Formalism of the Tersoff potential

Transcription

1 Originally written in December 000 Translated to English in June 014 Formalism of the Tersoff potential 1 The original version (PRB 38 p.990, PRB 37 p.6991) Potential energy Φ = 1 u ij i (1) u ij = f ij A ij exp(λ ij ) f ij b ij B ij exp(µ ij ) () b ij = [1 + (β i ζ ijk = f ik exp[µ m i ij ( ) m i 1 n i (3) 1 + c i d i c i d i + (h i cos θ kij ) θ kij is the angle between ik and ij. The first term of the right-hand side of Eq. () is denoted as u R ij, and the second term as u S ij. f is the cut-off function. A, B, λ, µ, m, c, d and h ara parameters. Calculation of force 1 (u R ij + u S ij) i (5) = 1 u R ij 1 u S ij i i (6) F i = Φ = The first term of the right-hand side is denoted as F R i, and the second as F S i. For the first term, it can be easily calculated like the case of a pairwise potential. F R i = = = u R ij f ij A ij (λ ij ) exp(λ ij ) j (8) f ij A ij λ ij exp(λ ij ) rij (9) } (4) (7) Here, j = (10) 1

2 is used (explained later). Caution must be drawn for F S i. To be exact, one should treat this as F S i = ( 1 ) (11) = ( 1 + ( 1 u S i j i j i u S i j ) + ( 1 r j i i i j =i i =i u S i j ) (1) u S i j (k =i) ). (13) i j i Especially, the third term must be treated carefully. The first, second and third terms are denoted as F S1 i, F S i and F S3 i. F S1 i = 1 (f ij b ij B ij exp(µ ij )) (14) = 1 f ij B ij b ij exp(µ ij )} (15) The underlined part can be written as b ij exp(µ ij )} = b ij exp(µ ij ) + exp(µ ij ) b ij (16) = µ ij b ij exp(µ ij ) + exp(µ ij ) b ij (17) The second term will be explained later. F S i = 1 u S ji (18) = 1 = 1 (f ji b ji B ji exp(µ ji r ji )) (19) f ji B ji b ji exp(µ ji r ji )} (0) The underlined part can be written as b ji exp(µ ji r ji )} = b ji exp(µ ji r ji ) + exp(µ ji r ji ) b ji (1) = µ ij b ji exp(µ ij ) + exp(µ ij ) b ji () The second term will be explained later. i = 1 f i r j B i j [1 + (β i i i j k =i 1 F S3 n j (β j l j,k f jk B jk exp(µ jk r jk )( 1 n j )[1 + (β j 1 ζ i j k )n i n i exp(µ i j r i j ) (3) l j,k ζ jkl ) n j 1 n j 1 (4) ζ jkl ) nj1 β j ζ jki (5)

3 The underlined part will be explained later. 1.1 The derivative of r (may be trivial..) j = (6) r ij = (7) 1. The derivative of b 1..1 b ij b ij = [1 + (β i = 1 n i [1 + (β i 1 n i (8) 1 1 n i 1 + (β i } (9) Thus, Similarly, 1 + (β i ζ ijk ) n i } = n i (β i 1 (β i ζ ijk ) (30) = n i (β i b ij = b ijβi n ( 1 [1 + (β i b ji = 1 n j [1 + (β j 1 + (β j 1 β i ζ ijk (31) ζ ijk (3) ζ jik ) n j 1 1 n j 1 + (β j ζ jik ) n j } = n j (β j ζ jik ) n j1 β j ζ jik ) n j } (33) ζ jik (34) 3

4 1.3 The derivative of ζ ζ jki is written as ζ jki = f ji exp[µ m j jk (r jk r ji ) m j 1 + c j d j c } j d j + (h j cos θ ijk ) (35) Thus, ζ jki = j ζ jki j = ζ jki j + (cos θ ijk) + (cos θ ijk) ζ jki (cos θ ijk ) ζ jki (cos θ ijk ) (36) (37) ζ jki j = µ m j jk m j(r jk ) m j1 ζ jki (38) Here, cos θ ijk = + r jk r ik r jk (39) Thus, (cos θ ijk ) = (cos θ ijk) + (cos θ ijk) (40) The derivative of cos is (cos θ ijk ) j j = (cos θ ijk) j j k (cos θ ijk) k k (41) = + r ik r jk r ij (4) and (cos θ ijk ) k = r jk (43) Therefore, Eq. (41) becomes (cos θ ijk ) = r jk r ik r ij r 3 ij + 1 r jk (44) And, from (cos θ ijk ) 1 + c j d j c } j d j + (h j cos θ ijk ) = c j(h j cos θ ijk ) d j + (h j cos θ ijk ) } (45) 4

5 we get ζ jki (cos θ ijk ) = f ji exp[µ m j jk (r jk ) m j c j(h j cos θ kij ) (46) d j + (h j cos θ ijk ) } Similarly, the derivative of ζ ijk is written as ζ ijk = j ζ ijk j = ζ ijk j + k ζ ijk k ζ ijk k + (cos θ kij) + (cos θ kij) ζ ijk (cos θ kij ) ζ ijk (cos θ kij ) (cos θ kij ) = 1 } r rik ij (r rik + r ij r kj) + (rik rij + r ij r kj) ik (47) (48) (49) ζ ijk j ζ ijk k = µ m i ij m i ( ) m i1 ζ ijk (50) = µ m i ij m i ( ) m i1 ζ ijk (51) ζ ijk (cos θ kij ) = f ik exp[µ m i ij ( ) m i c i (h i cos θ kij ) (5) d i + (h i cos θ kij ) } ζ jik is written as ζ jik = f jk exp[µ m j ji (r ji r jk ) m j 1 + c j d j c } j d j + (h j cos θ kji ) (53) Therefore we get ζ jik = ζ jik j + (cos θ kji) This can be calculated using the following. (cos θ kji ) = (cos θ ijk) ζ jik (cos θ kji ) (54) (55) ζ jik j = µ m j ji m j (r ji r jk ) m j1 ζ jik (56) ζ jik (cos θ kji ) = f jk exp[µ m j ji (r ji r jk ) m j c j(h j cos θ kji ) (57) d j + (h j cos θ kji ) } 5

6 New version for multicomponent system (PRB 39 p.5566) In the new version for multicomponent systems, b and zeta are somewhat different. b ij = χ ij [1 + (β i 1 n i (58) ζ ijk = f ik ω ik 1 + c i c } i d i d i + (h i cos θ kij ) For the derivative of β, the equations explained above must be just multiplied by χ. The derivative of ζ becomes simpler. For example, the derivative of ζ jki becomes (59) ζ jki = (cos θ ijk) ζ jki (cos θ ijk ) (60) ζ jki (cos θ ijk ) = f c j(h j cos θ kji ) jiω ji (61) d j + (h j cos θ kji ) }.1 f For R < r < S, f c = π Otherwise, sin π(r R) S R (6) f c = 0 (63) 6