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Formulas here have plenty of
indices, underlining etc. - and we will now give up the cursive font
normally used for variables because it gets too cumbersome. |
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Basic
Idea |
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The Coincidence Site lattice (CSL) provided a
relatively easy way to grasp the concept of special orientations between grains
that give cause for special grain boundaries. With the extension to grain
boundary dislocations in the DSC lattice, the CSL concept became
in principle applicable to all grain boundaries, because any arbitrary
orientation is "near" a CSL orientation. But yet, the
CSL concept is not powerful enough to allow the deduction of grain
boundary structures in all possible cases. The reasons for this are physical,
practical and mathematical: |
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The CSL by itself
is meaningless; meaningful is the
special grain boundary structure that is possible if there is a coincidence
orientation. The grain boundary structure is special, because it is periodic
(with the periodicity of the CSL) and contains coincidence points (cf.
the picture). |
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But we have no guarantee that periodic grain boundary
structures may not exist in cases where no CSL exists; i.e. by only
looking at CSL orientation, we may miss other
special orientations. That will be certainly true whenever we
consider boundaries between different lattices - be it that lattice constants
of the same materials changed ever so slightly because one grain has a somewhat
different impurity concentration, or that we look at phase boundaries between
different crystals. If the lattice constants are incommensurable, there will be
no CSL at all. |
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As we have seen, even a CSL with Σ = 41 is significant, even so it is virtually
unrecognizable as anything special in a drawing. This is an expression of the
mathematical condition, that you either have perfect coincidence or none. If
two points coincide almost, but not quite, no recognizable CSL will be
seen. If two lattice points coincide except for, lets say, 0,01 nm, we
certainly would say we have a physical coincidence, but mathematically we have
none. |
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The same is true if we rotate a lattice away from a
coincidence position by arbitrarily small angles. Mathematically, the coincidence is totally destroyed
and the situation has completely changed, whereas physically an arbitrarily small change of the
orientation would be expected to cause only small changes in properties. |
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Only a very small fraction of grain orientations have a
CSL. The "trick" we used to transform any orientation into a
coincidence orientation by introducing grain boundary dislocations in the
DSC lattice is somewhat questionable: the effect (= CSL) comes
before the cause (= dislocations in the DSC lattice), because at the
orientation that we want to change no CSL and therefore no DSC
lattice exists. |
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It becomes clear that the main problem lies in the
discreteness of the CSL. Any useful theory for special grain boundary
(and phase boundary) structures must be a continuum theory, i.e. give results for continuous
variations of the crystal orientation (and lattice type). |
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This theory exists, it is the so-called "O-lattice theory" of W. Bollmann;
comprehensively published in his opus magnus "Crystal Defects and Crystalline
Interfaces" in 1970. |
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The O-lattice theory is not
particularly easy to grasp. (Sorry, but it took me many hours, too).
It is well beyond the scope of this hyperscript to go into details. What will
be given is the basic concept, the big ideas; together with some formulas and a
few examples. You should first read just over it, trying to get the basic
ideas, than study it point by point. If you don't get it the first time - don't
despair, you are in good if not excellent company! |
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There are two basic ideas behind the
O-lattice theory: |
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1. Take a crystal lattice
I and transform it in any way you like. That means you can
not only rotate it into an arbitrary orientation relative to crystal I,
but also deform it by stretching, squeezing and shearing it. The
crystal lattice II generated in this
way from a simple cubic lattice I thus could even be an arbitrarily
oriented triclinic lattice. |
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2. Now look for coincidence points between lattice
I and lattice II. But do not restrict the search for coinciding
lattice points, but expand the concept of
coincidence to all "equivalence
points" within two overlapping unit cells. What
equivalence points are becomes clear in
the illustration. |
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Points in lattice I and lattice II
are called equivalent, if their space vectors are identical (always in their
respective lattice coordinate system). |
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Let's look at the example illustrated
above. Lattice I is deformed by first rotating it and then stretching
the axis x1; this produces lattice II |
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An arbitrary point within the elementary cell of
lattice I is described by a vector r(I) |
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r(I) transforms into a vector
r(II) by the transformation applied. The point reached within the
unit cell of lattice II by r(II) is then an equivalence
point to the one in crystal I. |
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Of course there is more than one
equivalence point; there is always an infinite set defined by one point plus
all points reachable by a lattice translation vector T from this
particular point. |
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Any point r'(II) in lattice
II belonging to the set as defined above can be described in the
coordinate system of lattice II (defined by the units vectors
x1(II) and x2(II)) by |
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With T(II) = any translation
vector of lattice II, or |
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T(II) |
= |
n · x1(II) + m · x2(II)
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And n, m = 0, ± 1, ± 2, ...
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All these points are by definition equivalence
points to the corresponding set of
points in lattice I given by |
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Let us designate the set of all equivalence points defined above in
lattice I by C1
and the corresponding set in lattice II by C2 and, for the sake of clarity, all
vectors pointing to equivalence points of the respective sets by
r(C1) and r(C2). |
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If we now look at a certain
equivalence point in lattice II, it always originated from lattice
I by the general transformation as shown in the picture below |
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The blue lattice was obtained from
the pink one by some transformation; in this case by a simple rotation. |
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The dark point with the red vector pointing to it
is an arbitrary point in lattice I (for the sake of easy recognition
about in the center of a lattice I cell). |
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After the transformation, it is now the red point
at the apex of the blue and pink arrows in lattice II. It is still about
in the center of a cell in lattice II, but for the particular transformation shown, it is now also
about in the center of a lattice I cell - there is (about) a coincidence of equivalence points. |
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Let's assume perfect coincidence, then the red
point denotes coinciding equivalence
points, i.e. equivalence points that are "on top of each other".
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We need a precise mathematical
formulation that gives us the conditions under which coincidence of equivalence points occcurs. |
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This is easy, we just have to consider that for
coinciding equivalence points the blue vector in lattice II can be
obtained in two ways:
- By the transformation equation from the corresponding red vector of lattice
I (valid for all equivalence points)
or, since the coincidence point belongs to both lattices at once, by
- adding some translation vector of lattice I to the red vector. This
is symbolically shown in the picture.
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In formulas we can write for any vector in lattice II pointing to some
equivalent point of the set C2: |
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r(C2) |
= |
A {r(C1)}
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r(C1) |
= |
A1
{r(C2)} |
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With A = transformation matrix (we will encounter examples for
A later; see also the basic
module for matrix calculus)
since this simply describes how lattice II originates from lattice
I. This was the first way mentioned above. |
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On the other hand, we can obtain new equivalence
points in lattice I, i.e. other elements of the set C1
designated by e(C1) quite generally by the
equation |
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We will now use these relations for coinciding
equivalence points: |
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We are looking for coincidences of
any one member of the set r(C1) with any one member of
the set r(C2); any coincidence point thus obtained
will be named r0. Since this point, describable in
lattice II by r(C2) must be reachable in lattice
I by first going down r(C1) and then adding a
translation vector of lattice I, we obtain |
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r(C2) =
r(C1) + T(I) = r0 |
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Using the transformation equation for lattice
I from above and substituting it into the above equation yields |
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We wrote r0 instead of
r(C2) because we do not need the distinction between
the sets C1 and C2 any more because
r0 belongs to both
sets. |
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Rearranging the terms following
matrix calculus by
using the identity or unit transformation matrix I, we obtain the
fundamental equation of O-lattice theory: |
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What does that equation mean? |
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For a given transformation, i.e. for given orientation
relationship between two grains, its solution for r0
defines all the coincidence points or O-points of
the lattices. The coincidence of lattice points is a subset of the general
solution for the coincidence of equivalence points. |
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The question comes up if there are any solution of this equation. Algebra tells us that
this requires that the determinant of the matrix, |I A1|, must be ¹ 0. This will be generally true (but not
always), so generally we must expect that solutions exist, i.e. that a
CSL (= O-lattice) for some equivalence points (= O-points)
exists - for any possible combination of lattice I and lattice
II. |
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How do we solve the O-lattice equation,
i.e. obtain the set of O-points for a given lattice and transformation?
Simply by inverting the matrix we obtain: |
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r0 = ( I
A1 )1 ·
T(I) |
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That is all there is to do; it looks easy. If we
have a given transformation matrix A,
the equation above gives us the set of vectors defining the equivalent points,
or as we are going to call them, the O-points of the two lattices. |
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However, the diffusion equations look easy, too, but are not
easy to solve. Also, we do not yet know what the solution, the
O-lattice, really means with respect to grain- or phase-boundaries.
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We will look at this more closely in the next
paragraph; but first we will discuss a simple example. |
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To keep the matter simple, we
look at a two-dimensional situation where a
square lattice rotates on top of another one. This will include our former
example of the Σ = 5 CSL case. (A word of warning: In Bollmanns
book are occasional mistakes when it comes to the Σ5 orientation (which is frequently used for
illustrations)). |
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The transformation matrix is a pure rotation matrix, for the
rotation angle α it writes |
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A |
= |
( |
cos α |
sin α |
sin α |
cos α |
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From this we
get |
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A1 |
= |
( |
cos α |
sin α |
sin α |
cos α |
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1
A1 |
= |
( |
1 cos α |
sin α |
sin α |
1 cos α |
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(1 A1
)1 |
= |
( |
½ |
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½ cotan α/2 |
½ cotan α/2 |
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½ |
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) |
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Now let's do an example. The base vectors of the
square lattice I are x1(I) = (1, 0),
x2(I) = (0, 1). |
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If we use them as the smallest possible translation vector
T(I) of lattice I, we obtain by multiplication with the
last matrix the smallest vectors of the
O-lattice which then must be the unit vectors of the O-lattice,
u1 and u2: |
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u1 |
= |
( |
½
½ · cotan (α/2) |
) |
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u2 |
= |
( |
½ · cotan (α/2)
½ |
) |
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This is easily graphically represented, but the pictures get
to be a bit complicated: |
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Lattice I is the blue lattice, lattice
II the red one; it has been rotated by the angle α. |
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The unit vectors of the O-lattice can be determined by
the intersection of the light- and dark-green lines (remember the definition of
tan and cotan!); they are depicted in black. |
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The O-lattice then can be constructed, its lattice
points are shown as orange blobs. |
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Note that a three dimensional expansion would not
produce much that is new. On any plane above or below the drawing plane, the
situation is exactly what we have drawn. This has one interesting consequence,
however: |
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In three dimension, we have no longer
O-points, but O-lines in this
case. |
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The O-lattice in this case thus is not a point lattice,
but a lattice of lines perpendicular to the
plane of rotation. This will come up naturally later, but it is good to keep it
in mind for what follows. However, since this is not the most general case, we
will keep talking of O-points. |
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The picture neatly helps to
overcome a possible misunderstanding: For
any O-point, a vector from the origin of either crystal to the
O-point (our vectors r(I) and r(II)) point
to a coinciding equivalence point or
O-point, but different points of the
O-lattice may be different
equivalence points. In the example we have O-points that are almost at
the center of both unit cells, or almost at a lattice point of
both unit cells - the O-lattice seems to be constructed of two
kinds of coinciding equivalence points; but: |
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If we would include more cells of the O-lattice, we
would see that equivalence points shift slightly for the example given. A few
O-lattice cells away, they would be more off-center or more distant to a
lattice point than close to the origin of the O-lattice. |
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Just how many equivalence
points of the set of equivalence points (which has an infinite number of
members) are needed for an O-lattice is an important (nontrivial)
question which we will take up later again. |
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We can rephrase this important question: |
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Is the pattern of equivalence points periodic (= finite number of equivalence point) or
non-periodic (infinite number)? In other
words: If any one point of the O-lattice defines a specific equivalence
point in the crystal lattices, does this specific point appear again at some
other point in the O-lattice (apart from the trivial symmetries of the
O-lattice)? |
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We will come back to this question
later; it is the decisive feature of the O-lattice for
defining the DSC-lattice. |
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How do we get the CSL from
the O-lattice? That is easy: It must be that particular subset of all
possible O-lattices where all O-points are also lattice points in
both lattices. |
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Looking at the unit vectors of the O-lattice, however,
there is no way of expressing them in integer values of the base vectors of
lattice I, because one component is always 1/2. How about
that? |
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This is not a real problem, best illustrated with
an example: If we chose α =
36o52,2', we have |
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u1 |
= |
( |
½
½ · cotan (α/2) |
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= |
( |
1/2
3/2 |
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u1 |
= |
( |
½ · cotan (α/2)
½ |
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= |
( |
3/2
1/2 |
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Thus every second point of
the O-lattice is a lattice point in both lattices (depicting O-
points of the equivalence class [0,0]), these points thus define the
Σ = 5 CSL. The other O-points
are of the equivalence class [1/2,1/2]. CSL lattices
(two-dimensional case) thus correspond to specific O-lattices, but with
lattice constant possibly larger by some integer value. This is quite important
so we will illustrate this in a
special module. |
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Note, too, that in this case the pattern of
equivalence point is obviously periodic, so we have a first specific answer to
the question asked above. |
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Before we delve deeper into the intricacies of
O-lattice theory, we shall first discuss some of its general
implications in the next paragraphs. |
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© H. Föll