applies only to cubic crystals which slip on {111} or {110}. To determine the operative slip system(s) in a specimen under uniaxial tension or compression: (i)
Ignoring the signs, identify the highest (H), intermediate (I) and lowest (L) valued indices of the tensile axis [UVW].
(ii)
The slip direction or {110} slip plane (whichever is appropriate) is the one with zerO in the position of the I index and the signs of the other two indices preserved.
(iii) The {111} slip plane or slip direction (whichever is appropriate) is the one with the L index Sign reversed and the signs of the other two indices preserved.
C6H12
-4-
C6H12
DIFFRACTION
Bragg Equation θ
θ
λ = 2 dhkl sin θ
d hkl
Spacings between lattice planes for crystals with orthogonal crystallographic axes (cubic, tetragonal and orthorhombic) 1 dhkl
2
= h a
2
+ k b
2
+ l c
2
a*. b = a*. c = 0 ; a*. a = 1 ; etc.
Reciprocal lattice
ghkl = ha* + kb* + lc* ; |ghkl | = l / dhkl
Fhkl = Σn fn exp 2 π i (hxn + kyn + lzn)
Structure factor
= Σn fn [cos 2 π (hxn + kyn + lzn) + i sin 2 π (hxn + kyn + lzn)] Intensity of diffraction maxima Ihkl ∝ Fhkl . F*hkl = [Σn fn cos 2 π (hxn + kyn + lzn)]2 + [Σn fn sin 2 π (hxn + kyn + lzn)]2 Systematic absences Lattice type Absent indices P None I (h + k + l) odd F h, k, l mixed odd and even
Lattice type A B C
Absent indices (k + l) odd (h + l) odd (h + k) odd
Electron wavelengths
λ = 0.388 V −1/2 (1 + 9.87 × 10− 4 V)−1/2 The wavelength of 100 kV electrons is 0.037 Å.
(λ in Å , V in kV)
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