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With the results of the elasticity
theory we can get approximate formulas for the line
energy of a dislocation and the elastic interaction with other defects,
i.e. the forces acting on dislocations. |
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The energy of a dislocation comes
from the elastic part that is contained in the elastically strained bonds
outside the radius r0 and from the energy stored in the core, which is of
course energy sitting in the distorted bonds, too, but is not amenable to
elasticity theory. |
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The total energy per unit length
Eul is the sum of the energy contained in the elastic
field, Eel, and the energy in the core,
Ecore. |
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Do not
confuse energies E with Youngs modulus Y which is often (possibly
here) written as E, too! From the context it is always clear what is
meant. |
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Using the
formula for the strain
energy for a volume element given before, integration over the total volume
will give the total elastic energy Eel of the
dislocation. The integration is easily done for the screw dislocation; in what
follows the equations are always normalized to a unit
of length. |
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dEel(screw) |
= |
π · r · dr ·
(σΘ z
· εΘ
z + σzΘ · εzQ ) |
= |
4π · r · dr ·
G · (εΘ
z)2 |
Eel(screw) |
= |
G · b2
4π |
· |
R
∫ ro |
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dr
r |
= |
G · b2
4π |
· ln |
R
ro |
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The integration runs from
r0, the core radius of the dislocation to
R, which is some as yet
undetermined external radius of the elastic cylinder containing the
dislocation. In principle, R should go to infinity, but this is
not sensible as we are going to see. |
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The integration for the edge
dislocation is much more difficult to do, but the result is rather simple,
too: |
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Eel(edge) |
= |
G · b2
4π(1 ν) |
· ln |
R
ro |
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So, apart from the factor (1
ν), this is the same result as for the screw
dislocation. |
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Let us examine these equations. There
are a number of interesting properties; moreover, we will see that there are
very simple approximations to be gained: |
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1. The total energy U of a dislocation is proportional to
its length L. |
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U |
= |
Eul · L = L ·
(Eel + Ecore) |
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Since we always have the principle of minimal
energy (entropy does not play a role in this case), we can draw a important
conclusion: |
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A dislocation tends to
be straight between its two "end points" (usually
dislocation knots). That is a first rule about the direction a dislocation
likes to assume. |
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2. The line energy of an edge dislocation is always larger than
that of a screw dislocation since (1 ν) < 1. With ν
» 1/3, we have Escrew
» 0,66 ·Eedge. |
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This means that a dislocation tends to have as
large a screw component as possible. This is a second rule about the direction a dislocation
likes to assume which may be in
contradiction to the first one. It is quite possible that a
dislocations needs to zig-zag to have as much screw character as geometrially
allowed - it then cannot be straight at the same time. |
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3. The elastic part of the energy depends (logarithmically) on the
crystal size (expressed in R), for an infinite crystal it is infinite (∞)! Does this make any sense? |
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Of course this doesn't make sense. Infinite
crystals, however, do not make sense either. And in finite crystals, even in
big finite crystals, the energy is
finite! |
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Moreover, in most real crystals it is not the outer dimension that counts, but the size of
the grains which are usually quite small. In addition, if there are many
dislocations with different signs of the Burgers vector, their strain fields
will (on average) tend to cancel each other. So for practical cases we have a finite energy. |
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4. The elastic part of the energy also depends (logarithmically)
on the core radius r0. |
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5. The energy is a weak function of the crystal (or grain size)
R. |
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Taking an extremely small value for
r0, e.g. 0,1 nm, we obtain for
ln (R/r0) an extreme range of 20,7 -
2,3 if we pick extreme values for R of 100 mm or 1
nm, respectively. A more realistic range for R would be
100 µm - 10 nm, giving ln (R/r0) =
13,8 - 4,6. Grain size variations, within reasonable limits, thus only
provide a factor of 2 - 3 for energy variations. |
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We will now deduce an approximation
for the line energy that is sufficiently good for most purposes. |
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We equate r0 with the
magnitude of the Burgers vector, |b|. This makes sense
because the Burgers vector is a direct
measure of the "strength" of a dislocation, i.e. the strength of
the displacement in the core region. |
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We need a value for the energy in the core of the
dislocations, which so far we have not dealt with. Since there is no easy way
of calculating that energy, we could equate it in a first approximation with
the energy of melting. That would make sense because the dislocation core is
comparable in its degree of distortion to the liquid state. More sophisticated
approaches end up with the best simple value: |
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There are, however, other approaches, too. |
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The total energy for r0 =
b then becomes |
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Etot |
= |
G · b2
4π |
· |
( |
ln |
R
b |
+ 2 |
) |
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More generally, the
following formula is often used |
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Etot |
= |
G · b2
4π(1 ν) |
· |
( |
ln |
R
b |
+ B |
) |
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with B = pure number best
approximating the core energy of the particular case. Often B = 1
is chosen, leading to
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Etot |
= |
G · b2
4π(1 ν) |
· |
( |
ln |
R
b |
+ 1 |
) |
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= |
G · b2
4π(1 ν ) |
· |
( |
ln |
e ·R
b |
) |
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For the last equation bear in mind that ln(e)
= 1. |
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The ln term is not very
important. To give an example; it is exactly 4π for e · R = 3,88 x 104
|b|, i.e. for R » 5 µm;
the total energy in this case would be Etot = 2G ·
b2. |
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In a very general way we can
write |
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And α (from
measurements) is found to be α
» 1,5 ....0,5. If we do
not care for factors in the order of unity, we get the final very simple formula for the line energy of a dislocation |
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With this expression for the line
energy of a dislocation, we can deduce more properties of dislocations. |
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6. Dislocations always tend to have the smallest possible Burgers
vector. |
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Since for Burgers vectors
bi larger than the smallest translation vector
of the lattice and thus expressible by bi =
b1 + b2 ;
b1,2 = some shorter vectors of the lattice, we
always have |
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A splitting into smaller Burgers vectors is therefore always
energetically favorable. |
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There are therefore no dislocations
with large Burgers vectors! |
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If a dislocation would have a large Burgers vector, it would
immediately split into two (or more) dislocations with smaller Burgers vectors.
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This is always possible,
because in the Volterra construction you can always replace one cut with the translation vector
b by two cuts with
b1 and b2 so
that b = b1 +
b2. |
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7. The line energy is in the order of 5 eV per Burgers
vector.. |
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This makes dislocations automatically non-equilibrium defects.
They will not come into being out of nothing like point defects for free
enthalpy reasons. |
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8. The line energy ( = energy per length) has the same dimension
as a a force, it expresses a line
tension, i.e. a force in the direction of the line vector which tries
to shorten the dislocation. |
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It actually is a force, as
we can see from the definition of such a line tension F: |
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We thus may imagine a dislocation as a stretched rubber band,
which tries to be as short as possible. But one should be careful not to
overreach this analogy. Ask yourself: What keeps dislocations loops
stable? |
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© H. Föll