Crystallography

C6H2

PART

II

and PART II

MATERIALS SCIENCE AND METALLURGY

Course C6: Crystallography

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:

b c a A B C (hkl) O a h b k c l Normal to (hkl)

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 + µ_AC

for suitable λ and µ. Hence, expressing AB and AC in terms of a, b and c, it follows that

r = λ b k – ah + µ cl – ah = – λ + µ h a + λ k b + µ l c

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 h , v = λk , w = µ 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.

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 k – ah × cl – ah , i.e., n || b ×c kl – b ×a hk – a ×c 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, d_hkl, it is convenient to consider the diagram below:

(hkl) a h O N A n ^ ψ

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| = d_hkl. Also,

OA.ON = OA ON cos ψ = ON2 = d_hkl2

since the angle ONA is a right angle. If we write the vector ON in the form ON = n = χ _{ha* + kb* + lc*}

for some χ, it follows that the dot product OA. ON = χ since OA = a/h. Hence,

χ = d_hkl2

Now,

C6H2 - 3 - C6H2 but we also know

ON2 = d_hkl2

and of course |ON|2 _{= ON.ON. Therefore,}

dhkl2 = dhkl4 ha* + kb* + lc* 2

and so we obtain the important result that

ha* + kb* + 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.

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.

Trace of (hkl) in the x-y plane

a₁ a₃ a₂ a k a h a i 60˚ θ O A B G 60˚ (60˚ – θ) (120˚ – θ)

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

a/k =

sin 60Þ – θ

a/h and sin

θ

a/p =

sin 120Þ – θ

a/h

respectively. Now sin (120° – θ) = sin (180° – (120° – θ)) = sin (60° + θ), and so these two equations can be rearranged in the forms

k

h sin θ = sin 60Þ – θ

and

p

h sin θ = sin 60Þ + θ

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

Three-index notation Four-index notation [100] 1 3 2110 [010] 1 3 1210 [110] ] 0 2 11 [ 3 1 [uv0] 1 3 2u – v, 2v – u, – u + v , 0 [uvw] 1 3 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 3h 2u – v + 1 3k 2v – u + 1 3 h + k u +v +lW = hu + kv + lw

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)),

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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> <111> <110> {100} and {110} Note that for any crystal system:

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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:

Course C6: Crystallography

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.

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-by-point 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.

sketch of sphere 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 A_O and B_O. It follows that the projection of a great circle intersects the primitive circle in two diametrically opposite points, C and C_O. 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

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Figure 8 (projection sphere) 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 CC_O (Figures 8 and 9), the pole must lie in the vertical plane perpendicular to the line CC_O; 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 C_O 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.

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.

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. 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.

5. 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. 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.

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.

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) = tan–1 2c

3 a = 65.0° and the angle (0001) : (0112) = tan–1

c

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 a b.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)].

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.

C6H4

PART

II

and PART II

MATERIALS SCIENCE AND METALLURGY

Course C6: Crystallography

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

A cos φ cos λ

The factor (cos φ cos λ) is called the Schmid factor.

Figure 2

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

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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 <110>-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.

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 } <111> slip (e.g., b.c.c. metals)

For b.c.c. metals slipping on {110} <111> 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 u2 _{+ 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.

Looking more closely at the Schmid factors for B and K, we have:

B: | cos φ cos λ| ( × 6 u2 _{+ v}2 _{+ w}2_{) = w}2 _{– u}2 _{+ v(w – u)} K: | cos φ cos λ| ( × 6 u2 _{+ v}2 _{+ w}2_{) = w}2 _{– u}2 _{+ 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 u2 _{+ v}2 _{+ w}2_{) = w}2 _{+ vu + vw – u}2 I: | cos φ cos λ| ( × 6 u2 _{+ v}2 _{+ w}2_{) = w}2 _{+ vu + uw – v}2

and since

vw + v2 _{> uw + u}2

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

II

and PART II

MATERIALS SCIENCE AND METALLURGY

Course C6: Crystallography

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:

rotation axes,

mirror planes,

centre of symmetry

and 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 I}A 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).

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

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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.

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 ;}

C6H5 - 5 - C6H5 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 <100> <111> <110>

Tetragonal, hexagonal, trigonal <001> <100> <110>

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.

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Course C6: Crystallography

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 n_p, 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

2, 0, z and 0, 1 2, z.

The screw axis at relates the Cu atom at 1

4, 1 4, 1 4 to those at (i) 1 4, – 1 4, 3 4 , (ii) 3 4, – 1 4, 1 4 and (iii) 3 4, 1 4, 3 4. Oxygen Copper x y 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 1 2 1 2 1 2 1 2 A

C6H6 - 2 - C6H6

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.

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 Magnitude Symbol

Axial glide || to a axis 1_{/2a a}

Axial glide || to b axis 1_{/2b b}

Axial glide || to c axis 1_{/2c c}

Diagonal glide || to face diagonal 1_{/2a +} 1_{/2 b,} 1_{/2b +} 1_{/2 c,} 1_{/2c +} 1_{/2 a}

n

Diamond glide || to face diagonal for a face centred cell

1_{/4a +} 1_{/4 b,} 1_{/4b +} 1_{/4 c,} 1_{/4c +} 1_{/4 a}

d

Diamond glide || to face diagonal for a body centred cell

1_{/4a +} 1_{/4 b +} 1_{/4 c d}

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_{/2b +} 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_{/2a +} 1_{/2 c.}

Thus, the copper atom at 1

4, 1 4, 1 4 is related to that at 3 4, 1 4, 3

4. and the oxygen atom at 0, 0, 0 is related

to the oxygen atom at 1

2, 1 2,

1

2 by the (010) glide plane at y = 1_/4.

C6H6 - 4 - C6H6

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 <100>, <111> and <110> respectively. This is space group no. 224, more commonly referred to as Pn3m

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.

C6H6 - 6 - C6H6

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.

Course C6: Crystallography

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.

C6H7 - 2 - C6H7 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.

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.

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.

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.

C6H7 - 8 - C6H7

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].

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

II

and PART II

MATERIALS SCIENCE AND METALLURGY

Course C6: Crystallography

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 Σ Σ 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.

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. <100> or <110>. 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.

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.

C6H8 - 5 - C6H8

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

S = λ1 0 0 0 λ₂ 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.

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

II

and PART II

MATERIALS SCIENCE AND METALLURGY

Course C6: Crystallography

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, F_hkl, is given by

F_hkl = f_n exp 2

{

(

)

}

n atoms

∑

where f_n is the scattering factor of atom n. Hence, it follows from this definition that for a face-centred lattice, systematic absences occur when h, k and l are mixed odd and even, and for a body-centred 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 _{+ k}2 _{+ l}2_{) 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 g_hkl. This is 1/d_hkl, as in the diagram below:

Hence, it follows that sinθ = 1/ 2dhkl

1/λ , i.e., we derive the Bragg equation λ = 2dhklsinθ.

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.

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 dynamicalnature 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.

000 002

111

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.

C6H9 - 3 - C6H9 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 <111> 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].

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, r_n*, is given by the formula

rn* = k2 – 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*.