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A perfect dislocation may dissociate
into two partial dislocations because this lowers the total energy. |
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The Burgers vector b =
a/2[110] may, e.g., decompose into the two
Shockley partials a/6[121]
and a/6[2,1,1] as shown below. |
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Of necessity, a stacking fault between the two partial dislocations
must also be generated. |
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You can think of this as doing two Volterra cuts in the same plane, each on with
the Burgers vector of one of the Shockley partials, but keeping the cut line
apart by the distance d. Each cut by itself makes a stacking
fault, but the superposition of both creates a perfect lattice. |
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Lets balance the energy of this
reaction: |
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Energy of the
perfect dislocation |
= G · b2 =
G · (a/2<110>)2 |
= |
G · a2
2 |
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Energy of the
two partial dislocations |
= 2G ·
(a/6<112>)2 = 2G ·
a2/36 · (12 + 12 + 22)
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= |
G · a2
3 |
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We thus have a clear energy gain Esplit =
G · a2 by having smaller Burgers vectors. This
energy gain does not depend on the distance d between the
dislocations. |
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But we are not done yet; we have two
more energy terms to consider: |
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1. The energy of
interaction +Einter; it will be large at
short distances. The dislocations repulse each other and the energy going with
this interaction is proportional to 1/d. Based on this alone, the partial dislocations thus would tend to
maximize d. |
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2. The energy of the stacking
fault +ESF stretched out by
necessity between the two partial dislocations. This
stacking fault
energy is always ESF = γ · area, or, taken per per unit of length as for the dislocations,
E'SF =γ ·
d. Based on this alone,
the partial dislocations thus would tend to minimize d. |
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In total we have some energy gain by just forming partial dislocations in the
first place, but energy losses if we keep
them too close together, or if we move them too far apart. |
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We thus must expect that there is an
equilibrium
distance
deq which gives a minimum energy for the total defect
which consists of a split dislocation
and a stacking
fault. This equilibrium distance deq will
depend mostly on the stacking fault energy γ;
for small γ's we expect a larger distance
between the partials. |
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In principle, we can calculate deq by
writing down the total energy, i.e. the sum of the energy gain by forming
partial dislocations plus the energy of the interaction plus the stacking fault
energy, then find the minimum with respect to d by
differentiation. This is a basic execise, what you will get is
deq µ
γ½ |
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Instead of a a pure one-dimensional
defect - our perfect dislocation - we have now something complicated, some kind
of ribbon stretching through the crystal. Moreover, this stacking fault ribbon
may be constricted at some knots or jogs, and may look like this: |
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How would this look in cross-section?
We take a picture after "Hull and Bacon" |
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It is clear that a dislocation split into Shockley partials is still able to glide on the same
glide plane as the perfect dislocation; the stacking fault just moves along. It
can also change its length without any problems. |
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For Frank type partials
this is not true. The loop it usually
bounds could only move on its
glide cylinder. Changing
the length would involve the absorption or emission of point defects. |
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Reactions between dislocation now
tend to become messy. You must consider the reaction between the partials and
taking into account the stacking fault. However, processes now become possible
that could not have occurred before. Lets look at some examples. |
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A small dislocation loop formed by the agglomeration of
vacancies, that in its pure form cannot add much to plastic deformation, may
transmutate into a dislocation loop bounded by a perfect Burgers vector (which
in turn may split into Shockley partials) - it is now glissile and can increase
its length ad libitum. How does that happen? |
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As shown below, the Frank partial bounding
the vacancy disc defining the stacking fault has a Burgers vector of the type
b = a/3<111>. It then may split into a perfect
dislocation with b = a/2<110> and a Shockley partial with
b = a/6<112> (which must lie in the loop plane). The
Shockley partial moves across the loop, removing the stacking fault - we have
an "unfaulting"
process. A loop bounded by a perfect dislocation, free to move, is left. The
glide plane of the perfect dislocation is not the plane of the loop; the
Burgers vector of the perfect dislocation, after all, must have a sizeable
component perpendicular to the loop plane in order for the sum of the Burgers
vectors to be zero. |
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The Shockley dislocation, once formed, will move quickly over
the loop - pulled by the stacking fault like by a tense rubber sheet. The
driving force for the reaction is the stacking fault energy: As the loop
increases in size because more and more vacancies are added and the radius
r grows, the energy of the loop increases with
r2 due to the stacking fault. However, the line energy
of the dislocation only increases with r no matter what kind of
dislocation is bounding the loop. |
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There is therefore always a critical radius
rcrit where a perfect loop becomes energetically
favorable. |
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The perfect loop now feels the
Peierls potential, it may try to align
the dislocation into the <110> directions, always favorable in
fcc lattices the loop then assumes a hexagonal shape. |
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Now all segments are able to glide. If the resolved shear
stress for some segments is large enough, they are going to move, pulling out
long dislocation dipoles in the direction of the movement. The beginning of
this process may look look this: |
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What we have, in summary, is one of
the problems of Si materials technology: |
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We have an efficient source for dislocation
generation by vacancy (or, in Si, interstitial) agglomeration in
formerly dislocation free crystals! And this is not a theoretical possibility,
but reality if you are not very careful in growing your crystals.
Many examples are shown in
the link. |
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As we have seen, there are now many
possible dislocation reactions. In writing down reaction equations, you must
use the specific Burgers vector (e.g. a/6[1, -2, 1]) and not the
general type (a/6<112> for the example). This can be
cumbersome and is prone to produce errors. |
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Fortunately there is a extremely useful tool for fcc
lattices to keep the vectors in line: The Thompson
tetrahedron. |
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The Thompson tetrahedron is simply the tetrahedron formed by
the {111} planes with consistently indexed planes and edges. |
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If we look at the {111}-planes
tetrahedron, we see the following connections
- The edges are <110>
directions, they may be used to represent the Burgers vectors of the perfect
dislocations and the preferred direction for the line vectors because of the
Peierls potential (red lines).
- The faces are {111} planes, they
show the positions of potential stacking faults.
- The Burgers vector of the Shockley partials that may bound a stacking fault
of the given {111} plane are the vectors running from the center
of the triangular faces to the corners (blue lines)
- The Frank dislocations that also can bound a stacking fault, run from the
center of the triangular faces to the center of the tetrahedron (not shown).
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For a "short-hand" description, it is conventional,
to enumerate the edges by A,B,C,D and the centers of their triangles by
α, β,
γ and δ. The
relevant vectors than become, e.g., AB or Aγ. |
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It is a good idea (really!) to really build a Thompson tetrahedron -
maybe from some stiff cardboard; the link gives the
detailed net. |
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© H. Föll