F
T
B
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
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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.
I
Also, depending on the potential, there may be symmetries
which can be used to reduce calculations.
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 Winter '12
 Kotakoski
 Atom, Energy, Force, Kinetic Energy, Potential Energy, uij

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