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Unformatted text preview: F TB P I For a pair potential, U ij = U ji since the potential only depends on r ij =  r ij  =  r ji  . This simplifies the force calculation. I In the case of a threebody potential, things are more difficult, since U ij 6 = U ji . I When we have both twobody terms U ij = U 2 ( r i , r j ) and threebody terms U ijk = U 3 ( r i , r j , r k ) , the force (on atom i ) becomes f i =  ∇ i X j ( U ij + U ji ) + X j , k U ijk =  X j ( ∇ i U ij + ∇ i U ji ) + X j , k ∇ i U ijk (22) Notes I Often the only threebody dependency is implemented through a cosine term: U 3 ( r i , r j , cos θ ijk ) . I When this is the case, one can utilize the following equalities: cos θ ijk = r ij · r ik r ij r ik ⇒ (23) ∇ i cos θ ijk = ∇ i r ij · r ik r ij r ik = . . . = " cos θ ijk r 2 ij 1 r ij r ik r ij # + cos θ ijk r 2 ik 1 r ij r ik (24) thus, there’s no need to actually evaluate the cos function, which is computationally expensive....
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This note was uploaded on 02/14/2012 for the course CSE 6590 taught by Professor Kotakoski during the Winter '12 term at York University.
 Winter '12
 Kotakoski

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