A Polygonal Dislocation Loop in MD++

Additional Manual for MD++
Creating a Polygonal Dislocation Loop

Keonwook Kang and Wei Cai

Jan 19 , 2009

This short summary explains how we create a polygonal dislocation loop in the atomistic simulation, using the displacement field of a triagonal dislocation loop.

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Displacement Field of a Triangular Dislocation Loop

Fig.1.

While studying a dislocation nucleation problem by atomistic simulations, the authors needed to create a polygonal dislocation loop, which requires the knowledge of the displacement field of a polygonal dislocation loop in the infinite isotropic medium.

In elasticity, the displacement field of a polygonal dislocation loop is merely the sum of fields of triangular loops that constitute the polygon loop. The coordinate-free form of analytic expression for the elastic displacement field u of a triangular loop ABC is given as

$\mathbf {u(x)} =-{\frac {\Omega }{4\pi }}\mathbf {b} -{\frac {1-2\nu }{8\pi (1-\nu )}}\left[\mathbf {f} _{AB}+\mathbf {f} _{BC}+\mathbf {f} _{CA}\right]+{\frac {1}{8\pi (1-\nu )}}\left[\mathbf {g} _{AB}+\mathbf {g} _{BC}+\mathbf {g} _{CA}\right]$

(1)

where

$\mathbf {f} _{AB}=\left(\mathbf {b} \times \mathbf {t} _{AB}\right)\ln \left({\frac {R_{B}}{R_{A}}}\cdot {\frac {1+{\boldsymbol {\lambda }}_{B}\cdot \mathbf {t} _{AB}}{1+{\boldsymbol {\lambda }}_{A}\cdot \mathbf {t} _{AB}}}\right)$

(2)

and

$\mathbf {g} _{AB}=({\boldsymbol {\lambda }}_{A}+{\boldsymbol {\lambda }}_{B}){\frac {\mathbf {b} \cdot ({\boldsymbol {\lambda }}_{A}\times {\boldsymbol {\lambda }}_{B})}{1+{\boldsymbol {\lambda }}_{A}\cdot {\boldsymbol {\lambda }}_{B}}}.$

(3)

f_BC and f_CA have the same expression as Eqn.(2) except cyclic permutation of A, B and C, and so do g_BC and g_CA with Eqn.(3). R_A, R_B and R_C are the magnitudes of R_A, R_B and R_C, respectively. (See Fig.1.) ${\boldsymbol {\lambda }}_{A}$, ${\boldsymbol {\lambda }}_{B}$ and ${\boldsymbol {\lambda }}_{C}$ are the unit vectors of R_A, R_B and R_C, respectively, e.g. ${\boldsymbol {\lambda }}_{A}=\mathbf {R} _{A}/R_{A}$. And t_AB, t_BC and t_CA are the unit tangent vectors of the segments AB, BC and CA, respectively. b is the loop Burgers vector.

The solid angle $\Omega$ in Eqn.(1) is expressed as

$\Omega =-2\tan ^{-1}{\frac {\mathbf {R} _{\textrm {A}}\cdot (\mathbf {R} _{\textrm {B}}\times \mathbf {R} _{\textrm {C}})}{R_{\textrm {A}}R_{\textrm {B}}R_{\textrm {C}}+(\mathbf {R} _{\textrm {A}}\cdot \mathbf {R} _{\textrm {B}})R_{\textrm {C}}+(\mathbf {R} _{\textrm {A}}\cdot \mathbf {R} _{\textrm {C}})R_{\textrm {B}}+(\mathbf {R} _{\textrm {B}}\cdot \mathbf {R} _{\textrm {C}})R_{\textrm {A}}}}.$

(4)

The sign of $\Omega$ is reversed from the original expression to be in accordance with the convention of Barnett (1985).

By displacing all the atoms according to Eqn.(1), we can create a triangular dislocation loop in the atomistic simulation as shown in Fig.2(a).

(a)	(b)
Fig.2 (a) A triangular and (b) a tetragonal dislocation loop in DC Si. $\mathbf {C} _{1}=8a[1\,1\,{\bar {2}}]$, $\mathbf {C} _{2}=20a[1\,1\,1]$ and $\mathbf {C} _{3}=12a[1\,{\bar {1}}\,0]$, where a is the lattice constant and $a=5.431\mathrm {\AA}$ in Si. The loop Bergurs vector $\mathbf {b} =a/2[1\,{\bar {1}}\,0]$ in both (a) and (b).

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Construction of a Polygonal Dislocation Loop

For the displacement field of a polygonal dislocation loop, we first decompose the polygonal loop into a sum of triangular loops, calculate the fields of each loops according to Eqn.(1)-(4) and finally sum them up. For example, Fig.3 shows how a tetragonal dislocation loop can be decomposed of multiple triangular loops.

(a)	(b)
Fig.3 A tetragonal dislocation loop P1P2P4P3 can be decomposed into two different sets of triangular loops. In either case, the net displacement field is identical, or $\mathbf {u} _{\Box 1243}=\mathbf {u} _{\triangle 124}+\mathbf {u} _{\triangle 143}=\mathbf {u} _{\triangle 123}+\mathbf {u} _{\triangle 243}$. The inside arrows designate the order of vertex indices, which are cyclically permutatable.

Note that the displacement given in Eqn.(1) changes its sign depending on the circular direction of the loop, or $\mathbf {u} _{\triangle 123}=-\mathbf {u} _{\triangle 321}$. Hence, when we sum fields of two or more triangular loops, we need to carefully decide the circular direction of each loop such that the displacement fields due to the shared segment of neighboring triangular loops can be canceled each other. (See the direction of circular arrows in Fig.3.)

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Is b given in lab coordinate or crytal coordinate?

The Burgers vector b is conventionally given in crystal coordinate such as $\mathbf {b} =a/2[1\,{\bar {1}}\,0]$. However, to use Eqn.(1), b should be given in lab coordinate. In the example of Fig.(2), the three main axes are chosen as $\mathbf {C} _{1}=8a[1\,1\,{\bar {2}}]$, $\mathbf {C} _{2}=20a[1\,1\,1]$ and $\mathbf {C} _{3}=12a[1\,{\bar {1}}\,0]$, where a is the lattice constant. Then the Burgers vector b in lab coordinate is

$a\left[\mathbf {b} \cdot {\frac {\mathbf {C} _{1}}{|\mathbf {C} _{1}|}}\quad \mathbf {b} \cdot {\frac {\mathbf {C} _{2}}{|\mathbf {C} _{2}|}}\quad \mathbf {b} \cdot {\frac {\mathbf {C} _{3}}{|\mathbf {C} _{3}|}}\right]=a[0\quad 0\quad {\frac {1}{\surd {2}}}].$

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