Molekula dinamika - Számítógépes szimulációk ff1n4i11/1

Θ

•
• 6 × 10 23
•
N N 2
•

• N
•
•
〈K〉 =
1
2 mv2
= 3
2 kT
k = 1.38 × 10 −23 J/K;
〈K〉
•
T p
CV

•
T = 300K
3
2 kT ∼ = 6.2 × 10 −21 J = 0.039eV
11.6 m = 6.69 × 10 −26
2π
p
= 2π
√ 3mkT
∼= 2.2 × 10 −11 m,
r0 = 3.4 × 10 −10 m

•
r0
∼= 7.9 × 10
3kT/m
−13 s,
N = 10 3
N(N − 1)/2

•
•

• R(t) = (r1, . . . , rN)
V(t) A(t)
Rn+1 = 2Rn − Rn−1 + τ 2 An + O(τ 4 )
Vn = Rn+1 − Rn−1
+ O(τ
2τ
2 )
•
•
•
O(τ 5 )
•
•

Rn+1 = 2Rn − Rn−1 + τ 2 An + O(τ 4 )
Vn = Rn+1 − Rn−1
2τ
+ O(τ 2 )
•
•
• O(τ 2 )
•
•
Rn+1 = Rn + τVn +
τ 2
2 An + O(τ 3 )
Vn+1 = Vn + τ
2 (An+1 + An) + O(τ 3 )
• R O(τ 3 )

• N
•
r0 12
r0
6
V (r) = 4V0 − ,
r r
r
V0 = 1.65 × 10 −21 J V0/kB = 119.8
r0 = 3.41 × 10 −10
•
F(r) = − ∇(r) = 24V0
r 2
r0
2
r
12
−
r0
6
r
r

0 = 1 V0 = 1

• r = r0 0 r = 2 1/6 r0
−V0
•
•
•
E =
N 1
2 mv2 i +
V (rij),
i=1
rij 〈ij〉
•
P(v)dv =
m
kT
〈ij〉
mv2
−
e 2kT vdv.

•
c = 1
m = 1
r0 = 1 V0 = 1
τ0 =
mr 2 0
V0
= 2.17 × 10 −12 s,
τ0 = 1
N = 16
τ = 0.01
•
Fi =
j = 1
j = i
Fj hat i−re

•
L 3
•
rij = |ri − ri|
rij ′ = |ri − r ′ i|
•
i j ′ j
•

•
•
•
3(N − 1) × 1
2 kBT
N m
= v
2
2
i .
〈. . .〉 3(N − 1)
¯v = 0 v 2 i
(vi − ¯v) 2
i=1

•
•

•
•

•
N = 4M 3 , M = 1, 2, 3, . . . 32 = 4 × 2 3 108 = 4 × 3 3
256, 500, 864, . . .
• 1
(0, 0, 0) (0.5, 0.5, 0) (0.5, 0, 0.5) (0, 0.5, 0.5)
(0.5, 0.5, 0.5)

• T
P(v) =
m
2πkBT
3/2
e − m(v2 x +v2 y +v2 z )
2k B T .
0 √ T
•
0 1
• 0
0
vi := vi − vCM
vCM =
N
i=1 mvi
N
i=1 m

• 1
T
λ =
vi → λvi
2(N − 1)kBT
N
i=1 mv2 i
•

•
N(N − 1)/2 O(N 2 )
•
•
r > r0
rcutoff
0 rcutoff K
N = n × K n
O(n × K 2 )
n
rcutoff

•
rij = |ri − rj| < rcutoff
O(N 2 )
L >> rmax > rcutoff
rcutoff = 2.5r0 rmax = 3.2r0
rmax
•
•
Ucorr(r) = U(r) − d
dr U(rcutoff )(r − rcutoff )

•
•

•
E = m
2
N
i=1
v 2 i +
i=j
U(|ri − rj|)
•
CV =
∂E
∂T V
= 1
kBT 2
E2 − 〈E〉 2
T 2 − 〈T 〉 2 = 3
2 N(kBT ) 2
1 − 3NkB
2CV

•
P V = NkBT + 1
3
i 1
Z < 1
Z = 1
i

cutoff
rmax
N

pV = NkBT